Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 11 Chemical Coordination and Integration Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter Chapter 11 Chemical Coordination and Integration

11th Bio Zoology Guide Chemical Coordination and Integration Text Book Back Questions and Answers

Part I

Question 1.
The maintenance of a constant internal environment is referred as
a) Regulation
b) homeostasis
c) Co-ordination
d) hormonal control
Answer:
b) homeostasis

Question 2.
Which of the following are exclusive endocrine glands?
a) Thymus and testis
b) adrenal and ovary
c) parathyroid and adrenal
d) pancreas and parathyroid
Answer:
c) parathyroid and adrenal

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 3.
Which of the following hormone is not secreted under the influence of the pituitary gland?
a) thyroxine
b) insulin
c) oestrogen
d) glucocorticoids
Answer:
b) insulin

Question 4.
Spermatogenesis in mammalian testes is controlled by
a) Luteinising hormone
b) Follicle-stimulating Hormone
c) FSH and prolactin
d) GH and prolactin
Answer:
b) Follicle-stimulating Hormone

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 5.
Serum calcium level is regulated by
a) Thyroxine
b) FSH
c) Pancreas
d) Thyroid and parathyroid
Answer:
d) Thyroid and parathyroid

Question 6.
Iodised salt is essential to prevent
a) rickets
b) scurvy
c) goiter
d) acromegaly
Answer:
c) goiter

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 7.
Which of the following gland is related to immunity?
a) Pineal gland
b) adrenal gland
c) thymus
d) parathyroid gland
Answer:
c) thymus

Question 8.
Which of the following statement about sex hormones is correct?
a) Testosterone is produced by Leydig ceils under the influence of the luteinizing hormone
b) Progesterone is secreted by corpus luteum and softens pelvic ligaments during childbirth
c) Oestrogen is secreted by both sertoli cells and corpus luteum
d) Progesterone produced by corpus luteum is biologically different from the one produced by the placenta.
Answer:
a) Testosterone is produced by Leydig cells under the influence of luteinizing hormone

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 9.
Hypersecretion of GH in children leads to
a) Cretinism
b) Gigantism
c) Graves disease
d) Tetany
Answer:
b) Gigantism

Question 10.
A pregnant female delivers a baby who suffers from stunted growth, mental retardation, low intelligence quotient, and abnormal skin. This is the result of
a) Low secretion of growth hormone
b) Cancer of the thyroid gland
c) Over secretion of pars distalis
d) Deficiency of iodine in the diet
Answer:
a) Low secretion of growth hormone

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 11.
The structure which connects the hypothalamus with the anterior lobe of the pituitary gland is the
a) Dendrites of the neurohypophysis
b) Axons of neuro hypophysis
c) Bands of white fibers from the cerebellar region
d) Hypophysial portal system
Answer:
d) Hypophysial portal system

Question 12.
Comment on homeostasis.
Answer:

  • Maintenance of constant internal environment of the body by the different co-ordinating system.
  • The maintenance of the constant internal environment of the body is due to the functioning of the endocrine system and nervous system.
    Endocrine gland.

1. a) If the calcium level in the blood decreases the parathyroid gland secretes parathyroid hormones and increases the calcium level in the blood.
b) If there is more calcium in the blood the other hormone secreted by the parathyroid gland called calcitonin acts against the parathormone and reduces calcium level.

2. a) The adrenalin secreted by the adrenal gland in emergency increases heartbeat rate and blood pressure.
b) The other hormone secreted by the adrenal is nor-adrenalin which reduces heartbeat rate and blood pressure.

Nervous system:
1. a) The sympathetic nervous system at the time of shock induces tear glands to secrete tears,
b) The parasympathetic nerve at the time of shock and emotion reduces the tear secretion.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 13.
Which one of the following statement is correct
a) Calcitonin and thymosin are thyroid hormones
b) Pepsin and prolactin are secreted in the stomach
c) Secretin and rhodopsin are polypeptide hormones
d) Cortisol and aldosterone are steroid hormones
Answer:
d) Cortisol and aldosterone are steroid hormones

Question 14.
Which of the given option shows all wrong statements for thyroid gland Statements?
i) It inhibits the process of RBC formation
ii) It helps in the maintenance of water and electrolytes
iii) It’s more secretion can reduce blood pressure
iv) It stimulates osteoblast
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iv)
(d) (i) and (iii)
Answer:
(d) (i) and (iii)

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 15.
Hormones are known as a chemical messengers. Justify.
Answer:
Hormones are released into the blood stream and circulated as chemical signals. These act specifically on certain organs or tissues called target organs or tissues. These speed up or slow down or alter the activity of target tissues or organs. Hence they are known as chemaical messengers.

Question 16.
Write the role of t-estrogen in ovulation.
Answer:

  • Oestrogen hormones promote the growth of the ovarian follicle
  • While the follicles are growing it secretes the hormone oestrogen which in turn promotes the ovum to develop.
  • The shedding up of ovum at the Luteal phase takes place by the influence of the Luteinizing hormone.

Question 17.
Comment on Acini of the thyroid gland.
Answer:
The thyroid gland is a bilobed endocrine gland. Each lobe is made up of many lobules. The lobules consist of follicles called acini. Each acinus is lined with glandular, cuboidal, or squamous epithelial cells. The lumen of acinus is filled with colloid, a thick glycoprotein mixture consisting of thyroglobulin molecules.

Question 18.
Write the causes for diabetes mellitus and diabetes insipidus.
Answer:

  • Diabetes emeritus: If the insulin is not secreted sufficiently the liver and muscles are unable to convert the glucose into glycogen. As a result, more glucose enters into the bloodstream raising the blood sugar level leads to diabetes mellitus.
  • Diabetes Insipidus: The hormone antidiuretic hormone is secreted by the neurohypophysis, which promotes reabsorption of water and thus reduces the loss of water through urine ADH deficiency induces the production of large amounts of urine leads to diabetes insipidus.

Question 19.
Specify the symptoms of acromegaly.
Answer:
Acromegaly is caused due to excessive secretion of growth hormone in adults. The symptoms of acromegaly are an overgrowth of hand bones, feet bones, jawbones malfunctioning of gonads. enlargement of viscera, tongue, lungs, heart, liver, spleen, and endocrine glands like thyroid, or adrenal glands.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 20.
Write the symptoms of cretinism.
Answer:
Cretinism is caused due to hypothyroidism in infants. A cretin child shows the following symptoms:-

  1. Retarded skeletal growth.
  2. Absence of sexual maturity
  3. Retarded mental ability
  4. Thick and short limbs
  5. Thick wrinkled skin
  6. Bloated face
  7. Protruded enlarged tongue
  8. Low BMR, slow pulse rate, subnormal body temperature, and elevated blood cholesterol levels

Question 21.
Briefly explain the structure of the thyroid gland.
Answer:
The thyroid gland is butterfly-shaped, bilobed situated below the larynx on each side of the upper trachea. The two lobes are connected by a median tissue mass called isthmus. Each lobe is made up of many lobules. The lobules consist of follicles called acini. Each acinus is lined with glandular, cuboidal, or squamous epithelial cells.

The lumen of acini is filled with colloid, a thick glycoprotein mixture consisting of thyroglobulin molecules. The thyroid gland secretes Tri-iodothyronine (T,) and tetra-iodothyronine (T4) or thyroxine hormones. These are concerned with metabolism.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 1

Question 22.
Name the layers of the adrenal cortex and mention their secretions?
Answer:
A pair of adrenal glands are located at the anterior end of the kidneys. Elence, they are called suprarenal glands. The outer region is called cortex and the inner region is medulla. The adrenal cortex has three distinct zones,

  1. Zona Glomerulosa
  2. Zona fasciculata
  3. Zona reticularis

1. Zona Glomerulosa:

  • It is the outer thin layer. It constitutes about 15% of the cortex.
  • It secretes mineralocorticoids.

2. Zona fasciculata:

  • It is the middle wide layer constituting about 15% of the adrenal cortex.
  • It secretes glucocorticoids such as cortisol, corticosterone and trace amounts of adrenal androgen and oestrogen.

3. Zona reticularis:

  • It is the inner zone of adrenal cortex.
  • It constitutes about 10% of adrenal cortex.
  • It constitutes about 10% of adrenal cortex.
  • It secretes the adrenal androgen, trace amount of oestrogen and glucocorticoids.

Question 23.
Differentiate hyperglycemia from hypoglycemia.
Answer:
Hyper glycemia:
Hyperglycemia is a condition in which there is increased blood glucose level, it may be due to the reduced secretion of insulin.

Hypo glycemia:
Hypoglycemia is a condition in which the blood glucose level decreases. It may be due to increased secretion of insulin thereby reducing blood glucose level.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 24.
Write the functions of (CCK)- Cholecystokinin.
Answer:
CholecystokininCholecystokinin is secreted by duodenum in response to the presence of fat and acid in the diet. It acts on the gall bladder to release bile into duodenum and stimulates the secretion of pancreatic enzymes and its discharge.

Question 25.
Growth hormone is important for normal growth. Justify the statement.
Answer:
Growth hormone promotes growth of all the tissues and metabolic process of the cell. The growth hormone stimulates the growth in the following ways

  • It influences the metabolism of carbohydrate proteins and lipids.
  • It increases the rate of protein biosynthesis.
  • It stimulates cartilage formation of chondrogenesis.
  • It stimulates bone formation osteogenesis.
  • It helps in the retention of minerals like nitrogen, potassium, phosphorous sodium in the body.
  • It increases the release of fatty acid from adipose tissue.
  • It decreases the rate of glucose utilization for energy by the cell, by which it conserves glucose for glucose-dependent tissues such as the brain.

Question 26.
The pineal gland is an endocrine gland, write its role.
Answer:
The pineal gland or epiphysis cerebri or conarium is located behind the third ventricle of the brain. It is formed of parenchymal cells and interstitial cells. It secretes the hormone, melatonin.

It plays a central role in the regulation of the circadian rhythm of our body and maintains a normal sleep-wake cycle. It also regulates the timing of the sexual maturation of gonads. It also influences metabolism, pigmentation, menstrual cycle and defence mechanism of our body.

Question 27.
Comment on the functions of adrenalin.
Answer:

  • Adrenalin increases liver glycogen breakdown into glucose and increases the release of fatty acids from fat cells.
  • During emergency, it increases heartbeat rate and blood pressure.
  • It stimulates the smooth muscles of cutaneous and visceral arteries to decrease blood flow.
  • It increases blood flow to the skeletal muscles thereby increases the metabolic rate of skeletal muscles cardiac muscles and nervous tissues.

Question 28.
Predict the effects of removal of the Pancreas from the human body.
Answer:
Pancreas is both exocrine and endocrine gland. It is located just below the stomach as a leaf-like structure. It secretes digestive enzymes and hormones like insulin and glucogon.

The digestive enzymes digest carbohydrates, proteins, and fats. Insulin and glucose regulate blood sugar levels. If pancreas is removed from the body, digestion and maintenance of blood sugar level get affected.

Question 29.
Enumerate the role of the kidney as an endocrine gland.
Answer:
Kidney has endocrine tissues which act as a partial endocrine gland. It secretes renin, erythropoietin, and calcitriol. Renin is secreted by juxtaglomerular cells. It increases blood pressure when angiotensin is formed in the blood.

Erythropoietin is also secreted by the juxtaglomerular cells of the kidney and stimulates erythropoiesis in the bone marrow. Calcitriol is secreted by proximal tubes of nephrons. It is an active form of vitamin D3 which promotes calcium and phosphorus absorption from the intestine and accelerates bone formation.

Part – II.

11th Bio Zoology Guide Chemical Coordination and Integration Additional Important Questions and Answers

I. Choose The Correct Answer

Question 1.
Which of the following is not related to hormones?
(a) They are secreted by ductless glands
(b) They act on the target
(c) They act as chemical messengers
(d) They play an important role in indigestion
Answer:
(d) They play an important role indigestion

Question 2.
Find out the endocrine glands from the following.
a) Sebaceous gland
b) Sweat glands
c) Salivary glands
d) Thyroid gland
Answer:
d) Thyroid gland

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 3.
Which of the following is the function of Growth Hormone?
(a) It increases blood pressure
(b) It influences the metabolism of carbohydrates, proteins, and lipids
(c) It stimulates melanin synthesis in melanocytes
(d) It promotes reabsorption of water by nephrons
Answer:
(b) It influences the metabolism of carbohydrates, proteins, and lipids

Question 4.
Where is the pituitary situated?
a) On the ethmoid bone
b) On sella turcica
c) On the foramen of Monro
d) On the Rathke’s packet
Answer:
b) On sella turcica

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 5.
The retention of nitrogen is due to this hormone.
a) TSH
b) GH
c) FSH
d) ACTH
Answer:
b) GH

Question 6.
A person is unable to sleep normally. He may not get proper secretion of?
(a) Thyroxine
(b) Vasopressin
(c) Melatonin
(d) Oxytocin
Answer:
(c) Melatonin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 7.
A male child is born to a parent as he grew if FSH and LH are not properly secreted what happens to the male child.
a) He grows like a normal male child
b) He grows and he shows stunted growth
c) Secondary sexual characters are not developed properly
d) He grows as a mentally depressed male
Answer:
c) Secondary sexual characters are not developed properly

Question 8.
These hormones are collectively known as Gonadotropins?
a) Oxytocin and ADH
b) Oestrogen and Progestron
c) Testosterone and androgens
d) FSH and LH
Answer:
d) FSH and LH

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 9.
Which adrenal hormone is concerned with maintaining electrolytes, osmotic pressure and blood pressure?
(a) Cortisol
(b) Glucocorticoids
(c) Aldosterone
(d) Adrenalin
Answer:
(c) Aldosterone

Question 10.
Prabhu is working in a MNC Company. He has to do night shift once in every fifteen days.
Aftern a few years he is suffering from sleeplessness (in somnia) What may be the cause of his problem.
a) It may be becauses of his nature of work
b) More work pressure
c) The sleepwake cycle is disturbed due to the irregular synthesis of melotonin hormone
d) Due to the metabolic disturbances
Answer:
c) The sleepwake cycle is disturbed due to the irregular synthesis of melotonin hormone

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 11.
Name the structure that connects the lobes of thyroid gland?
a) Acinus
b) Ischium
c) Isthmus
d) bridge tissue
Answer:
c) Isthmus

Question 12.
Which of the following is the function of Glucogon?
(a) It increases the cellular utilization of glycogen into glucose
(b) It promotes the breakdown of glycogen into glucose
(c) It inhibits the breakdown of glycogen into glucose
(d) It inhibits the production of glucose from non-carbohydrate sources
Answer:
(b) It promotes the breakdown of glycogen into glucose

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 13.
Assertion: The Oxyphil cells of parathyroid secrete parathormone
Reason: The chief cells of parathyroid regulates the synthesis of parathormone
a) Assertion and Reason are correct
b) Assertion incorrect and the reason is wrong
c) The assertion is true. The reason is false
d) Both the assertion and reason are false.
Answer:
d) Both the assertion and reason are false.

Question 14.
Which of the following is not the function of progesterone?
(a) Implantation of the zygote in the uterus
(b) Decreasing the contraction of the uterus
(c) Formation of placenta
(d) Maturation of reproductive organs
Answer:
(d) Maturation of reproductive organs

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 15.
Which of the following option shows the with its action matched hormones
(given in column I) (given in column II)

Column I Column II
A Pituitary i) Partial endocrine gland
B Thyroid ii) Secretes steroid hormone oestrogen
C Ovary iii) Thyroxine regulates basal metabolic rate
D Thymus iv) Anterior lobe of pituitary originate from Rathke’s pouch

a) A – ii; B – iv; C – i; D – iii
b) A – iv ; B – iii; C – ii; D – i
c) A – iii; B – ii; C – iv; D – i
d) A – iv; B – i; C – iii; D – ii
Answer:
b) A – iv ; B – iii; C – ii; D – i

Question 16.
These are known as suprarenal glands.
a) Thymus gland
b) Pancreas
c) Adrenal gland
d) Ovary
Answer:
c) Adrenal gland

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 17.
Find out the hormone which are catecholamines.
a) Thyroxine
b) Insulin
c) Adrenalin
d) Glucagon
Answer:
c) Adrenalin

Question 18.
Why does cretenism occur?
(a) Hypothyroidism in adults
(b) Hypothyroidism in infants
(c) Hyperthroidism in adults
(d) lower level of iodine in the blood
Answer:
(b) Hypothyroidism in infants

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 19.
This hormone promotes the activation of Vitamin D thus by absorbing calcium?
a) Calcitonin
b) Thymosin
c) Parathormone
d) Thyroxine
Answer:
c) Parathormone

Question 20.
Does hypocalcemia occur due to?
(a) Hypersecretion of parathyroid hormone
(b) Hyposecretion of parathyroid hormone
(c) Hypersecretion of cortisol
(d) Hyposecretion of cortisol
Answer:
(b) Hyposecretion of parathyroid hormone

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 21.
Name the cells that secrete somatostatin
a) a cells
b) b cells
c) Lampda cells
d) delta cells
Answer:
d) delta cells

Question 22.
Why is insulin tablet not advisable for chronic diabetic Mellitus patients?
a) It takes more time to act
b) The insulin find it difficult to reach the substrate
c) Insulin is easily digested by the digestive enzymes
d) The insulin is not an effective one.
Answer:
c) Insulin is easily digested by the digestive enzymes

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 23.
Excess secretion of cortisol leads to
(a) Addison’s disease
(b) Grave’s disease
(c) Cushing’s syndrome
(d) Gull’s disease
Answer:
(c) Cushing’s syndrome

Question 24.
Contraction of the uterus leads to miscarriages in pregnant women.
What would be administered to evacuate the embryo?
a) ADH
b) Vasopressin
c) Androgen
d) Oxytocin
Answer:
d) Oxytocin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 25.
In the case of thyrotecdomic patients, what is advisable to give along with the thyroxine?
a) Insulin
b) Thyroglobulin
c) Parathormone
d) Adrenalin
Answer:
c) Parathormone

Question 26.
The hyposecretion of vasopressin results in ……………..?
(a) Diabetes mellitus
(b) Diabetes insipidus
(c) Ketosis
(d) Hyperglycemia
Answer:
(b) Diabetes insipidus

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 27.
Name the hormone that stimulates the synthesis of HCl.
a) Enterokinase
b) Pepsinogen
c) Gastrin
d) Secretin
Answer:
c) Gastrin

Question 28.
What is the other name for type I diabetes?
a) Insulin-dependent diabetes
b) Non-insulin-dependent diabetes
c) Sensitive diabetes
d) Dependent types
Answer:
a) Insulin-dependent diabetes

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 29.
Which is called the second messenger.
a) Receptor cells
b) CAMP
c) Adenylate
d) Substrate
Answer:
b) CAMP

Question 30.
In the following diagram what are the parts A, B, C, and D represent?
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 2
I) A) Pineal body B) Anterior tube C) Capillary bed D) Infundibulum
II) A) Hypophyseal vein B) Optic chiasma C) Endocrine cells D) Artery
III) A) Infundibuim B) Posterior C) Hypophyseal vein D) Endocrine cells
IV) A) Endocrine cells B) Infundibulum C) Paradisaical D) Pineal body
Answer:
III) A) Infundibuim B) Posterior C) Hypophyseal vein D) Endocrine cells

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 31.
Find out the true and false statements from the following and on that basis find the correct answer.
i) ADH causes blood vessels to dialate
ii) The ‘c’ cells of the thyroid gland secretes calcitonin
iii) Zona fasciculate secretes cortisol
iv) Minerals corticoids regulate water balance
i) a) True b) False c) True
ii) a) False b) True c) False
iii) a) False b) True c) False
iv) a) False b) False c) True
Answer:
iv) a) False b) False c) True

Question 32.
Which part interlinks both the nervous system and endocrine system?
a) Receptor molecules
b) Target tissue
c) Hypothalamus
d) Infundibulum
Answer:
c) Hypothalamus

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 33.
Which of the following disease is not caused by iodine or thyroxine deficiency?
a) Sporodic goitre
b) Exophthalmic goitre
c) Simple goitre
d) Myxodema
Answer:
a) Sporodic goitre

Question 34.
Which is the correct location of the receptors of the hormones?
a) Extracellular matrix
b) Blood
c) Plasma membrane
d) Nucleus
Answer:
c) Plasma membrane

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 35.
The primary target of the hormones of the hypothalamus is
a) pineal gland
b) thymus
c) testis
d) pituitary
Answer:
d) pituitary

Question 36.
Functionally the adenohypophysis of the pituitary gland includes
a) anterior lobe
b) Posterior lobe
c) intermediate lobe
d) both (a) and (c)
Answer:
d) both (a) and (c)

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 37.
An outgrowth of the hypothalamus from the base of the brain develops into
a) pars nervosa
b) pars intermedia
c) pars distalis
d) pars intermedia
Answer:
a) pars nervosa

Question 38.
The hormone secreted by neurohypophysis is
a) ACTH
b) ADH
c) GH
d) TSH
Answer:
b) ADH

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 39.
Among the following hormones is the peptide hormone.
a) GH
b) TSH
c) FSH
d) LH
Answer:
a) GH

Question 40.
Among the following is not a glycoprotein hormone.
a) TSH
b) LTH
c) LH
d) FSH
Answer:
b) LTH

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 41.
Find out the protein hormone from the following.
a) LTH
b) FSH
c) TSH
d) ACTH
Answer:
a) LTH

Question 42.
Which is the pituitary hormone that is present in other vertebrates?
a) MSH
b) ACTH
c) TSH
d) GH
Answer:
a) MSH

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 43.
Find out the wrong statement about vasopressin or ADH
a) It promotes reabsorption of water and electrolytes by distal tubules of the nephron.
b) Causes constriction of blood vessels
c) It stimulates vigorous contraction of smooth muscles of the uterus during childbirth
d) It is a peptide hormone
Answer:
c) It stimulates vigorous contraction of smooth muscles of the uterus during childbirth

Question 44.
The growth hormone releasing hormone stimulates the
a) Thymus to release the hormone
b) Pituitary for synthesis and release of growth hormone
c) Testis to release gonadotropin
d) Adrenal to release growth hormone
Answer:
b) Pituitary for synthesis and release of growth hormone

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 45.
Posterior pituitary secretion is controlled by
a) hypothalamic hypophyseal portal blood vessel
b) neuroendocrine gland
c) hypophysis
d) hypothalamic hypophyseal axis
Answer:
d) hypothalamic hypophyseal axis

Assertion and Reason:- Directions: In the following questions a statement of Assertion is followed by a statement of reason. Mark the correct choice as;
a) If both Assertion and reason are true and reason is the correct explanation of assertion
b) If both Assertion and reason and true but reason is not the correct explanation of Assertion.
c) If assertion is true but reason is false
d) If both Assertion and reason are false

Question 46.
Assertion: Adrenalin nor-adrenalin, melatonin, and thyroid hormones are proteins
Reason: Derived from cholesterol mostly water-soluble
Answer:
d) If both Assertion and reason are false

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 47.
Assertion: Growth hormone stimulates chondrogenesis and osteogenesis
Reason: Growth hormones promote growth of all tissues and metabolic process of the body.
Answer:
a) If both Assertion and reason are true and reason is the correct explanation of assertion

Question 48.
Assertion: Adrenal medulla hormones are referred to as 3F hormones fight, flight, and hormones.
Reason: During emergency it increases heartbeat rate and blood pressure
Answer:
a) If both Assertion and reason are true and reason is the correct explanation of assertion

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 49.
Assertion: The effects of aldosterone, oestrogen FSH are long-lived as they alter the amount of m- RNA and protein in a cell
Reason: The action of cAMP are terminated by phosphodiesterases
Answer:
b) If both Assertion and reason and true but the reason is not the correct explanation of Assertion.

Question 50.
Assertion: The immunity of old age people, becomes week and causes sickness.
Reason: Due to degeneration of pineal gland thymosin level decreases
Answer:
c) If assertion is true but reason is false

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 51.
Assertion: JGA cells of the kidney stimulates erythropoiesis or formation of RBC
Reason: JGA cells also secrete Erythropoietin
Answer:
a) If both Assertion and reason are true and reason is the correct explanation of assertion

Question 52.
Assertion: Diabetes inspidus is marked by production of large amount of urine.
Reason: Hypo secretion of ADH leads to a condition of reduced water absorption. Thus the urine becomes diluted without sugar or glucose
Answer:
a) If both Assertion and reason are true and reason is the correct explanation of assertion

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 53.
The Luteinizing hormone in males is
a) Testosterone
b) ICSH
c) FSH
d) LTH
Answer:
b) ICSH

Question 54.
The cells that secrete parathyroid hormone is known as
a) chief cells
b) oxyphil cells
c) goblet
d) both b and c cells
Answer:
a) chief cells

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 55.
Find out the correct statement about pineal gland.
a) Stimulates the production and release of sperms.
b) Increases blood pressure.
c) It plays a central role and the regulation of circadian rhythm of our body and maintains the normal sleep wake cycle.
d) It is about 1cm in diameter and 0.5gm.
Answer:
c) It plays a central role and the regulation of circadian rhythm of our body and maintains the normal sleep wake cycle.

Question 56.
The BMR is regulated by
a) Parathormone
b) Thyrocalcitonin
c) Adrenalin
d) Thyroxine
Answer:
d) Thyroxine

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 57.
The primary function of the thymus gland is
a) regulation of body temperature
b) regulation of body growth
c) immunological functions
d) Secretion of thyrotropin
Answer:
c) immunological functions

Question 58.
A hormone secreted by adrenal gland and called life-saving hormone is
a) adrenaline
b) cortisone
c) aldosterone
d) Cortisol
Answer:
d) Cortisol

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 59.
Which of the following condition is not linked to a deficiency of thyroid hormone?
a) cretinism
b) goitre
c) Myxoedema
d) Exophthalmia
Answer:
d) Exophthalmia

Question 60.
Androgens are sex corticoids secreted from
a) Zona reticulate
b) Adrenal medulla
c) Zona glomerulosa
d) Acini
Answer:
a) Zona reticulate

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 61.
A gland which is well developed in newborn child and produce lymphocytes
a) Thyroid gland
b) Thymus
c) Parathyroid gland
d) Pineal gland
Answer:
b) Thymus

Question 62.
Which one reduces the blood calcium level and shows an opposite effect to parathormone?
a) ADH
b) Insulin
c) Thyrocalcitonin
d) Thyroxine
Answer:
c) Thyrocalcitonin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 63.
Secretion of HCl and pepsinogen is controlled by
a) Cholecystokinin
b) Gastrin
c) Calcitriol
d) Renin
Answer:
b) Gastrin

Question 64.
A man has an IQ equivalent to that of a boy 5 years old this is due to the deficiency of which hormone?
a) Thyroxine
b) Adrenaline
c) Aldosterone
d) Somatotropin
Answer:
a) Thyroxine

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 65.
Which of the following option shows the correct matching of disorders and causes.
A) Addison’s disease -(i) Hypo secretion of thyroid
B) Cushing’s syndrome – (ii) Hypersecretion of parathyroid hormone
C) Softening of bone – (iii) Hypo secretion of glucocorticoids
D) Gull’s disease (in adults) – (iv) Excess secretion of cortisol
Codes:
a) A – (ii); B – (i); C – (iii); D – (iv)
b) A – (iii); B – (iv); C – (ii); D – (i)
c) A – (iii); B – (ii); C – (i); D – (iv)
d) A – (iv); B – (iii); C – (iv); D – (ii)
Answer:
b) A – (iii); B – (iv); C – (ii); D – (i)

Question 66.
A person is having problems with calcium and phosphorous metabolism in his body.
Which one of the following glands may not be functioning.
a) Parathyroid
c) Pancreas
b) Parotid
d) Thyroid
Answer:
a) Parathyroid

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 67.
The presence of fat and acid in the diet induces the secretion of
a) Gastrin
b) Secretion
c) Cholecystokinin
d) Calcitriol
Answer:
c) Cholecystokinin

Question 68.
Which is an amino acid derivative hormone?
a) Epinephrine
c) Progesterone
b) Oestrogen
d) Relaxin
Answer:
a) Epinephrine

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 69.
Which of the following is a mineralocorticoid?
a) Testosterone
b) Cortisol
c) Adrenalin
d) Aldosterone
Answer:
d) Aldosterone

Question 70.
Which hormone increases the rate of protein biosynthesis chondrogenesis and osteogenesis and helps in the retention of minerals?
a) Prolactin
b) Somatotrophic hormone
c) Thyrotropin
d) Glucagon
Answer:
b) Somatotrophic hormone

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 71.
The male sex hormones are secreted by
a) Zona glomerulosa
b) Fat cells
c) Lay-dig cells
d) Acini
Answer:
c) Lay-dig cells

Question 72.
Which of the following pituitary hormone is secreted without the involvement of a releasing hormone?
a) Thyrotropin
b) Follicle-stimulating hormone
c) Oxytocin
d) Prolactin
Answer:
c) Oxytocin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 73.
Match the following columns.

Column I Column II
A Melatonin 1. T-lymphocyte formation
B Thymus 2. Formation of RBC
C Insulin 3. Sleep-wake cycle
D Erythropoietin 4. Hypoglycemic hormone

Code:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 3
Answer:
b) A -3,B-1,C-4,D-2

Question 74.
Which hormones can easily pass through the cell membrane and bind to their receptors and alter gene function?
a) Peptide hormones
b) Amino acid-derived hormones
c) Neurohormones
d) Steroid hormones
Answer:
d) Steroid hormones

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 75.
The hormone which helps to find out conception is
a) FSH
b) Oestrogen
c) HCG
d) LH
Answer:
c) HCG

Question 76.
Gigantism and Acromegaly are due to
a) Hypothyroidism
b) Hyperthyroidism
c) Hypopituitarism
d) Hyper pitituitarism
Answer:
d) Hyper pitituitarism

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 77.
After performing the functions in the target organs how hormones are inactivated and excreted?
a) By intestine
b) By lungs and skin
c) By liver and Kidney
d) Both a and b
Answer:
c) By liver and Kidney

Question 78.
The half-life period of insulin is
a) 6 minutes
b) 8 minutes
c) 15 minutes
d) 7 minutes
Answer:
a) 6 minutes

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 79.
The time taken by insulin to be cleared from circulation is
a) 10-20 minutes
b) 10-15 minutes
c) 5-10 minutes
d) 2-10 minutes
Answer:
b) 10-15 minutes

(2 Marks)

II. Very Short Answer.

Question 1.
What are the causes of bone cyst formation?
Answer:
Hyper para thyrodism causes demineralization of bone causes bone cyst makes the bone fragile and soft.

Question 2.
Why is the hypothalamus considered a neuroendocrine gland?
Answer:
The hypothalamus alone with its neural function produces hormones. Hence it is considered a neuro-endocrine gland.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 3.
What is the meaning of the word hormone?
Answer:
Hormone means to excite.

Question 4.
Write on the origin of the pituitary gland?
Answer:
The anterior lobe of the pituitary gland originates from the embryonic invagination of pharyngeal epithelium called Rathke’s pouch. The posterior lobe originates from the base of the brain as an outgrowth of the hypothalamus.

Question 5.
What is meant by Homeostasis?
Answer:
Maintenance of the constant internal environment of the body by the different coordinating system is homeostasis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 6.
What is Pars nervosa?
Answer:
The neurohypophysis of the pituitary gland is known as pars nervosa.

Question 7.
Give example for partial endocrine glands.
Answer:
Pancreas, gastrointestinal tract epithelium kidney, heart gonads, and placenta.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 8.
Write a paragraph on the Growth Hormone?
Answer:
Growth hormone (GH):
It is also known as somatotropic hormone (STH) or Somatotropin. It is a peptide hormone Growth hormone promotes the growth of all the tissues and metabolic processes of the body. It influences the metabolism of carbohydrates, proteins, and lipids and increases the rate of protein biosynthesis in the cells.

It stimulates chondrogenesis (cartilage formation), osteogenesis (bone formation) and helps in the retention of minerals like nitrogen, potassium, phosphorus, sodium, etc., in the body. GH increases the release of fatty acid from adipose tissue and decreases the rate of glucose utilization for energy by the cells. Thus it conserves glucose for glucose-dependent tissues, such as the brain.

Question 9.
Name the hormone of peptide nature?
Answer:
Insulin, glucagon, and pituitary hormone are water-soluble.

Question 10.
What are steroid hormones?
Answer:
They are derived from cholesterol and are lipid-soluble.
Ex. Cortisol aldosterone testosterone oestrogen progesterone.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 11.
What is the role of the Follicle-stimulating hormone in a man?
Answer:
Follicle-stimulating hormone (FSH):
FSH is a glycoprotein hormone that regulates the functions of the gonads (ovary and testis). In males, FSH along with androgens acts on the germinal epithelium of seminiferous tubules and stimulates the production and release of sperms (spermatogenesis). In females, FSH acts on the ovaries and brings about the development and maturation of gratfran follicles.

Question 12.
What is the function of LH in females?
Answer:
LH induces ovulation maintains the corpus luteum and promotes the synthesis and release of ovarian hormones.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 13.
Why there is a short life span for hormone-like insulin?
Answer:
Insulin hormone is usually short-lived as it has to work through a second messenger (cAMP) system.

Question 14.
Why the hormones steroid may be long-lived?
Answer:
The effect of steroid hormones is long-lived as they alter the amount of mRNA and protein in a cell.
Ex. Aldosterone oestrogen FSH.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 15.
How do the hormones reach and act on the specific target organs?
Answer:

  • The hormones are released into the blood and circulated as chemical signals
  • The target organs contain receptor molecules either on the surface or within the cell.
  • The cells that contain the receptor molecules specific to the hormone are physiologically activated.
  • Whereas other hormones that come in contact can not be activated.

Question 16.
Name the three zones which are present in the Adrenal gland.
Answer:

  1. Zona glomerulosa
  2. Zona fasciculata
  3. Zona reticularis

Question 17.
Differentiate amines and steroid hormones
Answer:

Amines Steroids
A. Small water-soluble 1. Mostly lipid-soluble
B. Derived from tyrosine or tryptophan 2. Derived from Cholesterol
C. Examples are Adrenalin Noradrenalin, Melatonin, and thyroid hormone 3. Examples are cortisol, aldosterone, testosterone, oestrogen, and progesterone

Question 18.
Differentiate exocrine glands from endocrine glands.
Answer:

Exocrine glands Endocrine glands
1. They have ducts to carry their substances to the membrane surfaces 1. These lack ducts and so release their hormone to the surrounding tissue fluid
2. They secrete enzymes, saliva, and sweat 2. Their secretions are collectively termed hormones.
3. Eg. Salivary gland, Sweat gland, and Gastric glands 3. Eg. Pituitary, Thyroid Pineal, Adrenal Parathyroid

Question 19.
How endocrine glands be classified based on their secretary function?
Answer:

  • Exclusive endocrine glands, e.g. Pituitary thyroid, Parathyroid, Pineal, Adrenal
  • Neuroendocrine gland – Hypothalamus
  • Partial endocrine glands, e.g. Pancreas, gastrointestinal tract epithelium, kidney, heart gonads, and placenta – have endocrine tissues.

Question 20.
The pituitary gland is known as the master endocrine glands. Justify the statement.
Answer:

  • The pituitary gland is known as the master endocrine gland, but it is in turn controlled by the hypothalamus and so the statement may not be totally applicable.
  • The hypothalamus with its neurosecretory cells produces neurotransmitters which regulate the secretions of the pituitary.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 21.
Write about the location and dimension of the pituitary gland.
Answer:

  • The pituitary gland is located in the bony cavity of the sphenoid bone the sella tunica at the base of the brain so is also called the hypothalamus cerebri.
  • it is about one centimeter in diameter and 0.5gm in weight.

Question 22.
What is an infundibulum?
Answer:
The pituitary gland is connected to the hypothalamic region of the brain by a stalk called the infundibulum.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 23.
What is Rathke’s pouch?
Answer:
The anterior lobe of the pituitary originates the embryonic invagination of the pharyngeal epithelium called Rathke’s pouch.

Question 24.
Comment on MSH or Melanocyte stimulating hormone.
Answer:

  • In mammal’s role of Pass, intermedia is insignificant.
  • In other vertebrates pass intermedia secretes melanocyte-stimulating hormone or MSH.
  • It induces pigmentation in the skin.

Question 25.
What are the functions of the pineal gland?
Answer:
In humans, the pineal gland or epiphysis cerebri or conarium is located behind the third ventricle of the brain and is formed of parenchymal cells and interstitial cells.

It secretes the hormone, melatonin, which plays a central role in the regulation of the circadian rhythm of our body and maintains the normal sleep-wake cycle. It also regulates the timing of sexual maturation of gonads. In addition, melatonin also influences metabolism, pigmentation, menstrual cycle and defence mechanism of our body.

Question 26.
What are the other names of Luteotropic hormones?
Answer:

  1. Luteotropin
  2. Lactogenic hormone
  3. Protein
  4. Mammotropin

Question 27.
Name the hormones that their secretions are regulated by negative feedback mechanisms?
Answer:

  • Thyroid-stimulating hormone TSH or Thyrotropin
  • Adreno corticotropic hormone or ACTH

Question 28.
Give the meanings of the word pituitary oxytocin and Hormone?
Answer:

  1. Pituitary means “to grow under”.
  2. Oxytocin means “quick birth”.
  3. Hormone means “to excite”.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 29.
Mention the importance of Luteotropic hormone in females?
Answer:
It stimulates milk secretion after the childbirth.

Question 30.
Why prolactin is called luteotropic hormone?
Answer:

  • High prolactin secretion during lactation
  • Since it induces the corpus luteum hence named as luteotropic hormone.

Question 31.
Point out the role of oxytocin in females?
Answer:

  • It stimulates vigorous contraction of the smooth muscles of uterus during child birth.
  • Ejection of milk from the mammary glands after child birth.

Question 32.
Name the hormones that play a key role in milk secretion?
Answer:

  • Prolactin or lactogenic hormone or mammotropin or luteotropic hormone.
  • Oxytocin.

Question 33.
Name the hormone of pituitary that act on bloodvessels?
Answer:
The vasopressin causes constriction of blood vessels when released in large amount and in-creases blood pressure.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 34.
Which gland is located behind the third ventricle of brain?
Answer:
The pineal gland or epiphysis cerebri or conarium.

Question 35.
Which is the largest endocrine in the body and where it is located?
Answer:

  • The thyroid gland is the largest gland.
  • It is a bibbed gland located below the larynx on each side of upper trachea.

Question 36.
Name the hormones secreted by thyroid gland.
Answer:

  • Tri-iodo thyronine or T3
  • Tetra-iodo thyromne or T4 or Thyroxine
  • Thyrocalcitonin

Question 37.
What are parafollicular cells or ‘C’ cells?
Answer:

  • These are the cells in the thyroid gland.
  • These cells secrete a hormone called thyrocalcitonin.

Question 38.
What element is important for synthesis of thyroid hormones? In what quantity it is essential?
Answer:

  • Iodine is essential for the normal synthesis of thyroid hormones.
  • About 1m. a week of Iodine is required.

Question 39.
Mention the normal glucose level in prepandial and postpandial test for glucose?
Answer:

  1. Prepandial- 70-100 mg/dl (Fasting)
  2. Post pandial (About 2 hours after food) – 110 – 140mg/ dl.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 40.
Is it advisable to take alcohol frequently? What ill effects does it cause over the body?
Answer:

  • Alcohol consumption has widespread effect on endocrine system.
  • It impairs the regulation of blood glucose level.
  • More over it reduces the testosterone level
  • Increases the risk of osteoporosis.

Question 41.
What is the injection administered to diabetic patients? Why is it given as injection and not as oral pills?
Answer:

  • Humiline-N is administered to diabetic patients.
  • Human insulin is administered as injection and not by oral consumption.
  • The reason is if it is administered by oral consumption it may be digested by digestive enzymes.

Question 42.
What is sporadic goiter?
Answer:

  • It is a genetic disease
  • It is not caused by iodine or thyroxine deficiency.

Question 43.
Why laughing is good for health?
Answer:
It reduces the secretion of stress hormone, adrenalin and makes us to relax.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 44.
Define circadian rhythm.
Answer:
It is the 24-hour cycle of biological activities associated with natural periods of light and darkness.
Eg. Sleep wake cycle, body temperature, appetite etc.

(3 Marks)

III. Short Answer

Question 1.
What is simple goiter? What are its symptoms?
Answer:

  • Simple goitre is otherwise known as endemic goitre.
  • It is due to the hypo secretion of thyroxine.
  • The symptoms are Enlargement of thyroid gland, Fall in serum thyroxine level, Increased TSH secretion.

Question 2.
What is tetany?
Answer:

  • The hyposecretion of parathyroid hormones leads to a decrease in calcium level and increase in phosphate level, it causes.
  • convulsion in muscle, locking of jaws, increased heart beat rate increased body temperature muscular spasm lead to tetany.

Question 3.
What is cushing’s syndrome? What are its symptoms?
Answer:
The excess secretion of cortisol causes cushing’s syndrome. The symptoms are obesity of the face and trunk.

  1. Redness of face hand and feet.
  2. Thick skin.
  3. Excessive hair growth
  4. Loss of mineral from bone
  5. Systolic hypertension

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 4.
What is the cause of dwarfism?
Answer:
It is due to the hypo-secretion of growth hormone in children in which skeletal growth and sexual maturity is arrested. They attain a maximum height of 4 feet only.

Question 5.
What is Gigantism?
Answer:

  • Due to the hyper-secretion of growth hormone in children there is a over growth of skeletal
  • structure may occur up to 8 feet height. The visceral growth is not appropriate with that of limbs.

Question 6.
You are walking on the road, suddenly a man with a knife in his hand comes in front of you. What would be your reaction?
Answer:
I will get bimple goose due to the Surge of vasodielation on the arector pili muscle and though, i am very much frightened, i will fight with that man. i will get that energy by the action of the hormone adrenalin.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 7.
Why oxytocin is considered as quick or rapid birth?
Answer:
Oxytocin stimulates the contraction of the smooth muscles of uterus during child birth which helps in the expulsion of foetus, hence it is called as quick birth.

Question 8.
Why thymus is considered as a lymphoid organ?
Answer:
Thy lymphocytes that comes to thymus is educated to become immuno competent T lymphocytes and provides cell mediated immunity. Hence thymus is a lymphoid organ.

Question 9.
Why is gluco corticoids considered in the life saving activity?
Answer:
During the period of emergency the gluco corticoids stimulate the synthesis of glucose from lipid and protein by lipolysis and proteolysis and releases energy to meet that emergency situation.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 10.
Why cortisol works as a stress combat hormone?
Answer:

  • Cortisone involved in maintaining cardio vascular and kidney functions. It produces anti inflammatory reactions and suppresses the immune response, it stimulates the RBC production.
  • Hence it is known as stress combat hormone.

Question 11.
How do the hormones act in the target tissue?
Answer:

  • The hormones are released into the blood and circulated as chemical signals and act on specific target organs.
  • Hormones may speed up or slow down or alter the activity of the target organs.
  • The hormones secreted do not remain permanently in the blood.
  • They are converted by the liver into inactive compound and excreted by kidneys.

Question 12.
What is a limbic system?
Answer:

  • It is a collection of special structures located in the middle of the brain.
  • It is also known as paleo mammalian brain.
  • It controls emotions, behavior, motivation of long term memory and olfaction.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 13.
What are exocrine glands? Give examples
Answer:

  • The exocrine glands have ducts to carry their secretion to the membrane surfaces.
  • They secrete enzymes, saliva arid sweat.
  • Examples are salivary gland, gastric gland.

Question 14.
Write down the functions of hypothalamus.
Answer:

  1. Hypothalamic hormones control anterior pituitary secretion through hypothalamic hypophyseal portal blood vessel.
  2. The nerve signal produced by the hypothalamic hypophyseal axis control the posterior pituitary secretion.
  3. Hypothalamus maintains homeostasis.
  4. Blood pressure
  5. Body temperature.
  6. Cardio and fluid electrolyte balance of the
  7. As the part of limbic system it influences various emotional responses.

Question 15.
Draw the diagram of the hypothalamus and pituitary gland and label the following parts.
Answer:
A. Anterior lobe
B. Posterior lobe
C. Hypothalamus
D. Interior Hypophyseal artery
E. Endocrine cells
F. Hypophyseal vein
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 4

Question 16.
Draw the diagram of thyroid gland and label the following parts
Answer:
A Thyroid cartilage
B. Thyroid gland
C. Trachea
D.Isthumus
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 5

Question 17.
Draw the diagram of thyroid gland and label the following parts
A. Capsule
B. Cortex
C. Medulla
D. Blood Vessels
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 6

Question 18.
Explain the role of leuteinizing hormone (LH) in male and females.
Answer:

  • It is a glycoprotein hormone
  • It is also known as Interstitial cell stimulating hormone (ICSH) in males.
  • The ICSH hormones acts on the interstitial cells of testis to produce the male sex hormone testosterone.
  • In females along with FSH matures the ovarian follicles.
  • LH independently induces ovulation, maintains the corpus luteum and promotes synthesis and release of ovarian.
  • FSH and LH secretion starts only during pre pubertal period.
  • FSH and LH are collectively referred as gonadotropins.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 19.
Compare the structural difference of vasopressin and oxytocin. Vasopressin Oxytocin
Answer:

Exocrine glands Endocrine glands
1. Composed of a amino acids 1. Composed of a amino acids
2. Amino acid sequence is cysteine – tyrosine – Phenyl alanine – glutamine arspargine – cysteine – proline – argirune – glycine. 2. Amino acid sequence is – cysteine – tyrosine –
Isoleucine – Glutamine – aspargine – cysteine –
proline – leucine – glycine.

Question 20.
How does the posterior lobe of the pituitary help in osmoregulation?
Answer:

  • ADH is a peptide hormone
  • It promotes absorption of water and electrolytes by distal tubules of nephron and there by reduces loss of water through urine.
  • Hence it is called as anti diuretic hormone.
  • ADH deficiency causes Diabetes insipidus which induces the production of large amount of urine.
  • This is how it helps in osmoregulation.
  • ADH when released in large amount causes constriction of blood vessels and increases blood pressure.

Question 21.
Explain the role of melatonin in our body.
Answer:

  • It is secreted by pineal gland.
  • It plays a central role in the regulation of circadium rhythm of our body
  • Maintains the normal sleep wake cycle.
  • It also regulates the timing of sexual maturation of gonads.
  • In addition it also influences metabolism, pigmentation, menstrual cycle and defence mechanism of our body.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 22.
List out the role of thyroxine or tetra iodo thyronine (T4) in our body.
Answer:

  • It regulates the Basal Metabolic Rate (BMR) and body heat production.
  • It stimulates protein synthesis and promotes growth
  • It is essential for the development of skeletal and nervous system.
  • It plays an important role in maintaining blood pressure
  • It reduces serum cholesterol levels.
  • Optimum levels of thyroxine in the blood are necessary for gonadial functions.

Question 23.
Which hormone is called hypercalcemic hormone? Explain its role.
Answer:

  • The parathyroid hormone (PTH) is the hypercalcemic hormone.
  • It is a peptide hormone.
  • It involves in controlling the calcium and phosphate homeostasis.
  • The secretes or PTH is controlled by calcium level in the blood.
  • It increases the blood calcium level by stimulating osteoclasts to dissolve the bone matrix.
  • As a result calcium and phosphate are released into the blood.
  • PTH enhances the re-absorption of calcium and exertion of phosphates by the renal tubules.
  • It promotes activation of vitamin D to increase calcium absorption by intestinal mucosal cells.

Question 24.
Explain the role of heart as a partial endocrine gland.
Answer:
1. In the heart, cardiocytes on the atrial walls secretes an important peptide hormone called Atrial Natriuretic Factor (ANF)
2. When blood pressure is increased ANF is secreted and causes dilation of the blood vessels to reduce the blood pressure.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 25.
Is it good to take synthetic soft drinks? Why?
Answer:

  • It is not good to take synthetic soft drinks.
  • The branded soft drinks damage our endocrine system.
  • While consuming soft drinks the sugar level increases in blood. Which leads to elevated insulin secretion
    to reduce the blood glucose level.
  • The elevated insulin level diminishes immunity.
  • It causes obesity, cardio-vascular disorders etc.

Question 26.
The doctors avoid prescribing steroid tablets most often. Why?
Answer:
The abuse of steroids can cause serious health problems such as.

  • HighB.P.
  • Heart diseases.
  • Liver damage
  • Cancer
  • Stroke
  • Blood clotsa

Side effects such as

  1. Nausea
  2. V omiting
  3. Ligament and tendon injuries
  4. Head ache
  5. JointPain
  6. Muscle cramps
  7. Diarrhoea
  8. Sleep problem

Question 27.
if para-thyroid gland shows hyper-secretion. What will be the symptoms of this disorder?
Answer:

  • The excess secretion of parathyroid gland is known as hyperparathyroidism
  • Symptoms are Demineralisation of bone.
  • Cyst formation
  • Softening of bone
  • Loss of muscletone
  • General weakness
  • Renal disorders.

Question 28.
Differentiate Glycogenolysis from gluconeogenesis.
Answer:

  • Glycogenolysis is breakdown of glycogen to glucose
  • This process is carried out in the liver by the Glucagon hormone.
  • Glucagon thus releases glucose from the liver cells increasing Gluconeogenesis.
  • it is the synthesis of glucose from lactic acid and from non-carbohvdrate molecules is called Gluconeogenesis. this increases the blood glucose levels.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 29.
Mention the symptoms of diabetes mellitus.
Answer:

  • Poly urea – Excessive urination
  • Polyphagia – Excessive intake of food
  • Polydipsia – Excessive consumption of liquids due to thirst.
  • Ketosis – Breakdown of fat into glucose results in accumulation of ketone bodies.

Question 30.
Give a short account on hypothalamus.
Answer:

  • Hypothalamus is a small cone shaped structure that projects downward from the brain ending into the pituitary stalk.
  • It interlinks both the nervous system and endocrine system.
  • Pituitary gland is controlled by the hypothalamus.
  • It produces neuro transmitters which act either as a releasing hormone or as an inhibitory hormone.
  • Hypothalamus contains groups of neuro secretory cells which produces neuro transmitters which regulate the secretions of the pituitary gland.

Question 31.
Comment on pineal gland.
Answer:

  • It is located behind the third ventricle of brain.
  • In human it is called epiphysis cerebri conarium.
  • It is formed of parenchymal cells and interstitial cells.
  • It secretes the hormone melatonin.
  • It plays a central role in the regulation of circadian rhythm of our body and maintains the normal sleep wake cycle.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 32.
Write down the functions of thyrocalcitonin.
Answer:

  1. It is a polypeptide hormone.
  2. It regulates the blood calcium and phosphate levels.
  3. It reduces the blood calcium level and oppose the effects of parathyroid hormone.

Question 33.
Give a brief account on parathyroid gland.
Answer:

  • In man, four tiny parathyroid glands are found in the posterior wall of the thyroid glands.
  • It composed of two types of cells. The chief cells and oxyphil cells.
  • The chief cells secrete parathyroid hormone.
  • The functions of oxyphil cells are not known.

Question 34.
Write short notes on thymus gland.
Answer:

  • Thymus gland is partially an endocrine and partially a lymphoid gland.
  • It is a bi-lobed structure located just above the heart and aorta behind the sternum.
  • It is covered by a fibrous capsule.
  • Anatomically it is divisible into an outer cortex and an inner medulla.
  • It secretes four hormones such as thymulin, thymosin, thymopoietin and thymic humoral factor (THF).
  • The primary function is the production of immuno competent T lymphocytes which provides cell mediated immunity.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 7

Question 35.
Write about Acromegaly.
Answer:
Acromegaly is due to excessive secretion of growth hormone in adults.
Symptoms:-

  • Over growth of hand bones, feet bones, jaw bones
  • Malfunctioning of gonads
  • Enlargement of viscera, tongue, lungs, heart, liver, spleen and endocrine gland like thyroid adrenal.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 36.
What is hyperparathyroidism? Write down its symptoms.
Answer:
1. Hyperparathyroidism is caused due to excess parathyroid hormone in blood.
Symptoms:-

  • Demineralisation of bones
  • Cyst formation, softening of bone.
  • Loss of muscle tone
  • General weakness
  • Renal disorders.

Question 37.
Comment on Addison’s disease.
Answer:
Addison’s disease is caused due to hyposecretion of glucocorticoids and mineralocorticoids from the adrenal cortex.
Symptoms:-

  • Muscular weakness, Low BP
  • Loss of appetite, vomiting
  • Hyper pigmentation of the skin
  • Low metabolic rate, subnormal temperature.
  • Weight loss reduced blood volume.
  • Low aldosterone level increases urinary excretion of NaCl and water and decreases potassium excretion leading to dehydration.

Question 38.
Give reasons for Diabetes insipidus and point out its symptoms.
Answer:
Diabetes insipidus is caused due to hypo secretion of vasopressin.
Symptoms:-

  • Polyurea-frequent urination
  • Polydipsia – excessive consumption of liquids due to thirst.

Question 39.
Define BMR.
Answer:
The amount of energy needed to keep the body at rest.

Question 40.
Write down the general function of adrenalin hormone and its nature of secretion?
Answer:
Function:- The general function of nor adrenalin is to mobilize the brain and body for action. Nature of secretion: Its secretion is less during sleep, more during wakefulness and reaches much higher levels during stress situations. This response is known as ‘fight or flight’ response.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 41.
Old age people are sick often why?
Answer:
Due to degeneration of thymus gland, thymosin level decreases as a result the immunity of old age people becomes weak and causes sickness.

Question 42.
What is the role of pass intermedia in mammals and in other vertebrates?
Answer:

  • In mammals the role of pass intermedia is insignificant.
  • In other vertebrates it secretes melanocyte stimulating hormone (MSH)
  • It induces pigmentation in skin.

(5 Marks)

IV. Brief Answers

Question 1.
Describe the mechanism of peptide hormone action with a diagram.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 8

  • Peptide hormones cannot cross the phospholipid cell membrane and bind to the receptors on the exterior cell surface.
  • They are transported to the golgi which is the site of modification.
  • It acts as a first messenger in the cell.
  • Hormones generate a second messenger like cyclic AMP (cAMP) that regulates cellular metabolism. This cAMP is catalyzed by the adenylate cyclase ezyme.
  • The interaction between the hormone surface and the effect of cAMP within the cell is known as a signaling cascade. There may be amplication at each step.
  • One hormone molecule may bind to multiple receptor molecules before it is degraded.
  • Each receptor activates to form more cAMP and produces more signals.

Question 2.
Describe the mechanism of action of steroid hormone.
Answer:

  • Steroid hormones can easily cross the cell membrane and bind their receptors which may be intracellular or intercellular.
  • Then they pair up with another receptor and form a receptor-hormone complex. This can bind to DNA and alter its transcription.
  • As it changes the mRNA and protein the effect will be with stand for a long time. Eg. Estrogen

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 9

Question 3.
What is typical about Addison’s disease.
Answer:

  • There is a hyper pigmentation of the skin. It is caused due to hypo secretion of glucocorticoids from the adrenal cortex.
  • Muscular weakness low BP, loss of appetite vomiting, low metabolic rate subnormal temperature reduced blood volume and weight are the other symptoms of Addison’s disease.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 4.
What is the cause of cretinism? What are the symptoms?
Answer:

  • Hypothyroidism is the cause of cretinism in infants. There is a retarded skeletal growth absence of sexual maturity, retarded mental ability, thick wrinkled skin protruded enlarged tongue bloated face thick and short limbs.
  • There is low BMR, slow pulse rate subnormal body temperature and elevated blood cholesterol levels.

Question 5.
What are the hormones responsible for maintaining blood glucose levels?
Answer:
Insulin and glucagon are the hormones secreted by Islets of Langerhans of the pancreas:-
Insulin:
Insulin is a peptide hormone and plays an important role in glucose homeostasis. Its main effect is to lower blood glucose levels by increasing the uptake of glucose into the body cells, especially muscle and fat cells. Insulin also inhibits the breakdown of glycogen to glucose, the conversion of amino acids or fats to glucose, so insulin is rightly called a hypoglycemic hormone.

Glucagon:
Glucagon is a polypeptide hormone. It is a potent hyperglycaemic hormone that acts on the liver and promotes the breakdown of glycogen to glucose (Glycogenolysis), synthesis of glucose from lactic acid and from non-carbohydrate molecules (gluconeogenesis).

Releases glucose from the liver cells, increasing the blood glucose levels. Since glucagon reduces the cellular uptake and utilisation of glucose it is called a hyperglycemic hormone. Prolonged hyperglycemia leads to the disorder called diabetes mellitus.

Question 6.
What is an exophthalmic goiter? What are its symptoms?
Answer:
This disease is caused due to hyper secretion of the thyroid.
Symptoms:

  • Enlargement of the thyroid gland
  • Increased BMR.
  • Elevated respiratory and excretory rates
  • Increased heart beat
  • High blood pressure
  • Increased body temperature
  • Protection of eye ball (Exophthalmic)
  • The weakness of eye muscles
  • Weight loss.

Question 7.
What is meant by negative feedback mechanism? Explain with an example.
Answer:
When the thyroxine level in the blood decreases it is sensed by the hypothalamus to release the thyroid releasing factor that induces the pituitary to secrete thyroid stimulating hormone that stimulates the thyroid to release thyroxine – when the thyroxine level in the blood increases TTH acts on both the pituitary and hypothalamus to inhibit TSH secretion. This is meant by negative feedback mechanism.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 10

Question 8.
Why parathormone is considered as a hypercalcemic hormone.
Answer:

  • It increases the blood calcium level by stimulating osteo clasts to dissolve the bone matrix.
  • It enhances the reabsorption of calcium and also increases calcium adsorption by intestinal mucosal cells.
  • By all these action it increases the blood calcium level. Hence it is known as hypercalcemic hormone.

Question 9.
Write on Hyperthyroidism and Hypothyroidism?
Answer:
In infants, hypothyroidism causes cretinism. A cretin shows retarded skeletal growth, absence of sexual maturity, retarded mental ability, thick wrinkled skin, protruded enlarged tongue, bloated face, thick and short limbs occurs. The other symptoms are low BMR, slow pulse rate, subnormal body temperature and elevated blood cholesterol level.

Hyposecretion of the thyroid in adults causes myxoedema. It is otherwise called Gull’s disease. This disease is characterised by decreased mental activity, memory loss, slowness of movement, speech, and general weakness of body, dry coarse skin, scarce hair, puffy appearance, disturbed sexual function, low BMR, poor appetite, and subnormal body temperature.

Grave’s disease also called thyrotoxicosis or exophthalmic goiter. This disease is caused due to hypersecretion of the thyroid. It is characterised by enlargement of the thyroid gland, increases BMR (50% -100%), elevated respiratory and excretory rates, increased heartbeat, high BP, increases body temperature, protrusion of eyeball, and weakness of eye muscles and weight loss.

Simple goitre is also known as Endemic goitre. It is caused due to hyposecretion of thyroxine. The symptoms include enlargement of thyroid gland, fall in serum thyroxine level, increases TSH secretion.

Question 10.
Give an account of the actions of insulin.
Answer:

  • Insulin lowers the blood glucose level by increasing the uptake of glucose into the body cells.
  • It inhibits the conversion of amino acids or fat to glucose.
  • Insulin is called a hypoglycemic hormone.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 11.
Write on the disorders associated with parathryroid gland?
Answer:
Tetany is caused due to the hyposecretion of parathyroid hormone (PTH). Due to hyposecretion of PTH serum calcium level decreases (Hypocalcemia), as a result serum phosphate level increases. Calcium and phosphate excretion level decreses. Generalized convulsion, locking of jaws increased heart beat rate, increases body temperature, muscular spasm are the major symptoms of tetany. Hyperparathyroidism is caused due to excess PTH in blood. Demineralisation of bone, loss of muscle tone, general weakness, renal disorders are the symptoms of hyperparathyroidism.

Question 12.
a) What are the systems that regulate and co-ordinate the physiological functions of our body?
b) Give the meaning of the word “Hormone”.
c) What are the general characters of hormone?
Answer:
a) The systems that regulate and co-ordinate the physiological functions of our body are

  • Nervous or neural system
  • Endocrine system

b) The meaning of the word “Hormone”is “to excite”
c) General characters of hormones:

  • The endocrine system influences the metabolic activities through the hormones
  • These are chemical messengers released in to the blood and acts specifically on certain target organs or target tissues.
  • Hormones may speed up or slow down or alter the activity of the target organs.
  • Hormones secreted do not remain permanently in the blood.
  • After their function they are converted by the liver into inactive compounds and excreted by the kidneys.

Question 13.
Write a paragraph on different types of diabetes mellitus?
Answer:
Hyperglycaemia is otherwise known as Diabetes mellitus. Is caused due to reduced secretion of insulin. As the result, the blood glucose level is elevated. Diabetes mellitus is of two types, Type! Diabetes and Type II Diabetes. Type I diabetes is also known as Insulin-dependent diabetes, caused by the lack of insulin secretion due to illness or viral infections. Type II diabetes is also known as Non-Insulin dependent diabetes, caused due to reduced sensitivity to insulin, often called insulin resistance.

Symptoms of diabetes include polyuria (excessive urination), polyphagia (excessive intake of food), polydipsia (excessive consumption of liquids due to thirst), ketosis (the breakdown of fat into glucose results in accumulation of ketone bodies) in blood. Gluconeogenesis (Conversion of non-carbohydrate) also occurs in diabetes.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 14.
Give a diagrammatic sketch of the glandular system.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 11

Question 15.
Tabulate the major hypothalamic hormones and their functions.
Answer:

Hormones Functions
1. Thyrotropin-releasing hormone (TRH) Stimulates the secretion of TSH
2. Gonadotropin-releasing hormone(GnJRH) Stimulates the secretion of FSH
3. Corticotropin-releasing hormone (CRH) Stimulates the secretion of ACTH
4. Growth hormone-releasing hormone (GHRH) Stimulates the secretion of GH
5. Prolactin releasing hormone (PRH) Stimulates the secretion of Prolactin
6. Luteinizing hormone-releasing hormone (LHRH) Stimulates the secretion of LH
7. MSH releasing hormone Stimulates the secretion of MSH
8.Growth hormone-inhibiting hormone(GHIH) Inhibits the secretion of GH
9. Prolactin inhibiting hormone (PIEI) Inhibits the secretion of Prolactin
10. MSH inhibiting hormone Inhibits the secretion of MSH

Question 16.
Discuss the role of the hypothalamus and pituitary as a coordinated unit in maintaining the physiological process.
Answer:

  • Hypothalamus is a small cone-shaped structure that projects downward from the brain ending into the pituitary stalk.
  • It interlinks both the nervous system and endocrine system.
  • Hypothalamus contains a group of neuro secretary cells and it produces neuro transmitters which regulate the secretions of the pituitary.
  • The hormones produced by the hypothalamus act either as a releasing hormone or as an inhibitory hormone.
  • Though the pituitary gland is known as the master endocrine glands really it is in turn controlled by the hypothalamus.
  • The hypothalamus and pituitary gland are anatomically different they are interlinked and functioning as a co-ordinated unit in maintaining the physiological process. They can not functions as a separated unit.
  • The hypothalamic hypophyseal portal blood vessel allows hypothalamic hormones to control the anterior pituitary.
  • The nerve signal produced by the hypothalamic hypophyseal axis (nerve bundle) controls the posterior pituitary secretion.
  • Hypothalamus maintains homeostasis.
  • Maintains blood pressure, body temperature cardio, and fluid-electrolyte balance of the body it influences emotional responses.
  • The pituitary gland through its hormones performs various functions.

Question 17.
Give the classification of hormones based on their chemical nature?
Answer:

Class Chemical properties Example
Amines Small, water-soluble derived from tyrosine or tryptophan Adrenalin, nor adrenalin, melatonin, and thyroid hormone
Protein/ Peptides Water-soluble Insulin, glucagon, and pituitary hormones
Steroids Derived from cholesterol mostly lipid-soluble Cortisol, aldosterone, testosterone, oestrogen, progesterone.

Question 18.
Explain the structure of Testis?
Answer:

  • A pair of the testis is present in the scrotal sac of males.
  • The testis functions as a sex organ and also as an endocrine gland.
  • Testis composed of seminiferous tubules and Ley dig cells.
  • Lay dig cells secrete several male sex hormones collectively called Androgens.
  • The main male sex hormone is Testosterone.

Functions of Testosterone:

  1. Under the influence of FSH and LH, testosterone initiates the maturation of the male reproductive organ.
  2. The appearance of secondary sexual characters. Eg. Muscular growth, growth of facial and axillary hair, masculine voice, and male sexual behaviour.
  3. It enhances the total bone matrix and stimulating the process of spermatogenesis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 19.
Give a brief account of the ovary.
Answer:

  • Females have a pair of ovaries located in the pelvic region of the abdomen.
  • It is composed of ovarian follicles and stromal tissue.
  • It produces the egg or ova.
  • It secretes the steroid hormones oestrogen and progesterone.

Functions:

  1. Oestrogen is responsible for the maturation of reproductive organs and the development of secondary sexual characters at puberty.
  2. Along with progesterone, oestrogen promotes breast development and initiates the menstrual cycle.
  3. Progesterone prepares the uterus for implantation of the fertilized ovum.
  4. It decreases the uterine contraction during pregnancy and stimulates the development of mammary glands and milk secretion.
  5. It is responsible for premenstrual changes in the uterus and for the formation of the placenta.

Question 20.
Explain the functions of hormones of the heart and kidney.
Answer:
Heart: The cardiocytes on the atrial walls secrete an important peptide hormone called atrial natriuretic factor (ANF)

Function: When blood pressure is increased ANF is secreted and causes dilation of the blood vessels to reduce the blood pressure.

Kidney: Three hormones are secreted by the kidneys. They are Renin, erythropoietin, and calcitriol.

Renin:

  • It is secreted by Juxta glomerular Cells (JGA).
  • It increases blood pressure when angiotensin is formed in the blood.

Erythropoietin:

  • It is also secreted by JGA.
  • Stimulates erythropoiesis ie formation of RBC in bone mawow.

Calcitriol:

  • It is secreted by proximal tubules of the nephron.
  • It is an active form of vitamin D3.
  • It promotes calcium and phosphorus absorption from the intestine and accelerates bone formation.

Question 21.
If you happen to see a man/lady with short stature how will you identify him?
Differentiate and give a reason for that disorder.
Answer:
The person may be suffering from either cretinism or Dwarfism.
The following symptoms may be observed in cretinism. It is due to the hypo-secretion of Thyroxine.

  • A cretin shows retarded skeletal growth.
  • Absence of sexual maturity.
  • Retarded mental ability.
  • Thick wrinkled skin.
  • Protruded enlarged tongue.
  • Bloated face.
  • Thick and short limb occur
  • Low BMR
  • Slow pulse rate.
  • Subnormal body temperature
  • Elevated blood cholesterol levels.

If the person shows the following symptom, he may have Dwarfism, it is due to hyposecretion of growth hormone.

  1. Skeletal growth and sexual maturity is arrested.
  2. They attain a maximum height of 4 feet only.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 22.
(i) Describe the structure of the pancreas.
(ii) Draw the diagram and marced it parts
(iii) How insulin controls blood sugar?
(iv) What is the role of glucagon in our body?
Answer:
i) Structure of pancreas:

  • Pancreas is a composite gland which performs both endocrine and exocrine functions.
  • It is located just below the stomach as a leaf-like structure.
  • It is composed of two major tissues such as acme and islets of Langerhans
  • Acme secretes digestive enzymes and the islets of Langerhans secretes hormones like insulin and glucagon.
  • Human pancreas has 1 -2 million islets of Langerhans.
  • In each islet about 60% cells are beta cells 25% cells are alpha cells and 10% are delta cells.
  • The alpha cells secrete glucagon the beta cells secrete insulin and delta cells secrete somatostatin.

ii) Structure of Islets of Langerhans (pancreas)
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 12

iii) Insulin control-blood sugar:-

  • Insulin is a peptide hormone and plays an important role in glucose homeostasis.
  • Its main effect is to lower blood glucose levels by increasing the uptake of glucose into the body cells
  • especially muscle and fat cells.
  • Insulin also inhibits the breakdown of glycogen to glucose, the conversion of amino acids or fats to glucose.
  • So insulin is rightly called a hypoglycemic hormone.
  • Reduced secretion of insulin leads to diabetes mellitus. As a result glucose level is elevated.
  • It is of two types. Type I. Diabetes and Type II Diabetes. Type I Diabetes is caused by lack of secretion of insulin due to illness or viral infection. Type II or Non-insulin-dependent diabetes caused due to reduced sensitivity to insulin.

iv) Role of Glucagon:

  • Glucagon is a polypeptide hormone.
  • It is a potent hyper glycemic hormone.
  • It acts on the liver and promotes the break down of glycogen to glucose (glycogenolysis)
  • It also promotes synthesis of glucose from lactic acid and from non-carbohydrate molecules is called Gluconeogenesis.
  • Glucogon releases glucose from the liver cell increasing the blood glucose levels.
  • Since glucagon reduces the cellular uptake and utilisation of glucose it is called a hyperglycemic hormone. Prolonged hyperglycemia leads to the disorder called diabetes mellitus.

Question 23.
Symptoms of diabetes
Answer:

  • Polyurea – excessive urination
  • Polyphagia – excessive intake of food
  • Poly dispsia – excessive consumption of liquids due to thirst.
  • Ketosis – breakdown of fat into glucose results in accumulation of ketone bodies in blood.
  • Gluconeogenesis also occur ie conversion of the non-carbohydrate form like amino acids and fat into glucose.

Question 24.
a) Write down the location and dimension of the pituitary gland?
b) Explain the internal structure of the pituitary gland?
c) Draw the diagram of the pituitary gland and label the parts.
Answer:
a. Location of pituitary gland:-

  • It is located in a depression called sella tursica a bony cavity of the sphenoid bone below the brain.
  • The pituitary gland means “to grow under”.
  • It is connected to the brain by a stalk called the infundibulum
  • Dimension:- It is about 1cm in diameter and 0. 5gm in weight.

b. Internal structure:-

  • The pituitary consists of two lobes, anterior glandular adenohypophysis and posterior neural neuro hypophysis.
  • Anatomically the anterior lobe or adenohypophysis has three lebesor zones namely pass intermedia, pass distalis and pass tuberalis.
  • The neurohypophysis is otherwise known as pars nervosa.
  • Embryonic origin:- The anterior lobe originates from the Embryonic invagination of pharyngeal epithelia called Rathke’s pouch.
  • The posterior lobe originates from the base of the brain as an outgrowth of the hypothalamus.

c. Hypothalamus and pituitary gland
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 13

Question 25.
Give a brief account of the hormones of the adenohypophysis.
Answer:
It secretes six hormones.
1. Growth Hormone:-

  • It is also known as somatotropic hormone (STH) or somatotropin
  • It is a peptide hormone.
  • It promotes the growth of all the tissues and metabolic processes of the body.
  • It influences the metabolism of carbohydrates, proteins, and lipids and increases the rate of biosynthesis in the cells.
  • It stimulates chondrogenesis (cartilage formation), osteogenesis (bone formation).
  • It helps in the retention of minerals like nitrogen and potassium, phosphorous, sodium, etc. in the body.

2. Thyroid Stimulating Hormone TSH or Thyrotropin:-

  • It is a glycoprotein.
  • It stimulates the thyroid gland to secrete Tri-iodothyronine (T3) and Thyroxine (T4).
  • TSH secretion is regulated by a negative feedback mechanism.
  • Its release from the anterior pituitary is induced by the thyrotropin-releasing hormone (TRH).
  • When thyroxine level in the blood increases TRH acts on both the pituitary and hypothalamus to inhibit TSH secretion.

3. Adreno Corticotropic Hormone (ACTH):-

  • It is a peptide hormone.
  • It stimulates the adrenal cortex to secrete glucocorticoids and mineral corticoids.
  • It stimulates melanin synthesis in melanocytes induces the release of fatty acids from adipose tissues and stimulates insulin secretion.
  • its secretion is regulated by a negative feedback mechanism.

4. Follicle Stimulating Hormone (FSH):-

  • It is a glycoprotein hormone.
  • It regulates the functions of the gonads.
  • In males, FSH along with androgens act on the germinal epithelium of seminiferous tubules and stimulates the production and release of sperms (Spermatogenesis).
  • In females, it acts on the ovaries and brings about the development and maturation of graffian follicles.

5. Luteinizing hormone (LH):-

  • It is a glycoprotein hormone.
  • It is also known as interstitial cell-stimulating hormone.
  • In females LH, along with FSH matures the ovarian follicles.
  • LH independently induces ovulation, maintains the corpus luteum and promotes synthesis and release of ovarian hormones.
  • In males, ICSH acts on interstitial cells of testes to produce the male sex hormone testosterone.

6. Luteo Tropic Hormone (LTH):-

  • It is a protein hormone.
  • It is also called luteotropin or lactogenic hormone or prolactin or mammotropin.
  • It stimulates milk secretion after childbirth.
  • Since it induces the corpus luteum hence named as a luteotropic hormone.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 26.
a) Name the hormones secreted by the neuro hypophysis?
b) Give a brief account of its functions?
Answer:
a. Neuro hypophysis secretes two hormones.

  1. Vasopressin or anti-diuretic hormone
  2. Oxytocin.

Vasopressin or Anti diuretic Hormone (ADH):-

  • It is a peptide hormone.
  • it promotes reabsorption of water and electrolytes by distal tubules of the nephron and thereby reduces the loss of water through urine.
  • Hence it is called an anti-diuretic hormone.
  • When released in large amounts it causes constriction of blood vessels and increases blood pressure.
  • ADH deficiency causes Diabetes insipidus which induces the production of a large amount of urine.

Oxytocin:-

  • It means quick birth.
  • It is a peptide hormone.

3. It stimulates vigorous contraction of the smooth muscles of the uterus during childbirth.
4. And it also stimulates the ejection of milk from the mammary glands.

b. Chemical nature:

  • The insulin hormone is a peptide hormone with 51 amino acids.
  • The molecular weight of insulin hormone is 5734 Dalton.
  • It consists of two chains A and B which are linked together by disulphide bridges formed between cystine residues.

Role of Insulin:

  • It increases uptake of glucose into body cells especially muscle and fat cells.
  • It inhibits the breakdown of glycogen to glucose, the conversion of amino acids or fats to glucose.
    Thus it decreases the blood glucose level and so insulin is rightly called as a hypoglycemic hormone.

c. How can this condition be reversed?

  • maintenance of normal body weight through adoption of nutritional habits ie. balanced diet and physical exercise.
  • Correction of over nutrition and obesity may reduce the risk of diabetes and its complications.
  • Alcohol and smoking should be avoided.
  • Control of high blood pressure elevated cholesterol and high triglyceride levels.

Question 27.
a) Why hormones are called chemical messengers?
b) Explain how the target organs are specifically fit for the action of hormones?
Answer:
a) Hormones are chemical messengers because they act as organic catalysts and co-enzymes to perform specific functions in the target organs.

b) Special features of target organs:

  • The target organs contain receptor molecules either on the surface or within the cell.
  • Although different hormones come in contact, only the cells that contain receptor molecules specific to the hormone are physiologically activated.
  • A single hormone may have multiple effects on a single target tissue or on different target tissues.
  • Many hormones exhibit long-term changes like growth, puberty, and pregnancy.
  • Serious deficiency or excess secretion of hormones leads to disorders.
  • Hormones coordinate different physiological mental activities and maintain homeostasis.

Question 28.
Draw the Endocrine gland is our body.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration 14

Question 29.
Give an account on parathyroid hormones its hypo and hyper secretion deficiencies.
Answer:

  • Four tiny parathyroid glands are found in the posterior wall of the thyroid gland.
  • It is composed of Chief cells and oxyphil cell.
  • The chief cells secrete parathyroid hormone and the function of oxyphill cell are not known

Functions

  • It involves in calcium and phosphate.
  • It promotes the activation of vitamin D to increase calcium absorption by intestine.

Hyper secretion

  • Hyper parathyroidism causes demineral isation of calcium in bones.
  • Thus increasing the calcium and phosphate level in the blood.
  • PTH enhances the reabsorption of calcium and excretion of phosphate by the renal tubules.

Hypo secretion

  • Due to the hyposecretion of PTH serum calcium level decreases and the serum phosphate level increases.
  • Generalised convulsion occurs.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 11 Chemical Coordination and Integration

Question 30.
a. Give an account of position of adrenal gland and it’s internal structure, b. List out the function of adrenal hormone.
Answer:
a) Position
A pair of adrenal glands are located at the anterior end of the kidney. Hence called supra renal glands.
b) Structure
The adrenal gland is composed of the outer cortex and an inner medulla.
The cortex consists of

  1. zona glomerulosa
  2. zona fasciculata
  3. zona reticularis.

1. zona glomerulosa
– secretes minerals corticoids

2. zona fasciculata
– secretes gluco corticoids

3. zona reticularis.
– secretes androgen and oestrogen.

b) Functions

  • Gluco corticoids stimulate gluco neogenesis.
  • Cortisolies involved in maintaining cardio vascular and kidney functions.
  • Cortisol stimulates RBC production.
  • Mineralocorticoids regulated the water and electrolyte balance of our body.
  • Androgen plays a role in hair growth in the axial region pubis and face during Puberty.

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 13 Hydrocarbons Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

11th Chemistry Guide Hydrocarbons Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is
a) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
c) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
d) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Question 2.
C2H5 Br + 2Na Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 1 C4H10 + 2NaBr.
The above reaction is an example of which of the following
a) Reimer Tiemann reaction
b) Wurtz reaction
c) Aldol condensation
d) Hoffmann reaction
Answer:
b) Wurtz reaction

Question 3.
An alkyl bromide (A) reacts with sodium in ether to form 4, 5 – diethyloctane, the compound (A) is
a) CH3(CH2)3Br

b) CH3(CH2)5Br

c) CH3(CH2)3 CH(Br)CH3

d) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 2
Answer:
d) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 2

Question 4.
The C – H bond and C – C bond in ethane are formed by which of the following types of overlap
a) sp3 – s and sp3 – sp3
b) sp2 – s and sp2 – sp2
c) sp – sp and sp – sp
d) p – s and p – p
Answer:
a) sp3 – s and sp3 – sp3

Question 5.
In the following reaction,
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 3

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 4
Answer:
c) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 27

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 6.
Which of the following is optically active
a) 2 – methyl pentane
b) citric acid
c) Glycerol
d) none of these
Answer:
a) 2 – methyl pentane

Question 7.
The compound formed at anode in the electrolysis of an aqueous solution of potassium acetate are
a) CH4 and H2
b) CH4 and CO2
c) C2H6 and CO2
d) C2H4 and Cl2
Answer:
c) C2H6 and CO2

Question 8.
The general formula for cyclo alkanes
a) CnHn
b) CnH2n
C) CnH2n-2
d) CnH2n + 2
Answer:
b) CnH2n

Question 9.
The compound that will react most readily with gaseous bromine has the formula
a) C3H6
b) C2H2
c) C4H10
d) C2H4
Answer:
a) C3H6

Question 10.
Which of the following compounds shall not produce propene by reaction with HBr followed by elimination (or) only direct elimination reaction
a) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 5
b) CH3 – CH2 – CH2 – OH
c) H2C = C = O
d) CH3 – CH2 – CH2Br
Answer:
c) H2C = C = O

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 11.
Which among the following alkenes on reductive ozonolysis produces only propanone?
a) 2 – Methyl propene
b) 2 – Methyl but – 2 – ene
c) 2, 3 – Dimethyl but – 1- ene
d) 2, 3 – Dimethyl but – 2 – ene
Answer:
d) 2, 3 – Dimethyl but – 2 – ene

Question 12.
The major product formed when 2 – bromo – 2 – methyl butane is refluxed with ethanolic KOH is
a) 2 – methylbut – 2- ene
b) 2 – methyl butan – 1 – ol
c) 2 – methyl but – 1 – ene
d) 2 – methyl butan – 2- ol
Answer:
a) 2 – methylbut – 2- ene

Question 13.
Major product of the below mentioned reaction is, (CH3)2 C = CH2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 6
a) 2 – chloro – 1 – iode – 2 – methyl propane
b) 1 – chloro – 2 – iodo – 2 – methyl propane
c) 1, 2 – dichloro – 2 – methyl propane
d) 1, 2 – diiodo – 2 – methyl propane
Answer:
a) 2 – chloro – 1 – iode – 2 – methyl propane

Question 14.
The IUPAC name of the following compound is
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 7
a) trans – 2- chloro – 3- iodo – 2- pentane
b) cis – 3- iodo – 4 chloro – 3 – pentane
c) trans – 3 – iodo – 4 – chloro – 3 – pentene
d) cis – 2 – chloro – 3 iodo – 2 – pentene
Answer:
a) trans – 2- chloro – 3- iodo – 2- pentane

Question 15.
Cis – 2- butene and trans – 2 – butene are
a) conformational isomers
b) structural isomers
c) configurational isomers
d) optical isomers
Answer:
c) configurational isomers

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 16.
Identify the compound (A) in the following reaction

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 8

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 9
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 28

Question 17.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 11Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 10 CH ≡ CH , where A is,
a) Zn
b) conc. H2SO4
c) alc. KOH
d) dil. H2SO4
Answer:
c) alc. KOH

Question 18.
Consider the nitration of benzene using mixed con H2SO4 and HNO3 if a large quantity of KHSO4 is added to the mixture, the rate of nitration will be
a) unchanged
b) doubled
c) faster
d) slower
Answer:
d) slower

Question 19.
In which of the following molecules, all atoms are co-planar
a) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 12

b) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 13

c) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 14
d) both (a) and (b)
Answer:
d) both (a) and (b)

Question 20.
Propyne on passing through red hot iron tube gives
a) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 15

b) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 16

c) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 17
d) none of these
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 15

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 21.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 18 is

a) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 19

b) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 20

c) both (a) and (b)

d) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 21
Answer:
d) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 21

Question 22.
Which one of the following is non aromatic?
a) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 22

b) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 23

c) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 24

d) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 25
Answer:
d) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 25

Question 23.
Which of the following compounds will not undergo Friedal – crafts reaction easily?
a) Nitro benzene
b) Toluene
c) Cumene
d) Xylene
Answer:
a) Nitro benzene

Question 24.
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
a) – COOH
b) – NO2
c) -C ≡ N
d) -SO3H
Answer:
b) – NO2

Question 25.
Which of the following can be used as the halide component for friedal – crafts reaction?
a) Chloro benzene
b) Bromo benzene
c) Chloro ethene
d) Isopropyl chloride
Answer:
d) Isopropyl chloride

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 26.
An alkane isobtainedbydecarboxylation of sodium propionate. Same alkane can be prepared by
a) Catalytic hydrogenation of propene
b) action of sodium metal on iodomethane
c) reduction of 1 – chloro propane
d) reduction of bromomethane
Answer:
b) action of sodium metal on iodomethane

Question 27.
Which of the following is aliphatic saturated hydrocarbon
a) C8H18
b) C9H18
c) C8H14
d) All of these
Answer:
a) C8H18

Question 28.
Identify the compound ‘Z’ in the following reaction
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 26
a) Formaldehyde
b) Acetaldehyde
c) Formic acid
d) none of these
Answer:
a) Formaldehyde

Question 29.
Peroxide effect (Kharasch effect) can be studied in case of
a) Oct – 4 – ene
b) hex – 3 – ene
c) pent – 1 – ene
d) but – 2 – ene
Answer:
c) pent – 1 – ene

Question 30.
2 – butyne on chlorination gives
a) 1 – chloro butane
b) 1, 2 – dichloro butane
c) 1, 1, 2, 2 – tetrachlorobutane
d) 2, 2, 3, 3 – tetra chloro butane
Answer:
d) 2, 2, 3, 3 – tetra chloro butane

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

II. Write brief answer to the following questions:

Question 31.
Give IUPAC names for the following compounds
1) CH3 – CH = CH – CH = CH – C ≡ C – CH3

2) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 29

3) (CH3)3 C – C ≡ C – CH(CH3)2

4) ethyl isopropyl acetylene

5) CH ≡ C – C ≡ C – C ≡ CH
Answer:
1) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 30

2) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 31

3) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 32

4) ethyl isopropyl acetylene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 33

5) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 34

Question 32.
Identify the compound A, B, C and D in the following series of reactions
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 35
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 36

Question 33.
Write short notes on ortho, para directors in aromatic electrophilic substitution reactions.
Answer:
All the activating groups are ‘ortho-para’ directors.
Example: – OH, – NH2, -NHR, -NHCOCH3, -OCH3 -CH3 – C2H5 etc.
Let us consider the directive influences of phenolic (-OH) group. Phenol is the resonance hybrid of following structures.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 37

In these resonance structures, the (-) charge residue is present on ortho and para position of ring structure. It is quite evident that the lone pair of electron on the atom which is attached to the ring involves in resonance and makes the ring more electron rich than benzene. The electron density at ortho and para positions increases as compared to the meta position. Therefore phenolic group activates the benzene ring for electrophilic attack at ‘ortho’ and ‘para positions and hence -OH group is an ortho-para director and activator.

In aryl halides, the strong -I effect of the halogens (electron withdrawing tendency) decreases the electron density of benzene ring, thereby deactivating for electrophilic attack. However the presence of lone pair on halogens involved in the resonance with pi electrons of benzene ring, increases electron density at ortho and para position. Hence the halogen group is an ortho-para director and deactivator.

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 34.
How is propyne prepared from an alkene dihalide?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 38

Question 35.
An alkylhalide with molecular formula C6H13Br on dehydro halogenation gave two isomeric alkenes X and Y with molecular formula C6H12. On reductive ozonolysis, X and Y gave four compounds CH3COCH3, CH3CHO, CH3CH,CHO and (CH3)2 CHCHO. Find the alkyihalide.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 39

Question 36.
Describe the mechanism of Nitration of benzene.
Answer:
It is prepared by the action of a mixture of con. HNO3 and con. H2SO4 (nitrating mixture) on benzene maintaining the temperature below 333 K.
Nitration:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 40

Sulphuric acid generates the electrophile – NO2+, nitronium ion-from nitric acid. This is an example of aromatic electrophilic substitution reaction.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 41

The generation of nitronium ion
H2SO4 + HONO2Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 42 + HSO4
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 43

To the nitronium ion (being an electron deficient species) the π bond of benzene, donates a pair of electrons forming a 6-bond. A species with a + ve charge is formed as an intermediate. This is called ‘arenium ion’ and is stabilised by Resonance.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 44

In the last step, the hydrogen atom attached to the carbon carrying the nitro group is pulled out as a proton, by the Lewis base HSO4, so that stable aromatic system is formed.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 45

Question 37.
How does Huckel rule help to decide the aromatic character of a compound.
Answer:
In 1865, August Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds.
There were two objections:
Benzene forms only one ortho disubstituted products whereas the Kekule’s structure predicts two o-di substituted products as shown below.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 46

Kekule’s structure failed to explain why benzene with three double bonds did not give addition reactions like other alkenes.To overcome this objection, Kekule suggested that benzene was mixture of two forms (1 and 2) which are in rapid equilibrium.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 47

Resonance description of benzene:
The phenomenon in which two or more structures can be written for a substance which has identical position of atoms is called resonance. The actual structure of the molecule is said to be resonance hybrid of various possible alternative structures. In benzene, Kekule’s structures I & II represented the resonance structure, and structure III is the resonance hybrid of structure I & II.
The structures 1 and 2 exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 48

Spectroscopic measurements:
Spectroscopic measurements show that benzene is planar and all of its carbon- carbon bonds are of equal length 1.40 Å. This value lies between carbon-carbon single bond length 1.54 Å and carbon- carbon double bond length 1.34 Å.

Molecular orbital structure:
The structure of benzene is best de¬scribed in terms of the molecular orbital theory. All the six carbon atoms of benzene are sp2 hybridized. Six sp2 hybrid orbitals of carbon linearly overlap with six 1 s orbitals of hydrogen atoms to form six C – H sigma bonds. Overlap between the remaining sp2 hybrid orbitals of carbon forms six C – C sigma bonds.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 49
Formation of Sigma bond in benzene

All the σ bonds in benzene lie in one plane with bond angle 120°. Each carbon atom in benzene possess an un hybridized p-orbital containing one electron. The lateral overlap of their p-orbital produces 3 π- bond. The six electrons of the p-orbitals cover all the six carbon atoms and are said to be delocalised. Due to delocalization, strong π-bond is formed which makes the molecule stable. Hence unlike alkenes and alkynes benzene undergoes substitution reactions rather addition reactions under normal conditions.

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 50

All carbon atoms have the delocalized π MO is formed by p-orbitais the overlap of six p-orbitals.

Representation of benzene:
Hence, there are three ways in which benzene can be represented.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 51

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 38.
Suggest the route for the preparation of the following from benzene.
1) 3 – chloro nitrobenzene
2) 4 – chlorotoluene
3) Bromo benzene
4) m – dinitro benzene
Answer:
1) 3 – chloro nitrobenzene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 52

2) 4 – chlorotoluene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 53

3) Bromo benzene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 54

4) m – dinitro benzene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 55

Question 39.
Suggest a simple chemical test to distinguish propane and propene.
Answer:
Propene decolourises Br2/H2O it forms dibromo compound but propane does not react with Br2/ H2O.

Question 40.
What happens when Isobutylene is treated with acidified potassium permanganate?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 56

Question 41.
How will you convert ethyl chloride into
(i) ethane
(ii) n – butane
Answer:
(i) ethane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 57

(ii) n – butane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 58

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 42.
Describe the conformers of n – butane.
Answer:
Conformations of n-Butane:
n-Butane may be considered as a derivative of ethane, as one hydrogen on each carbon is replaced by a methyl group

Eclipsed conformation:
In this conformation, the distance between the two methyl group is minimum. So there is maximum repulsion between them and it is the least stable conformer.

Anti or staggered form:
In this conformation, the distance between the two methyl groups is maximum and so there is minimum repulsion between them. And it is the most stable conformer.

The following potentially energy diagram shows the relative stabilities of various conformers of n-butane.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 59

Question 43.
Write the chemical equations for combustion of propane.
Answer:
C3H8(g) + 5O2(g) ) → 3CO2(g) + 4H2O(l)
propane

Question 44.
Explain Markovnikoff’s rule with suitable example.
Answer:
Markovnikoff ‘s rule:
“When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen adds to the carbon that has more number of hydrogen and halogen add to the carbon having fewer hydrogen”. This rule can also be stated as in the addition reaction of alkene/alkyne, the most electro negative part of the reagent adds on to the least hydrogen attached doubly bonded carbon.

Addition of water: (Hydration of alkenes)
Normally, water does not react with alkenes. In the presence of concentrated sulphuric acid, alkenes react with water to form alcohols. This reaction follows carbocation mechanism and Markovnikoff’s rule.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 60

Question 45.
What happens when ethylene is passed ‘ through cold dilute alkaline potassium permanganate.
Answer:
CH2 = CH + H2O + (O) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 61 CH2OH – CH2OH
Ethylene                                                                Ethylene glycol

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 46.
Write the structures of following alkanes.
1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane
3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
Answer:
1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 62

2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 63

3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 64

Question 47.
How will you prepare propane from a sodium salt of fatty acid?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 65

Question 48.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 66
Identify A and B
Answer:
A) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 67

b) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 68

Question 49.
Complete the following :

i) 2 – butyne  Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 69

ii) CH2 = CH2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 70

iii) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 71

iv) CaC2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 72
Answer:
i) 2 – butyne Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 69
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 73

ii) CH2 = CH2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 70
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 74

iii) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 71
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 75

iv) CaC2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 72
CaC2 + 2H2O → Ca(OH)2 + C2H2
Calcium Carbide                  Acetylene

Question 50.
How will distinguish 1 – butyne and 2 – butyne?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 76
1 – butyne reacts with ammoniacal AgNO3 solution it forms white precipitate of silver acetylide but, 2 – butyne doesnot reacts with ammoniacal AgNO3 solution.

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

11th Chemistry Guide Hydrocarbons Additional Questions and Answers

I. Choose the best answer:

Question 1.
Statement-I: Methane. ethane, propène, and butane are alkane group compounds.
Statement-II: They are obeying C2H2n+2 + formula and each member differs from it proceeding member by a CH2 group.
(a) Statement-I and II are correct and Statement-II is the correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I
(e) Statëment-I is correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of statement-I.

Question 2.
Alkanes are also known as
a) olefins
b) unsaturated aliphatic hydrocarbons
c) saturated aromatic hydrocarbon
d) paraffin
Answer:
d) paraffin

Question 3.
Which one of the following shows three possible isomeric structures’?
(a) C4H10
(b) C5H12
(c) C6H12
(d) C3H4
Answer:
(b) C5H12

Question 4.
The gas supplied in cylinders for cooking is
a) marsh gas
b) LPG
c) mixture CH4 and C2H6
d) mixture of ethane and propane
Answer:
d) mixture of ethane and propane

Question 5.
Which of the following compound cannot be prepared by Kolbe’s electrolytic method’?
(a) CH3-CH3
(b) CH4
(c) CH2 = CH2
(d) CH = CH
Answer:
(b) CH4

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 6.
Soda-lime is
a) NaOH
b) NaOH + CaO
c) CaO
d) Na2CO3
Answer:
b) NaOH + CaO

Question 7.
Statement – I : Alkenes shows both structural and geometrical isomerism.
Statement – II : Because of the presence of double bond.
(a) Statement – I and II are correct and statement-II is correct explanation of statement – I.
(b) Statement – I and II are correct but statement-II is not correct explanation of statement – I
(c) Statement – I is correct but statement – II is wrong.
(d) Statement – II is wrong but statement – II is correct.
Answer:
(a) Statement – I and II are correct and statement – II is correct explanation of statement – I.

Question 8.
Hydrocarbon which is liquid at room temperature is
a) Pentane
b) Butane
c) Propane
d) Ethane
Answer:
a) Pentane

Question 9.
Pyrolysis of Methane and respectively are
a) Exothermic & Endothermic
b) Endothermic & Exothermic
c) Both are endothermic
d) Both are exothermic
Answer:
c) Both are endothermic

Question 10.
Peroxide effect is not observed in ………..
(a) HCl
(b) HI
(c) HBr
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 11.
Final products of complete oxidation of hydrocarbon is
a) Acid
b) Dihydric alcohol
c) Aldehyde
d) H2O + CO2
Answer:
d) H2O + CO2

Question 12.
Isomerisation in alkane can be brought by using
a) Al2O3
b) Fe2O3
c) Anh.AlCl3/ HCl at 200°C
d) Cone. H2SO4
Answer:
c) Anh.AlCl3/ HCl at 200°C

Question 13.
In aromatization of n – hexane, the catalyst used is
a) Cr2O3
b) V2O5
c) Mo2O3
d) All
Answer:
d) All

Question 14.
The most oxidized form of ethane is
a) CO2
b) HCHO
c) HCOOH
d) CH3COOH
Answer:
a) CO2

Question 15.
The following substance is used as anti-knocking compound
a) TEL
b) Lead tetrachloride
c) Lead acetate
d) C2H2PbCl
Answer:
a) TEL

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 16.
The molecular formula of benzene is ………….
(a) C6H6
(b) C6H5
(c) C7H8
(d) CH4
Answer:
(a) C6H6

Question 17.
The most stable conformation of ethane is
a) Eclipsed
b) Skew
c) Staggered
d) All are equally stable
Answer:
c) Staggered

Question 18.
IUPAC name of the following compound
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 77
a) 3 – methyl hexane
b) 3 – Ethyl pentane
c) 2, 3 – dimethyl pentane
d) 2, 2 – dimethyl pentane
Answer:
a) 3 – methyl hexane

Question 19.
The number of sigma bonds formed in ethane by the overlapping of sp3 – sp3 orbitals
a) 7
b) 5
c) 1
d) 4
Answer:
c) 1

Question 20.
Which one of the following is not an ortho-para director?
(a) – NO2
(b) – CH3
(c) – OH
(d) – C2H5
Answer:
(a) – NO2

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 21.
The correct IUPAC name of the following alkane is
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 78
a) 3, 6 – diethyl – 2- methyloctane
b) 5 – isopropyl – 3- ethyloctane
c) 3 – ethyl – 5- isopropyloctane
d) 3 – isopropyl – 6 – ethyloctane
Answer:
a) 3, 6 – diethyl – 2- methyloctane

Question 22.
In the Wurtz reaction, n-hexane is obtained from
a) n – propyl chloride
b) n – butyl chloride
c) Ethyl chloride
d) isopropyl chloride
Answer:
a) n – propyl chloride

Question 23.
When sodium acetate is heated with soda lime the reaction is called
a) Dehydration
b) Decarboxylation
c) Dehydrogenation
d) Dehydrohalogenation
Answer:
b) Decarboxylation

Question 24.
The following substance reacts with water to give ethane
a) CH4
b) C2H5MgBr
c) C2H4OH
d) C2H5OC2H5
Answer:
b) C2H5MgBr

Question 25.
Reaction of ROH with R’ MgX produces
a) RH
b) R’H
c) R – R
d) R’ – R”
Answer:
b) R’H

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 26.
The solvent used in Wurtz reacton is
a) C2H5OH(aq)
b) CH3COOH
c) H2O
d) C2H5OC2H5(dry)
Answer:
d) C2H5OC2H5(dry)

Question 27.
Which of the following is less reactive than benzene towards electrophilic substitution reactions?
(a) Nitrobenzene
(b) Aniline
(c) Bromobenzene
(d) Chlorobenzene
Answer:
(a) Nitrobenzene

Question 28.
The compound with the highest boiling point is
a) n-Hexane
b) n – Pentane
c) 2, 2 – dimethyl propane
d) 2 – methyl butane
Answer:
a) n-Hexane

Question 29.
The increasing order of reduction of alkyl halides with zinc and dilute HCl is
a) R – Cl < R – I < R – Br
b) R – Cl < R – Br < R – I
c) R – I < R – Br < R – Cl
d) R – Br < R – I < R – Cl
Answer:
b) R – Cl < R – Br < R – I

Question 30.
The volume of oxygen required for the complete combustion of 4 lit of ethane is
a) 4 lit
b) 8 lit
c) 12 lit
d) 14 lit
Answer:
d) 14 lit

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 31.
The dihedral angle between the hydrogen atoms of 2 methyl groups in the staggered conformation of ethane is
a) 0°
b) 60°
c) 120°
d) 240°
Answer:
b) 60°

Question 32.
The distances between the hydrogen nuclei in staggered and eclipsed form in ethane respectively are
a) 2.55 Å & 2.29 Å
b) 1.54 Å & 1.34 Å
c) 3.5 Å & 2.5 Å
d) 2.29 Å & 2.55 Å
Answer:
a) 2.55 Å & 2.29 Å

Question 33.
Energy barrier between staggered and eclipsed form in ethane is
a) 0.6 kcal / mole
b) 2.9 kcal / mole
c) 12 kcal / mole
d) 14 cal / mole
Answer:
b) 2.9 kcal / mole

Question 34.
IUPAC name of the following compound
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 79
a) 3 – Ethyl, 5 – methyl heptane
b) 5 – Ethyl, 3 – methyl heptane
c) 2 – Ethyl, 5 – methyl heptane
d) 4 – Ethyl, 5 – methyl heptane
Answer:
a) 3 – Ethyl, 5 – methyl heptane

Question 35.
Ethylene is converted to ethane in the presence of Ni at 300°C. In this reaction the hybridization of carbon changes from
a) sp to sp2
b) sp2 to sp3
c) sp3 to sp
d) sp to sp3
Answer:
b) sp2 to sp3

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 36.
In which of the following reactions in the preparation of ethane a new C – C bond is formed
a) Sabatier – Senderen’s reaction
b) Reduction of ethyl iodide
c) Decarboxylation
d) Kolbe’s electrolysis
Answer:
d) Kolbe’s electrolysis

Question 37.
Select the correct statements
a) eclipsed and staggered ethanes give different products on reaction with chlorine in presence of light
b) the conformational isomers can be isolated at room temperature
c) torsional strain is minimum in ethane at dihedral angles 60°, 180° and 300°
d) steric strain is minimum in gauche form of n – butane
Answer:
c) torsional strain is minimum in ethane at dihedral angles 60°, 180° and 300°

Question 38.
The fully eclipsed conformation of n-butane is least stable due to the presence of
a) bond opposition strain only
b) steric strain only
c) bond opposition strain as well as steric strain
d) no strain is present in the molecule
Answer:
c) bond opposition strain as well as steric strain

Question 39.
Vinyl group among the following is
a) (CH3)2 CH –
b) HC ≡ C –
c) H2C = CH – CH2
d) CH2 = CH –
Answer:
d) CH2 = CH –

Question 40.
The alkene that exhibits geometrical isomerism is
a) propene
b) 2 – methyl propene
c) 2 – butene
d) 2 – methyl – 2 – butene
Answer:
c) 2 – butene

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 41.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 80
a) 5 – methylocta – 1, 3, 5, 7 – tetraene
b) 4 – methylocta – 1, 3, 5, 7 – tetraene
c) 4 – butenylocta -1, 3- diene
d) octa – 1, 5 – diene
Answer:
b) 4 – methylocta – 1, 3, 5, 7 – tetraene

Question 42.
CH2 = C (CH2CH2CH3)2
a) 2 – Propyl pent – 1 – ene
b) 2 – Propyl pent – 2- ene
c) 2 – Propyl pent – 3 – ene
d) 3 – Propyl pent – 1- ene
Answer:
a) 2 – Propyl pent – 1 – ene

Question 43.
The number of sigma (σ) and pi (π) bonds in the following structure are
a) σ bonds – 33 π bonds – 2
b) σ bonds – 22 π bonds – 2
c) σ bonds – 42 π bonds – 2
d) σ bonds – 40 π bonds – 3
Answer:
a) σ bonds – 33 π bonds – 2

Question 44.
In dehydrohalogenation, hydrogen and halogen are removed from
a) the same carbon atom
b) from adjacent carbon atoms
c) from isolate carbon atoms
d) from any two carbon atoms
Answer:
b) from adjacent carbon atoms

Question 45.
Ethylene readily undergoes the following type of reaction.
a) Elimination
b) Addition
c) Rearrangement
d) Substitution
Answer:
b) Addition

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 46.
Baeyer’s reagent is
a) Aqueous bromine solution
b) Neutral permanganate solution
c) Acidified permanganate solution
d) Alkaline potassium permanganate solution
Answer:
d) Alkaline potassium permanganate solution

Question 47.
Baeyer’s reagent oxidizes ethylene to
a) Ethylene chlorohydrin
b) Ethyl alcohol
c) CO2 and H2O
d) Ethane – 1, 2 – diol
Answer:
d) Ethane – 1, 2 – diol

Question 48.
On reductive ozonolysis ethylene gives
a) Aldehyde
b) Ketone
c) Carboxylic acid
d) Ether
Answer:
a) Aldehyde

Question 49.
Polythene is obtained by the polymerization of
a) Styrene
b) A mixture of ethylene and styrene
c) Acetylene
d) Ethene
Answer:
c) Acetylene

Question 50.
Polytetrafluoroethylene is commercially known as
a) Teflon
b) Freon
c) Lewisite
d) Westron
Answer:
a) Teflon

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 51.
Polythens is
a) (- H2C = CH2 -)n
b) (- HC = CH – )n
c) (- H3C – CH3 – )n
d) (- H2C – CH2 -)n
Answer:
d) (- H2C – CH2 -)n

Question 52.
When ethanol vapours are passed over alumina heated at 350°C, the main product obtained is
a) C2H6
b) C2H4
c) C2H2
d) C2H5OC2H5
Answer:
b) C2H4

Question 53.
Dehydrohalogenation of ethyl chloride in presence of ale. KOH produces the following
a) HC ≡ CH + KCl + H2O
b) CH4 + KCl + H2O
c) CH2 = CH2 + KCl + H2O
d) C2H4 + HCl
Answer:
c) CH2 = CH2 + KCl + H2O

Question 54.
Ethylene is prepared by
a) Dehalogenation of chloroform
b) Pyrolysis of ethane at 450°C
c) Dehydration of methanol with Al2O3 / 350°C
d) Methyl chloride on reduction
Answer:
b) Pyrolysis of ethane at 450°C

Question 55.
The peroxide effect involves
a) Ionic mechanism
b) Free – radical mechanism
c) Heterolytic fission of double bond
d) Homolytic fission of double bond
Answer:
b) Free – radical mechanism

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 56.
In which of the following will Kharasch effect operate?
a) CH3 – CH2 – CH = CH2 + HCl
b) CH3 – CH2 – CH = CH2 + HBr
c) CH3 – CH = CH – CH3 + HBr
d) CH3 – CH2 – CH = CH2 + HI
Answer:
b) CH3 – CH2 – CH = CH2 + HBr

Question 57.
Anti Markownikoff addition of HBr is not observed in
a) Propene
b) Butene – 1
c) Butene – 2
d) pentene – 2
Answer:
c) Butene – 2

Question 58.
Conditions used for the formation of ethylene glycol from ethylene
a) bromine water
b) cold alkaline KMnO4
c) dil H2SO4, 60°C
d) Ag / 200°C
Answer:
b) cold alkaline KMnO4

Question 59.
The olefin which on ozonolysis gives CH3CH2CHO and CH3CHO is
a) 1 – butene
b) 2 – butene
c) 1 – pentene
d) 2 – pentene
Answer:
d) 2 – pentene

Question 60.
In the following reaction, A and B respectively are,
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 81
a) C2H4 and alcoholic KOH / ∆
b) C2H5Cl and aqueous KOH / ∆
c) C2H5OH and aq KOH / ∆
d) C2H2 and Br2
Answer:
a) C2H4 and alcoholic KOH / ∆

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 61.
The number of possible alkynes with molecular formula C5H8 is
a) 3
b) 4
c) 5
d) 6
Answer:
a) 3

Question 62.
The number of open chain structural isomers possible with molecular formula C5H8 is
a) 7
b) 6
c) 5
d) 4
Answer:
a) 7

Question 63.
Alkynes exhibit.
a) Chain isomerism
b) Position isomerism
c) Functional isomerism
d) All the above
Answer:
d) All the above

Question 64.
The IUPAC name of the compound having the formula CH ≡ C – CH = CH2
a) Butene – 2 – ye
b) But – 2- yne – 3 – ene
c) 3 – butane 1 – ene
d) But – 1 – ene – 3 – yne
Answer:
d) But – 1 – ene – 3 – yne

Question 65.
Acetylene can be obtained by the electrolysis of the following compound
a) Potassium fumerate
b) Potassium succinate
c) Potassium acetate
d) Potassium formate
Answer:
a) Potassium fumerate

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 66.
The gas obtained when ethylene chloride reacts with alcoholic potash and sodamide is
a) C2H4
b) C2H6
c) C2H2
d) C2H5Cl
Answer:
c) C2H2

Question 67.
PVC is the polymer of the following
a) Ethyl chloride
b) Vinyl Chloride
c) Allyl Chloride
d) Ethynyl chloride
Answer:
b) Vinyl Chloride

Question 68.
Which of the following possess acidic hydrogen
a) C2H6
b) C2H4
c) C2H2
d) CH4
Answer:
c) C2H2

Question 69.
Hydrocarbon which gives an oxyacetylene flame
a) ethane
b) ethene
c) ethyne
d) ethanol
Answer:
c) ethyne

Question 70.
The isomer of propyne
a) Allene
b) Propene
c) Cyclopropane
d) Propane
Answer:
a) Allene

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 71.
Bond angle between C – C in alkyne
a) 109°.28′
b) 120°
c) 180°
d) 60°
Answer:
c) 180°

Question 72.
The molecule having linear structure is
a) Methane
b) Ethylene
c) Acetylene
d) Water
Answer:
c) Acetylene

Question 73.
The C – C bond length is shortest in
a) C2H6
b) C3H8
c) C6H6
d) C2H4
Answer:
b) C3H8

Question 74.
The hydrolysis of Mg2C3 produces
a) acetylene
b) propyne
c) butyne
d) ethylene
Answer:
b) propyne

Question 75.
1 – pentyne and 2 – pentyne can be distinguished by
a) Silver mirror test
b) Iodoform test
c) Addition of H2
d) Baeyers test
Answer:
a) Silver mirror test

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 76.
Acetylene on reaction with silver nitrate shows
a) Oxidizing property
b) Reducing property
c) Basic nature
d) Acidic nature
Answer:
d) Acidic nature

Question 77.
The acidic nature of hydrogens in acetylene cannot be explained by the reaction with
a) Sodium metal
b) Ammonical cuprous chloride solution
c) Ammonical silver nitrate solution
d) HCN
Answer:
d) HCN

Question 78.
Westron is the solvent obtained by the reaction of chlorine with
a) Ethylene
b) Ethyne
c) Ethane
d) Methane
Answer:
b) Ethyne

Question 79.
CH ≡ CHSamacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 82 is
a) CH2 = CH – CH = CH2
b) HC ≡ C – CH3
c) CH2 = CH – CH3
d) CH3 – CH2 – CH3
Answer:
b) HC ≡ C – CH3

Question 80.
Hydration of ethyne to ethanal takes place through the formation of
a) CH3CH(OH)2
b) CH2 = CHOH
c) CH2 = CHO
d) CH ≡ C
Answer:
b) CH2 = CHOH

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 81.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 83 B is
a) Acetylene
b) Acetaldehyde
c) Acetone
d) Acetic acid
Answer:
b) Acetaldehyde

Question 82.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 84 is
a) Ethyl chloride
b) 1, 2 dichloro ethene
c) Vinyl chloride
d) Ethylidine chloride
Answer:
c) Vinyl chloride

Question 83.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 85 The Polymer ‘B’ is
a) orlon
b) PVC
c) nylon
d) Teflon
Answer:
b) PVC

Question 84.
Benzene is ______ molecule.
a) Tetrahedral
b) Planar
c) Trigonal
d) Square planar
Answer:
b) Planar

Question 85.
Bond length of C – C in benzene.
a) 1.34 Å
b) 1.39 Å
c) 1.54 Å
d) 1.20 Å
Answer:
c) 1.54 Å

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 86.
The resonance energy Benzene is
a) 36 kcal / mol
b) 85.8 kJ / mole
c) 150.48 kJ / mole
d) Both (a) and (c)
Answer:
d) Both (a) and (c)

Question 87.
In Huckel’s (4n + 2) π rule for aromaticity, ‘n’ represents
a) Number of carbon atoms
b) Number of rings
c) Whole number
d) Fractional number (or) integer (or) zero
Answer:
c) Whole number

Question 88.
Coal tar is obtained as a by product during
a) Destructive distillation of wood
b) Destructive distillation of coal
c) Destructive distillation of bones
d) steam distillation of light oil
Answer:
b) Destructive distillation of coal

Question 89.
Gammaxene is ________ isomer of benzene hexa chloride.
a) α
b) β
c) γ
d) δ
Answer:
c) γ

Question 90.
The empirical formula of benzene and acetylene is/are
a) CH2, CH
b) CH2, CH2
c) CH, CH
d) CH3, CH3
Answer:
c) CH, CH

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 91.
Chemical name of the insecticide gammaxene is
a) DDT
b) Benzene hexa chloride
c) Chloral
d) Hexa chloro ethane
Answer:
b) Benzene hexa chloride

Question 92.
Which is non benzenoidal aromatic compound?
a) Benzene
b) Pyridine
c) Toluene
d) Phenol
Answer:
b) Pyridine

Question 93.
Benzene is purified by
a) distillation
b) fractional distillation
c) Evaporation
d) sublimation
Answer:
b) fractional distillation

Question 94.
Preparation of benzene from phenol is
a) Reduction
b) Oxidation
c) Addition
d) Dehydrogenation
Answer:
a) Reduction

Question 95.
The true statement about benzene is
a) Because of unsaturation benzene easily undergoes addition reactions
b) There are two types C – C bonds in benzene molecule
c) There is a cyclic delocalization of pi – electrons in benzene
d) Mono substitution of benzene gives three isomeric products
Answer:
c) There is a cyclic delocalization of pi – electrons in benzene

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 96.
– COOH group in electrophilic substitution directs the incoming group to
a) o – position
b) p – position
c) m – position
d) o – and p – position
Answer:
c) m – position

Question 97.
Which of the following is not meta directing group?
a) -SO3H
b) – NO2
c) – CN
d) – NH2
Answer:
d) – NH2

Question 98.
Which among the following is very strong o – , p – directing group?
a) – Cl
b) – OR
c) – NH2
d) – NHR
Answer:
d) – NHR

Question 99.
Cyclo butadiene is said to be
a) Aromatic
b) Aliphatic
c) anti aromatic
d) heterocyclic
Answer:
c) anti aromatic

Question 100.
In the reaction Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 86
the attacking species is
a) Cl2
b) Cl+
c) Cl
d) FeCl4
Answer:
b) Cl+

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 101.
The electrophile in Acetylation of Benzene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 87
Answer:
b) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 88

Question 102.
The ratio of sigma and pi bonds in benzene is
a) 4 : 1
b) 2 : 3
c) 6 : 1
d) 1 : 1
Answer:
a) 4 : 1

Question 103.
Benzene reacts with _______ to yield acetophenone.
a) CH3COCl + AlCl3
b) C6H5COCl + AlCl3
c) RCOCl + AlCl3
d) C2H5COCl + AlCl3
Answer:
a) CH3COCl + AlCl3

Question 104.
The order of activites of the various O and P director is
a) – O > – OH > – OCOCH3 > – COCH3
b) – OH > – O > – OCOCH3 > – COCH3
c) – OH > – O > – COCH3 > – OCOCH3
d) -O > – COCH3 > -OCOCH3 > -OH
Answer:
a) – O > – OH > – OCOCH3 > – COCH3

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

II. Very short question and answers (2 Marks):

Question 1.
What are alkanes? Give example.
Answer:
Alkanes are saturated hydrocarbons represented by the general formula CnH2n + 2 where ‘n’ is the number of carbon atoms in the molecule. Methane CH4, is- the first member of alkane family. The successive members are ethane C2H6, propane C3H8,butane C4H10, pentane C5H12 and so on. It is evident that each member differs from its proceeding or succeeding member by a -CH2 group.

Question 2.
Give the IUPAC name of the following alkane.
a) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 89

b) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 90
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 91

b) eer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 92

Question 3.
Draw the structural formula for 4,5 – diethyl – 3, 4, 5 – trimethyl octane.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 93

Question 4.
How is methane prepared from sodium acetate?
Answer:
When a mixture of sodium salt of carboxylic acid and soda lime (sodium hydroxide + calcium oxide) is heated, alkane is formed. The alkane formed has one carbon atom less than carboxylic acid. This process of eliminating carboxylic
group is known as decarboxylation.
CH3COONa + NaOH Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 94 CH4 + Na2CO3
Sodium acetate                            Methane

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 5.
Write a short note on Kolbe’s Electrolytic method.
Answer:
When sodium or potassium salt of carboxylic acid is electrolyzed, a higher alkane is formed. The decarboxylative dimerization of two carboxylic acid occurs. This method is suitable for preparing symmetrical alkanes(R-R).
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 95

Question 6.
How will you convert chloro propane into propane?
Answer:
When chloro propane is reduced with Zn / HCl it gives propane.
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 96

Question 7.
Explain the combustion reaction of alkane with suitable example.
Answer:
A combustion reaction is a chemical reaction between a substances and oxygen with evolution of heat and light (usually as a flame). In the presence of sufficient oxygen, alkanes undergoes combustion when ignited and produces carbondioxide and water.

The combustion reaction is expressed as follows.
Example:
CH4 + 2O2 → CO2 + 2H2O
∆H° = -890.4 KJ

When alkanes burn in insufficient supply of oxygen, they form carbonmonoxide and carbon black.
2CH4 + 3O2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 97 2CO + 4H2O
CH4 + O2 → C + 2H2O

Question 8.
How are the following compound prepared by the Aromatisation method?
(i) Benzene
(ii) Toluene
Answer:
(i) Benzene
n-Hexane passed over Cr2O3 supported on alumina at 873 k gives benzene.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 98

(ii) Toluene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 99Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 100

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 9.
What happens when methane reacts with steam?
Answer:
Methane reacts with steam at 1273K in the presence of Nickel and decomposes to form carbon monoxide and hydrogen gas.
CH4(g) + H2O(g) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 100 CO(g) + 3H2
Production of H2 gas from methane is known as steam reforming process and it is a well-established industrial process for the production of H2 gas from hydrocarbons.

Question 10.
Explain the Isomerisation reaction with suitable example.
Answer:
Isomerisation is a chemical process by which a compound is transformed into any its isomeric forms. Normal alkanes can be converted into branched alkanes in the presence of AlCl3 and HCl at 298 k.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 101
This process is of great industrial importance. The quality of gasoline is improved by isomerizing its components.

Question 11.
What is isomerism? Mention the types of isomerism?
Answer:
The phenomenon in which the same molecular formula may exhibit different structural arrangements is called isomerism.
There are two types of isomerism. namely.

  • Structural isomerism
  • Stereoisomerism

Question 12.
Explain the Geometrical isomerism of alkene with suitable example.
Answer:
It is a type of stereoisomerism and it is also called cis-trans isomerism. Such type of isomerism results due to the restricted rotation of doubly bounded carbon atoms.
If the similar groups lie on the same side, then the geometrical isomers are called Cis-isomers. When the similar groups lie on the opposite side, it is called a Trans isomer.
Example:
The geometrical isomers of 2-Butene is expressed as follows
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 102

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 13.
Write the test for Alkene.
Answer:
Bromine in water is reddish brown colour. When small about of bromine water is added to an alkene, the solution is decolourised as it forms dibromo compound. So, this is the characteristic test for unsaturated compounds.
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 103

Question 14.
Mention uses of Alkenes.
Answer:

  • Alkenes find many diverse applications in industry. They are used as starting materials in the synthesis of alcohols, plastics, liquors, detergents and fuels
  • Ethene is the most important organic feed stock in the polymer industry. E.g,’ PVC, Sarans and polyethylene. These polymer are used in the manufacture of floor tiles, shoe soles, synthetic fibres,raincoats, pipes etc.

Question 15.
How is propyne prepared from propylidine chloride?
Answer:

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 104

Question 16.
How is ethyne prepared from CaC2?
Answer:
Ethyne can be manufactured in large scale
by action of calcium carbide with water.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 105
Calcium carbide needed for this reaction is prepared by heating quick lime and coke in an electric furance at 3273 K
CaO + 3C Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 106 CaC2 + CO

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 17.
Write the reaction of propyne using Br2.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 107
When Br2 in CCl4 (Reddishbrown) is added to an aikyne, the colour is decolounsed. This is the test for unsaturation.

Question 18.
Mention the uses of alkynes.
Answer:

  1. Acetylene is used in oxy acetylene torch used for welding and cutting metals.
  2. It is used for manufacture of PVC, polyvinyl acetate, polyvinyl ether, orlon and neoprene rubbers.

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

III. Short question and answers (3 Marks):

Question 1.
How are the following conversions carried out?
(i) propene → propane
(ii) ethene → ethane
(iii) prop – 1 – yne → propane
Answer:
(i) propene → propane
CH3 – CH = CH2 + H2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 108 CH3 – CH2 – CH3
propene                                     propane

(ii) ethene → ethane
CH2 = CH2 + H2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 109 CH3 – CH3
ethene                               ethane

(iii) prop – 1 – yne → propane
CH3 – C ≡ CH + 2H2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 108 CH3 – CH2 – CH3
prop – 1 – yne                            propane

Question 2.
write the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 110
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 111

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 3.
Write the structural formula for the following compounds.
(i) 3 – Ethyl – 4, 5 – dipropyl octane
(ii) 2, 3 – Dimethyl pentane
(iii) 4 – Ethyl – 2,7 – Dimethyl octane
Answer:
(i) 3 – Ethyl – 4, 5 – dipropyl octane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 112

(ii) 2, 3 – Dimethyl pentane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 113

(iii) 4 – Ethyl – 2,7 – Dimethyl octane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 114

Question 4.
Write a short note on
(i) Wurtz reaction
(ii) Corey – House Mechanism
Answer:
(i) Wurtz reaction:
When a solution of halo alkanes in dry ether is treated with sodium metal, higher alkanes are produced. This reaction is used to prepare higher alkanes with even number of carbon atoms.
Example:
CH3 – Br + 2Na + Br – CH3 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 115 CH3 – CH3 + NaBr
methyl bromide                                        ethane

(ii)Corey-House Mechanism:
An alkyl halide and lithium di alkyl cuprate are reacted to give higher alkane.
Example:
CH3CH2Br + (CH3)2LiCu → CH3CH2CH3 + CH3Cu + LiBr
Ethyl bromide

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 5.
How is methane prepared from Grignard reagent?
Answer:
Methyl chloride reacts with magnesium in presence of dry ether gives methyl magnesium chloride. Then methyl magnesium chloride reacts with water to give methane.
CH3 – Cl + Mg Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 116 CH3MgCl
chloromethane                 methyl magnesium chloride

CH3MgCl + H2O → CH4 + Mg(OH)Cl
methane

Question 6.
What is meant by pyrolysis? Explain the pyrolysis reaction of Ethane and propane.
Answer:
Pyrolysis is defined as the thermal decomposition of organic compound into smaller fragments in the absence of air through the application of heat. ‘Pyro’ means ‘fire’ and ‘lysis’ means ‘separating’. Pyrolysis of alkanes also named as cracking.

In the absence of air, when alkane vapours are passed through red-hot metal it breaks down into simpler hydrocarbons.

1) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 204

2) 2CH3 – CH3 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 117  CH2 = CH2 + 2CH4
Ethane                                 Ethylene

Question 7.
Mention the uses of alkanes.
Answer:
The exothermic nature of alkane combustion reaction explains the extensive use of alkanes as fuels. Methane present in natural gas is used in home heating. Mixture of propane and butane are known as LPG gas which is used for domestic cooking purpose. GASOLINE is a complex mixture of many hydrocarbons used as a fuel for internal combustion engines.

Carbon black is used in the manufacture of ink, printer ink and black pigments. It is also used s fillers.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 118

Question 8.
Write all possible structural isomers with the molecular formula C4H10 and name them.
Answer:
(i) CH3 – CH = CH – CH3 2 – butene
(ii) CH2 = CH – CH2 – CH3 1 – butene
(iii) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 119 2 – mehyl – 1 – propene
structures (i) 8s (ii) are position isomers, structures (i) & (iii), (ii) 8s (iii) are chain isomers.

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 9.
How is ethene prepared by Kolbe’s electrolytic method?
Answer:
When an aqueous solution of potassium succinate is electrolyzed between two platinum electrodes, ethene is produced at the anode.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 120

Question 10.
How are the following compounds prepared by ozonolysis method?
(i) Formaldehyde
(ii) Acetaldehyde
Answer:
(i) Formaldehyde:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 121

(ii) Acetaldehyde:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 122

Question 11.
How ozone reacts with 2 – methyl propene?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 123
When 2 – methyl propene reacts with ozone to give ozonoid. Ozonide is treated with Zn/H2O gives acetone.

Question 12.
How is acetylene prepared from ethylene?
Answer:
This process involves two steps:
(i) Halogenation of alkenes to form vicinal dihalides
(ii) Dehalogenation of vicinal dihalides to form alkynes.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 124

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 13.
How is acetylene prepared by Kolbe’s electrolytic method?
Answer:
Electrolysis of sodium or potassium salt of maleic or fumaric acid yields alkynes.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 125

Question 14.
Write Ozonolysis reaction of Propyne?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 126

Question 15.
How is BHC prepared? Give its uses.
Answer:
Chlorination of Benzene:
Benzene reacts with three molecules of Cl2 in the presence of sun light or UV light to yield Benzene Hexa Chloride (BHC) C6H6Cl6. This is known as gammaxane or Lindane which is a powerful insecticide.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 127

Question 16.
How propane is prepared form 1, 2 – dichloro propane?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 128

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 17.
Explain the polymerization reaction of alkenes.
Answer:
A polymer is a large molecule formed by the combination of larger number of small molecules. The process is known as polymensation. Alkenes undergo polymerisation at high temperature and pressure, in the presence of a catalyst.
Example:
red hot
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 129

Question 18.
Write a note on acidic nature of Alkynes.
Answer:
An alkyne shows acidic nature only if it contains terminal hydrogen. This can be explained by considering sp hybrid orbitals of carbon atom in alkynes. The percentage of S-character of sp hybrid orbital (50%) is more than sp2 hybrid orbital of alkene (33%) and sp3 hybrid orbital of alkane (25%). Because of this, Carbon becomes more electronegative facilitating donation of H+ ions to bases. So hydrogen attached to triply bonded carbon atoms is acidic but not the others.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 130

Question 19.
Write a short note on
(i) Wurtz – Fittig reactions
(ii) Friedel Craft’s reaction
Answer:
(i) Wurtz – Fittig reactions:
When a solution of bromo benzene and iodo methane in dry ether is treated with metallic sodium, toluene is formed.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 131

(ii) Friedel Craft’s reaction:
When benzene is treated with methyl chloride in the presence of anhydrous aluminium chloride, toluene is formed.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 132

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

IV. Long Question and answers (5 Marks):

Question 1.
How to draw structural formula for given IUPAC name with suitable example.
Answer:
After you learn the rules for naming alkanes, it is relatively easy to reverse the procedure and translate the name of an alkane into a structural formula. The example below show how this is done.
Let us draw the structural formula for
3 – ethyl – 2, 3 – dimethyl pentane
Step 1:
The parent hydrocarbon is pentane. Draw the chain of five carbon atoms and number it. .
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 133

Step 2:
Complete the carbon skeleton by attaching the alkyl group as they are specified in the name. An ethyl group is attached to carbon 3 and two methyl groups are attached to carbon 2 and 3.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 134

Step 3:
Add hydrogen atoms to the carbon skeleton so that each carbon atoms has four bonds.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 135

Question 2.
Explain the conformations of ethane.
Answer:
The two tetrahedral methyl groups can rotate about the carbon – carbon bond axis yielding several arrangements called conformers. The extreme conformations are staggered and eclipsed conformation. There can be number of other arrangements between staggered and eclipsed forms and their arrangements are known as skew forms.
Eclipsed conformation:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 136

In this conformation, the hydrogen’s of one carbon are directly behind those of the other. The repulsion between the atoms is maximum and it is the least stable conformer.

Staggered conformation:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 137

In this conformation, the hydrogen’s of both the carbon atoms are far apart from each other. The repulsion between the atoms is minimum and it is the most stable conformation.

Skew conformation:

The infinite numbers of possible intermediate conformations between the two extreme conformations are referred as skew conformations. The stabilities of various conformations of ethane are:
Staggered > Skew > Eclipsed
The potential energy difference between the staggered and eclipsed conformation of ethane is around 12.5 kJmol-1. The various conformations can be represented by new man projection formula.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 138

Eclipsed Skew Staggered newmann projection formula for Ethane.

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 3.
Explain the steps involved in the mechanism of halogenations of alkane.
Halogenation:
A Halogenation reaction is the chemical reaction between an alkane and halogen in which one or more hydrogen atoms are substituted by the halogens.

Chlorination and Bromination are two widely used halogenation reactions. Fluorination is too quick and iodination too slow. Methane reacts with chlorine in the presence of light or when heated as follows.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 139

Mechanism:
The reaction proceeds through the free radical chain mechanism. This mechanism is characterized by three steps initiation, propagation and termination.

(i) Chain Initiation:
The chain is initiated by UV light leading to homolytic fission of chlorine molecules into free radicals (chlorine atoms).
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 140

Here we choose Cl – Cl bond for fission because C – C & C – H bonds are stronger than Cl – Cl.

ii) Propagation:
It proceeds as follows
a) Chlorine free radial attacks the methane molecule and breaks the C – H bond resulting in the generation of methyl free radical
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 141

b) The methyl free radical thus obtained attacks the second molecule of chlorine to give chloromethane (CH3Cl) and a chlorine free radical as follows.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 142

c) This chlorine free radical then cycles back to step a) and both step a) and b) are repeated many times and thus chain of reaction is set up.

iii) Chain termination:
After sometimes, the reactions stops due to consumption of reactant and the chain is terminated by the combination of free radicals.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 143

Question 4.
Write the IUPAC name of the following compounds.
1) CH3 – CH = CH2
2) CH3 – CH = CH – CH3
3) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 144
4) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 145
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 146

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 5.
Write the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 147
Answer:
1) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 148

2) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 149

3) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 150

4) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 151

Question 6.
Draw the structures for the following alkenes.
(i) 6 – Bromo – 2, 3 – dimethyl – 2 – hexene
(ii) 5 – Bromo – 4 – chloro – 1 -heptene
(iii) 2, 5 – dimethyl – 4 – octene
(iv) 4 – Methyl – 2 – pentene
Answer:
(i) 6- Bromo – 2, 3-dimethyl – 2 -hexene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 152

(ii) 5 – Bromo – 4 – chloro – 1 -heptene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 153

(iii) 2, 5 – dimethyl – 4 – octene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 154

(iv) 4 – Methyl – 2 – pentene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 155

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 7.
Draw the structure and write down the IUPAC name for the isomerism exhibited by the molecular formulae:
(i) C5H12 – Pentene (3 isomers)
(ii) C6H14 – Hexene (5 isomers)
(iii) C5H12 – Pentene (3 isomers)
Answer:
(i) C5H12 – Pentene
a) CH3 – CH2 – CH2 – CH = CH2
1 – Pentene

b) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 156
2 – methyl – 2 – butene

c) CH3CH = CH – CH2CH3
2 – pentene

(ii) C6H14 – Hexene
a) CH3 – CH2 – CH2 – CH2 – CH = CH2
1 – hexane

b) CH3 – CH2 – CH2 – CH = CH – CH3
2 – hexane

c) CH3 – CH2 – CH = CH – CH3 – CH2
3 – hexane

d) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 157

e) Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 158

Question 8.
How are the following conversions carried out?
(i) ethanol → ethene
(ii) 2- butyne cis → 2 – butene and trans – 2 – butane
(iii) 1 – bromopropane → prop – 1 – ene
(iv) 1 – 2 – dibromoethane – ethene
Answer:
1) ethanol → ethene
C2H5OH Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 159 H2C = H2C
ethanol ethane

(ii) 2 – butyne → cis – 2 – butene
H3C – C ≡ C – CH3 + H2 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 160

2 – butane → trans – 2 – butane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 161

(iii) 1 – bromopropane → prop – 1 – ene
H3C – CH2 – CH2 – Br Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 162 H3C – CH = CH2 + KBr + H2O
1 – bromopropane → prop – 1 – ene

(iv) 1 – 2 – dibromoethane – ethene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 163
1 – 2 – dibromoethane – ethene

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 9.
Write the mechanism involved in the addition of HBr to alkene in the presence of organic peroxide.
Answer:
Anti-Markovnikoff’s Rule (Or) Peroxide Effect (Or) KharaschAddition:
The addition of HBr to an alkene in the presence of organic peroxide takes place in the direction opposite to that predicted by Markovnikoff’s rule.
CH3 – CH = CH2 + HBr Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 164 CH3 – CH2 – CH3 – CH2 – CH2 – Br
prop – 1 – ene                                                          1 – bromopropane

Mechanism:
The reaction proceeds via free radical mechanism.

Step 1:
The week O – O single bond linkages of peroxides undergoes homolytic cleavage to generate free radical.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 165

Step 2:
The radicals abstract H from HBr thus generating bromine radical.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 166

Step 3:
The Bromine radical adds to the double bond in the way to form more stable alkyl free radical.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 167

Step 4:
Addition of HBr to secondary free radical
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 168

Step 5:
Addition of HBr to primary free radical
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 169

Question 10.
How are the following conversions carried out?
Answer:
(i) Propene → 2 – Propanol
(ii) ethene → ethylene glycol
(iii) 2 – butene → Acetic acid
(iv) 2 – methyl – 1 – propene → Acetone
Answer:
(i) Propene → 2 – Propanol
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 170

(ii) ethene → ethylene glycol
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 171

(iii) 2 – butene → Acetic acid
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 172

(iv) 2 – methyl – 1 – propene → Acetone
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 173 Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 174

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 11.
An organic compound (A) C2H4 decolourises bromine water. (A) on reaction with chlorine gives (B) A reacts with HBr to give (C). Identity (A), (B), (C). Explain th. reactions.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 175
Result:
(A) – ethylene
(B) – 1, 2 – dichloroethane
(C) – ethyl bromide

Question 12.
How are the following conversions carried out?
(i) 2 – butyne → 2, 2 – dichlorobutane
(ii) 1 – butyne → 2, 2 – dibromobutane
(iii) ethyne → acetaldehyde
(iv) Propyne → propan – 2 – one
Answer:
(i) 2 – butyne → 2, 2 – dichlorobutane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 176

(ii) 1 – butyne → 2, 2 – dibromobutane
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 177

(iii) ethyne → acetaldehyde
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 178

(iv) Propyne → propan – 2 – one
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 179

Question 13.
How are the following compounds prepared from ethyne?
(i) Formic acid
(ii) Vinyl ethylene
(iii) Benzene
Answer:
(i) Formic acid
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 180

(ii) Vinyl ethylene
2CH ≡ CH Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 181 CH2 = CH – C ≡ CH (Vinyl ethylene)

(iii) Benzene
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 182

Question 14.
Explain the Huckel’s rule of Aromaticity.
Answer:
Huckel proposed that aromaticity is a function of electronic structure. A compound may be aromatic, if it obeys the following rules (Huckel’s rule)
i) The molecule must be co-planar
ii) Complete delocalization of n electron in the ring
iii) Presence of (4n+2) n electrons in the ring where n is an integer (n = 0,1,2….)
This is known as Huckle’s rule.
Some of the examples which obey Huckel rule.

1. The benzene ring is planar with delocalized π electrons (6)
It obeys Huckel’s rule because 4n + 2 = (4 × 1) + 2 = 6π electrons.
Hence, it is an aromatic compound
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 183

2. It has planar ring structure with delocalized π electrons (10).
Applying Huckle’s rule (4 × 2) + 2 = 10π electrons.
Hence it is an aromatic compound
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 184
Naphthalene

3. It has a planar ring structure with delocalized π electrons(14).
Applying Huckle’s rule (4 × 3) + 2 = 14π electrons.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 185

4. It has planar ring structure but the π electrons are not delocalized.
Applying Huckle’s rule (4 × 1) + 2 = 6π electrons.
But it has only 4electrons. So it is not an aromatic compound
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 186

5. It has a tub shaped ring structure. It does not obey Huckel’s rule.
Applying Huckle’s rule 4n + 2 = 8π electrons if n = 2
So it is not an aromatic compound.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 187

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 15.
Explain the Kekule’s structure of benzene.
Answer:
In 1865, August Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds.
There were two objections:
(i) Benzene forms only one orthodisubstituted products whereas the Kekule’s structure predicts two o-di substituted products as shown below.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 188
Presence of double bond between the substituents

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 189
Presence of single bond between the substituents

(ii) Kekule’s structure failed to explain why benzene with three double bonds did not give addition reactions like other alkenes.To overcome this objection, Kekule suggested that benzene was mixture of two forms (1 and 2) in rapid equilibrium.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 190

Question 16.
Explain the resonance in benzene.
Answer:
The phenomenon in which two or more structures can be written for a substance which has identical portions of atoms in called resonance. The actual structure of the molecule is said to be resonance hybrid of various possible alter nature structures. In benzene, Kekule’s structures I & II represented the resonance structure, and structure III is the resonance hybrid of structure I & II.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 191

The structures 1 and 2 exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 17.
Write the industrial preparation of benzene from coal tar.
Answer:
Coal tar is a viscous liquid obtained by the pyrolysis of coal. During fractional distillation, coal tar is heated and distills away its volatile compounds namely benzene, toluene, xylene in the temperature range of 350 to 443 K. These vapours are collected at the upper part of the fractionating column
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 192

Question 18.
How is benzene prepared from
(i) Acetylene
(ii) Decarboxylation
(iii) Phenol
Answer:
(i) From acetylene
Acetylene on passing through a red -hot tube trimerises to give benzene. We havealready studied this concept in polymerization of alkynes.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 193

(ii) Decarboxylation of aromatic acid.
When sodium benzonate is heated with sodalime, benzene vapours distil over.
C6H5COONa + NaOH Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 194 C6H6 + Na2CO3

(iii) Preparation of benzene from Phenol
When phenol vapours are passed over zinc dust, then it is reduced to benzene.
C6H6OH + Zn Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 195 C6H6 + ZnO

Question 19.
Explain Electrophilic substitution reaction of benzene.
Answer:
(i) nitration:
When benzene is heated at 330K with a nitrating mixture (Con. HNO3 + Con. H2SO4), nitro benzene is formed by replacing of hydrogen atom by nitronium ion NO2+ (electrophile).
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 196
Concentrated H2SO4 is added to produce nitronium ion NO2+.

(ii) Halogenation:
Benzene reacts with halogens (X2 = Cl2, Br2,) in the presence of Lewis acid such as FeCl3, FeBr3 orAlCl3 and give corresponding halo benzene.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 197

(iii) Sulphonation:
Benzene reacts with fuming sulphuric acid (Con. H2SO4 + SO3) and gives benzene sulphonic acid. The electrophile SO3 is a molecule.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 198

(iv) Friedel craft’s alkylation: (methylation)
When benzene is treated with a alkyl halide in the presence of only AlCl3, alkyl benzene is formed.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 199

(v) Friedel craft’s acylatlon: (Acetylation)
When benzene is treated with acetyichioride in the presence of AlCl3, acyl benzene is formed.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 200

Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons

Question 20.
How is benzene converted into
a) Cyclo hezane
b) maleic anhydride
c) 1, 4 cyclo hexadiene
Answer:
a) Cyclo hexane:
Benzene reacts with hydrogen in the presence of Platinum or Palladium to yield Cyclohexane. This is known as hydrogenation.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 201

b) maleic anhydride:
Although benzene is very stable to strong oxidizing agents, it quickly undergoes vapour phase oxidation by passing its vapour mixed with oxygen over V2O5 at 773k. The ring breaks to give maleic anhydride.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 202

c) 1, 4 cyclo hexadiene:
Benzene can be reduced to non- conjugated dienes 1, 4 – cyclohexadiene by treatment with Na or Li in a mixture of liquid ammonia and alcohol. It is the convenient method to prepare cyclicdienes.
Samacheer Kalvi 11th Chemistry Guide Chapter 13 Hydrocarbons 203

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 8 Excretion Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

11th Bio Zoology Guide Excretion Text Book Back Questions and Answers

Part – I

Question 1.
Arrange the following structures in the order that a drop of water entering the nephron would encounter them.
a) Afferent arteriole
b) Bowman’s Capsule
c) Collecting duct
d) Distal tubule
e) Glomerulus
f) Loop of Henle
g) Proximal tubule
h) Renal Pelvis
Answer:
(a), (e), (b), (g), (f), (d), (c), (h)

Question 2.
Name the three filtration barriers that solutes must come across as they move from plasma to the lumen of Bowman’s capsule. What components of the blood are usually excluded by these layers.
Answer:
The three filtration barriers that solutes must come across as they move from plasma to the lumen of Bowman’s capsule are

  1. Glomerular capillary endothelium
  2. Basal lamina or basement membrane
  3. Epithelium of Bowman’s capsule. Blood corpuscles and plasma protein are excluded by these layers.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 3.
What forces promote glomerular filtration? What forces oppose them? What is meant by net filtration pressure?
Answer:

  • Glomerulus hydrostatic pressure
  • Glomerulus pressure
  • Opposing pressure: Colloidal osmotic pressure, Capsular hydrostatic pressure
  • Net filtration pressure = Glomerular hydrostatic pressure – (Colloidal osmotic pressure + capsular hydrostatic pressure.

Question 4.
Identify the following structures and explain their significance in renal physiology?
Answer:
1. Juxtaglomerular apparatus:
Juxtaglomerular apparatus is a specialized tissue in the afferent arteriole of the nephron that consists of macula densa and granular cells. The macula densa cells sense distal tubular flow and affect afferent arteriole diameter. The granular cells secrete an enzyme called renin. It plays an important role in reabsorption of water, Na+ and excretion of K+.

2. Podocytes:
The visceral layer of the Bowman’s capsule is made up of epithelial cells called podocytes. The podocytes end in foot processes which cling to the basement membrane of the glomerulus. The openings between the foot processes are called filtration slits. It is important for glomerular filtration.

3. Sphincters in the bladders:
Sphincter muscles in the bladder controls the flow of urine from the bladder. When urinary bladder is filled with urine, it stretches and stimulates the central nervous system through the sensory neurons of the parasympathetic nervous system and brings about contraction of the bladder.

Simultaneously, somatic motor neurons induce the sphincters to close. Smooth muscles contracts resulting in the opening of the internal sphincters passively and relaxing the external sphincter. When the stimulatory and inhibitory controls exceed the threshold, the sphincter opens and the urine is expelled out.

4. Renal cortex:
The outer portion of the kidney is the renal cortex. It contains renal corpuscles and the proximal and distal tubules. It is thin and fibrous.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 5.
In which segment of the nephron most of the re-absorption of substances takes place?
Answer:
In the proximal convoluted tubule.

Question 6.
When a molecule or ion is reabsorbed from the lumen of the nephron, Where does it go?
Answer:
When a molecule or ion is reabsorbed from the lumen of the nephron, it goes to the bloodstream through an efferent arteriole which carries blood away from the glomerulus. If a solute is filtered and not reabsorbed from the tubule, it goes along with urine.

Question 7.
Which segment is the site of secretion and regulated reabsorption of ions and pH homeostasis?
Answer:

  • 1. Distal tubule; 2. Collecting duct
  • In the renal tubules H+ and NH4 are secreted and liberated into the tubules and excreted through
    urine thus maintaining acid base balance.
  • For each H+ ions from the liberated filtrate one Na+ is reabsorbed in the tubules.
  • The secreted HCO3, PO3 and NH3 combines to form carbonic acid and phosphoric acids.
  • Thus the H+ are maintained and their reabsorption is prevented.

Question 8.
What solute is normally present in the body to estimate GFR in humans?
Answer:

  • The glomerulus filtrate consists of water glucose amino acids creatinine protein salts and urea.
  • These solutes decide the glomerular filtration rate.

Question 9.
Which part of the autonomic nervous system is involved in the micturition process?
Answer:
Parasympathetic nervous system.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 10.
If the afferent arteriole of the nephron constricts, what happens to the GFR in that nephron? If the efferent arteriole constricts what happens to the GFR in that nephron? Assume that no autoregulation takes place.
Answer:

  • Blood enters the glomerulus faster with greater force through the afferent arteriole and leaves the glomerulus through the efferent arterioles much slower.
  • Because afferent arteriole is wider than efferent arteriole.
  • The constriction of this does not affect the rate of filtration.

Question 11.
The concentration of urine depends upon which part of the nephron.
Answer:
a) Bowman’s capsule
b) Length of Henle’s loop
c) PCT
d) network of capillaries arising from the glomerulus.
Answer:
b) Length of Henle’s loop

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 12.
If Henle’s loop were absent from mammalian nephron, which one of the following is to be expected?
a) There will be no urine formation
b) There will be hardly any change in the quality and quantity of urine formed.
c) The urine will be more concentrated
d) The urine will be more dilute
Answer:
d) The urine will be more dilute

Question 13.
What will happen if the stretch receptors of the urinary bladder wall are totally removed?
a) Micturition will continue
b) Urine will continue to collect normally in the bladder
c) There will be micturition
d) Urine will not collect in the bladder
Answer:
a) Micturition will continue

Question 14.
The end product of the Ornithine cycle is
a) Carbon dioxide
b) Uric acid
c) Urea
d) Ammonia
Answer:
c) Urea

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 15.
Identify the wrong match.

a) Bowman’s capsule Glomerular filtration
b) DCT Absorption of glucose
c) Henle’s loop Concentration of urine
d) PCT Absorption of Na+ and K+ ions

Answer:
b) DCT – Absorption of glucose

Question 16.
Podocytes are the cells present on the
a) Outer wall of Bowman’s capsule
b) Inner wall of Bowman’s capsule
c) Neck of the nephron
d) Wall glomerular capillaries
Answer:
b) Inner wall of Bowman’s capsule

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 17.
Glomerular filtrate contains
a) Blood without blood cells and proteins
b) Plasma without sugar
c) Blood with proteins but without cells
d) Blood without urea
Answer:
c) Blood with proteins but without cells

Question 18.
Kidney stones are produced due to the deposition of uric acid and Silicates
b) Minerals
c) Calcium Carbonate
d) Calcium oxalate
Answer:
d) Calcium oxalate

Question 19.
Animal requiring the minimum amount of water to produce urine is
a) Ureotelic
b) Ammonotelic
c) Uricotelic
d) Chemotelic
Answer:
c) Uricotelic

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 20.
Aldosterone acts at the distal convoluted tubule and collecting duct resulting in the absorption of water through
a) Aquaporins
b) Spectrins
c) GLUT
d) Chloride channels
Answer:
a) Aquaporins

Question 21.
The hormone which helps in the reabsorption of water in kidney tubules is
a) Cholecystokinin
b) Angiotensin Il
c) Antidiuretic hormone
d) Pancreozymin
Answer:
c) Antidiuretic hormone

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 22.
Malpighian tubules remove excretory products from
a) Mouth
b) Oesophagus
c) Haemolymph
d) Alimentary canal
Answer:
c) Haemolymph

Question 23.
Identify the biological term excretion, glomerulus, urinary bladder, glomerular filtrate, ureters, urine, Bowman’s capsule, urinary system, reabsorption, micturition, osmosis, proteins.
Answer:

  • A liquid which gathers in the bladder – Urine
  • Produced when blood is filtered in a Bowman’s capsule – Glomerular filtrate
  • The temporary storage of urine – Urinary bladder
  • A ball of intertwined capillaries – Glomerulus
  • Removal of unwanted substances from the body – Excretion
  • Each contains a glomerulus – Bowman’s Capsule
  • Carry urine from the kidneys to the bladder – Ureter
  • The scientific term for urination – Micturition
  • Regulation of water and dissolved substances in the blood and tissue fluid – Osmoregulation
  • Consists of the kidneys ureters and bladder Excretory system
  • Removal of useful substances from glomerular filtrate reabsorption
  • What solute does blood contain that is not present in the glomerular filtrate? – Plasma Protein

Question 24.
With regards to toxicity and the need for dilution in water how different are ureotelic and ureocotelic excretions? Give examples of animals that use these types of excretions?
Answer:

  • The type of nitrogenous end product (Urea or Uric acid or ammonia) of an animal excretion depends upon the habitat of the animal.
  • For the Excretion of Ammonia, more water is needed.
  • Animals that excrete most of their nitrogen in the form of ammonia are called ammonites.
  • Fishes Amphibians. Uric acid can be eliminated with a minimum loss of water and is called Uricoteles.
  • Reptiles, birds insects, landrails. Mammals and terrestrial amphibians excrete urea and are called ureoteles.
    Nitrogenous

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 1

Question 25.
Differentiate protonephridia from metanephridia.
Answer:

Protonephridia Metanephridia
1. Proto – first it is a network of dead-end tubules lacking internal opening. (E.g) Platyhelminthes Meta – after They have a tubular internal opening called nephrostome. (E.g) Earthworm
2. Flame cells are excretory structures. The excretory products are filters and selectively reabsorbed.
3. Excretory product excretes through nephridiopore The excretory product excretes through the nephridiopore.
4. Excretory structure are usually osmo regulators Excretory structures are osmo regulators and helps in excretion.

Question 26.
What is the nitrogenous waste produced by amphibian larvae and by adult animals?
Answer:
The adult amphibian’s excretory product is urea. The larval form of amphibian’s excretory product is ammonia.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 27.
How is urea formed in the human body?
Answer:
Urine formation involves three main processes namely Glomerular Alteration, tubular reabsorption and tubular secretion.

I. Glomerular filtration:

  • Blood enters the kidney from the renal artery into the glomerulus.
  • Blood is composed of water colloidal proteins, sugars and nitrogenous end product.
  • The filteration is started in the glomerulus. The fluid that leaves the glomerular capillaries and enters the Bowman’s capsule is called the glomerular filtrate. It formed 170-180 litre within 24 hours.
  • Glomerular membrane has large surface area. Blood enters the glomerulus faster with greater force through the afferent arteriole and leaves the glomerulus through the efferent arterioles much slower.
  • This is because afferent arteriole is wider than efferent arteriole and the glomerular hydrostatic pressure is around 55 mm Hg.
  • Molecules larger than 5mm are barred from entering the tubule.

The two opposing forces are contributed by the plasma proteins in the capillaries.

  1. Colloidal osmotic pressure – 30 mm Hg
  2. Hydrostatic pressure -15 mm Hg

Both pressures combine. 30 mm Hg + 15mm Hg = 45mm Hg

  • The net filtration pressure of 10mm Hg is responsible for the renal filtration.
  • Net filtration pressure – Glomerular
  • Hydrostatic Pressure – Colloidal Osmotic pressure + Capsular hydrostatic pressure.

Net filteration pressure = 55 mm Hg – (30mm Hg +15mm Hg) = 10mm Hg

Substance Concentration in blood Plasm / g dm-3 Concentration in glomerular filtrate / g dm-3
Water 900 900
Proteins 80. 0.05
Aminoacids 0.5 0.5
Glucose 1.0 1.0
Substance Concentration in blood PIasm /gdm-3 Concentration glomerular filtrate/gdm-3
Urea 0.3 0.3
Uric Acid 0.04 0.04
Creatinine 0.01 0.01
Inorganic ions (mainly (Na+, K+and Cl) 7.2

 

7.2

II. Tubular reabsorption:

  • The volume of filtrate formed per day is around 170 – 180l.
  • Urine released is around 1.51 per day. 99% of the glomerular filtrate is reabsorbed by the renal tubules.

Reabsorption in proximal convoluted tubule:

  • Glucose lactate amino acids Na in the filtrate are reabsorbed in the PCT.
  • Sodium is reabsorbed by active transport through a sodium-potassium pump in the PCT.
  • Small amounts of urea and uric acid are reabsorbed.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 2

Reabsorption in Henle Loop:

  • Descending limb of Henle’s loop is permeable to water due to the presence of aqua porlns but not permeable to salts.
  • Water is lost in the descending limb. Hence Na and Cl get concentrated in the filtrate.
  • Ascending limb of Henle’s loop is impermeable to water but permeable to Na+, Cl andK+.

Distal Convoluted Tubule:

  • It recovers water and secretes potassium into tubule.
  • Na+, Cl-, and water remain in the DCT.
  • Reabsorption of HCO3 takes place to regulate the blood pH.
  • Collecting the tubule is permeable to water potassium ions are actively transported into the tubule and Na+ to produce concentrated urine.

Tubular Secretion:

  • Once it enters the collecting duct water is absorbed and concentrated hypertonic urine is formed.
  • For every H secreted into the tubular filtrate, a Na+ is absorbed by the tubular cell.
  • The H+ secreted combines with HCO+3, HPO3, and NH+ and gets fixed as H2CO3, H2PO3, and NH4+.
  • Since H+ gets fixed in the fluid reabsorption of H+is prevented.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 28.
Differentiate cortical from medullary nephrons.
Answer:
In the renal tubules, proximal convoluted tubule and distal convoluted tubule are situated in the cortical region of the kidney and the loop of Henle is in the medullary region.

Cortical Nephron Medullary Nephron
1. The loop of Henle is too short and extends only very little into the medulla. Nephrons have a long loop of Henle that run deep in to the medulla.
2. There is no vasa recta. Vasa recta is present.

Question 29.
What vessels carry blood to the kidneys? Is this blood arterial or venous?
Answer:
Renal artery right and left artery. The blood is arterial.

Question 30.
Which vessels drain filtered blood from the kidneys?
Answer:
Renal vein. The filtered blood is taken from the kidney to the inferior vena cava through the renal vein.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 31.
What is tubular secretion? Name the substances secreted through the renal tubules?
Answer:
The movement of substances such as H+, K+, NH4+, creatinine and organic acids from the peritubular capillaries into the tubular fluid, the filtrate is called Tubular secretion.

Question 32.
How are the kidneys involved in controlling blood volume? How is the volume of blood in the body related to arterial pressure?
Answer:

  • When the volume of blood decreases the flow pressure, decreases.
  • This can be sensed by the hypothalamus and osmoreceptors are stimulated and the antidiuretic hormone is secreted from the neurohypophysis.
  • The aquaporins in the proximal convoluted tubules and collecting tubule reabsorb water. Hence the blood volume increases and the blood pressure increases.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 33.
Name the three main hormones that are involved in the regulation of the renal function?
Answer:

  • Renin
  • Angio Tensin I
  • Angiotensin II

Question 34.
What is the function of the antidiuretic hormone? Where is it produced and What stimuli increases or decreases its secretion?
Answer:

  • ADH which is also called vasopressin or Antidiuretic hormone is secreted from the neurohypophysis.
  • Fluid loss or if blood pressure increases the osmoreceptors of the hypothalamus is stimulated and hence neurohypophysis is stimulated and secretes ADH.
  • When the fluid level and pressure are maintained due to the negative feedback mechanism ADH secretion stops.

Question 35.
What is the effect of aldosterone on kidneys and where is it produced?
Answer:

  • Due to the stimulation of Angiotensin II the adrenal cortex secretes aldosterone. That causes reabsorption of Na+, K+ excretion, and absorption of water from distal convoluted tubule and collecting tubule.
  • This mechanism is known as Renin – Angiotensin Aldosterone System.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 36.
Explain the heart’s role in secreting a hormone that regulates renal function? What hormone is this?
Answer:

  • Excessive stretch of cardiac atrial cells cause an increase in blood flow to the atria of the heart and release Atrial Natriuretic Peptide or factor (ANF) travels to the kidney where H increases Na+ excretion and increases the blood flow to the glomerulus.
  • Acting on the afferent glomerular arterioles as a vasodilator or an efferent arterioles as a vasoconstrictor.
  • The first identical natri uretic hormone is atrial natriuretic horome of heart.

Part – II.

11th Bio Zoology Guide Excretion Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
The elimination of ……………………….. requires a large amount of water.
(a) Urea
(b) Uric acid
(c) Ammonia
(d) creatinine
Answer:
(c) ammonia

Question 2.
Whether the following statements are true or false find the correct sequence. Find the correct sequence.
I) According to the environmental changes organism change their osmotic concentration and are called osmo confirmers.
II) Osmo regulators maintain their internal osmotic concentration irrespective of their external osmotic environment.
III) The Euryhaline animals are able to tolerate only narrow fluctuations in the salt concentration?
IV) The steno haline animals can tolerate wide fluctuations in the salt concentration.
Series:
a) I – False, II – True, III – False, IV – True
b) I – True, II – True, III – False, IV- False
c) I – False, II – True, III – False, IV – True
d) I – False, II – False, III – True, IV – True
Answer:
b) I – True, II – True, III – False, IV- False

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 3.
Solenocytes are the specialized cells for excretion in
(a) Flatworms
(b) Molluscs
(c) Insects
(d) Amphioxus
Answer:
(d) amphioxus

Question 4.
Find out the odd one out.
a) Ammonoteles – Aquatic Amphibians
b) Urico teles – Birds
c) Ureoteles – Land amphibians
d) Ureoteles – Earthworm (When it is in water)
Answer:
d) Ureoteles – Earthworm (When it is in water)

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 5.
…………………… have antennal glands or green glands which perform an excretory function.
(a) Insects
(b) Annelids
(c) Crustaceans
(d) Flatworms
Answer:
(c) Crustaceans

Question 6.
What type of Urine is formed in organism with long Henle’s loop?
a) Isotonic
b) less concentrated
c) Concentrated
d) None of the above
Answer:
c) Concentrated

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 7.
In the marine organism the kidney with glomerulus produce ………………. type of concentrated urine.
a) Concentrated than body fluid
b) Equal to the concentration of body
c) Less than the concentration of body fluids
d) None of the above
Answer:
b) Equal to the concentration of body

Question 8.
Aglomerlar kidneys of marine fishes produce little urine that is ………………………. to the body fluid.
(a) Hypotonic
(b) Hyperosmotic
(c) Isoosmotic
(d) None of the above
Answer:
(c) Isoosmotic

Question 9.
What is the outer covering of the kidney?
a) Renal fascia perirenal fat capsule pleura
b) Renal fascia perirenal fat capsule peri cardial membrane
c) Renal fascia, Peri renal fat capsule meninges
d) Renal fascia perirenal fat capsule fibrous capsule
Answer:
d) Renal fascia perirenal fat capsule fibrous capsule

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 10.
What is meant by renal corpuscle?
a) Glomerulus and Bowman’s Capsule
b) Glomerulus and Malpighian capsule
c) Glomerulus and nephron
d) Glomerulus and Henle’s loop
Answer:
d) Glomerulus and Henle’s loop

Question 11.
Regarding renal tubules find the correct sequences.
a) Glomerulus Bowman’s capsule Malpighian capsule uriniferous tubules.
b) Proximal convoluted tubules thick descending loop thin ascending limb distal convoluted tubule.
c) Proximal convoluted tubule, Henle’s loop. Distal convoluted tubule.
d) Proximal convoluted tubule, thin descending limb thick ascending limb distal convoluted tubule.
Answer:
d) Proximal convoluted tubule, thin descending limb thick ascending limb distal convoluted tubule.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 12.
Glucose, amino acids, Na+, and water in the filtrate are reabsorbed in the
(a) descending limb of Henle’s loop
(b) ascending limb of Henle’s loop
(c) proximal convoluted tubule
(d) distal convoluted tubule
Answer:
(c) Proximal convoluted tubule

Question 13.
What is the pressure in the afferent arthery of Glomerulus?
a) 55 mm Hg
b) 50 mm Hg
c) 57 mm Hg
d) 58 mm Hg
Answer:
a) 55 mm Hg

Question 14.
What is the amount of filteration of the glomerulus in 24 hours?
a) 200l
b) 180l
c)190l
d) 170l
Answer:
b) 180l

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 15.
The pH value of human urine is ……………
(a) 7.5
(b) 6.0
(c) 4.3
(d) 9.5
Answer:
(a) 6.0

Question 16.
What is the pressure in glomerulus?
a) 50 mm Hg
b) 60 mm Hg
c) 55 mm Hg
d) 62 mm Hg
Answer:
c) 55 mm Hg

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 17.
The glomerular pressure encounter ………….. of colloidal osmotic pressure and …………… of hydrostatic pressure.
a) 30mmHg;15mmHg
b) 15mmHg;30mmHg
c) 40 mm Hg; 30 mm Hg
d) 30mmHg;40mmHg
Answer:
a) 30mmHg;15mmHg

Question 18.
What is the net filteration pressure?
a) 55mmHg – (15mmHg + 30mmHg) = 10mm Hg
b) 55 mm Hg – (30 mm Hg +15 mm Hg) =10 mm Hg
c) 55 mm Hg – (35 mm Hg + 5 mm Hg) = 15 mm Hg
d) 55 mm Hg – (5 mm Hg + 35 mm Hg) = 15 mm Hg
Answer:
b) 55 mm Hg – (30 mm Hg + 15 mm Hg) =10 mm Hg

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 19.
What is the amount of filtrate formed in one minute?
a) 100 ml -125 ml
b) 100 ml -150 ml
c) 120 ml -125 ml
d) 130 ml -140 ml
Answer:
c) 120 ml -125 ml

Question 20.
Through hemodialysis, ……………………….. can be removed from the blood.
(a) Ketone bodies
(b) Glucose
(c) Amino acids
(d) Urea
Answer:
(d) Urea

Question 21.
What is the amount of urine excreted in a day?
a) 1.5l
b) 1l
c) 2l
d) 2.5l
Answer:
a) 1.5l

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 22.
Which of the following hormones are secreted by the kidney?
a) Renin
b) Gastrin
c) Calcitriol
d) secretin
a) (ii) and (iii)
b) (i) and (iii)
c) (ii) and (iv)
d) (i) and (ii)
Answer:
b) (i) and (iii)

Question 23.
Where is aquaporins present?
a) Proximal convoluted tubule
b) Distal convoluted tubule and ascending limb of Henle
c) The descending limb of Henle and distal convoluted tubule
d) All the above
Answer:
b) Distal convoluted tubule and ascending limb of Henle

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 24.
Name the hormone that helpsin the reabsorption of water in distal convoluted tubule.
a) Vasopressin
b) Serotonin
c) Oxytocin
d) All the above
Answer:
a) Vasopressin

Question 25.
Where is Renin synthesized in Kidney?
a) Afferent arteriole
b) Efferent arteriole
c) Vasarecta
d) Collecting duct
Answer:
a) Afferent arteriole

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 26.
Name the hormone that helpsin the conversion of plasma protein angioten- sinogen into angiotensin I
a) Renin
b) Vasopressin
c) ADH
d) STH
Answer:
a) Renin

Question 27.
What is micturition?
a) Excretion of urine from the urinary bladder
b) The formation of urine in the glomerulus
c) Urine formation in distal convoluted tubule
d) the absorption of urine in Bowman’s capsule
Answer:
a) Excretion of urine from the urinary bladder

Question 28.
According to the food that maneats the pH of urine ……………… to ……………. can be altered.
a) pH 4.8-7.5
b) PH 4.9-7.9
c) pH 4.5-8.0
d) pH 4.4 -7.4
Answer:
c) pH 4.5-8.0

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 29.
What is the reason for the yellow colour of the urine?
a) Urochrome
b) Haemoerythrin
c) Haemocyanin
d) None
Answer:
a) Urochrome

Question 30.
What is the normal pH of urine?
a) pH 6.0
b) pH 5.3
c) pH 6.4
d) pH 5.9
Answer:
a) pH 6.0

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 31.
What is the amount of CO2 released from lungs in a day?
a) 20l
b) 19l
c) 18l
d) 119l
Answer:
c) 18l

Question 32.
Confirm
Statement A: The important function of sweat gland is to cool the body.
Statement B: The sweat glands excrete sodi¬um chloride urea and lactic acid.
a) Statement A-True B-True
b) Statement A-True B-False
c) Statement A-FalseB-True
d) Statement A – False B – False
Answer:
a) Statement A-True B-True

Question 33.
Confirmation:
Statement S: When kidney fails suddenly there is more chance of recovery.
Statement T: In the chronic kidney failure there may not be any chance of recovery.
a) Statement S-True T-True
b) Statement S-True T-False
c) Statement S-True T-True
d) Statement S – True T – False
Answer:
c) Statement S-True T-True

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 34.
Confirmation:
Statement A: The kidney infection leads to inflammation of bladder and kidney
Statement B: Urination with pain, urinary urgency bloodytinged urine.
a) Statement A – True Statement B explain the symptom of A.
b) A – True – The statement B does not explains the statement A
c) A and B are false
d) A False B – True
Answer:
a) Statement A – True Statement B explain the symptom of A.

Question 35.
What is the normal urea level in the blood
a) 17-30mg/100ml
b) 30-35mg/ 100ml
c) 10-15 mg/100 ml
d) 5-10 mg/100 ml
Answer:
a) 17-30mg/100ml

Question 36.
Find the wrong pair.
a) Renal stone – nephrolithiasis
b) Urine – Urochrome
c) Deficiency of ADH – Urine output decreases
d) Skin- Lactic acid excretion
Answer:
c) Deficiency of ADH – Urine output decreases

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 37.
Confirmation:
Statement A: Bright’s disease is due to the infection of streptococcus in children
statement B: There is inflammation of glomerulus.
a) Statement A and Bare false
b) Statement A True B explain the A
c) A True B False
d) A False B – True
Answer:
b) Statement A True B explain the A

Question 38.
Find the wrong pair.
a) Heporin – Anticoagulating factor
b) Glomerulonephritis – Accumulation of water in the body
c) Primary kidney – Meso nephridia
d) Uremia – Increase in the blood urea level
Answer:
c) Primary kidney – Meso nephridia

Question 39.
How much urine can be stored up in the bladder?
a) 300-600 ml
b) 200-300 ml
c) 400-700 ml
d) 500-800 ml
Answer:
a) 300-600 ml

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 40.
For how much time urine can be held in the bladder?
a) 6 hours
b) 2 hours
c) 5 hours
d) 3 hours
Answer:
c) 5 hours

Question 41.
The urinary bladder is made up this muscle ……………………….
a) Detrusor muscle
b) Striated muscle
c) Sphincter muscle
d) None of the above
Answer:
a) Detrusor muscle

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 42.
Match the following:

1. Steno haline I Shark
2. Eurihaline II Otter
3. Osmo regulators III Goldfish
4. Osrno confirmers IV Salmon

a) 1-IV 2-III 3-II 4-1
b) 1 -II 2-II 3-IV 4-1
c) 1 -III 2-IV 3-II 4-1
d) 1-1 2-II 3-IV 4-III
Answer:
c) 1 – III, 2- IV, 3 – II, 4 – I

Question 43.
Find out the less toxic waste among the following:
a) Urea
b) Uric acid
c) Ammonia
d) Creatinine
Answer:
a) Urea

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 44.
Find out the ammonotelic organisms?
a) Reptiles
b) Birds
c) Aquatic amphibians
d) Frog
Answer:
c) Aquatic amphibians

Question 45.
Find out the wrong pair.
a) Rennette cells – Annelida
b) Molluscs – Metanephridia
c) Amphioxus – Mesonephridia
d) Tapeworm – Flame cells
Answer:
c) Amphioxus – Mesonephridia

Question 46.
Find the excretory structure of prawn?
a) Malpighian tubules
b) Green glands
c) Rennette cells
d) Peyer gland
Answer:
b) Green glands

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 47.
Kidney with noglomerulus form this types of urine?
a) Hypertonic
b) Isotonic
c) Hypotonic
d) Neutrotonic
Answer:
c) Hypotonic

Question 48.
Match the following:

1. Conical tissue masses a) Columns of Bertini
2. Extension in renal b) Renal pelvis
3. Broad part of Hilum c) Calyces
4. Projection of pelvis d) Medullary pyramids

a) 1-d 2-a 3-b 4-c
b) 1-a 2-b 3-c 4-d
c) 1-b 2-a 3-c 4-d
d) 1-a 2-c 3-b 4-d
Answer:
a) 1-d 2-a 3-b 4-c

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 49.
How much urine can be held in the urinary bladder?
a) 300-800ml
b)300-600ml
c) 100-400ml
d)200-300ml
Answer:
b)300-600ml

Question 50.
The muscles of urinary bladder is called by this name?
a) sphincter muscle
b) Detrusor muscle
c) constricting muscle
d) Peristaltic muscle
Answer:
b) Detrusor muscle

Question 51.
Name the nephron where the Henle’s loop is short?
a) Medullary nephron
b) Cortical nephron
c) Juxta medullary nephron
d) Medulla nephron
Answer:
b) Cortical nephron

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 52.
The vessel which runs parallel to the loop of Henle is called by this name.
a) Efferent artery
b) Afferent artery
c) Vasarecta
d) Vasanervosa
Answer:
c) Vasarecta

Question 53.
What is the process of urea formation?
a) Ross cycle
b) Ornithine cycle
c) Urea cycle
d) b and c
Answer:
b) Ornithine cycle

Question 54.
What is the pressure in the afferent arteriole?
a) 35 mm Hg
b) 55 mm Hg
c) 20 mm Hg
d) 40 mm Hg
Answer:
b) 55 mm Hg

Question 55.
What is the filtrate entered into the Bowman’s capsule from glomerulus?
a) Secondary urine
b) Primary urine
c) Tertiary urine
d) Quarternary urine
Answer:
b) Primary urine

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 56.
Find the correct option
Assertion: The glomerular filtrate resembles the blood
Reason: The glomerulus filtered the blood received from efferent artery
a) Assertion True Reason True
b) Assertion False Reason False
c) Assertion True The Reason does not explains the statement
d) Assertion True Reason explains A
Answer:
c) Assertion True The Reason does not explains the statement

Question 57.
What is the fate of water in the descending limb of Henle?
a) The concentrations of Na reduces
b) The concentration of Cl ions reduces
c) Na and Cl gets concentrated in the filtrate.
d) Formation of Hypotonic solution.
Answer:
c) Na and Cl gets concentrated in the filtrate.

Question 58.
How can the pH of blood be regulated?
a) Due to glucose reabsorption
b) Due to the reabsorption of HCO3
c) Due to the reabsorption of Na+
d) Due to the reabsorption of Cl
a) 1-2 II-4 III-3 IV-1
b) 1-1 II-2 III-3 IV-4
c) 1-4 II-3 III-2 IV-1
d) 1-2 II-l III-3 IV-4
Answer:
b) 1-1 II-2 III-3 IV-4

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 59.
Match the following
Answer:

I. Potassium 1. Descending limb of Renie
II. Water 2. Active transport
III. Glucose 3. Proximal convoluted tubule
IV. Na, cl, lc 4. Aqua porin

Question 60.
How is osma regulation in medulla maintained?
a) Juxtamedullary
b) Counter current
c) Positive current exchanger
d) Negative control
Answer:
b) Counter current

Question 61.
Find the wrong pair.
a) Vasa recta – Proximal convoluted tubule
b) ADH – Neuro hypophysis
c) Specialized nephron tissue – Juxtaglomerular apparatus
d) Renin – Glanularcell
Answer:
a) Vasa recta – Proximal convoluted tubule

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 62.
This synthesizes angio tenginogen.
a) Kidney
b) Liver
c) Malpighian tubule
d) Glomerular
Answer:
b) Liver

Question 63.
This increases the absorption of sodium ions.
a) Renin
b) Angiotensin I
c) Angio tensin II
d) Angio tensinogen
Answer:
c) Angiotensin II

Question 64.
Name the hormone that decreases the blood pressure
a) Angiotensin I
b) Atrial natriuretic peptide
c) Vasopressin
d) ADH
Answer:
b) Atrial natriuretic peptide

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 65.
This decreases the excretion of Renin?
a) Aldosterone
b) vasopressin
c) Atrial natriuretic peptide
d) Ventrical natriuretic peptide
Answer:
c) Atrial natriuretic peptide

Question 66.
What is the pH of Urine?
a) 6.5
b) 7.0
c) 6.0
d) 8
Answer:
c) 6.0

Question 67.
What is the reason for the yellow colour of urine?
a) Uro chrome
b) Cytochrome
c) Chlorochrome
d) Phytochrome
Answer:
a) Uro chrome

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 68.
The increase in the level of urea
a) Uremia
b) ureomia
c) Uricmia
d) Ketosis
Answer:
a) Uremia

Question 69.
The accumulation of salt in the blood.
a) Poly urea
b) Oligo urea
c) Polydipsia
d) Polyphagia
Answer:
b) Oligo urea

Two marks

II. Very short answer 

Question 1.
What is osmotic regulation?
Answer:
Osmotic regulation is the control of tissue osmotic pressure which acts as a driving force for the movement of water across biological membranes.

Question 2.
What is meant by Eury haline animals?
Answer:
Eury haline animals are able to tolerate wide fluctuations in the salt concentrations. (eg) Salmons Tilapia.

Question 3.
Define excretion?
Answer:
The process by which the body gets rid of the nitrogenous waste products of protein metabolism is called excretion.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 4.
What is meant by Ionic regulation?
Answer:
It is the control of ionic composition of body fluids.

Question 5.
What are the nitrogenous waste formed due to the degeneration of amino acid?
Answer:

  • Ammonia
  • Urea
  • Uric acid

Question 6.
Name some nitrogenous waste product produced by various animals?
Answer:
Some of the nitrogenous wastes produced by various animals other than ammonia, urea and uric acid are:
Jrimethyl amine oxide (TMO) in marine teleosts, guanine in spiders, hippuric acid in mammals, reptiles and other nitrogenous wastes include allantonin, allantoic acid, omithuric acid, creatinine, creatine, purines, pyramidines and pterines.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 7.
What is meant by renal hilum?
Answer:
The centre of the inner concave surface of the kidney has a notch called the renal hilum through which ureter blood vessels and nerves innervate.

Question 8.
What is meant by malpighian capsule or renal corpuscle?
Answer:
The Bowman’s capsule and the glomerulus together constitute the renal corpuscle.

Question 9.
What is the difference between nephron present in reptiles and mammals?
Answer:
Reptiles have reduced glomerulus or lack glomerulus and Henle’s loop. Mammals have a long Henle’s loop. Reptiles produce hypotonic urine whereas mammals produce hypertonic urine.

Question 10.
What are cortical nephrons?
Answer:
The loop of Henle is too short and extends only very little into the medulla and are called cortical nephrons.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 11.
What is meant by Juxta medullary nephrons?
Answer:
Some nephrons have very long loop of Henle that run deep in to the medulla and are called Juxta medullary nephrons.

Question 12.
What is vasa recta?
Answer:
The efferent arteriole serving the juxtamedullary nephron forms bundles of long straight vessel called vasa recta and runs parallel to the loop of Henle.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 13.
List the three important process of urine formation?
Answer:

  1. Glomerulus filtration
  2. Tubular reabsorption
  3. Tubular secretion

Question 14.
What is meant by Glomerulus filtrate? What is its composition?
Answer:
The blood comes to the glomerulus are filtered and enters the Bowman’s capsule is called glomerular filtrate
Composition Water glucose amino acids and nitrogenous wastes.

Question 15.
What is meant by Net filtration pressure?
Answer:
The two opposing forces against the glomerular blood pressure

  1. Collodial osmotic pressure is 30 mm Hg
  2. Capsular hydrostatic pressure is 15mm Hg

Net filtration pressure- Glomerular
Hydrostatic pressure – Colloidal
Osmotic pressure + Capsular hydrostatic pressure = 55 mm Hg – 30 mm Hg + 15 mm Hg = 10 mm Hg

Question 16.
What is meant by Glomerular filtration rate?
Answer:
It is the volume of filtrate formed in a minute in all nephrons of both the kidneys.

Question 17.
What is the amount of glomerular filtrate?
Answer:
In adults the GFR is 120 -125 ml per minute.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 18.
What is meant by primary filtrate?
Answer:
The filtrate enters from glomerulus to the Bowman’s capsule is called as primary filtrate.

Question 19.
Why glomercular filtrate resembles blood plasma?
Answer:
In the glomerular filtrate all the contents that present in the blood except the plasma protein is present.

Question 20.
How much filtrate is formed in one day?
Answer:
The amount of filtrate formed in a day is 170 to 180l.

Question 21.
What is meant by selective permeability?
Answer:

  • Some substances present in the glomerular filtrate is essential for our body.
  • Hence these molecules are reabsorbed in a tubules. This process is called as selective reabsorption.

Question 22.
Name the process in which selective reabsorption is taking place?
Answer:

  • Passive transport
  • Active transport
  • Diffusion
  • Osmosis

Question 23.
What are aqua porins?
Answer:
Aquaporins are membrane transport proteins that allow water to move across the epithelial cells.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 24.
What is Micturition?
Answer:
The process of release of urine from the bladder is called micturition or urination.

Question 25.
What is meant by Isotonic solution?
Answer:
Isotonic condition of a solution indicates no passage of water across the membrane separating two such solution.

Question 26.
What is meant by Hypotonic solution?
Answer:
The solution in which there is a loss of water then that solution is a hypotonic solution.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 27.
What is meant by hypertonic solution?
Answer:
When two solutions A and B are separated by a semi permeable membrane when water move from solution A to B across the membrane then the B solution is hypertonic and the solution A where the water loses is known as hypotonic solution.

Question 28.
Name the organ that the angiotensin II acton?
Answer:

  • Heart
  • Kidney
  • Brain
  • Adrenal cortex

Question 29.
What are the symptoms of diabetes mellitus?
Answer:

  • Excess glucose and ketone bodies in the urine
  • Poly dipsia – Excessive drinking of water
  • Polyurea – Excretion of large quantities of urea
  • Polyphagia – Excessive appetite

Question 30.
Name the organ that excrete nitrogen other than kidney?
Answer:

  • Lungs
  • Liver
  • Skin

Question 31.
What are the significance of sweat glands?
Answer:

  • Sweat produced by the sweat glands helps to cool the body.
  • It excretes Na+ and Cl small quantities of urea and lactate.

Question 32.
What is meant by nephrolithiasis?
Answer:
It is the formation of hard stone like masses in the renal tubules of renal pelvis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 33.
What is meant by phleothotomy or lithotripsy?
Answer:
Renal stones can be removed by the technique pyleothotomy or lithotripsy.

Question 34.
What is meant by Bright’s disease?
Answer:
The inflammation of the glomeruli of both kidney of children due to the streptococcal infection is called as Bright’s disease.

Question 35.
What is meant by haemodialysis?
Answer:
The process of removing toxic urea from the blood of renal failure patients is known as haemodialysis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 36.
Why females are prone to urinary tract infection than men?
Answer:
Females are prone to recurring urinary tract infection as they have shorter urethra.

Question 37.
Why men are finding difficult to urinate in their old age?
Answer:
With age prostate in males may enlarge which forces urethra to tighten restricting a normal urinary flow.

Question 38.
What is the change in Urine formation when there is a deficiency of ADH?
Answer:
When there is a deficiency of ADG the reabsorption of water from the proximal convoluted tubule decreases leads to dilute urine formation.

Question 39.
Why there is a increase in the body fluid when we drink large volume of water with out eating anything salty?
Answer:

  • When we drink or eat salty products the Na+ enters into the body fluids.
  • The sodium ions helps in the reabsorption of water
  • But when we drink only water as there is no sodium ions the tubules cannot reabsorb water.
  • Hence there is an increase in the urine output.

Question 40.
Write a short note on Haemodialysis?
Answer:
Malfunction of the kidneys can lead to accumulation of urea and other toxic substances, leading to kidney failure. In such patients, toxic urea can be removed from the blood by a process called hemodialysis. A dialyzing machine or an artificial kidney is connected to the patient’s body. A dialyzing machine consists of a long cellulose tube surrounded by the dialyzing fluid in a water bath.

The patient’s blood is drawn from a convenient artery and pumped into the dialyzing unit after adding an anticoagulant like heparin. The tiny pores in the dialysis tube allow small molecules such as glucose, salts, and urea to enter the water bath, whereas blood cells and protein molecules do not enter these pores.

This stage is similar to the filtration process in the glomerulus. The dialyzing liquid in the water bath consists of a solution of salt and sugar in the correct proportion in order to prevent loss of glucose and essential salts from the blood. The cleared blood is then pumped back to the body through a vein.

Question 41.
Name the different process that maintains water level? When there is a severe loss of water in the body?
Answer:

  • The blood vessels supplies to skin constricts and thus there is a decrease in the secretion of sweat prevents loss of water.
  • There is a reduction in the glomerular blood pressure and the rate of filtration decreases.
  • The reaborption of water in the proximal distal convoluted tubules increase.
  • There is absorption of water from the small intestine and large intestine and thus increases the water content in the blood.

Question 42.
What is meant by Osmolarity?
Answer:

  • The solute concentration of a solution of water is known as osmolarity,
  • The unit is millosmoles / litre (mOsm /l)

Question 43.
What is meant by aquaporins? What are its functions?
Answer:
Aquaporins are water permeable channels.
Functions
It helps in allowing water to move across the epithelial cells in relation to the osmotic difference from the lumen to the interstitial fluid.

Question 44.
How can we measure that there is a efficient glomerular filtration?
Answer:
If the renal clearance is equal to the glomerular filtration rate with little reabsoption and secretion. Then we know the kidney is functioning efficiently.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 45.
What is the Significance of having long and short Henle’s loop of Nephrons?
Answer:

  • The main function of Henle’s loop is to reabsorb water from filtrate.
  • If the length of the loop is longer then there is more reabsorption of water and if the lengh of Henle’s loop is shorter then the reabsorption of water is less.

Question 46.
Give notes on Capillary to capsule.
Answer:
Bloodcells and most blood proteins are too big to cross the capsular membrane into the capsule space. But the membrane’s slits and pores allow through water mineral salts polypeptides and other small molecules including waste such as urea ammonia and creatinine.

Question 47.
Give short notes on Blood enters the glomerulus.
Answer:
Blood flows from renal arteriole into the knot of capillaries. It enters at pressure which will force water and other out of the capillaries into the capsular space.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 48.
Give notes on filteration in proximal convoluted tubule?
Answer:
Proximal tubule is nearer to the Bowman’s capsule. This region allows much water to be reabsorbed into the capillaries and surronding fluids as well as glucose mineral salts and other useful substances.

Question 49.
Give notes on filteration in peritubular capillaries.
Answer:
It is also called the vasarecta this network reabsorbs upto 99 percent of the water in the tubule as well as various other substance using active pumps it also moves sodium from the blood to the tubule.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 50.
Give notes on filteration in Henle’s loop (Ascending)
Answer:

  • As the loop of the Henle dipsin to the renal medcula more water moves from the tubule into the blood as well as small amounts of salts and some urea and creatinine.
  • Some acids and amines may move into the tubule in which ammonia cango in both the direction.

Question 51.
Give notes on filteration in distal tubule?
Answer:

  • Distal tubule is far from capsule. This region may see water go in or out of the tubule depending on the concentration of water already in the tubule/ while hydrogen and potassium ions move to regulate both blood and urine pH.
  • Acids amines and ammonia compounds may also transported into the tubule.

Question 52.
Give notes on filteration in collecing duct.
Answer:

  • Fine adjustment of urine composition continues into the collecting duct system.
  • About 5 percent of all the water and sodium being reabsorbed into the blood is recovered here.

Question 53.
Give notes on venousflow?
Answer:

  • Blood flowing away from the nephrons carries 99 % of its orginal water.
  • 98% of its sodium calcium and cholrides and about 40% of its urea.

Question 54.
We are not consuming urea. But in our body area is produced. Why?
Answer:
Through Arnithine cycle in the lives the nitrogenous waste formed due to the breakdown of amino acid creates urea.

Question 55.
What is meant by Ionic regulation?
Answer:
It is the control of the ionic composition of body fluids.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 56.
What are stenohaline animals?
Answer:
They can tolerate only narrow fluctuations in the salt concentration. Ex: Goldfish

Question 57.
What are Euryhaline animals?
Answer:

  • They are able to tolerate wide fluctuations in the salt concentrations. Ex: Artemia Salmons.
  • Acids amines and ammonia compounds may also transported into the tubule.

Question 58.
List the nitrogenous wastes
Answer:
Ammonia urea uricacid.

Question 59.
What are the other nitrogenous wastes of protein metabolism?
Answer:
Alantonin Alantoic acid, Ornithuriacid creatinine creatine purines pyramidines and pterines.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 60.
Which are called ammoneteles?
Answer:
Animals that excreate most of its nitrogen inthe form of ammonia are called ammonoteles.

Question 61.
What are called as uricoteles?
Answer:
Animals that excrete uric acid crystals with minimum loss of water are called uricoteles.

Question 62.
What are ureoteles?
Answer:
Mammals and terrestrial amphibians mainly excrete urea are called ureoteles.

Question 63.
How is Reptiles produced lesshypotonic urine?
Answer:
Reptiles have reduced glomerulus or lack glomerulus and Henles loop and hence produce very little hypotonic urine.

Question 64.
How is mammals produced concentrated urine?
Answer:
Mammalian kidneys produce concentrated urine due to the presence of Henle’s loop.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 65.
What are the three coverings of kidney?
Answer:

  • Renal facia
  • Perirenal fat capsule
  • Fibrous capsule

Question 66.
What are medullary pyramids?
Answer:
The medulla is divided into a few conical tissue masses called medullary pyramids.

Question 67.
What is meant by renal columns of Bertini?
Answer:
The part of cortex that extends in between the medullary pyramids is the renal columns of Bertini.

Question 68.
What is meant by renal pelvis?
Answer:
A broad funnel shaped space inner to the hilum is called renal pelvis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 69.
What is calyces?
Answer:
The projection in the pelvis is called calyces.

Question 70.
What are cortical nephrons?
Answer:
The loop of Henle is too short and extends only very little in to the medulla and are called cortical nephrons.

Question 71.
What is meant by juxta medullary nephron?
Answer:
Some nephrons have very long loop of Henle that run deep into the medulla and are called Juxta medullary nephrons.

Question 72.
What is the functions of aquaporins?
Answer:
This helps in allow water to move across the epithelial cells in relation to the osmotic difference from the human to the interstitial fluid.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 73.
What is meant by juxtaglomerular apparatus?
Answer:
It is a specialized tissue in the afferent arteriole of the nephron that consists of maculadensa and granular cells.

Question 74.
What is meant by micturition?
Answer:
The process of release of urine from the bladder is called micturition.

Question 75.
What are the symptoms of diabetes mellitus?
Answer:
Presence of glucose and ketone bodies in the urine.

Question 76.
How is lung acting as a excretory organ?
Answer:
Lungs remove large quantities of carbondixide 181 / day and significant quantities of water every day

Question 77.
What is meant by renal clearnace?
Answer:
The amount of solute passing from the urine in a given period of time is renal clearance.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 78.
How can we estimate the efficiency of kidney?
Answer:

  • The renal clearance is equal to the glomerular filtrate then there is efficient filtration with little reabsorption and secretion.
  • Thus we can estimate the clearance is equal.

Three Marks

Short Answer –

Question 1.
a) List the ma j or nitrogenous wastes.
b) What are other wastes formed during protein metabolism?
Answer:
a) 1. Major nitrogenous wastes.

  • ammonia
  • Urea
  • Uricacid

b) Other Wastes:

  • Trimethy lamine oxide TMO
  • Quanine
  • Allantonin
  • Creatinine
  • Creatine
  • Purines

Question 2.
Give notes on ammonoteles uricoteles and ureoteles.
Answer:
Ammonoteles:

  • Animals that excrete most of its nitrogen in the form of ammonia are called ammonoteles.
    (e.g) fishes Amphibians aquatic insects.
  • In bony fishes ammonia diffuses out acrossthe body surface.

Uricoteles:

  • Animals which excrete uricacid crystals with a minimum loss of water is called.
  • Uricoteles (e.g) Reptiles Birds land snails and insects.

Ureoteles:

  • Animals which excrete urea as a nitrogenous wastes are called ureoteles
  • (e.g) Mammals, terrestrial amphibians.

Question 3.
Nephrons are the functional and structural unit of kidney’s. What is the relationship between glomerulus, Henle’s loop and urine formation?
Answer:

  • Reptiles have reduced glomerulus or lack glomerulus and Henle’s loop and produce hypotonic urine (dilute)
  • Mammalian kidneys produce concentrated urine due to the presence of long Henle’s loop.
  • Aglomerular kidneys of marine fishes produce little urine that is iso osmotic to the body fluid.
  • Amphibians and fresh water fishlack Henle’s loop hence produce dilute urine.

Question 4.
Differentiate the cortical nephron fron juxta medullary nephron
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 3

Question 5.
Give Short notes on capillary bed of the nephron:
Answer:
The first capillary bed of the nephron is the glomerulus.
The other is peritubular capillaries.
1. Glomerulus:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 4
Blood enters into the glomerulus through afferent arteriole and drained by the efferent arteriole.

2. Peritubular capillaries:

  • The efferent arteriole forms a fine cappillary network around the renal tubule called the peritubular capillaries.
  • The efferent arteriole serving the juxta medullary nephrons forms bundles of long straight vessel called vasa recta.
  • Vasa recta is absent in cortical nephrons.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 5

Question 6.
What happens to the filtrate that comes to the proximal convoluted tubule? (or) Explain about reabsorption?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 6

  • In the proximal convoluted tubule glucose lacticacid aminoacid sodium ions are reabsorbed.
  • Sodium is reabsorbed – potassium pump in the proximal convoluted tubule.
  • Small amounts of urea and uric acid are also reabsorbed.

Question 7.
What happen to the filtrate that comes to the Henle’s loop? (or) Explain the reabsorptionin the Henle’s loop?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 7

Descending Loop:
The aquaporin present in the descending limb of Henle permeable to water but not permeable to salts.
Hence Na+ and cl gets concentrated in the filtrate.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 8

Ascending Limb:
It is impermeable to water but permeable to solutes like Na+cl and K+.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 8.
Give an account of tubular reabsorption?
Answer:
The volume of filterate formed perday is 170-180 is and the urine released in a day is 1.5l

  • Nearly 99 % of the glomerular filtrate is reabsorbed by the renal tubules. It is called selective reabsorption.
  • Reabsorption is taken place by the tubular epithelial cells in different segments of the nephron by active transport or passive transport diffusion and osmosis.

Question 9.
What is happenning to the filtrate in distal convoluted tubule (or) Reabsorption taking place here?
Answer:

  • Depending on the body’s need the reabsorption taking place here and is regulated by hormones.
  • Reabsorption of bicarbonate HCO3 takes place to regulate the blood pH.
  • Homestasis of K+ and Na+ in the blood is also regulated in this region.

Question 10.
Name the structures that regulate the functioning of kidney?
Answer:

  • Hypothalamus
  • Juxta glomerular apparatus
  • Heart

Question 11.
What is meant by diabetes incipidus?
Answer:
If there is deficiency or absence of ADH that leads to dilute urine called diabetes incipidus.
Symptoms.

  • Excessive thirst
  • Excretion of large quantities of dilute urine.
  • FaIl in blood pressure.

Question 12.
Give notes on juxta glomerular apparatus?
Answer:

  • Specialized tissue in the afferent arteriole of ncphron is the juxta glomerular apparatus.
  • It consists of macula densa and granular cells.
  • The macula densa cells sense distal tubular flow and affect afferent alteriole diameter.
  • The granular cells secrete renin.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 13.
Why female are prone to urinary tract infections? (Urethritis)
Answer:

  • Female’s urethra is very short and its external opening is close to the analopening.
  • Hence improper toilet habits can easily carry faecal bacteria into the urethra.
  • The urethral mucusa is continuous with the urianary tract and the inflammation of the urethra is called urethriti’s.

Question 14.
What is cystitis?
Answer:
The urinary tract infection leads to inflammation of bladder called cystiti’s.
Symptoms:

  • Painful urination
  • Urinary Urgency
  • Cloudy or bloodtingedurine
  • Back pain head ache offen occurs

Question 15.
What is meant by renal failure? What are its types?
Answer:
When the kidney fails to excrete wastes may lead to accumulation of urea with marked reduction in the out put called renal failure.
Types
Acute renal failure
Chronic renal failure

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 16.
Why the chronic renal failure is dangerous than acute renal failure?
Answer:

  • Though the kidney stops its function abruptly there are chances for recovery of kidney function in acute renal failure.
  • But in chronic failure there is a progressive loss of function of the nephron which gradually decreases the function of kidney.

Question 17.
What is meant by glomerulo nephritis or Bright’s disease? What are its symptoms?
Answer:
Inflammation of the glomerulus of both the kidneys due to the strepto coccal infection in children is called as Bright’s disease Symptoms

  • Haematuria
  • Proteinuria
  • Salt and water retention – Oligouria (Low urine out put)
  • Hypertension and Pulmonaryoedema

Question 18.
a) What is meant by kidney transplantations.
b) Where is graft kidney received from?
c) What are the steps to be taken to avoid graft rejection?
Answer:

  • Transfer of healthy kidney from one person (donor) to another person with kidney failure is called kidney transplantation.
  • The donated kidney may be taken from a healthy person who is declared brain death or from sibling or close relatives.
  • Immuno supressive drugs are administered to the patient to avoid tissue rejection.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 19.
a) Name the hormone that the dipsticks contain which tests urine?
b) Which colour indicates the presence of glucose in the urine?
Answer:

  • Glucose oxidase and peroxidase.
  • Brown coloured compound is produced.

Question 20.
What are Osmo confirmers?
Answer:
Osmo confirmers are able to change their internal osmotic concentration with change in external environments asin marine and sharks. molluscs

Question 21.
What are Osmo regulators?
Answer:
They maintain their internal osmotic concentration irrespective of their external osmotic environment, (eg) Otters.

Question 22.
List the excretory structures of different organisms.
Answer:

  • Protonephridia
  • Meta nephridia
  • Flame cells – Platy helminthes
  • Rennette cells – Nematodes
  • Malpighian tubules – Insects
  • Greenglands – Prawns

Question 23.
What is meant by filtration slits?
Answer:

  • The visceral layer of glomerulus is made of epithelial cells called podocytes and ends in foot processes which cling to the basement membrane of the glomerulus.
  • The openings between the foot processes are called filtration slits.

Question 24.
Draw the diagram of ornithine cycle?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 9

Question 25.
Why there is a pressure reduction when the blood goes through efferent arteriole?
Answer:

  • Blood enters the glomerulus faster with greater force through afferent arteriole.
  • Because the afferent arteriole is broader than efferent arteriole that is why the pressure reduces when it goes through the efferent arteriole.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 26.
What are the changes taking place in our body when there is a fluid loss?
Answer:

  • The osmo receptors in the hypothalamus is stimulated.
  • The neurohypophysis is stimulated and anti diuretic hormone is liberated.
  • The aquaporins in the tubuler are increased and water is reabsorbed and enters into the interstitial cell and the water loss is rectified.

Question 27.
How is skin acted as a excretory organ?
Answer:

  • Skin excretes Na+ and Cl small quantities of urea and lactate.
  • Sebaceous glands eliminated certain substances like steroids, hydrocarbons and waxes.

Question 28.
What is meant by urethritis?
Answer:
The urethral mucosa is continuous with the urinary tract. The infection in the urethra is called urethritis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 29.
What is meant by Cystitis?
Answer:
The infection in the urethra can ascend the tract to cause bladder inflammation called cystitis.

Question 30.
What is meant by Phelitis or Pyelone phritis?
Answer:
The bladder infection ascend to the renal inflammation called pyelitis or pyelonephritis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 31.
What are the two types of renal failure?
Answer:

  1. Acute renal failure
  2. Chronic renal failure

1. Acute renal failure

  • In acute renal failure the kidney stops its function abruptly.
  • There are chances for recovery of kidney function.

2. Chronic renal failure
In chronic renal failure there is a progressive loss of function of the nephron which gradually decreases the function of kidneys.

Question 32.
What is meant by Uremia?
Answer:
Uremia is characterized by increase in Urea uric acid and creatinine in blood.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 33.
What is meant by Nephrolithiasis?
Answer:
It is a formation of hard stone like masses in the renal tubules of renal pelvis.

Question 34.
How is water excess taken through drinking too much fruit juice regulated?
Answer:

  • When we drink much fruit juice the osmo receptors in hypothalamus is not stimulated and hence the secretion of vaso pressin from neuro hypophysis is reduced.
  • The aquaporin escapes from collecting duct to cytoplasm and hence water reabsorption is prevented and formed dilute urine.

Question 35.
What is uremia?
Answer:
Uremia is a condition in which there is a increase level of urea uric acid and creatinine in blood.

Question 36.
What is the amount of Urea present in the blood?
Answer:

  • The level of urea in the blood is 17 -30 mg /100ml.
  • In chronic kidney failure there is 10 times increase in urea level.

Five marks

IV. Detailed Answers –

Question 1.
Name the different excretory structure and different organisms.
Answer:
1. Invertebrate – Protonephridia / Meta nephridia
2. Platvhelminthes – Flamecells
3.  Amphioxues – Solenocytes
4 Nematodes – Rennette cells
5. Annelida – Metanephridia
6. Insects – Malpighian tubules
7. Prawn / Crustaceans – Green glands / Antenna! glands

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 2.
The concentration of urine depends on the structure of nephron? Explain?
Answer:

  1. In the reptiles the glomerulus is reduced or there may be no glomerulus and 1-lenle’s loop and hence produces dilute urine (chypotonic).
  2. In the mammals the long Henle’s loop produces concentrated urine (hypertonic)
  3. A glomerular kidneys of marine fishes produce little urine that is isoosmotic to the body fluid.
  4. Amphibians and freshwater fish lack Henle’s loop hence produce dilute urine.

Question 3.
Draw the following diagram and mark the following parts.
Answer:
A) Aorta
B) Renal vein
C) Ureter
D) Urinary bladder
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 10

Question 4.
a. What is the weight of kidneys? What are its outer coverings?
b. Draw the L.S of kidney and name the parts.
c. Explain the internal structure of kidney
Answer:
a. Each kidney weighs an average of 120 – 170 gms.
The outer layer of the kidney is covered by three layers of supportive tissue namely renal fascia perirenal fat capsule fibrous capsule.

b. Draw the LS of kidney and name the parts.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 18

c. Internal Structure of kidney

  • The longitudinal section of kidney shows an outer cortex inner medulla and pelvis.
  • The inner concave surface of the kidney is renal hilum through which ureter blood vessels and nerves enter.
  • Inner to the hium is a funnel shaped renal pelvis with projection called calyces.
  • The calyces collect the urine and empties in to the ureter.
  • The medulla consists of conical tissues called medullary pyramids or renal pyramids.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 5.
a) What is the structural and functional unit of kidney.
b)Draw the diagram of nephron and name the parts.
c) Give short notes on Malpighian body or Renal Corpuscle.
Answer:
a) Hie structural and functional unit of kidney is nephron. It is composed of Malpighian body or renal corpuscle and Urine ferous tubule.
b) Structure of nephron
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 11
c. Malpighian body/Renal Corpuscle.

  • The Bowman’s capsule and the glomerulus together constitutes Malpighian corpuscle.
  • Bowman’s capsule is made up of two layers. It contains blood vessels called glomerules.
  • The endothelial of glomerulus has many pores. The viscral layers of glomerulus is made of epithelial cells called podocytes.
  • Tire podocytes end in foot processes which cling to the basement membrance of the glomerulus.
  • The openings between the foot processes are called filtration slits.

Question 6.
a) Give the different regions of renal tubule.
b) Where is renal tubule present in the kidney?
c) How is renal tubules differentiated depending on the Henle’s loop?
Answer:
a)
1. proximal convoluted tubule.
2. Henle’s loop
a. Thindescending limb of Henle’s loop.
b. Thick ascending limb.

2. Distal convoluted tubules
The distal convoluced tubules opens in to acollecting duct.
Several collecting ducts fuse to form papillary duct that delivers urine in to the calvces which opens into the renal pelvis.
b) The PCT and DCT are situated in the cortical region of the kidney.
The loop of Henle is in the medulla region.
c) The loop of Henle is too short and extends only very little into the medulla and are called cortical nephron.
Some nephrons have very long loops of Henle that run deep into the medulla and are called Juxta medullary nephrons.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 7.
a) Give notes on capillaries of nephron
b) Give an account of blood vessels of glomerulus.
c) What is vasa recta? Where is it seen?
Answer:
Capillaries of nephron
1. Glomerulus capillary bed
2. Peritubular capillaries

1. Glomerular capillarybed

  • It consists of afferent and efferent arteriole.
  • The afferent arteriole is broader than efferent arteriole.
  • The efferent arteriole that comes out of the glomerulus forms a fine capillary network around the renal tubule called the peritubular capillaries.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 12
Vasa recta
The efferent arteriole serving the juxta medullary nephrons form bundles of long straight vessel called vasa recta and runs parallel to the loop of Henle.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 8.
a) Describe the three important process in the urine formation?
b) Give an account on glomerular filtration
Answer:
a. Main Processes

  • Clornerular Filtration
  • Tubular reabsorption
  • Tubular secretion

b. Glomerular filtration

  • Blood enters the kidney from the renal artery.
  • Blood enters the glomerulus faster with greater force through the afferent arterioles because the afferent arteriole is wider than efferent arteriole.
  • The glomerular hydrostatic pressure is around 55 mm Hg.
  • The colloidal osmotic pressure is 30 mm Hg. Capsular hydrostatic pressure is 15 mm Hg. The net pressure of 10 mm Hg is responsible for the renal filtration.
  • Next filtration pressure 55 mm Hg – (30 mm Hg +15 mm Hg) 10 mm Hg.
  • The kidneys produce 1801 of glomerular filtrate 24 hours and 120 -125 ml/min.
  • This filtrate will enter into the Bowman’s capsule and called primary urine. It contains more water colloidal protein, glucose salts and nitrogenous waste.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 13

Question 9.
a) Why glomerular filtrate resembles blood plasma?
b) Tabulate the concentration of substances in the blood plasma and in the glomerular filtrate.
Answer:
As the glomerular filtrate forms it contain all the substances except plasma protein of blood. Hence it resembles blood.

Substance Concentration in blood Plasma / g dm-3 Concentration in glomerular filtrate / g dm-3
Water 900 900
Proteins 80.0 0.05
Aminoacids 0.5 0.5
Glucose 1.0 1.0
Urea 0.3 0.3
Uric Acid 0.04 0.04
Creatinine 0.01 0.01
Inorganic ions (mainly (Na+, K+ and Cl ) 7.2 7.2

Question 10.
a) What is meant by Tubular Secretion?
b) Give an account of tubular secretion of nephron.
Answer:
Tubular Secretion:

  • The collecting tubule of nephron secrete H+ NH4 , Creatinine and Organic acid and liberated into the tubules and excreted through urine.
  • Most of the water is absorbed in the proximal convoluted tubule.
  • Na is exchanged for water in the loop of Henle.
  • The hypotonic fluid enters the distal convoluted tubule.
  • Substances such as urea and salts pass from peritubular blood into the cells as distal convoluted tubule and then to collecting duct.
  • Water is absorbed and concentrated hypertonic urine is form ed.
  • For every H+ secreted into the tubular filtrate a Na+ is absorbed by the tubular cell.
  • The H secreted combines with HCO3, HPO3 and NH3 and gets fixed as carbonic acid CH2CO3 and Phosphoric acid CH2PO4
  • Since H+ gets fixed in the fluid reabsorption of H+ is prevented.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 11.
a) In which process concentrated urine is formed?
b) How is concentrated urine formed in the Hen le’s loop.
Answer:
a) The major function of Henle’s loop is to concentrate Na+ and Cl.

  • There is low osmolarity near the cortex and high osmolarity towards the medulla.
  • This osmolarity in the medulla is due to the presence of the solutes transporters and is maintained by
    the arrangement of the loop of Henle collecting duct and vasa recta.
  • The osmolarity of interstitial fluid is 300 m Osm.
  • The Henle’s loop create a countercurrent multiplier.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 14

  • As the fluid enters the descending limb water moves from the lumen into the inter stitial fluid the osmolarity reduces.
  • To counteract this dilution the region of the ascending limb actively pumps solutes from the lumen into the interstitial fluid and the osmolarity increases to about 1200 m OSM in medula.

b) The vasa recta maintains the medullary osmotic gradient via counter current exchanger.

  • The counter current exchanger of vasa recta preserves the medullary gradient while removing reabsorbed water and solutes.
  • The vasa recta leaves the kidney at the junction between the cortex and medulla.
  • When the blood leaves the efferent arteriole and enters vasa recta the osmolarity in the medulla increases (1200 rnOsm) and result in passive up take of solutes and loss of water.
  • As the blood enters the cortex the osmolarity in the blood decreases and the blood loses solutes and gain water to form concentrated urine.

Question 12.
How is vasa recta helps in producing concentrated urine?
Answer:
Vasa recta maintains the medullary osmotic gradient via counter current exchanger.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 15

  • Vasa recta preserves the medullary gradient while removing reabsorbed water and solutes through counter current exchanges.
  • The vasa recta leave the kidney at the junction between the cortex and medulla.
  • The interstitial fluid at this point is iso – osmotic to blood.
  • When the blood leaves the efferent arteriole and enters the vasa recta the osmalarity in the medulla increases (1200 mOsm) and results in passive up take of solutes and loss of water.
  • As the blood enters the cortex the osmolarity in the blood decreases (300mOsm) and the blood loses solutes and gains water to form concentrated urine.
  • Human kidneys can produce urine nearly four times concentrated than the initial filtrate formed.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 13.
a) What are the structures that regulate kidney functioning?
b) What is the role of ADH in Urine formation.
c) What are the symptoms of diabetes insipidus.
Answer:
a) The structures that regulate kidney functioning

  • Hypothalamus
  • Juxtaglomerular apparatus
  • Heart

b) The functions of ADH

  • When there is excessive loss of fluid from the body or when there is an increase in the blood pressure the osmoreceptors of the hypothalamus is stimulated.
  • The osmoreceptors stimulate the neurohypophysis to secrete an antidiuretic hormone (AOH) or vasopressin.
  • ADH facilitates reabsorption of water by increasing the number of aquaporins on the cell surface of the distal convoluted tubule and collecting duct and prevents excessloss of water.

Question 14.
a) Name the cell that secretes the enzyme renin.
b) Where is granular cell present?
c) What is the role of renin in Osmoregulation.
Answer:
a) Renin is secreted by granular cells.
b) Granular cells are present in the afferent arteriole.
c) The role of renin

  • A fall in glomerular blood flow blood pressure and filtration rate can activate granular cells of juxtaglomerular cells to release renin.
  • Renin converts the plasma protein angiotensinogen into angiotensis I and angiotensin II.
  • Angiotensis II stimulates Na+ reabsorption in the proximal convoluted tubule by vasoconstriction of the blood vessels and increases the glomerular blood pressure.
  • Angiotensis II stimulates adrenal cortex to secrete aldosterone that causes reabsorption of Na+ ,K+ excretion and absorption of water.
  • This increases the glomerular blood pressure and glomerular filtration rate.
  • Hence renin regulates osmoregulation.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 15.
a) Where is atrial natriuretic peptide liberated from?
b) Write its significance in short.
Answer:
a) This is liberated from atrium of heart.
b) Use of Atrial natriuretic peptide

  • It increases Na excretion and increases the blood flow to the glomerulus.
  • It acts on the afferent glomerular arteriole as a vaso dilator or an efferent arteriole as a vaso constrictor.
  • It reduces aldosterone from adrenal cortex and renin secretion.
  • Thus decreases the angiotensin II.
  • The atrial natri uretic factor acts antagonistically to renin angiotensin system aldosterone and vasopressin.

Question 16.
a) What is micturition?
b) How is central nervous system regulates urination?
Answer:
a) The process of release of Urine from the bladder is called micturition or urination.
b) Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored till it receives a signal from the central nervous system.

  • The stretch receptors present inthe urinary bladder are stimulated when it gets filled with urine.
  • At the same time the internal sphincters opens and relaxing the external sphincter.
  • The sphincter opens and the urine is expelled out.

Question 17.
Answer for the following questions.
a) What is the average excretion of an adult human?
b) What is the pH of Urine
c) How much pH is differed due to the food we eat?
d) What is the reason for the yellow colour of urine?
e) How much urea is excreted in a day?
f) If there is more glucose, and ketone what does it indicates?
Answer:

  • 1.5l
  • pH = 6
  • pH = 4.5-8
  • Uro chrome
  • 25 – 30 g
  • Diabetes melitus

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 18.
Answer for the following question in excretion by other organs.
a) What are the other structures that excrete nitrogenous wastes rather than kidney?
b) How much CO2 is excreted through lungs?
c) What are the wastes excreted by digestive systems? ’
d) What are the glands that excrete waste through skin?
e) What is the main function of sweat glands?
f) What is the second important function of sweat gland?
g) Name the substance excreted by sebaceous glands.
h) Name the substances excreted by sebaceous glands.
i) Name the waste excreted through saliva.
Answer:

  • Lungs, Liver and skin
  • 18 Litre, much Water
  • Bilirubin, Biliverdin, Cholesterol, Vitamins and drugs.
  • Sweat glands and sebaceous glands
  • To cool the body
  • Na+ and Cl, small quantities of Urea and Lactate excretion.
  • Sebum
  • Steroid Hydrocarbon and wax.
  • Nitrogenous wastes

Question 19.
Draw the schematic representations of renin hormone in the regulation of body fluid concentration.
Answer:

Question 20.
a) What is meant by Haemodialysis?
b) Why is it called an artifical kidney?
c) Give an account of Haemodialysis.
Answer:
a) in the patients of kidney failure toxic urea can be removed from the blood by a process called haemodialysis.
b) The dialyzing machine is a artificial kidney.
c) Hemodialysis
Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion 17

  • A dialyzing machine consists of a long cellulose tube surrounded by the dialysing fluid in a water bath.
  • A patient’s blood is drawn from a convenient artery and pumped into the dialysing unit after adding an anticoagulant like heparin.
  • The tiny pores in the dialysis tube allow small molecules such as glucose salts and urea to enter into the water bath.
  • Whereas blood cells and protein molecules do not enter these pores the cleared blood is then pumped back to the body through a vein.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 8 Excretion

Question 21.
Give notes on kidney transplantation?
Answer:

  • This involves transfer of healthy kidney from one person (donor) to another person who is with kidney failure.
  • The donated kidney may be taken from a healthy person who is declared brain dead or from sibling or close relatives to minimize the chances of rejection by the immune system of the host.
  • Immuno suppressive drugs are usually administered to the patient to avoid tissue rejection.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 11 Fundamentals of Organic Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

11th Chemistry Guide Fundamentals of Organic Chemistry Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
Select the molecule which has only one π bond.
a) CH3 – CH = CH – CH3
b) CH3 – CH = CH – CHO
c) CH3 – CH = CH – COOH
d) All of these
Answer:
a) CH3 – CH = CH – CH3

Question 2.
In the hydrocarbon Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 1 the state of hybridization of carbon 1, 2, 3, 4 and 7 are in the following sequence.
a) sp, sp, sp3, sp2, sp3
b) sp2, sp, sp3, sp2, sp3
c) sp, sp, sp2, sp, sp3
d) none of these
Answer:
a) sp, sp, sp3, sp2, sp3

Question 3.
The general formula for alkadiene is
a) CnH2n
b) CnH2n – 1
c) CnH2n – 2
d) CnHn – 2
Answer:
c) CnH2n – 2

Question 4.
Structure of the compound whose IUPAC name is 5, 6 – dimethylhept – 2 – ene is
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 2
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 14

Question 5.
The IUPAC name of the compound is
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 3
a) 2, 3 – Dimethylheptane
b) 3 – methyl – 4 – ethyloctane
c) 5 – ethyl – 6- methyloctane
d) 4 – Ethyl – 3 methyloctane
Answer:
d) 4 – Ethyl – 3 methyloctane

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 6.
Which one of the following names does not fit a real name?
a) 3 – Methyl – 3- hexanone
b) 4 – Methyl – 3 – hexanone
c) 3 – Methyl – 3 hexanol
d) 2 – Methyl cyclo hexanone
Answer:
a) 3 – Methyl – 3- hexanone

Question 7.
The IUPAC name of the compound CH3 – CH = CH – C ≡ CH is
a) Pent – 4- yn – 2 – ene
b) Pent – 3- en – 1- yne
c) Pent – 2 – en – 4 – yne
d) Pent – 1 yn – 3 – ene
Answer:
b) Pent – 3- en – 1- yne

Question 8.
IUPAC name of  Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 4 is
a) 3, 4, 4 – Trimethylheptane
b) 2 – Ethyl – 3, 3, – dimethyl heptane
c) 3, 4, 4 – Trimethyloctane
d) 2 – Butyl – 2 – methyl – 3 ethyl – butane
Answer:
c) 3, 4, 4 – Trimethyloctane

Question 9.
The IUPAC name of
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 5 is
a) 2,4,4 – Trimethylpent – 2 – ene
b) 2,4,4 – Trimethylpent – 3 – ene
c) 2,2,4 – Trimethylpent – 3 – ene
d) 2,2,4 – Trirnethylpent – 2 – ene
Answer:
c) 2,2,4 – Trimethylpent – 3 – ene

Question 10.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 6 is
a) 3 – Ethyl – 2 – hexene
b) 3 – Propyl – 3 – hexene
c) 4 – Ethyl – 4 – hexene
d) 3 – Propyl – 2 – hexene
Answer:
a) 3 – Ethyl – 2 – hexene

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
The IUPAC name of the compound
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 7 is
a) 2 – Hydroxypropionic acid
b) 2 – Hydroxy Propanoic acid
c) Propan – 2 – ol – 1 – oic acid
d) 1 – Carboxyethanol
Answer:
b) 2 – Hydroxy Propanoic acid

Question 12.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 8 is
a) 2 – Bromo – 3- methyl butanoic acid
b) 2 – methyl – 3 bromo butanoic acid
c) 3 – Bromo – 2 – methylbutanoic acid
d) 3 – Bromo – 2, 3 – dimethyl propanoic acid
Answer:
c) 3 – Bromo – 2 – methylbutanoic acid

Question 13.
The structure of isobutyl group in an organic compound is
a) CH3 – CH2 – CH2 – CH2

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 9

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 10

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 11
Answer:
c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 10

Question 14.
The number of stereoisomers of 1, 2 – dihydroxy cyclopentane
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

Question 15.
Which of the following is optically active?
a) 3 – Chloropentane
b) 2- Chloro propane
c) Meso – tartaric acid
d) Glucose
Answer:
d) Glucose

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 16.
The isomer of ethanol is
a) acetaldehyde
b) dimethyl ether
c) acetone
d) methyl carbinol
Answer:
b) dimethyl ether

Question 17.
How many cyclic and acyclic isomers are possible for the molecular formula C3H6O?
a) 4
b) 5
c) 9
d) 10
Answer:
c) 9

Question 18.
Which one of the following shows functional isomerism?
a) ethylene
b) Propane
c) ethanol
d) CH2Cl2
Answer:
c) ethanol

Question 19.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 12 are
a) resonating structure
b) tautomers
c) Optical isomers
d) Conformers
Answer:
b) tautomers

Question 20.
Nitrogen detection in an organic compound is carried out by Lassaigne’s test. The blue colour formed is due to the formation of
a) Fe3[Fe(CN)6]2
b) Fe4 [ Fe(CN)6]3
c) Fe4[Fe(CN)6]2
d) Fe3[Fe(CN)6]3
Answer:
b) Fe4 [ Fe(CN)6]3

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Lassaigne’s test for the detection of nitrogen fails in
a) H2N – CO – NH.NH2.HCl
b) NH2 – NH2.HCl
c) C6H5 – NH – NH2.HCl
d) C6H5CONH2
Answer:
c) C6H5 – NH – NH2.HCl

Question 22.
Connect pair of compounds which give blue colouration / precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
a) NH2 NH2 HCl and ClCH2 – CHO
b) NH2 CS NH2 and CH3 – CH2Cl
c) NH2 CH2 COOH and NH2 CONH2
d) C6H5NH2 and ClCH2 – CHO
Answer:
d) C6H5NH2 and ClCH2 – CHO

Question 23.
Sodium nitropruside reacts with sulphide ion to give a purple colour due to the formation of
a) [Fe (CN)5 NO]3-
b) [Fe (NO)5 CN]+
c) [Fe (CN)5 NOS]4-
d) [Fe (CN)5 NOS]3-
Answer:
c) [Fe (CN)5 NOS]4-

Question 24.
An organic Compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the Compound will be close to
a) 46 %
b) 34 %
c) 3.4 %
d) 4.6 %
Answer:
b) 34 %

Question 25.
A sample of 0.5 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50mL of 0.5 M H2SO4 The remaining acid after neutralization by ammonia consumed 80 mL of 0.5 M NaOH. The percentage of nitrogen in the organic compound is.
a) 14 %
b) 28 %
c) 42 %
d) 56 %
Answer:
b) 28 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 26.
In an organic compound, phosphorus is estimated as
a) Mg2P2O7
b) Mg3(PO4)2
c) H3PO4
d) P2O 5
Answer:
a) Mg2P2O7

Question 27.
Ortho and para – nitro phenol can be separated by
a) azeotropic distillation
b) destructive distillation
c) steam distillation
d) cannot be separated
Answer:
c) steam distillation

Question 28.
The purity of an organic – compound is determined by
a) Chromatography
b) Crystallization
c) melting or boiling point
d) both (a) and (c)
Answer:
d) both (a) and (c)

Question 29.
A liquid which decomposes at its boiling point can be purified by
a) distillation at atmospheric pressure
b) distillation under reduced pressure
c) fractional distillation
d) steam distillation
Answer:
b) distillation under reduced pressure

Question 30.
Assertion:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 13
Reason:
The principal functional group gets lowest number followed by double bond (or) triple bond.
a) both the assertion and reason are true and the reason is the correct explanation of assertion.
b) both assertion and reason are true and the reason is not the correct explanation of assertion.
c) assertion is true but reason is false.
d) both the assertion and reason are false
Answer:
a) both the assertion and reason are true and the reason is the correct explanation of assertion.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

II. Write brief answers to the following questions:

Question 31.
Give the general characteristics of organic compounds.
Answer:
All organic compounds have the following characteristic properties.

  1. They are covalent compounds of carbon and generally insoluble in water and readily soluble in organic solvent such as benzene, toluene, ether, chloroform etc…
  2. Many of the organic compounds are inflammable (except CCl4). They possess low boiling and melting points due to their covalent nature.
  3. Organic compounds are characterized by functional groups. A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present. In almost all the cases, the reaction of an organic compound takes place at the functional group. They exhibit isomerism which is a unique phenomenon.

Question 32.
Describe the classification of organic compounds based on their structure.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 15

Question 33.
Write a note on homologous series.
Answer:
Homologous series:
A series of organic compounds each containing a characteric functional group and the successive members differ from each other in molecular formula by a CH2 group is called homologous series.
Example:
Alkanes:
Methane (CH4), Ethane (C2H6), Propane (C3H8) etc..
Alcohols:
Methanol (CH3OH), Ethanol (C2H5OH) Propanol (C3H7OH) etc…

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 34.
What is meant by a functional group?
Identify the functional group in the following compounds.
a) acetaldehyde
b) oxalic acid
e) dimethyl ether
d) methylamine
Answer:
Functional group:
An atom or group of atoms within a molecule that shows characteristics set of physical and chemical properties.
a) acetaldehyde → – CHO
b) oxalic acid → – COOH
c) di methyl ether → – O –
d) methyiamine → – NH2

Question 35.
Give the general formula for the following classes of organic compounds
a) Aliphatic monohydric alcohol
b) Aliphatic ketones
c) Aliphatic amines
Answer:
a) Aliphatic monohydric alcohol → CnH2n + 2O
b) Aliphatic ketones → CnH2nO
c) Aliphatic amines → CnH3n + 2N

Question 36.
Write the molecular formula of the first six members of homologous series of nitre alkanes.
Answer:

  • CH3NO2
  • C2H5NO2
  • C3H7NO2
  • C4H9NO2
  • C5H11NO2
  • C6H13NO2

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 37.
Write the molecular formula and possible structural formula of the first four members of homologous series of carboxylic acids.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 16

Question 38.
Give the IUPAC names of the following compounds.
i) (CH3)2CH – CH2 – CH(CH3) – CH(CH3)2

ii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 17

iii) CH3 – O – CH3

iv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 18

v) CH2 = CH – CH – CH2

vi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 19

vii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 20

viii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 21

ix) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 22

x) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 23

xi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 24

xii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 25

xiii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 26

xiv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 27

xv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 28

xvi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 29
Answer:
(i) 2, 3, 5 – trimethyihexane
(ii) 2 – bromo – 3 – methylbutane
(iii) methoxymethane
(iv) 2 – hydroxybutanal
(y) buta – 1 ,3 – diene
(vi) 4 – chloropent – 2 – yne
(vii) 1 – bromobut – 2 – ene
(viii) 5 – oxohexanoic acid
(ix) 3 – ethyl – 4 – ethenylheptane
(x) 2, 4, 4 – trimethylpent – 2 – ene
(xi) 2 – methyl -1 – phenyipropan – 1 -amine
(xii) 2, 2 – dimethyl – 4oxopentanenitrile
(xiii) 2 – ethoxypropane
(xiv) 1 – fluoro – 4 – methyl – 2 -nitrobenzene
(xv) 3 – bromo – 2 – methylpentanal
(xvi) Acetophenone

Question 39.
Give the structure for the following compound
(i) 3 – ethyl – 2 methyl – 1 – pentene
(ii) 1, 3, 5 – Trimethyl cyclohex – 1 – ene
(iii) tetry butyl iodide
(iv) 3 – Chlorobutanal
(V) 3 – Chlorobutanol
(vi) 2 – Chloro – 2 – methyl propane
(vii) 2, 2 – dimethyl – 1 – chloropropane
(viii) 3 – methylbut -1- ene
(ix) Butan – 2, 2 – diol
(x) Octane – 1, 3 – diene
(xi) 1, 3 – Dimethylcyclohexane
(xii) 3 – Chlorobut – 1 – ene
(xiii) 2 – methylbutan – 3 – ol
(xiv) acetaldehyde
Answer:
(i) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 30

(ii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 31

(iii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 32

(iv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 33

(v) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 34

(vi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 35

(vii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 36

(viii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 37

(ix) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 38

(x) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 39

(xi) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 40

(xii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 41

(xiii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 42

(xiv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 43

Question 40.
Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
This method involves the conversion of covalently bonded N, S or halogen present in the organic compounds to corresponding water soluble ions in the form of sodium salts For this purpose a freshly cut piece of Na of the size of a paper, dried it by pressing between the folds of a filter paper is taken in a fusion tube in an iron tand clamping it just near the upper end and it is gently heated.

When it melts to a shiñing globule, put a pinch of the organic compound on it. Heat the tube with the tip of the flame till all reaction ceases and it becomes red hot. Now plunges it in about 50 mL of distilled water taken in a china dish and break the bottom of the tube by striding against the dish. Boil the contents of the dish for about 10 mts and filter. This filtrate is known as lassaignes extract or sodium fusion extract and it used for qualitative analysis of nitrogen, sulfur and halogens present in organic compounds.

Test for Nitrogen:
If nitrogen is present it gets converted to sodium cyanide which on reaction with freshly prepared ferrous sulphate and ferric ion followed by cone. HCl gives a Prussian blue color or green color or precipitate. It confirms the presence of nitrogen. HCl is added to dissolve the greenish precipitate of ferrous hydroxide produced by the excess of NaOH on FeSO4 which would otherwise mark the Prussian blue precipitate. The following reaction takes part in the formation of Prussian blue.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 44

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 41.
Give the principle involved in the estimation of halogen in an organic compound by Carius method.
Answer:
Estimation of halogens (Carius method):
A known mass of the organic compound is heated with fuming HNO3 along with AgNO3. C, H & S gets oxidized to CO2, H2O, SO2 and halogen combines with AgNO3 to form a precipitate of silver halide.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 45

The ppt of AgX is filtered, washed, dried and weighed. From the mass of AgX and the mass of the organic compound taken, percentage of halogens are calculated.

A known mass of the substance is taken along with fuming HNO3 and AgNO3 is taken in a clean carius tube. The open end of the Carius tube is sealed and placed in a iron tube for 5 hours in the range at 530 – 540 K Then the tube is allowed to cool and a small hole is made in the tube to allow gases produced to escape. The tube is broken and the ppt is filtered, washed, dried and weighed. From the mass of AgX obtained, calculations are made.

Calculation:
Weight of the organic compound: w g
Weight of AgCl precipitate = a g
143. 5 g of AgCl contains 35.5 g of Cl
∴ a g of AgCl contains \(\frac{35.5}{143.5}\) × a
w g Organic compound gives a g AgCl
Percentage of Cl in w g organic compound = \(\left(\frac{35.5}{143.5} \times \frac{\mathrm{a}}{\mathrm{w}} \times 100\right) \%\)
Let Weight of silver Bromide be ‘b’ g
188 g of AgBr contains 80 g of Br
∴ b g of AgBr contains \(\frac{80}{180} \times \frac{b}{w}\) of Br
w g Organic compound gives b g AgBr
Percentage of Br in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)
Let Weight of silver Iodide be ‘c’ g
235 g of AgI contains 127 g of I
∴ C g of AgI contains \(\left(\frac{127}{235} \times \frac{c}{w}\right)\) of I
w g Organic compound gives c g AgI
percentage of I in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)

Question 42.
Give a brief description of the principles of
1. Fractional distillation
2. Column Chromatography
Answer:
1. Fractional distillation:
This method is used to purify and separate liquids present in the mixture having their boiling point close to each other. The process of separation of the components in liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. This method is applied in distillation of petroleum. coal tar and crude oil.

2. Column chromatography:
(a) The principle behind chromatography is selective distribution of mixture of organic substances between two phases a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. when the stationary phase is a solid, the moving phase is a liquid or gas.

(b) If the stationary phase is solid, the basis is adsorption, and when it is a liquid, the basis is partition.

(c) Chromatography is defined as a technique for the separation of a mixture brought about by differential movement of the individual component through porous medium under the influence of moving solvent.

(d) In column chromatography, the above principle is carried out in a long glass column.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 43.
Explain paper chromatography.
Answer:
Paper chromatography (PC) is an example of partition chromatography. The same procedure is followed as in thin layer chromatography except that a strip of ‘ paper acts as an adsorbent. This method involves continues differential portioning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatography paper is used. This paper act as a stationary phase.

A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which act as the mobile phase. The solvent rises up and flows over the spot. The paper selectively retains different components according to their different partition in the two phases where a chromatogram is developed. The spots of the separated colored compounds are visible at different heights from the position of initial spots . on the chromatogram. The spots of the separated colorless compounds may be observed either under ultraviolent light or by the use of an appropriate spray reagent.

Question 44.
Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:
Structural isomerism:
This type of isomers have same molecular formula but differ in their bonding sequence.

(a) Chain or nuclear or skeletal isomerism:
These isomers differ in the way in which the carbon atoms are bonded to each other in a carbon chain or in other words isomers have similar molecular formula but differ in the nature of the carbon skeleton (ie. Straight or branched).
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 46

(b) Position isomerism:
If different compounds belonging to same homologous series with the same molecular formula and carbon skeleton, but differ in the position of substituent or functional group or an unsaturated linkage are said to exhibit position isomerism.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 47

(c) Functional isomerism:
Different compounds having same molecular formula but different functional groups are said to exhibit functional isomerism.
Molecular formula C3H6O
CH3 – CH2 – CHO
propanal (aldehyde group)
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 48

Question 45.
Describe optical isomerism with suitable example.
Answer:
Optical isomerism:
1. Compounds having same physical and chemical property but differ only in the rotation of plane of polarised light are known as optical isomers and the phenomenon is known as optical isomerism.

2. Glucose have the ability to rotate the plane of plane polarised light and it is said to be an optically active compound and this property of any compound is called optical activity.

3. The optical isomer which rotates the plane of plane polarised light to the right or in clockwise direction is said to be dextrorotatory and is denoted by the sign (+).

4. The optical isomer which rotates the plane of plane polarised light to the left or in anti- clockwise direction is said to be laveo rotatory and is denoted by the sign (-).

5. Dextrorotatory compounds are represented as ‘d’ (or) by (+) sign and leave rotatory compounds are represented as l (or) by (-) sign.

6. The optical isomers which are non-superimposible mirror images of each other are called enantiomers.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 46.
Briefly explain geametrical isomerism in alkene by considering 2-butene as an example.
Answer:
Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid frame work of double bonds. This type of isomerism occurs due to restricted rotation of double bonds, or about single bonds in cyclic compounds.

In alkenes, the carbon-carbon double bond is sp2 hybridized. The carbon-carbon double bond consists of a σ bond and a π bond. The π bond is formed by the head on overlap of sp2 hybrid orbitals. The n bond is formed by the side-wise overlap of ‘p’ orbitals. The presence of the π bond lock the molecule in one position. Hence, rotation around C = C bond is not possible. This restriction of rotation about C – C double bond is responsible for geometrical isomerism in alkenes.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 49

These two compounds are termed as geometrical isomers and are distinguished from each other by the terms c is and trans. The c is isomer is one in which two similar groups are on the same side of the double bond. The trans isomers is that in which the two similar groups are on the opposite side of the double bond, hence this type of isomerism is often called cis- trans isomerism.

The cis-isomer can be converted to trans isomer or vice versa is only if either isomer is heated to a high temperature or absorbs light. The heat supplies the energy (about 62kcal/ mole) to break the n bond so that rotation about a bond becomes possible. Upon cooling, the reformation of the n bond can take place in two ways giving a mixture both cis and trans forms of trans-2-butene and cis-2-butene.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 50

Question 47.
0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
Solution:
Weight of organic compound = 0.30 g
Weight of carbon dioxide = 0.88 g
Weight of water = 0.54 g

Percentage of carbon:
44 g of carbondioxide contains, carbon = 12 g
0.88 g of carbon dioxide contains, carbon = \(\frac{12 \times 0.88}{44}\) g
0.30 g substance contains,
carbon = \(\frac{12 \times 0.88}{44}\) g

100 g substance Contains \(\frac{12 \times 0.88}{44}\) × \(\frac{100}{0.30}\) = 80 g of carbon
Percentage of carbon = 80 %

Percentage of hydrogen:
18 g of water contains, hydrogen = 2 g
0.54 g of water contains, hydrogen = \(\frac{2 \times 0.54}{18}\) g
0.30 g of substance contains hydrogen = \(\frac{2 \times 0.54}{18 \times 0.30}\) g
100 g of substance contains = \(\frac{2 \times 0.54}{18 \times 0.30}\) × 100 g = 20 g of hydrogen
Percentage of hydrogen = 20 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 48.
The ammonia evolved form 0.20 g of an organic compound by Kjeldahl method neutralized 15 ml of N / 20 sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer:
Weight of organic compound = 0.20 g
Volume of sulphuric acid taken = 15 ml
Strength of sulphuric acid taken = \(\frac {N}{20}\) = 0.05 N
Percentage of nitrogen = \(\frac{14 \times \mathrm{NV}}{1000 \times 10} \times 100\)
= \(\frac{14 \times 0.05 \times 15}{1000 \times 0.20} \times 100\)
= \(\frac {1050}{200}\) = 5.25
% of nitrogen = 5.25%

Question 49.
0.32 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals is a sealed tube gave 0. 466 g of barium sulphate. Determine the percentage of sulphur in the compound.
Solution:
Mass of the substance taken = 0.32 g
Mass of BaSO4 formed = 0.466 g
Molecular mass of BaSO4 = 137 + 32 + 64 = 233
Then, mass of S in 0.466 g of BaSO4 = \(\frac{0.466 \times 32}{233}\)

Percentage of S in compound= \(\frac{0.466 \times 32 \times 100}{233 \times 0.32}\) = 20 %

Question 50.
0.24 g of an organic compound gave 0.287 g of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Solution:
Mass of organic compound = 0.24 g
Mass of silver chloride = 0.287 g
143. 5 g AgCl contains = 35.5 g chlorine
0.287 g of AgCl contains = \(\frac{35.5}{143.5}\) × 0.287
Percentage of chlorine = \(\frac{35.5}{143.5} \times \frac{0.287}{0.24}\) × 100 = 29.58 %

Question 51.
In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 15° C and 760 mm pressure. Calculate the percentage of nitrogen in the compound.
Solution:
Volume of N2 at NTP = \(\frac{V \times P}{t+273} \times \frac{273}{760}\)
= V0 ml
Substituting the various values in the above equation,
V0 = \(\frac{20.7 \times 760}{288} \times \frac{273}{760}\) = 19.62 ml

weight of 19.62 ml of Nitrogen = \(\frac{28}{22400}\) × 19.62 g

∴ Percentage of Nitrogen = \(\frac{28}{22400}\) × 19.62 × \(\frac{100}{0.35}\)
= 4.9 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

11th Chemistry Guide Fundamentals of Organic Chemistry Additional Questions and Answers

I. Choose the correct answer:

Question 1.
Organic compounds can be formed by
a) Plants only
b) Animals only
c) Plants and Animals
d) Plants, animals and can be synthesized in the laboratory
Answer:
d) Plants, animals and can be synthesized in the laboratory

Question 2.
Which of the following is not an organic compound?
(a) DNA
(b) Lipid
(c) Glycogen
(d) Bronze
Answer:
(d) Bronze
Solution:
It is an alloy and a mixture of metals and all others are organic compounds.

Question 3.
The vital force theory was proposed by
a) Wohler
b) Berthlot
c) Berzelius
d) Kolbe
Answer:
c) Berzelius

Question 4.
Which of the following is an example of heterocyclic aromatic compound?
(a) THF
(b) Cyclopropane
(c) Pyridine
(d) Azulene
Answer:
(c) Pyridine

Question 5.
The first organic compound was synthesized in the laboratory by
a) Wohler
b) Kolbe
c) Berzelius
d) Neil Barthlot
Answer:
a) Wohler

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 6.
Which of the following pair is an example of aromatic compounds?
(a) Benzene, Toluene
(b) Cyclopropane, Cyclobuane
(c) Pyridine, Pyrrole
(d) Propane, Butane
Answer:
(a) Benzene, Toluene

Question 7.
Marsh gas mainly contains
a) C2H2
b) C2H4
c) CH4
d) C2H6
Answer:
c) CH4

Question 8.
Hybridization at 2nd carbon in CH2 = CH – CH3 is
a) sp
b) sp2
c) sp3
d) sp3d
Answer:
b) sp2

Question 9.
Number of possible position isomers for Dichlorobenzene is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 10.
Which of the following represents thiol?
(a) -SH
(b) -OH
(c) -SR
(d) -SCN
Answer:
(a) -SH

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
Alkanols and Alkoxyalkanes are
a) Functional isomers
b) Keto – enol tautomers
c) Geometrical isomers
d) Not isomers at all
Answer:
a) Functional isomers

Question 12.
n – propyl alcohol and isopropyl alcohol are examples of
a) Position isomerism
b) Chain isomerism
c) Tautomerism
d) Geometrical isomerism
Answer:
a) Position isomerism

Question 13.
The number of structural alcoholic isomers for C4H10O is
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4

Question 14.
Cycloalkanes are isomeric with
a) Alkadienes
b) Alkynes
c) Aromatic compounds
d) Olefins
Answer:
d) Olefins

Question 15.
Which one of the following is called benzylchloride?
(a) C6H5CH2Cl
(b) C6H5CHCl2
(c) C6H5CCl3
(d) C6H5Cl
Answer:
(a) C6H5CH2Cl

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 16.
Diethyl ether and n – propyl methyl ether are
a) Metamers
b) Chain isomers
c) Geometrical isomers
d) Position isomers
Answer:
a) Metamers

Question 17.
The total number of structural isomers for the compound of the formula C4H10O is
a) 7
b) 6
c) 4
d) 3
Answer:
a) 7

Question 18.
The number of primary alcoholic isomers with the formula C4H10O is
a) 1
b) 2
c) 3
d) 4
Answer:
b) 2

Question 19.
Which colour is formed in Lassaigne’s test for nitrogen?
(a) Purple
(b) Black
(c) Prussian blue
(d) Violet
Answer:
(c) Prussian blue

Question 20.
The number of isomeric amines possible for the formula C3H9N
a) 4
b) 3
c) 5
d) 6
Answer:
a) 4

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Which hybrid orbitals are involved in the CH3 – CH = CH – CH3 compound
a) sp and sp3
b) sp2 and sp3
c) sp and sp2
d) only sp3
Answer:
b) sp2 and sp3

Question 22.
Which of the following bonds is strongest?
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 51

b) > C = C <

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 52

d) – C – C –
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 51

Question 23.
Identify the colour formed when Lassigne’s extract of sulphur containing organic compound is mixed with sodium nitroprusside solution?
(a) Prussian blue
(b) Black
(c) Green
(d) Purple
Answer:
(d) Purple

Question 24.
Which of the following is an aromatic compound
a) Phenol
b) Naphthalene
c) Pyridine
d) All
Answer:
d) All

Question 25.
Which one of the following is not identified by Lassaigne’s test?
(a) nitrogen
(b) sulphur
(c) halogens
(d) phosphorous
Answer:
(d) phosphorous

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 26.
Which is an alicyclic compound?
a) benzene
b) cyclohexane
c) pyridine
d) pyrrole
Answer:
b) cyclohexane

Question 27.
Which of the following is not a cyclic compound?
a) Anthracene
b) Pyrrole
c) Phenol
d) Isobutylene
Answer:
d) Isobutylene

Question 28.
Functional group present in amides is
a) – COOH
b) – NH2
c) – CONH2
d) – COO –
Answer:
c) – CONH2

Question 29.
Which of the following is used as moisture absorbent?
(a) Potash
(b) Soda
(c) Conc. H2SO4
(d) Na3CO4
Answer:
(c) Conc. H2SO4

Question 30.
IUPAC name of methyl cyanide is
a) Cyano methane
b) Ethanenitrile
c) Methane nitrile
d) Methyl – n – butylamine
Answer:
b) Ethanenitrile

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 31.
The correct IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 53 is
a) 1, 2 – diethyl butene
b) 2 – ethyl – 3- methyl pentene
c) 3, 4 – dimethyl hex – 3 – ene
d) 2, 3 – dimethyl pent – 2 – ene
Answer:
d) 2, 3 – dimethyl pent – 2 – ene

Question 32.
Which of the following is not purified by the sublimation method?
(a) Camphor
(b) Benzoic acid
(c) Naphthalene
(d) Nitrobenzene
Answer:
(d) nitrobenzene

Question 33.
IUPAC name of CH ≡ C – CH = CH2 is
a) but – 3 – ene – 1 – yne
b) but – 1 – ene – 3 – yne
c) but – 1 – yne – 3 – ene
d) but – 3 – yne – 1 – ene
Answer:
b) but – 1 – ene – 3 – yne

Question 34.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 54 is
a) 4 – hydroxy – 2 – methyl pentanal
b) 2 – hydroxy – 4 – methyl pentanal
c) 4 – hydroxy – 2 – methyl pentanol
d) 2 – hydroxy – 4 – methyl pentanol
Answer:
a) 4 – hydroxy – 2 – methyl pentanal

Question 35.
3 – methyl penta -1, 3- diene is
a) CH2 = CH (CH2)2 CH3
b) CH2 = CHCH (CH3) CH2CH3
c) CH3CH = C(CH3)CH = CH2
d) CH3 = C = CH (CH3)2
Answer:
c) CH3CH = C(CH3)CH = CH2

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 36.
The structural formula of methyl amino methane is
a) (CH3)2 CH NH2
b) (CH3)3 N
c) (CH3)2 NH
d) CH3NH2
Answer:
c) (CH3)2 NH

Question 37.
Which of the following compounds gives curdy white precipitate in Lassaigne’s test?
(a) CH3Br
(b) C2H5I
(c) CH3Cl
(d) C6H5NO2
Answer:
(c) CH3Cl

Question 38.
The compound which is not isomeric with diethyl ether is
a) n – propyl methyl ether
b) 1 – Butanol
c) 2 – Methyl – 2 – propanol
d) Butanone
Answer:
d) Butanone

Question 39.
Which of the following pairs of compounds are tautomers?
a) Propanol & propanone
b) Ethanol & vinyl alcohol
c) Ethanol & allyl alcohol
d) Vinyl alcohol & ethanal
Answer:
d) Vinyl alcohol & ethanal

Question 40.
Which of the following compounds does not have any tertiary hydrogen atoms?
a) (CH3)3 CCH2 CH3
b) (CH3)2 CHCH2 CH3
c) (CH3)2 CHCH (CH3)2
d) (CH3)3 CH
Answer:
a) (CH3)3 CCH2 CH3

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 41.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 55 is
a) 2 – Methyl – 2 – butenoic acid
b) 3 – Methyl – 3 – butenoic acid
c) 3 – Methyl – 2 – butenoic acid
d) 2 – Methyl – 3 – butenoic acid
Answer:
c) 3 – Methyl – 2 – butenoic acid

Question 42.
Which of the following reagent is used to distinguish between halogens (Cl, Br, I) in an organic compound?
(a) NaOH
(b) FeCl3
(c) H2SO4
(d) NH4OH
Answer:
(d) NH4OH

Question 43.
The IUPAC name of the Compound CH3 – CH(OH) – COOH is
a) Lactic acid
b) 2 – Hydroxy propanoic acid
c) 3 – Hydroxy propanoic acid
d) Carboxy propanol
Answer:
b) 2 – Hydroxy propanoic acid

Question 44.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 56 is
a) 2 – Ethyl – ethyl acetate
b) Ethyl – 3 – methy1butanoate
c) Ethyl – 2 – methyl butanoate
d) 2- methyl butanoic acid
Answer:
c) Ethyl – 2 – methyl butanoate

Question 45.
The IUPAC name of the given compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 57 is
a) 2,2 — Dimethyl butane
b) lsohexane
c) 2, 3 – Dimethyl butane
d) Di isohexane
Answer:
c) 2, 3 – Dimethyl butane

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 46.
The IUPAC name of the given compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 58 is
a) Octyl cyclopentane
b) 3 – cyclopentyl octane
c) Cyclopentane octane
d) 6 – cyclopentyl octane
Answer:
b) 3 – cyclopentyl octane

Question 47.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 59 is
a) but – 2- ene – 2,3 – diol
b) pent – 2- ene – 2,3 – diol
c) 2 – methyl but – 2 – ene – 2,3 – diol
d) hex – 2- ene – 2,3 – diol
Answer:
b) pent – 2- ene – 2,3 – diol

Question 48.
The structure of 3-bromoprop-1-ene is
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 60

b) CH3 – CH = CH – Br

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 61

d) Br – CH2 – CH ≡ CH2
Answer:
d) Br – CH2 – CH ≡ CH2

Question 49.
Neo-heptyl alcohol is correctly represented as
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 62

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 63

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 64

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 65
Answer:
c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 64

Question 50.
The number of dibromo derivatives possible for propane is
a) 2
b) 3
c) 1
d) 4
Answer:
d) 4

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 51.
The number of aromatic isomers possible for C7H8O is
a) 2
b) 3
c) 4
d) 5
Answer:
d) 5

Question 52.
Isomers of propanoic acid are
a) HCOOC2H5 and CH3COOCH3
b) H – COOC2H5 and C3H7COOH
c) CH3COOCH3 and C3H7OH
d) C3H7OH and CH3COCH3
Answer:
a) HCOOC2H5 and CH3COOCH3

Question 53.
IUPAC name of CH3 – CH (OCH3) – CH2 – NH2
a) 2-methoxy propanamine
b) 1-amino – 2-methoxy propane
c) 1-amino – 2-methyl – 2-methoxy ethane
d) 1 – methoxy- 2-amino propane
Answer:
a) 2-methoxy propanamine

Question 54.
IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 66 is
a) 3 – cyanopentane – 1, 5 – dinitrile
b) Propane – 1, 2, 3-tri nitrile
c) 1, 2, 3-tri cyano propane
d) Propane 1, 2, 3-tricarbonitrile
Answer:
d) Propane 1, 2, 3-tricarbonitrile

Question 55.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 67 is
a) 3 – Carboxylic pentane – 1,5 -dioic acid
b) Propane – 1, 2, 3 – trioic acid
c) 1, 2, 3- tricarboxylic propane
d) Propane – 1,2, 3 – tricarboxylic acid
Answer:
b) Propane – 1, 2, 3 – trioic acid

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 56.
The IUPAC name of the following compound
CH3 – C(CH3)2 – CH2 – CH = CH2 is
a) 2, 2 – Dimethyl – 4 – pentene
b) 4, 4 – Dimethyl – 1 – pentene
c) 1, 1, 1 – trimethyl – 3 – butene
d) 4, 4, 4 – trimethyl – 1 – butene
Answer:
b) 4, 4 – Dimethyl – 1 – pentene

Question 57.
The IUPAC name of Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 68 is
a) 2, 4 – Dimethyl pentan – 2 – ol
b) 2, 4 – Dimethyl pentan – 4 – ol
c) 2,2 – Dimethyl butan – 2- ol
d) Butan – 2 – ol
Answer:
a) 2, 4 – Dimethyl pentan – 2 – ol

Question 58.
The IUPAC name the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 69 is
a) Butane – 2, 3, 4 – triol
b) Butane – 1,2, 3 – triol
c) Pentane – 1, 2, 3 – triol
d) 2, 3 dihydroxy butanol
Answer:
b) Butane – 1,2, 3 – triol

Question 59.
The IUPAC name the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 70 is
a) 3, 3 – Dimethyl – 1 – cyclohexanol
b) 1, 1- Dimethyl – 3 – hydroxy cyclohexane
c) 3, 3 – Dimethyl – 1 – hydroxy cyclohexane
d) 1, 1 – Dimethyl – 3 – cyclohexanol
Answer:
a) 3, 3 – Dimethyl – 1 – cyclohexanol

Question 60.
The IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 71 is
a) 2 – Ethylprop – 2 – en – 1 – ol
b) 2 – Hydroxymethyl butan – 1- ol
c) 2 – Methylene butan – 1 – ol
d) 2 – Ethyl -3 hydroxyprop – 1 – ene
Answer:
a) 2 – Ethylprop – 2 – en – 1 – ol

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 61.
The number of possible alkynes with molecular formula C5H8 is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 62.
What is the IUPAC name of the following Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 72 is
a) 3 – chloro cyclo hexa – 1, 5 – diene
b) 5 – chloro cyclo hexa – 1, 3 – diene
c) 1 – chloro cyclo hexa – 2, 5 – diene
d) 2 – chloro cyclo hexa – 1, 4 – diene
Answer:
b) 5 – chloro cyclo hexa – 1, 3 – diene

Question 63.
What is the IUPAC name of the following Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 73 is
a) 6 – hydroxy cyclohex – 2 – ene – 1 – al
b) 4 – hydroxy cyclohex – 1 – ene – 3 – al
C) 2 – hydroxy cyclohex – 5 – ene – 1 – al
d) 1 – formyl cyclohex – 5 – ene – 2 – ol
Answer:
a) 6 – hydroxy cyclohex – 2 – ene – 1 – al

Question 64.
What is the IUPAC name of the following?
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 74
a) 1 – chloro – 2 – bromo – 4 – nitrobenzene
b) 1 – bromo – 2- chloro – 4 – mtrobenzene
c) 3 – bromo – 4 – chloro – nitrobenzene
d) 2 – bromo – 1 – chloro – 4- nitrobenzene
Answer:
d) 2 – bromo – 1 – chloro – 4- nitrobenzene

Question 65.
What is the IUPAC name of the following?
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 75
a) Ethenyl cyclo pentane
b) cyclopentyl ethene
c) cyclopentyl ethylene
d) vinyl cyclopentane
Answer:
b) cyclopentyl ethene

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 66.
The hybridization of carbon atoms in C – C single bond is HC ≡ C – CH = CH2 is
a) sp3 – sp3
b) sp2 – sp3
c) sp – sp2
d) sp3 – sp
Answer:
c) sp – sp2

Question 67.
The correct IUPAC name of the compound Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 76 is
a) 1, 4 – Butane dioicacid
b) Ethane – 1, 2 – dicarboxylic acid
c) Succinic acid
d) 1, 2 – Ethane dioic acid
Answer:
a) 1, 4 – Butane dioic acid

Question 68.
Correct statements about CH3 – CH2 – CN is
a) common name of the compound is ethylcyanide.
b) IUPAC name of the compound propane – 1 – nitrile.
c) secondary suffix of the compound is nitrile.
d) IUPAC name of the compound is ethane nitrile.
Answer:
a) common name of the compound is ethylcyanide.

Question 69.
Tautomerism is shown by
a) R – C ≡ N
b) R – NO2
c) R – OH
d) R – COOH
Answer:
b) R – NO2

Question 70.
Stereo isomers have different
a) Molecular mass
b) Molecular formula
c) Structural formula
d) Configuration
Answer:
d) Configuration

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 71.
Geometrical isomerism may be exhibited by compounds having atleast.
a) One double bond
b) One triple bond
c) One asymmetric carbon
d) One polar bond
Answer:
a) One double bond

Question 72.
The prefixes syn – and anti – are used to denote
a) structural isomers
b) conformational isomers
c) geometrical isomers
d) optical isomers
Answer:
c) geometrical isomers

Question 73.
d – tartaric acid and l – tartaric acid are
a) geometrical isomers
b) conformers
c) enantiomers
d) diastereomers
Answer:
c) enantiomers

Question 74.
The method of separation of enantiomers from racemic mixture is known as
a) inversion
b) recemisation
c) resolution
d) asymmetric synthesis
Answer:
c) resolution

Question 75.
Racemic mixture is optically inactive due to
a) internal compensation
b) external compensation
c) inversion
d) plane of symmetry
Answer:
c) inversion

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 76.
Meso isomers are possible when the organic compound contains
a) one asymmetric carbon
b) two or more dissimilar asymmetric carbons
c) similar asymmetric carbons
d) unsaturation
Answer:
c) similar asymmetric carbons

Question 77.
Optical inactivity of meso isomer is due to
a) element of symmetry and element of asymmetry
b) internal compensation
c) due to lack of asymmetric carbon
d) External compensation
Answer:
b) internal compensation

Question 78.
A racemic mixture is a mixture of
a) meso and its isomers
b) d and l isomers of same compound in equimolar proportions
c) d and l isomers of same compound in different proportions
d) mixture of d and meso isomers
Answer:
b) d and 1 isomers of same compound in equimolar proportions

Question 79.
Which of the following is optically active?
a) HOOC – CH2 – COOH
b) CH3 – CO – COOH
c) CH3 – CH(OH) – COOH
d) CH3 – CH2 – COOH
Answer:
c) CH3 – CH(OH) – COOH

Question 80.
Which of the following is optically active?
a) n – propanal
b) 2 – chlorobutane
c) n – butanal
d) 3 – pentanol
Answer:
b) 2 – chlorobutane

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 81.
The number of optical enantiomers of tartaric acid
a) 3
b) 2
c) 4
d) 1
Answer:
b) 2

Question 82.
Geometrical isomers differ in
a) position of substituents
b) Position of double bond
c) C – C bond length
d) Spatial arrangement of groups
Answer:
d) Spatial arrangement of groups

Question 83.
Which of the following exhibit cis – trans isomerism
a) propene
b) 1 – butene
c) 2 – butene
d) benzene
Answer:
c) 2 – butene

Question 84.
Which of the following show geometrical isomerism?
a) CH3CH = CHCH3
b) (CH3)2C = CH2
c) C2H5CH = CH2
d) CH3CH = CH2
Answer:
a) CH3CH = CHCH3

Question 85.
Which of the following does not show geometrical isomerism?
a) 1, 2 – dichloro – 1 – pentene
b) 1, 3 – dichloro – 2 – pentene
c) 1, 1 – dichloro – 1 – pentene
d) 1, 4 – dichloro – 2 – pentene
Answer:
c) 1, 1 – dichloro – 1 – pentene

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 86.
Geometrical isomerism is not shown by
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 77

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 78

c) CH2 = C(CI) CH3

d) CH3 – CH = CH – CH = CH2
Answer:
c) CH2 = C(CI) CH3

Question 87.
Which of the following is optically active?
a) Glycerine
b) Acetaldehyde
c) Glyceraldehyde
d) Acetone
Answer:
c) Glyceraldehyde

Question 88.
The minimum number of C atoms for a hydrocarbon to exhibit optical isomerism
a) 4
b) 5
c) 6
d) 7
Answer:
d) 7

Question 89.
Which of the following can form cis – trans isomers?
a) C2H5Br
b) (CH)2(COOH)2
c) CH3CHO
d) (CH2)2COOH
Answer:
b) (CH)2(COOH)2

Question 90.
No.of geometrical isomers possible for the compound CH3 – CH = CH – CH = CH – C2H5
a) 2
b) 3
c) 4
d) 5
Answer:
c) 4

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 91.
The number of geometrical isomers of CH3 – CH = CH – CH = CH – CH = CHCl is
a) 2
b) 4
c) 6
d) 8
Answer:
d) 8

Question 92.
Minimum number of C atoms for an alkene hydrocarbon, that shows geometrical & optical isomerism both
a) 5
b) 6
c) 7
d) 8
Answer:
c) 7

Question 93.
Among the following compounds which exhibits optical isomerism?
a) propanol
b) 2 – propanol
c) 1 – butanol
d) 2 – butanol
Answer:
d) 2 – butanol

Question 94.
Which of the mesoisomer?
a) CH2OHCHOHCHO
b) CH2OHCHOHCHOHCHO
c) HOOCCHOHCHOHCOOH
d) HOH2CCHOHCHOHCOOH
Answer:
c) HOOCCHOHCHOHCOOH

Question 95.
d – tartaric acid and l – tartaric acid can be separated by
a) Salt formation
b) Fractional distillation
c) Fractional crystallization
d) Chromatography
Answer:
a) Salt formation

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 96.
Paper chromatography is
a) Adsorption chromatography
b) partition chromatography
c) Ion exchange chromatography
d) all of these
Answer:
b) partition chromatography

Question 97.
Simple distillation can be used to separate liquids which differ in their boiling points at least by
a) 5°C
b) 10°C
c) 40 – 50°C
d) 100°C
Answer:
c) 40 – 50°C

Question 98.
In adsorption chromatography mobile phase will be
a) Only solid
b) Only liquid
c) Only gas
d) Liquid as well as gas
Answer:
d) Liquid as well as gas

Question 99.
Which of the following can be used as adsorbent in adsorption chromatography?
a) Silica gel
b) Alumina
c) Cellulose powder
d) All of these
Answer:
d) All of these

Question 100.
Two substances when separated out on the basis of their extent of adsorption by one material, the phenomenon is called
a) Chromatography
b) Crystallization
c) Sublimation
d) Steam distillation
Answer:
a) Chromatography

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 101.
Chromatographic technique is used for the separation of
a) Camphor
b) Alcohol & Water
c) Acetone and Methanol
d) Plant pigments
Answer:
d) Plant pigments

Question 102.
In column chromatography stationary phase is
a) only solid
b) only liquid
c) only gas
d) All of these
Answer:
a) only solid

Question 103.
Which of the following method is used for the purification of solids?
a) Distillation under reduced pressure
b) Distillation
c) Strain distillation
d) Sublimation
Answer:
d) Sublimation

Question 104.
Vacuum distillation is used to purify liquids which
a) are highly volatile
b) are explosive in nature
c) soluble in water
d) decomposes below their B.P’s
Answer:
c) soluble in water

Question 105.
Impure Napthalene is purified by
a) Fractional crystallization
b) Fractional distillation
C) solvent extraction
d) sublimation
Answer:
d) sublimation

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 106.
A very common adsorbent used in coloun^n chromatography is
a) Powdered charcoal
b) Alumina
c) Chalk
d) Sodium carbonate
Answer:
b) Alumina

Question 107.
Simple distillation of liquids involves simultaneously
a) Vapourisation and condensation
b) Condensation and vapourisation
c) Vapourisation and sublimation
d) Sublimation and condensation
Answer:
a) Vapourisation and condensation

Question 108.
The latest technique for the purification of organic compounds is
a) Fractional distillation
b) Chromatography
c) Vacuum distillation
d) Crystallization
Answer:
b) Chromatography

Question 109.
Fixed melting point of an organic compound informs
a) Purity of an organic compound
b) Conductivity of compound
c) Chemical nature of compound
d) Whether the compound is liquid or gas
Answer:
a) Purity of an organic compound

Question 110.
Lassaigne’s test is used in qualitative analysis to detect
a) Nitrogen
b) Sulphur
c) Chlorine
d) All of these
Answer:
d) All of these

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 111.
In Lassaigne’s method organic compound is fused with
a) Sodium metal
b) Zinc dust
c) Sodium carbonate and Zinc dust
d) Calcium metal
Answer:
a) Sodium metal

Question 112.
Presence of nitrogen in organic compound in Lassaigne’s extract as
a) Nitrogen gas
b) NH3
c) NO
d) CN
Answer:
d) CN

Question 113.
Medium of Sodium extract is
a) Neutral
b) Basic
c) Acidic
d) Depends on organic compound
Answer:
b) Basic

Question 114.
H2O vapours on passing through anhydrous CuSO4 turns it to
a) Green
b) Blue
c) Violet
d) White
Answer:
b) Blue

Question 115.
When a nitrogenous organic compound is fused with sodium, the nitrogen present in the compound is converted into
a) Sodium Nitrate
b) Sodium nitrite
c) Sodamide
d) Sodium cyanide
Answer:
d) Sodium cyanide

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 116.
In the Lassainge’s test the Sulphur present in the organic compound first changes into
a) Na2SO3
b) CS2
c) Na2SO4
d) Na2S
Answer:
d) Na2S

Question 117.
Which of the following elements in an organic compound cannot be detected by Lassaigne’s test?
a) N
b) S
c) Cl
d) H
Answer:
d) H

Question 118.
A compound which does not give a positive result in the Lassaigne’s test for nitrogen is
a) Urea
b) Hydroxyl amine
c) Glycine
d) Phenylhydrazine
Answer:
b) Hydroxyl amine

Question 119.
Lassaigne’s test gives a violet colouration with sodium nitroprusside, it indicates presence of
a) N
b) S
c) O
d) Cl
Answer:
b) S

Question 120.
The presence of halogen in an organic compound is detected by
a) Iodoform test
b) Silver nitrate test
c) Beilstein’s test
d) Million’s test
Answer:
c) Beilstein’s test

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 121.
The Beilstein’s test in a rapid test used for . organic compound^ to^ detect
a) Phosphorous
b) Sulphur
c) Halogens
d) Nitrogen
Answer:
c) Halogens

Question 122.
Liebig’s method is used for the estimation of
a) Nitrogen
b) Sulphur
c) Carbon and hydrogen
d) Halogens
Answer:
c) Carbon and hydrogen

Question 123.
In Kjeldahl’s method of estimation of nitrogen, copper sulphate act as
a) Oxidizing agent
b) reducing agent
c) Catalytic agent
d) Hydrolysing agent
Answer:
c) Catalytic agent

Question 124.
Percentage of carbon in an organic compound is determined by
a) Duma’s method
b) Kjeldahl’s method
c) Carius method
d) Liebig’s method
Answer:
d) Liebig’s method

Question 125.
Halogen can be estimated by
a) Duma’s method
b) Carius method
c) Leibig’s method
d) All of these
Answer:
b) Carius method

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 126.
In Garius method halogens are estimated
a) X2
b) BaX2
c) PbX2
d) AgX
Answer:
d) AgX

Question 127.
In Duma’s method nitrogen in organic compound is estimated in the form of
a) N2
b) NO
c) NH3
d) N2O5
Answer:
a) N2

Question 128.
In Kjeldahl’s method to estimate nitrogen, compound is heated with conc.H2SO4 in presence of
a) CaSO4
b) (NH4)2SO4
c) CuSO4
d) P2O5
Answer:
c) CuSO4

Question 129.
In organic compounds, Sulphur is estimated as
a) BaSO4
b) BaCl2
c) Ba3(PO4)2
d) H2SO4
Answer:
a) BaSO4

Question 130.
In the Liebig’s method, if ‘w’ is the mass of compound taken and ‘x’ is the amount of C0„ formed then
a) %C = \(\frac{12 \times x}{16 \times w}\)

b) %C = \(\frac{12}{44} \times \frac{\mathrm{w}}{\mathrm{x}} \times 100\)

c) %C = \(\frac{12}{44} \times \frac{x}{w} \times 100\)

d) %C = \(\frac{12}{44} \times \frac{x}{w}\)
Answer:
c) %C = \(\frac{12}{44} \times \frac{x}{w} \times 100\)

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 131.
In Dumas method for estimating nitrogen in organic compound, the gas finally collected is
a) N2
b) NO
c) NH3
d) H2
Answer:
a) N2

Question 132.
In Dumas method, the gas which is collected in Nitrometer is
a) N2
b) NO
c) NH3
d) H2
Answer:
a) N2

Question 133.
In Kjeldahl’s method, the nitrogen presence is estimated as
a) N2
b) NH3
c) NO2
d) N2O3
Answer:
b) NH3

Question 134.
In Kjeldahl’s method, nitrogen present in the organic compound is first converted into
a) NH3
b) (NH4)2SO4
c) N2
d) NO
Answer:
b) (NH4)2SO4

Question 135.
In Liebig’s method for the estimation of C and H, the combustion tube is passed over
a) CuO pellets
b) Copper turnings
c) Iron fillings
d) Zinc – copper couple
Answer:
a) CuO pellets

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 136.
Which gas is introduced into the combustion tube in Liebig’s method?
a) Pure and dry CO2
b) Pure and dry N2
c) Pure and dry O2
d) Pure and dry He
Answer:
c) Pure and dry O2

Question 137.
Chromatographic techniques of purification can be used for
a) Coloured compounds
b) Liquids
c) Solids
d) All of these
Answer:
d) All of these

Question 138.
Two substances when separated on the basis of partition co – efficient between two liquid phase, then the technique is known as
a) column chromatography
b) Paper chromatography
c) GLC
d) TLC
Answer:
b) Paper chromatography

Question 139.
Ortho and para nitro phenols can be separated by
a) crystallization
b) distillation
c) sublimation
d) solvent extraction
Answer:
b) distillation

Question 140.
In steam distillation, the sum of the vapour pressure of the volatile compound and that of water is
a) Equal to atmospheric pressure
b) Less than atmospheric pressure
c) More than atmospheric pressure
d) Exactly half of the atmospheric pressure
Answer:
a) Equal to atmospheric pressure

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 141.
Organic compound is fused with metallic sodium for testing nitrogen, sulphur, and halogens because
a) To make the solution alkaline
b) To convert into elemental state of nitrogen, sulphur, and halogens
c) To convert covalent compound into ionic compound
d) To decrease fusion temperature
Answer:
c) To convert covalent compound into ionic compound

Question 142.
Sodium extract gives blood red colour when treated with FeCl3, Formation of blood-red colour confirms the presence of
a) Only nitrogen
b) Only sulphur
c) Only halogens
d) Both Nitrogen and Sulphur
Answer:
d) Both Nitrogen and Sulphur

Question 143.
The compound not formed in the positive test for nitrogen with the Lussaigne’s solution of an organic compound is
a) Fe4[Fe(CN)6]3
b) Na3[Fe(CN)6]
c) Fe(CN)3
d) Na3[Fe(CN)5NOS]
a) b, c, d
b) a, b
c) a, b, c
d) a only
Answer:
a) b, c, d

Question 144.
The Lassaigne’s solution when heated with ferrous sulphate and acidified with sulphuric acid gave intense blue colour indicating the presence of nitrogen. The blue colour is due to the formation of
a) Na4[Fe(CN)6]
b) Fe3[Fe(CN)6]2
c) Fe2[Fe(CN)6]
d) Fe4[Fe(CN)6]3
Answer:
d) Fe4[Fe(CN)6]3

Question 145.
Which of the following compounds will answer Lassaigne’s test for nitrogen?
a) NH2NH2
b) NH2OH
c) NaCN
d) NaNO3
Answer:
c) NaCN

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 146.
In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is
a) 28
b) 38
c) 18
d) 48
Answer:
a) 28

Question 147.
0.73 g of organic compound on oxidation gave 1.32 g of carbon dioxide. The percentage of carbon in the given compound will be
a) 49.32
b) 59.32
c) 29.32
d) 98.64
Answer:
a) 49.32

Question 148.
In an estimation of S by Carius method 0.217 g of the compound gave 0.5825 g of BaSO4. Percentage of S is
a) 36.78 %
b) 35.50 %
c) 36.48 %
d) 35.69 %
Answer:
a) 36.78 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

II. Very short question and answers (2 Marks):

Question 1.
What are Acyclic compounds? Give suitable example.
Answer:
These are the compounds in which carbon atoms are linked to form open chain (straight or branched). These compounds may be saturated (all single bonds) or unsaturated (multiple bonds).
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 79

These acyclic compounds are known as open chain or aliphatic compounds.

Question 2.
What arc Alicyclic Compounds? Give suitable example.
Answer:
These are saturated or unsaturated carbocyclic compounds which resemble the corresponding acylic compounds in their properties.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 80

Question 3.
What are Aromatic heterocyclic Compounds? Give example.
Answer:
These are the heterocyclic compounds which possess aromaticity and resemble, the corresponding aromatic compounds in most of their properties. These are also called non-benzenoid aromatic compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 81

Question 4.
What is functional group? Give example.
Answer:
An atom or group of atoms within a molecule that shows a characteristics set of physical and chemical properties.
Example:
(i) – NH2 – amines
(ii) = NH – Imines
(iii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 82

Question 5.
Almost all compounds of carbon form covalent bonds. Give reason.
Answer:
Carbon (Z = 6) have electronic configuration of is2 2s2 2p2. It is not possible for the carbon to form either C4+ or C4- ions to attain the nearest noble gas configuration as it requires large amount of energy. This implies that carbon cannot form ionic bond. So almost in all compounds of carbon, form four covalent bonds.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 6.
What is Metamerism? With suitable examples.
Answer:
Metamerism:
This type of isomerism is a special kind of structural isomerism arises due to the unequal distribution of carbon atoms on either side of the functional group or different alkyl groups attached to the either side of the same functional group and having same molecular formula. This isomerism is shown by compounds having functional group such as ethers, ketones, esters and secondary amines between two alkyl groups.
C4H10O
CH3 – O – C3H7 – Methyl propyl ether 1 – methoxypropane

C2H5 – O – C2H5 – diethyl ether ethoxyethane

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 83 – methyl isopropyl ether 2 – methoxypropane.

Question 7.
What is meant by stereochemistry?
Answer:
The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This branch of chemistry dealing with the study of the three-dimensional nature (spactial arrangement) of molecules is known as stereochemistry. The metabolic activities in living organisms, natural synthesis and drug synthesis involve various stereoisomers.

Question 8.
What is enantiomerism?
Answer:
An optically active substance may exist in two or more isomeric forms which have same physical & chemical properties but differ in terms of direction of rotation of plane-polarized light, such optical isomers which rotate the plane of polarized light with equal angle but in opposite direction are known as enantiomers and the phenomenon is known as enantiomerism.
Example: d and l lactic acid.

Question 9.
What are the conditions for an organic compound is said to be optically active?
Answer:
(1) The molecule must contains at least one chiral or Asymmetric carbon atom.
(2) The object molecule should not be super impossable with its own mirror image.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 84

Question 10.
How will you detect phosphorus present in the given organic compound?
Answer:
Test for phosphorous:
A solid compound is strongly heated with a mixture of Na2CO3 & KNO3. Phosphorous present in the compound is oxidized to sodium phosphate. The residue is extracted with water and boiled with Conc. HNO3. A solution of ammonium molybdate is added to the above solution. A canary yellow coloration or precipitate shows the presence of phosphorous.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
Explain the sublimation process for the purification of organic compounds.
Answer:
Few substances like benzoic acid, naphthalene and camphor when heated pass directly from solid to vapor without melting (ie liquid). On cooling the vapours will give back solids. Such phenomenon is called sublimation. It is a useful technique to separate volatile and non-volatile solid. It has limited application because only a few substance will sublime.
Example:
naphthalene, benzoic acid.

Question 12.
What is the need for purifying an organic compound?
Answer:
In order to study the structure, physical properties, chemical properties and biological properties of organic compounds they must be in a pure state.

III. Short question and answers (3 Marks):

Question 1.
How are naphthalene and camphor purified?
Answer:
1. Naphthalcne, camphor and benzoic acid when heated, pass directly from solid to vapour without melting. On cooling the vapours will give back solid. This phenomenon is known as sublimation. This technique is used to purify naphthalene, camphor from non-volatile impurities.

2. Substances to he purified is taken in a beaker. It is covered with a watch glass. The beaker is heated for a while and the resuJting vapours condense on the bottom of the watch glass. Then the watch glass is removed and the crystals are collected.

Question 2.
What are Alicycic heterocyclic Compounds? Give example.
Answer:
These heterocyclic compounds resemble the corresponding aliphatic compounds in most of their properties.
Example:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 85

Question 3.
Write the IUPAC name of the following compounds.
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 86

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 87

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 88

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 89
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 90

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 91

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 92

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 93

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 4.
Write the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 94
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 95

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 96

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 97

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 98

Question 5.
Write the IUPAC name of the following compounds.
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 99
b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 100

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 101

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 102
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 103

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 104

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 105

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 106

Question 6.
Write the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 112
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 113

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 114

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 115

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 116

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 7.
Write the IUPAC name of the following compounds.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 107
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 108

b) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 109

c) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 110

d) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 111

Question 8.
Classify the following compounds based on the structure.
i) CH ≡ C – CH2 – C ≡ CH

ii) CH3 – CH2 – CH2 – CH2 – CH3

iii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 117

iv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 118
Answer:
i) CH ≡ C – CH2 – C ≡ CH is unsaturated open chain compound

ii) CH3 – CH2 – CH2 – CH2 – CH3 is saturated open chain compound

iii) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 117 is aromatic benzenoid compound

iv) Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 118 is alicyclic compound

Question 9.
Give two examples for each of the following type of organic compounds.
(i) non-benzenoid aromatic
(ii) aromatic heterocyclic
(iii) alicyclic
(iv) aliphatic open chain
Answer:
(i) non-benzenoid aromatic
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 118

(ii) aromatic heterocyclic
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 119

(iii) alicyclic
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 120

(iv) aliphatic open chain
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 121

Question 10.
Explain how nitrobenzene and benzene arc separated and purified. (or) How will you separate the mixture of diethyl ether and ethanol?
Answer:
Distillation:

  • The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver.
  • This method ¡s to purify liquids from non-volatile impurities and used for separating the constituents of a liquid mixture which differ in their boiling points.
  • In this simple distillation process, liquids with large difference in boiling point (about 40K) and do not decompose under ordinary pressure can be purified.
  • e.g., the mixture of C6H5NO2 nitrobenzene (h.p. 484 K) and C6H6 benzene (b.p. 354 K) can be urified and separated. Similarly the mixture of diethyL ether (b.p. 308K) and ethyl alcohol (h.p. 351 K) can he purified and separated.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
In an estimation of sulphur by carius method. 0.2175 g of the substance gave 0.5825 g of BaSO4 calculate the percentage composition of S in the compound.
Solution:
Weight of organic compound = 0.2175 g
Weight of BaSO4 = 0.5825 g
233 g of BaSO4 contains = \(\left(\frac{32}{233} \times \frac{0.5825}{0.2175}\right)\)
0.5825 g of BaSO4 contains
Percentage of S = \(\left(\frac{32}{233} \times \frac{0.5825}{0.2175} \times 100\right)\)
= 36.78 %

Question 12.
0.16 g of an organic compound was heated in a carius tube and H2SO4 acid formed
was precipitated with BaCl2. The mass of BaSO4 was 0.35 g. Find the percentage of
sulphur [30.04]
Answer:
Weight of organic substance (w) = 0.16 g
Weight of Barium sulphate (x) = 0.35 g
Percentage of sulphur = \(\left(\frac{32}{233} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{32}{233} \times \frac{0.35}{0.16} \times 100\right)\)
= 30.04 %

Question 13.
0.284 g of an organic substance gave 0.287 g AgCl in a Carius method for the estimation of halogen. Find the percentage of Cl in the compound.
Solution:
Weight of the organic substance = 0.284 g
Weight of AgCl is = 0.287 g
143.5 g of AgCl contains = 35.5 g of chlorine
0.287 g of AgCl contains = \(\left(\frac{35.5}{143.5} \times \frac{0.287}{0.284}\right)\)

% of chlorine is = \(\left(\frac{35.5}{143.5} \times \frac{0.287}{0.284} \times 100\right)\)
= 24.56 %

Question 14.
0.185 g of an organic compound when treated with Conc. HNO3 and silver nitrate gave 0.320 g of silver bromide. Calculate the % of bromine in the compound.
(Ag = 108, Br = 80)
Answer:
Weight of organic substance (w) = 0.185 g
Weight of silver bromide (x) = 0.320 g
Percentage of bromine = \(\left(\frac{80}{188} \times \frac{x}{w} \times 100\right)\)
= \(\left(\frac{80}{188} \times \frac{0.32}{0.185} \times 100\right)\)
= 73.6%

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 15.
0.40 g of a iodo – substituted organic compound gave 0.235 g of AgI by carlus method. Calculate the percentage of iodine in the compound. (Ag = 108 I = 127).
Solution:
Weight of organic substance (w) = 0.40 g
Weight of silver iodide (x) = 0.235 g
Percentage of iodine = \(\left(\frac{127}{235} \times \frac{x}{w} \times 100\right)\)

= \(\left(\frac{127}{235} \times \frac{0.235}{0.40} \times 100\right)\)
= 31.75%

Question 16.
0.24 g of organic compound containing phosphorous gave 0.66 g of Mg2P2O7 by the usual analysis. Calculate the percentage of phosphorous in the compound.
Solution:
Weight of an organic compound = 0.24 g
Weight of Mg2P2O7 = 0.66 g
222 g of Mg2P2O7 contains = 62 g of P
0.66 g contains = \(\frac{62}{222} \times \frac{0.66}{0.24}\)

Percentage of p = \(\frac{80}{180} \times \frac{0.66}{0.24}\) = 76.80%

Question 17.
0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg2P2O7 by the analysis. Calculate the percentage of P in the compound.
Solution:
Weight of organic substance (w) = 0.33 g
Weight of Mg2P2O7 (x) = 0.397 g
Percentage of phosphorous = \(\frac{62}{222} \times \frac{x}{w} \times 100\)

= \(\frac{62}{222} \times \frac{0.397}{0.33} \times 100\) = 33. 59 %

Question 18.
Explain simple distillation process with suitable example.
Answer:
The process of distillation involves the impure liquid when boiled gives out vapour and the vapour so formed is collected and condensed to give back the pure liquid in the receiver. This method is called simple distillation. Liquids with large difference in boiling point(about 40k) and do not decompose under ordinary pressure can be purified by simple distillation.
Example:
The mixture of C6H5NO2 (b.p 484K) &
C6H6 (354 K) and mixture of diethyl ether (b.p 308K) and ethyl alcohol (b.p 35 K).

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 19.
How will you purify an organic compound by differential extraction process?
Answer:
The process of removing a substance from its aqueous solution by shaking with a suitable organic solvent is termed extraction. When an organic substance present as solution in water can be recovered from the solution by means of a separating funnel. The aqueous solution is taken in a separating funnel with little quantity of ether or chloroform (CHCl3).

The organic solvent immiscible with water will form a separate layer and the contents are shaken gently. The solute being more soluble in the organic solvent is transfered to it. The solvent layer is then separated by opening the tap of the separating funnel, and the substance is recovered.

Question 20.
Explain the steam distillation process for purifying organic compound.
Answer:
This method is applicable for solids and liquids. If the compound to be steam distilled the compound should not decompose at the steam temperature, should have a fairly high vapour pressure at 373 K, it should be insoluble in water; the impurities present should be non-volatile.

The impure liquid along with little water is taken in a round-bottom flask which is connected to a boiler on one side and water condenser on the other side, the flask is kept in a slanting position so that no droplets of the mixture will enter into the condenser on the brisk boiling and bubbling of steam. The mixture in the flask is heated and then a current of steam passed in to it. The vapours of the compound mix up with steam and escape into the condenser.

The condensate obtained is a mixture of water and organic compound which can be separated. This method is used to recover essential oils from plants and flowers, also in the manufacture of aniline and turpentine oil.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 21.
Write a note on azeotropic distillation with suitable example.
Answer:
These are the mixture of liquids that cannot be separated by fractional distillation. The mixtures that can be purified only by azeotropic distillation are called as azeotropes. These azeotropes are constant boiling mixtures, which distil as a single compound at a fixed temperature. Ethanol and water are the most common examples of azeotropic mixture in the ratio of 95.87 : 4.13.

In this method apart from azeotropic mixture a third component like C6H6, CCl4, ether, glycerol, glycol which act as a dehydrating agent depress the partial pressure of one component so that the boiling point of that component is raised sufficiently and thus other component will distill over.

Dehydrating agents like C6H6, CCl4 have low boiling points and reduce the partial vapour pressure of alcohol more than that of water whereas glycerol & glycol, etc. have high boiling point and reduce the partial vapour pressure of water more than that of alcohol.

Question 22.
What is Ring chain isomerism? Give an example.
Answer:
In this type of isomerism, compounds having same molecular formula but differ terms of bonding of carbon atom to form open-chain and cyclic structures.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 123

Question 23.
Briefly explain geometrical isomerism in oximes with suitable example.
Ans:
Restricted rotation around C = N (oximes) gives rise to geometrical isomerism in oximes. Here ‘syn’ and ‘anti’ are used instead of cis and trans respectively. In the syn isomer the H atom of a doubly bonded carbon and -OH group of doubly bonded nitrogen lie on the same side of the double bond, while in the anti isomer, they lie on the opposite side of the double bond.
For eg:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 124

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

IV. Long Question and answers (5 Marks):

Question 1.
Write the structure of the following compounds.
a) Cinnamic acid
b) Lactic acid
c) Phthalic acid
d) Tartaric acid
d) Benzoic acid
Answer:
a) Cinnamic acid
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 125

b) Lactic acid
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 126

c) Phthalic acid
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 127

d) Tartaric acid
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 128

d) Benzoic acid
C6H5COOH

Question 2.
Explain the methods for the representation of structure of organic compounds with suitable example.
Answer:
The structure of an organic compound can he represented using any one of the below mentioned methods.

  1. Lewis structure or dot structure,
  2. Dash structure or line bond structure,
  3. Condensed structure
  4. Bond line structure

We know how to draw the Lewis structure for a molecule. The line bond structure is obtained by representing the two electron covalent bond by a dash or line (-) in a Lewis structure. A single line or dash represents single a covalent bond, double line represents double bond (1 σ bond, 2 π bond) and a triple line represents triple bond (1 σ bond, 2 π bond). Lone pair of electrons on hetero atoms may or may not be shown. This represents the complete structural formula.

This structural formula can be further abbreviated by omitting some or all of the dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.

For further simplification, organic chemists use another way of representing the structures in which only lines are used. In this type of representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bonds are shown in a zigzag fashion. The only atoms specifically written are oxygen, chlorine, nitrogen etc. These representations can be easily understood by the following illustration.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 129

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 3.
Explain the molecular model method for the representation of structure of organic compounds.
Answer:
Molecular models:
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 130

Molecular models are physical devices that are used for a better visualisation and perception of three dimensional shapes of organic molecules. These are made of wood, plastic or metal and are commercially available.
(i) Frame work model
(ii) Ball and stick model &
(iii) space filling model.

In the frame work model only the bonds connecting the atoms themselves are shown. This model emphasizes the pattern of bonds of a molecule while ignoring the size of the atom.

In the ball and stick model, both the atoms and the bonds are shown. Ball represent atoms and the stick a bond. Compounds containing C = C can be best represented by using springs in place of sticks and this model is termed as ball and spring model.

The space filling model emphasizes the relative size of each atom based on its Vander Waals radius.

Question 4.
Explain briefly
(i) Fisher projection
(ii) sawhorse projection
(iii) Newman projection formula with neat example.
Answer:
(i) Fisher projection formula:
This is a method of representing three dimensional structures in two dimension. In this method, the chiral atom(s) lies in the plane of paper. The horizontal substituents are pointing towards the observer and the vertical substituents are away from the observer. Fisher projection formula for tartaric acid is given below.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 131

Sawhorse projection formula:
Here the bond between two carbon atoms is drawn diagonally and slightly elongated. The lower left hand carbon is considered lying towards the front and the upper right hand carbon towards the back. The Fischer projection inadequately portrays the spatial relationship between ligands attached to adjacent atoms. The sawhorse projection attempts to clarify the relative location of the groups.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 132

Newman projection formula:
In this method the molecules are viewed from the front along the carbon-carbon bond axis. The two carbon atom forming a bond is represented by two circles. One behind the other so that only the front carbon is seen. The front carbon atom is shown by a point whereas the carbon lying further from the eye is represented by the origin of the circle. Therefore, the C-H bonds of the front carbon are depicted from the circle while C-H bonds of the back carbon are drawn from the circumference of the circle with an angle of 120° to each other.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 133

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 5.
What is Tautomerism? Explain different types of Tautomerism with suitable examples.
Answer:
Tautomerism:
It is a special type of functional isomerism in which a single compound exists in two readily inter convertible structures that differ markedly in the relative position of atleast one atomic nucleus, generally hydrogen. The two different structures are known as tautomers. There are several types of tautomerism and the two important types are dyad and triad systems.

Dyad system:
In this system, hydrogen atom oscillates between two directly linked polyvalent atoms.
Example:

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 134

In this example, the hydrogen atom oscillates between carbon & nitrogen atom.

Triad system:
In this system, the hydrogen atom oscillates between three polyvalent atoms. It involves 1, 3 migration of hydrogen atom from one polyvalent atom to other within the molecule. The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are keto form and enol-form. The polyvalent atoms involved are one oxygen and two carbon atoms. Enolisation is a process in which keto-form is converted to enol form. Both tautomeric forms are not equally stable. The less stable form is known as the labile form.
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 135

Question 6.
How is sulphur detected by Lassaigne test?
Answer:
a) To a portion of the lassaigne’s extract, add freshly prepared sodium nitroprusside solution. A deep violet colouration is obtained. This test is also used to detect S2- in inorganic salt analysis.
Na2S + Na2 [Fe(CN5) NO] → Na4 [Fe (CN5) NOS]
sodium nitroprusside

b) Acidify another portion of lassaigne’s extract with acetic acid and add lead acetate solution. A black precipitate is obtained.
(CH3COO)2Pb + Na2S → PbS↓ + 2CH3COONa
(black ppt)

c) Oxidation test:
The organic substances are fused with a mixture of KNO3 and Na2CO3. The sulphur, if present is oxidized to sulphate.
Na2CO3 + S + 3O + Na2SO4 + CO2
The fused mass is extracted with water, acidified with HCl and then BaCl2 solution is added to it. A white precipitate indicates the pressure of sulphur.
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 7.
How is halogen detected by using the Lassaigne test?
Answer:
To another portion of the Lassaigne’s filtrate add dil HNO3 warm gently and add AgNO3
Solution:
a) Appearance of curdy white precipitate soluble in ammonia solution indicates the presence of chlorine.
b) Appearance of pale yellow precipitate sparingly soluble in ammonia solution indicates the presence of bromine.
c) Appearance of a yellow precipitate insoluble in ammonia solution indicates the presence of iodine.

Na + Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 136 NaX ( where X = Cl, Br, I)
from organic compound
NaX + AgNO3 → AgX + NaNO3

If N or S is present in the compound along with the halogen, we might obtain NaCN and Na2S in the solution, which interfere with the detection of the halogen in the AgNO3 test Therefore we boil the lassaignes extract with HNO3 which decomposes NaCN and Na2S as

NaCN + HNO3 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 137 NaNO3 + HCN T
Na2S + 2HNO3 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 137 2NaNO3 + H2S
NaCN + AgNO3 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 138 AgCN + NaNO3
white ppt confusing with AgCl
Na2S + AgNO3 → Ag2S ↓ + NaNO3
black ppt

Question 8.
Explain the various steps involved in the crystallization method.
Answer:
Most solid organic compounds are purified by the crystallization method. This process is carried out by the following steps.
1. Selection of solvent:
rganic substances being covalent do not dissolve in water, hence selection of suitable solvent becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is sufficient to dissolve the organic compound.

If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with benzene, ether, acetone and alcohol various solvent. till the most suitable one is sorted out.

2. Preparation of solution:
The organic compound is dissolved in minimum quantity of suitable solvent small amount of animal charcoal can he added to decolonize any colored substance. The solution may be prepared by heating over a wire gauze or water bath.

3. Filtration of hot solution:
The hot solution so obtained is filtered through a fluted filter paper placed in a funnel.

4. Crystallization:
The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper. the pure solid substance separate as crystal. If the rate of crystallization slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of pure compounds to the solution.

5. Isolation and drying of crystals:
The crystals are separated from the mother liquor by filtration is done under reduced pressure using a Buchner funnel. Finally the crystals are washed with small amount of pure cold solvent and dried.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 9.
Explain the Carius method for the estimation of sulphur in an organic compound.
Answer:
Estimation of sulphur Carius method:
A known mass of the organic substance is heated strongly with fuming HNO3, C & H get oxidized to CO2 & H2O while sulphur is oxidized to sulphuric acid as per the following reaction.

C Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 139 CO2
2H Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 139 H2O
S → SO2 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 140 H2SO4

The resulting solution is treated with excess of BaCl2 solution H2SO4 present in the solution in quantitatively converted into BaSO4, from the mass of BaSO4, the mass of sulphur and hence the percentage of sulphur in the compound can be calculated.

Procedure:
A known mass of the organic compound is taken in clean carius tube and added a few mL of fuming HNO3. The tube is the sealed. It is then placed in an iron tube and heated for about 5 hours. The tube is allowed to cool to temperature and a small hole is made to allow gases produced inside to escape.

The carius tube is broken and the content collected in a beaker. Excess of BaCl2 is added to the beaker H2SO4 acid formed as a result of the reaction is converted to BaSO4. The precipitate of BaSO4 is filtered, washed, dried and weighed. From the mass of BaSO4, percentage of S is found.

Mass of the organic compound = w g
Mass of the BaSO4 formed = x g
233 g of BaSO4 contains = 32 g of sulphur

∴ x g of BaSO4 contain = \(\left(\frac{32}{233} \times \frac{x}{w}\right)\) g of S

Percentage of sulphur = (\(\left(\frac{32}{233} \times \frac{x}{w}\right)\) × 100) %

Question 10.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO2 = 0.245 g
Percentage of hydrogen
18 g of water contain 2 g of hydrogen
0.039 of water contain = \(\left(\frac{2}{18} \times \frac{0.039}{0.26}\right)\) of hydrogen

% of hydrogen = \(\left(\frac{0.039}{0.26} \times \frac{2}{18} \times 100\right)\) = 1.66%
Percentage of Carbon
44 g of CO2 contain 12 g of C
0.245 g of CO2 contains \(\left(\frac{12}{44} \times \frac{0.245}{0.26}\right)\) g of C

% of Carbon = \(\left(\frac{12}{44} \times \frac{0.245}{0.26} \times 100\right)\) = 25.69 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 11.
0.2346 of an organic compound yielded C, H & 0 0.2754 g of H2O and 0.4488 g CO2. Calculate the % composition.
Solution:
Weight of organic substance w = 0.2346 g
Weight of water (x) = 0.2754 g
Weight of CO2 (y), = 0.4488 g
Percentage of carbon = \(\left(\frac{12}{44} \times \frac{y}{w} \times 100\right)\)

= \(\left(\frac{12}{44} \times \frac{0.4488}{0.2346} \times 100\right)\) = 52.17 %

Percentage of hydrogen = \(\left(\frac{2}{18} \times \frac{x}{w} \times 100\right)\)
= \(\left(\frac{2}{18} \times \frac{0.2754}{0.2346} \times 100\right)\) = 13. 04%
Percentage of oxygen= [100 – (52.17 + 13.04)]
= 100 – 65.21 = 34.79 %

Question 12.
Explain the estimation of phosphours by Carius method.
Answer:
Carius method:
A known mass of the organic compound (w) containing phosphorous is heated with fuming HNO in a sealed tube where C is converted into CO2 and H to H2O. phosphorous present in organic compound is oxidized to phosphoric acid which is precipitated, as ammonium phosphomolybdate by heating with Cone. HNO3 and then adding ammonium molybdate.

H3PO4 + 12(NH4)2MoO4 + 21 HNO3 Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 141 (NH4)3PO4 + 21 NH4NO3 + 12HNO3

The precipitate of ammonium phosphomolybdate thus formed is filtered washed, dried, and weighed. In an alternative method, the phosphoric acid is precipitated as magnesium-ammonium phosphate by adding magnesia mixture (a mixture containing MgCl2, NH4Cl, and ammonia) This ppt is washed, dried, and ignited to get magnesium pyrophosphate which is washed, dried a weighed. The following are the reaction that takes place.

By knowing the mass of the organic compound and the, mass of ammonium phosphomolybdate or magnesium pyrophosphate formed, the percentage of P is calculated.

Mass of organic compound = w g
Weight of ammonium phosphomolybdate = x g
Weight of magnesium pyrophosphate = y g
Mole mass of (NH4)3PO4. 12MoO3 is = 1877 g
[3 × (14 + 4) + 31 + 4(16)]+ 12(96 + 3 × 16)
Molar mass of Mg3P2O7 is 222 g
(2 × 24) + (31 × 2) + (7 × 16)
1877 g of (NH4)3PO4. 12 MoO3 contains 31 g of P
X g of (NH4)3PO4. 12 MoO3 in w g of organic
compound contains \(\frac{31}{1877} \times \frac{x}{w}\) of phosphorous
Percentage of phosphorous = \(\frac{31}{1877} \times \frac{x}{w}\) (or)
227 of Mg2P2O7 contains 62 g of P
y g of Mg2P2O2 in w g of Organic compound contains = \(\frac{62}{227} \times \frac{y}{w}\) of P
Percentage of phosphorous = \(\frac{62}{227} \times \frac{y}{w}\)%

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 13.
Explain the estimation of nitrogen by the Dumas method.
Answer:
1. Dumas method:
This method is based upon the fact that nitrogenous compound when heated with cupric oxide in an atmosphere of CO2 yields free nitrogen. Thus
Cx Hy Nz + (2x + y/2) CuO → x CO2 + y/2 H2O Z/2 N2 +(2X + y/2)Cu.

Traces of oxide of nitrogen, which may be formed in some cases, are reduced to elemental nitrogen by passing overheated copper spiral. The apparatus used in the Dumas method consists of a CO2 generator, combustion tube, Schiff’s nitrometer.

CO2 generator:
CO2 needed in this process is prepared by heating magnetite or sodium bicarbonate contained in a hard glass tube or by the action of dil. HCl on marble in a Kipps apparatus. The gas is passed through the combustion tube after being dried by bubbling through Cone. H2SO4 contained in a Drechel bottle.

Combustion Tube:
The combustion tube is heated in a furnace is charged with
a) A roll of oxidized copper gauze to prevent the back diffusion of the products of combustion and to heat the organic substance mixed with CuO by radiation
b) a weighed amount of the organic substance mixed with an excess of CuO,
c) a layer, of course, CuO packed in about 2/3 of the entire length of the tube and kept in position by a loose asbestos plug on either side; this oxidizes the organic vapors passing through it, and
d) a reduced copper spiral which reduces any oxides of nitrogen formed during combustion to nitrogen.

Schiff’s nitro meter:
The nitrogen gas obtained by the decomposition of the substance in the combustion tube is mixed with a considerable excess of CO2 It is estimated by passing a nitrometer when CO2 is absorbed by KOH and the nitrogen gets collected in the upper part of the graduated tube.

Procedure:
To start with the tap of the nitrometer is left open. CO2 is passed through the combustion tube to expels the air in it. When the gas bubbles rising through, the potash solution fails to reach the top of it and is completely absorbed it shows that only CO2 is coming and that all air has been expelled from the combustion tube. The nitrometer is then filled with KOH solution by lowering the reservoir and the tap is closed. The combustion tube is now heated in the furnace and the temperature rises gradually.

The nitrogen set free from the compound collects in the nitrometer. When the combustion is complete a strong current of CO2 is sent through, the apparatus in order to sweep the last trace of nitrogen from it. The volume of the gas gets collected is noted after adjusting the reservoir so that the solution in it and the graduated tube is the same. The atmospheric pressure and the temperature are also recorded.

Calculation:
Weight of the substance taken = w g
Volume of nitrogen = V1 L
Room Temperature = T1 K
Atmospheric Pressure = P mm of Hg
Aqueous tension at room temperature = P’ mm of Hg
Pressure of diy nitrogen = (P – P’) = P’ mm of Hg.
Let P0V0 and T0 be the pressure, volume, and temperature respectively of dry nitrogen at STP,
Then,

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 142

Calculation of percentage of nitrogen, 22.4 L of N2 at STP weight 28 g of N2
∴ V0 L of N2 at S.T.P weigh = \(\frac{28}{22.4}\) × V0
Wg Organic compound contain = \(\left(\frac{28}{22.4} \times \frac{\mathrm{V}_{\mathrm{o}}}{\mathrm{W}}\right)\) of nitrogen

∴ percentage of nitrogen = \(\left(\frac{28}{22.4} \times \frac{\mathrm{V}_{\mathrm{o}}}{\mathrm{W}}\right)\) × 100

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 14.
0.1688 g when analyzed by the Dumas method yield 31.7 ml of moist nitrogen measured at 14°C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14°C = 12 mm)
Solution:
Weight of Organic compound = 0.168 g
Volume of moist nitrogen (V1) = 31.7 mL
= 31.7 × 10-3 L
Temperature (T1) = 14°C = 14 + 273 = 287 K
Pressure of Moist nitrogen (P) = 758 mm Hg
Aqueous tension at 14°C =12 mm of Hg
∴ Pressure of dry nitrogen = (P – P1)
= 758 – 12 = 746 mm of Hg
Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry 143

Question 15.
Explain the estimation of nitrogen by Kjeldahl’s method.
Answer:
This method is carried much more easily than the Dumas method. It is used largely in the analysis of foods and fertilizers. Kjeldahl’s method is based on the fact that when an organic compound containing nitrogen is heated with Conc. H2SO4, the nitrogen in it is quantitatively converted to ammonium sulphate. The resultant liquid is then treated with excess alkali and then liberated ammonia gas absorbed in excess of standard acid. The amount of ammonia (and hence nitrogen) is determined by finding the amount of acid neutralized by back titration with same standard alkali.

Procedure:
A weighed quantity of the substance (0.3 to 0.5 g) is placed in a special long-necked Kjeldahl flask made of pyrex glass. About 25 mL of Cone. H2SO4 together with a little K2SO4 and CuSO 4 (catalyst) is added to it the flask is loosely stoppered by a glass bulb and heated gently in an inclined position. The heating is continued till the brown color of the liquid first produced, disappears leaving the contents clear as before. At this point all the nitrogen in the substance is converted to (NH4)2SO4.

The Kjeldahl flask is then cooled and its contents are diluted with some distilled water and then carefully transferred into a 1 lit round bottom flask. An excess NaOH solution is poured down the side of the flask and it is fitted with a Kjeldahl trap and a water condenser. The lower end of the condenser dips in a measured volume of excess the H2SO4 solution. The liquid in the round bottom flask is then heated and the liberated ammonia is distilled into sulphuric acid. The Kjeldahl trap serves to retain any alkali splashed up on vigorous boiling.

When no more ammonia passes over (test the distillate with red litmus) the receiver is removed. The excess of acid is then determined by titration with alkali, using phenolphthalein as the indicator.

Calculation:
Weight of the substance = w g
Volume of H2SO4 required for the complete neutralization of evolved NH3 = V ml.
Strength of H2S04 used to neutralize NH3 = N
Let the Volume and the strength of NH3 formed are V1 and N1 respectively.
We know that V1N1 = VN
The amount of nitrogen present in the = \(\frac{14 \times \mathrm{NV}}{1 \times 1000 \times \mathrm{w}}\)
Percentage of Nitrogen = \(\left(\frac{14 \times \mathrm{NV}}{1000 \times \mathrm{w}}\right)\) × 100 %

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 16.
0.6 g of an organic compound was kjeldalised and NH3 evolved was absorbed into 50 mL of semi-normal solution of H2SO4. The residual acid solution was diluted with distilled water and the volume made up to 150 mL. 20 mL of this diluted solution required 35 mL of \(\frac{\mathbf{N}}{20}\) NaOH solution for complete neutralization. Calculate the % of N in the compound.
Solution:
Weight of Organic compound = 0.6 g
Volume of sulphuric acid taken = 50 mL
Strength of sulphuric acid taken = 0.5 N
20ml of diluted solution of unreacted sulphuric acid was neutralized by 35 mL of 0.05 N Sodium hydroxide
Strength of the diluted sulphuric aicd = \(\frac{35 \times 0.05}{20}\) = 0.0875 N
Volume of the sulphuric acid remaining after reaction with Organic compound = V1 mL
Strength of the diluted H2SO4 = 0.5 N
Volume of the diluted H2SO4 = 150 mL
Strength of the diluted H2SO4 = 0.0875 N
V1 = \(\frac{150 \times 0.0875}{0.5}\) = 26.25 ml
Volume of H2SO4 consumed by ammonia = 50 – 26.25 = 23.75 ml
23.75 ml of 0.5 N H2SO4 = 23.75 ml of 0.5 N NH3
The amount of Nitrogen present in the 0.6 g of organic compound = \(\frac{14 g}{1000 m L \times 1 N}\) × 23.75 × 0.5 N = 0.166 g
Percentage of Nitrogen = \(\frac{0.166}{0.6}\) × 100 = 27.66%

Question 17.
Explain the various steps involved in the purification of organic compounds by crystallization process.
Answer:
(i) Selection of solvent:
Most of the organic substances being covalent do not dissolve in polar solvents like water, hence selection of solvent (suitable) becomes necessary. Hence the powdered organic substance is taken in a test tube and the solvent is added little by little with constant stirring and heating, till the amount added is just sufficient to dissolve the solute (ie) organic compound. If the solid dissolves upon heating and throws out maximum crystals on cooling, then the solvent is suitable. This process is repeated with other solvents like benzene, ether, acetone, and alcohol till the most suitably one is sorted out.

(ii) Preparation of solution:
The quantity to be purified is taken in a conical flash fitted with a reflux condenser. The solvent selected in first stage is also taken along with solute and the quantity of the solvent just enough to dissolve the whole solid on boiling. Small amount of are animal char-coal can be added before boiling to decolorize any colored substance. The heating may be done over a wire gauze or water bath depending upon the nature of liquid (ie) whether the solvent is low boiling or high boiling.

(iii) Filtration of hot solution:
The solution so obtained is filtered through a fluted filter paper placed in a hot water funnel.

(iv) Crystallization:
The hot filtrate is then allowed to cool. Most of the impurities are removed on the filter paper, the pure solid substance separate as crystal. When copious amount of crystal has been obtained, then the crystallization is complete. If the rate of crystallization is slow, it is induced either by scratching the walls of the beaker with a glass rod or by adding a few crystals of the pure compounds to the solution.

(v) Isolation and drying of crystals:
The crystals are separated from the mother liquor by filtration. Filtration is done under reduced pressure using a Bucher funnel. When the whole of the mother liquor has been drained into the filtration flask, the crystals are washed with small quantities of the pure cold solvent and then dried.

Samacheer Kalvi 11th Chemistry Guide Chapter 11 Fundamentals of Organic Chemistry

Question 18.
How will you purify an organic compound by Thin layer chromatography?
Answer:
This method is another type of adsorption chromatography with this method it is possible to separate even minute quantities of mixtures. A sheet of a glass is coated with a thin layer of adsorbent (cellulose, silica gel or alumina). This sheet of glass is called chromoplate or thin layer chromatography plate. After drying the plate, a drop of the mixture is placed just above one edge and the plate is then placed in a closed jar containing eluent (solvent). The eluent is drawn up to the adsorbent layer by capillary action.

The components of the mixture move up along with the eluent to different distances depending upon their degree of adsorption of each component of the mixture. It is expressed in terms of its retardation factors (ie) Rf value
Rf = \(\frac{\text { Distance moved by the substance from baseline }(\mathrm{x})}{\text { Distance moved by the solvent from baseline }(\mathrm{y})}\)

The spots of colored compounds are visible on TLC plate due to their original color. The colorless compounds are viewed under UV light or in another method using iodine crystals or by using an appropriate reagent.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 10 Chemical Bonding Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding

11th Chemistry Guide Chemical Bonding Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
In which of the following compound does the central atom obey the octet rule?
a) XeF4
b) AlCl3
c) SF6
d) SCl2
Answer:
d) SCl2

Question 2.
In the molecule OA = C = OB, the formal charge on OA, C and OB are respectively.
a) -1, 0, +1
b) +1, 0, -1
c) -2, 0, +2
d) 0, 0, 0
Answer:
d) 0, 0, 0

Question 3.
Which of the following is electron deficient?
a) PH3
b) (CH3)2
c) BH3
d) NH3
Answer:
c) BH3

Question 4.
Which of the following molecule contain no π bond?
a) SO2
b) NO2
c) CO2
d) H2O
Answer:
d) H2O

Question 5.
The ratio of number of sigma (σ) bond and pi (π) bonds in 2 – butynal is
a) 8/3
b) 5/3
c) 8/2
d) 9/2
Answer:
a) 8/3

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 6.
Which one of the following is the likely bond angles of sulphur tetrafluo-ride molecule?
a) 120°, 80°
b) 109°28’
c) 90°
d) 89°, 117°
Answer:
d) 89°, 117°

Question 7.
Assertion:
Oxygen molecule is paramagnetic.
Reason :
It has two unpaired electron in its bonding molecular orbital
a) both assertion and reason are true and reason is the correct explanation of assertion.
b) both assertion and reason are true but reason is not the correct explanation of assertion.
c) assertion is true but reason is false.
d) both assertion and reason are false.
Answer:
c) assertion is true but reason is false.

Question 8.
According to Valence bond theory, a bond between two atoms is formed when
a) fully filled atomic orbitals overlap
b) half filled atomic orbitals overlap
c) non – bonding atomic orbitals overlap
d) empty atomic orbitals overlap
Answer:
b) half filled atomic orbitals overlap

Question 9.
In ClF3, NF3 and BF3 molecules the chlorine, nitrogen and boron atoms are
a) sp3 hybridised
b) sp3, sp3 and sp2 respectively
c) sp3 hybridised
d) sp3d, sp3 and sp hybridised respectively
Answer:
d) sp3d, sp3 and sp hybridised respectively

Question 10.
When one s and three p orbitals hybridise,
a) four equivalent orbitals at 90° to each other will be formed
b) four equivalent orbitals at 109°28’ to each other will be formed
c) four equivalent orbitals, that are lying the same plane will be formed
d) none of these
Answer:
b) four equivalent orbitals at 109°28’ to each other will be formed

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 11.
Which of these represents the correct order of their increasing bond order.
a) C2+ < C22- < O22- < O2
b) C22- < C2+ < O2 < O22-
c) O22- < O2 < C22- < C2+
d) O22- < C2+ < O2 < C22-
Answer:
d) O22- < C2+ < O2 < C22-

Question 12.
Hybridisation of central atom in PCl5 involves the mixing of orbitals.
a) s, Px, Py, dx2, dx2 – y2
b) s, px, py, pxy, dx2 – y2
c) s, px, py, pz, dx2 – y2
d) s, px, Py, dxy, dx2 – y2
Answer:
c) s, px, py, pz, dx2 – y2

Question 13.
The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is
a) H2O2 > O3 > O2
b) O2 > O3 > H2O2
c) O2 > H2O2 > O3
d) O3 > O2 > H2O2
Answer:
b) O2 > O3 > H2O2

Question 14.
Which one of the following is diamagnetic?
a) O2
b) O22-
c) O2+
d) None of these
Answer:
b) O22-

Question 15.
Bond order of a species is 2.5 and the number of electrons in its bonding molecular orbital is formed to be 8. The no. of electrons in its antibonding molecular orbital is
a) three
b) four
c) Zero
d) can not be calculated from the given information
Answer:
a) three

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 16.
Shape and hybridisation of IF5 are
a) Trigonal bipyramidal, sp3d2
b) Trigonal bipyramidal, sp3d
c) Square pyramidal, sp3d2
d) Octahedral, sp3d2
Answer:
c) Square pyramidal, sp3d2

Question 17.
Pick out the incorrect statement from the following:
a) sp3 hybrid orbitals are equivalent and are at an angle of 109°28’ with each other
b) dsp2 hybrid orbitals are equivalent and bond angle between any two of them is 90°
c) All five sp3d hybrid orbitals are not equivalent out of these five sp3d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three
d) none of these
Answer:
c) All five sp3d hybrid orbitals are not equivalent out of these five sp3d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three

Question 18.
The molecules having same hybridisation, shape and number of lone pairs of electrons are
a) SeF4, XeO2F2
b) SF4, XeF2
c) XeOF4, TeF4
d) SeCl4, XeF4
Answer:
a) SeF4, XeO2F2

Question 19.
In which of the following molecules / ions BF3, NO2 H20 the central atom is sp2 hybridised?
a) NH2 and H2O
b) NO2 and H2O
c) BF3 and NO2
d) BF3 and NH2
Answer:
c) BF3 and NO2

Question 20.
Some of the following properties of two species, NO3 and H3O+ are described below. Which one of them is correct?
a) dissimilar in hybridisation for the central atom with different structure
b) isostructural with same hybridisation for the central atom.
c) different hybridisation for the central atom with same structure.
d) none of these
Answer:
a) dissimilar in hybridisation for the central atom with different structure

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 21.
The types of hybridisation on the five-carbon atom from right to left in the, 2,3 pentadiene.
a) sp3, sp2, sp, sp2, sp3
b) sp3, sp, sp, sp, sp3
c) sp2, sp, sp2, sp2, sp3
d) sp3, sp3, sp2, sp3, sp3
Answer:
a) sp3, sp2, sp, sp2, sp3

Question 22.
XeF2 is isostructural with
a) SbCl2
b) BaCl2
c) TeF2
d) ICl2
Answer:
d) ICl2

Question 23.
The percentage of s-character of the hybrid orbitals in methane, ethane, ethene, and ethyne are respectively
a) 25, 25, 33.3, 50
b) 50, 50, 33.3, 25
c) 50, 25, 33.3, 50
d) 50, 25, 25, 50
Answer:
a) 25, 25, 33.3, 50

Question 24.
Of the following molecules, which have shape similar to carbondioxide?
a) SnCl2
b) NO2
c) C2H2
d) All of these
Answer:
c) C2H2

Question 25.
According to VSEPR theory, the repulsion between different parts of electrons obey the order
a) 1. p – 1. p > b. p – b. p > 1. p – b. p
b) b. p – b. p > b. p – 1. p > 1. p – b. p
c) 1. p – 1. p > b. p – 1. p > b. p – b. p
d) b. p – b. p > 1. p – 1. p > b. p – 1. p
Answer:
c) 1. p – 1. p > b. p – 1. p > b. p – b. p

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 26.
Shape of ClF3 is
a) Planar triangular
b) Pyramidal
c) “T” Shaped
d) none of these
Answer:
c) “T” Shaped

Question 27.
Non – Zero dipole moment is shown by
a) CO2
b) p – dichlorobenzene
c) carbontetrachloride
d) water
Answer:
d) water

Question 28.
Which of the following conditions is not correct for resonating structures?
a) the contributing structure must have the same number of unpaired electrons
b) the contributing structures should have similar energies
c) the resonance hybrid should have higher energy than any of the contributing structure.
d) none of these
Answer:
c) the resonance hybrid should have higher energy than any of the contributing structure.

Question 29.
Among the following, the compound that contains, ionic, covalent, and Coordinate linkage is
a) NH4Cl
b) NH3
c) NaCl
d) none of these
Answer:
a) NH4Cl

Question 30.
CaO and NaCl have the same crystal structure and approximately the same radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is
a) U
b) 2U
c) U /2
d) 4U
Answer:
d) 4U

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

II. Write brief answer to the following questions:

Question 31.
Define the following:
i) Bond order
ii) Hybridisation
iii) σ – bond
Answer:
i) Bond order:
The number of bonds formed between the two bonded atoms in a molecule is called the bond order.

ii) Hybridisation:
Hybridisation is the process of mixing of atomic orbitals of the same atom with comparable energy to form an equal number of new equivalent orbitals with the same energy.

iii) σ – bond:
When two atomic orbitals overlap linearly along the axis, the resultant bond is called a sigma (σ) bond.

Question 32.
What is a pi – bond?
Answer:
When two atomic orbitals overlap sideways, the resultant covalent bond is called a pi (π) bond. When we consider x-axis as the molecular axis, the py – py and pz – pz overlaps will result in the formation of a π – bond.

Question 33.
In CH4, NH3, and H2O, the central atom undergoes sp3 hybridization – yet their bond angles are different. Why?
Answer:
According to VSEPR theory, as H2O has two lone pairs so it repels the bond pairs much more and makes the bond angle shorter of 104.5 degrees, and as NH3 has one lone pair that repels the three bond pairs but not much effectively and strongly as two lone pairs of water repel one bond pair.

So the bond angle between the Hydrogen atom of ammonia is 107.5 greater than that of water. Similarly, methane molecules have no lone pair and bond pair repels each other with an equal bond angle between two adjacent hydrogen atoms becomes 109°.28′.

Question 34.
Explain Sp2 hybridization in BF3.
Answer:
Consider boron trifluoride molecule. The valence shell electronic configuration of the boron atom is [He]2s2 2p1.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 1

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 2

In the ground, state boron has only one unpaired electron in the valence shell. In order to form three covalent bonds with fluorine atoms, three unpaired electrons are required. To achieve this, one of the paired electrons in the 2s orbital is promoted to the 2py orbital in the excited state. In boron, the s orbital and two p orbitals (px and py) in the valence shell hybridises, to generate three equivalent sp2 orbitals as shown in the Figure. These three orbitals lie in the same xy plane and the angle between any two orbitals is equal to 120°.

Overlap with 2pz orbitals of fluorine:
The three sp2 hybridised orbitals of boron now overlap with the 2pz orbitals of fluorine (3 atoms). This overlap takes place along the axis as shown below.
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 3

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 35.
Draw the M.O diagram for oxygen molecule calculate its bond order and show that O2 is paramagnetic.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 4
Electronic configuration of O atom:
1s2 2s2 2p4

Electronic configuration of O2 molecule:
σ1s2, σ1s*2, σ2s2, σ2s*2, σ2px2, π2py2 π2pz2 π2py*1 π2pz*1

Bond order = \(\frac{N_{b}-N_{a}}{2}\)
= \(\frac{10-6}{2}\)
= 2
The molecule has two unpaired electrons hence it is paramagnetic.

Question 36.
Draw MO diagram of CO and calculate its bond order.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 5

Bonding in some hetero nuclear di-atomic molecules:
Molecular orbital diagram of Carbon monoxide molecule (CO)
Electronic configuration of C atom: 1s2 2s2 2p2
Electronic configuration of O atom: 1s2 2s2 2p4
Electronic configuration of CO molecule :
σ1s2, σ1s*2, σ2s2, σ2s*2, π2py2, π2pz2 σ2px2

Bond order = \(\frac{N_{b}-N_{a}}{2}\)
= \(\frac{10-4}{2}\)
= 3
The molecule has no unpaired electrons hence it is diamagnetic.

Question 37.
What do you understand by the Linear combination of atomic orbitals in MO theory?
Answer:
The wave functions for the molecular orbitals can be obtained by solving the Schrodinger wave equation for the molecule. Since solving the Schrodinger equation is too complex, approximation methods are used to obtain the wave function for molecular orbitals. The most common method is the linear combination of atomic orbitals (LCAO).

We know that the atomic orbitals are represented by the wave function ψ. Let us consider two atomic orbitals represented by the wave function ψA and ψB with comparable energy, combines to form two molecular orbitals. One is bonding molecular orbital(ψbonding) and the other is antibonding molecular orbital (ψantibonding) The wave functions for these two molecular orbitals can be obtained by the linear combination of the atomic orbitals ψA and ψB as below,

ψbonding = ψA + ψB;
ψantibonding = ψA – ψB

The formation of bonding molecular orbital can be considered as the result of constructive interference of the atomic orbitals and the formation of anti-bonding molecular orbital can be the result of the destructive interference of the atomic orbitals. The formation of the two molecular orbitals from two is orbitals is shown below.

Constructive interaction:
The two 1s orbitals are in phase and have the same sign,
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 6

Destructive interaction:
The two orbitals are out phase
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 7

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 38.
Discuss the formation of N2 molecule using MO theory.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 8
Molecular orbital diagram of nitrogen molecule (N2)
Electronic configuration of N atom 1s2 2s2 2p3
Electronic configuration of N2 molecule:
σ1s2, σ1s*2, σ2s2, σ2s*2, π2py2, π2pz2 σ2px2
Bondorder = \(\frac{N_{b}-N_{a}}{2}\)
= \(\frac{10-4}{2}\) = 3
Molecule has no unpaired electrons hence it is diamagnetic.

Question 39.
What is dipole moment?
Answer:
The polarity of a covalent bond can be measured in terms of dipole moment which is defined as
μ = q × 2d
Where μ is the dipole moment, q is the charge and 2d is the distance between the two charges. The dipole moment is a vector and the direction of the dipole moment vector points from the negative charge to positive charge.
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 9
representation of dipole

The unit of dipole moment is coloumb meter (C m). It is usually expressed in Debye unit (d). The conversion factor is 1 Debye = 3.336 × 10-30 C m.

Question 40.
Linear form of carbondioxide molecule has two polar bonds. Yet the molecule has Zero dipole moment. Why?
Answer:
The linear form of carbon dioxide has zero dipole moment, even though it has two polar bonds. In CO2, the dipole moments of two polar bonds (CO) are equal in magnitude but have opposite direction. Hence, the net dipole moment of the CO2 is,
μ = μ1 + μ2 = μ1 + (-μ1) = 0
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 10

Question 41.
Draw the Lewis structures for the following species.
(i) NO3
(ii) SO42-
(iii) HNO3
(iv) O3
Answer:
(i) NO3
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 12

(ii) SO42-
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 13

(iii) HNO3
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 11

(iv) O3
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 14

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 42.
Explain the bond formation if BeCl2 and MgCl2.
Answer:
Bond formation of BeCl2:
Be = 4;
Electronic configuration of Be atom is = 1s2 2s2
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 15

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 16
Bond formation of MgCl2:
Mg (Z = 12), 1s2 2s2 2p6 3s2 it loses two of its valence electron and became Mg2+ with inert gas configuration Neon.
The chlorine accepts one electron in its valence shell and because Cl ion with Ar electron configuration.
Mg + Cl2 → Mg2+ + 2 Cl → MgCl2
Magnesium cation and two chlorides are attracted by strong electrostatic force to form MgCl2 crystals.
Mg = 12,
Electronic configuration: 1s2 2s2 2p6 3s2
Mg+2:
Electronic configuration: 1s2 2s2 2p6 3s0
Cl = 17, 1s2 2s2 2p6 3s2 3p5
Cl:
Electronic configuration: 1s2 2s2 2p6 3s2 3p6

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 17

Question 43.
Which bond is stronger σ or π? Why?
Answer:
1. Sigma bonds (σ) are stronger than Pi bonds (π). Because, sigma bonds are formed from bonding orbitals directly between the nuclei of the bonding atoms, resulting in greater overlap and a strong sigma bond (axial overlapping).

2. π bonds result from the overlap of atomic orbitals that are in contact through two areas of overlap (lateral overlapping). Pi bonds are more diffused bonds than sigma bonds.

Question 44.
Define bond energy.
Answer:
The bond enthalpy is defined as the minimum amount of energy required to break one mole of a particular bond in molecules in their gaseous state. The unit of bond enthalpy is kJ mol-1.

Question 45.
Hydrogen gas is diatomic whereas inert gases are monoatomic – Explain on the basis of MO theory.
Answer:
The molecular orbital electronic configuration of the hydrogen molecule is (σ1s2). The molecular orbital energy level diagram of the H2 molecule is given in
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 18
Here, N2 = 2, Na = 0
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{2-0}{2}\) = 1

He2: σ1s2 σ1s*2
The molecular orbital energy level diagram of He2 (hypothetical) is given in
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 19

Here, Nb = 2 and Na = 2
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{2-2}{2}\) = 0

As the bond order for He2 comes out to form between two hydrogen atoms. But as the bond order of helium is zero, there is no bond between helium atoms and hence it is monoatomic.

Result:
As the bond order of the H2 molecule is one, it is diatomic and a single bond is formed between two hydrogen atoms. But as the bond order of helium is zero, there is no bond between helium atoms and hence it is monoatomic.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 46.
What is the Polar Covalent bond? Explain with example.
Answer:
Polar covalent bond is a chemical bond in which the electrons required to form a bond is unequally shared between two atoms. The atom which is more electronegative attracts more electrons from the bonded pair than the other atom. As a result there is a slight separation of charges in a molecule in which more electronegative atom (comparatively) carries a slight negative charge and less electronegative atom carries a positive charge. The bonds formed between two atoms have a permanent electric dipole.

In water (H2O) molecule, covalent bond exists between Hydrogen and Oxygen. Being more electronegative Oxygen carries negative charge and Hydrogen carries positive charge. In a water molecule Oxygen carries a negative charge (anionic in nature) and hydrogen’s carries positive charge (cationic in nature) which forms a polar covalent bond. The size of an oxygen atom is comparatively higher than a hydrogen atom, hence it polarizes the molecule towards itself i.e., it attracts shared pair of bonding electrons towards itself and makes the bond to be more polarized. Hydrogen which is comparatively a small-sized cation makes the bond to be polarized better.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 20

Question 47.
considering x-axis as the molecular axis which out of the following will form a sigma bond.
i) 1s and 2py
ii) 2px and 2py
iii) 2px and 2pz
iv) 1s and 2pz
Answer:
i) 1s and 2py: No sigma bond
ii) 2px and 2py: sigma bond
iii) 2px and 2pz: No sigma bond
iv) 1s and 2pz: No sigma bond

Question 48.
Explain resonance with reference to a carbonate ion.
Answer:
It is evident from the experimental results that all carbon-oxygen bonds in carbonate ions are equivalent. The actual structure of the molecules is said to be resonance hybrid, an average of these three resonance forms. It is important to note that carbonate ion does not change from one structure to another and vice versa. is not possible to picturise the resonance hybrid by drawing a single Lewis structure. However, the following structure gives a qualitative idea about the correct structure.
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 21

(b) Resonance structure of CO32-:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 22

Resonance Hybrid structure of CO32-:
It is found that the energy of the resonance hybrid (structure 4) is lower than that of all possible canonical structures (Structure 1, 2 & 3). The difference in energy between structure 1 or 2 or 3, (most stable canonical structure) and structure 4 (resonance hybrid) is called resonance energy.

Question 49.
Explain the bond formation in ethylene and acetylene.
Answer:
Bonding in ethylene:
The bonding in ethylene can be explained using the hybridization concept. The molecular formula of ethylene is C2H4. The valency of carbon is 4. The electronic configuration of valence shell of carbon in ground state is [He] 2s2 2px1 2py1 2pz0. To satisfy the valency of carbon promote an electron from 2s orbital to 2pz orbital in the excited state.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 23

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 24

In ethylene both the carbon atoms undergo sp2 hybridization involving 2s, 2px, and 2py, orbitals, resulting in three equivalent sp2 hybridised orbitals lying in the XY plane at an angle of 120° to each other. The unhybridized 2pz orbital lies perpendicular to the XY plane.

Formation of sigma bond:
One of the sp2 hybridised orbitals of each carbon lying on the molecular axis (x-axis) linearly overlaps with each other resulting in the formation of a C-C sigma bond. The other two sp2 hybridised orbitals of both carbons linearly overlap with the four is orbitals of four hydrogen atoms leading to the formation of two C-H sigma bonds on each carbon.

Formation of sigma bond:
The unhybridized 2pz orbital of both carbon atoms can overlap only sideways as they are not in the molecular axis. This lateral overlap results in the formation of a pi bond between the two carbon atoms as shown in the figure.
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 25

Bonding in acetylene :
Similar to ethylene, the bonding in acetylene can also be explained using hybridization concept. The molecular formula of acetylene is C2H2. The electronic configuration of valence shell of carbon in ground state is [He] 2s2 2px1 2py1 2pz0. To satisfy the valency of carbon promote an electron from 2s orbital to 2pz orbital in the excited state.

In acetylene molecule, both the carbon atoms are in sp hybridized state. The 2s and 2px orbitals, resulting in two equivalent sp hybridized orbitals lying in a straight line along the molecular axis (x-axis). The unhybridized 2py and 2pz orbitals lie perpendicular to the molecular axis.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 26

Formation of sigma bond:
One of the two sp hybridized orbitals of each carbon linearly overlaps with each other resulting in the formation a C – C sigma bond. The other sp hybridized orbitals of both carbons linearly overlap with the two 1s orbitals of two hydrogen atoms leading to the formation of one C – H sigma bond on each carbon.

Formation of pi bond:
The unhybridized 2py and 2pz orbitals of each carbon overlap sideways. This lateral overlap results in the formation of two pi bonds (py – py and pz – pz) between the two carbon atoms as shown in the figure.
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 27

Question 50.
What type of hybridization is possible in the following geometries?

  1. octahedral
  2. tetrahedral
  3. square planar

Answer:

  1. octahedral: sp3d2
  2. tetrahedral: sp3
  3. square planar: dsp2

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 51.
Explain VSEPR theory. Applying this theory to predict the shapes of IF7 and SF6.
Answer:
Lewis’s concept of the structure of molecules deals with the relative position of atoms in the molecules and sharing of electron pairs between them. However, we cannot predict the shape of the molecule using Lewis concept. Lewis theory in combination with VSEPR theory will be useful in predicting the shape of molecules.

IF7:
Iodine has 7 valence electrons in the valence shell. In an excited state it has 6 valency electrons by 1st and 2nd excited state & under sp3d2, hybridization and combines with 7 Fluorine to get pentagonal bipyramidal shape. There are no lone pairs.
I = ns2 np5 nd0
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 29

SF6:
Sulphar attain six valence electron in 1st and 2nd excited states undergoes sp3d2 hybridization and combines with six ‘F’ atoms, as there no lone pair electrons it geometry in octahedral.
S = 16 = ns2 np4 nd0
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 28

Question 52.
CO2 and H2O both are triatomic molecule but their dipole moment values are different. Why?
Answer:

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 30
Sum of the dipole moment are cancelled.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 31
Water is ‘v’ shape sum of the dipole moments are not equal to zero.

Question 53.
Which one of the following has highest bond order?
(i) N2
(ii) N2+
(iii) N2
Answer:
(i) N2
N2 = 14
σ1s2 σ1s*2 σ2s2 σ2s*2 π2py2 π2pz2 π2px2
Bond order = \(\frac{6}{2}\) = 3

(ii) N2+
σ1s2 σ1s*2 σ2s2 σ2s*2 π2py2 π2pz2 π2px1
Bond order = \(\frac{5}{2}\) = 2.5

(iii) N2
σ1s2 σ1s*2 σ2s2 σ2s*2 π2py2 π2pz2 π2px*1
Bond order = \(\frac{5}{2}\) = 2.5
The highest bond order is N2

Question 54.
Explain the covalent character in ionic bond.
Answer:
Like the partial ionic character in covalent compounds, ionic compounds show partial covalent character. For example, the ionic compound, lithium chloride shows covalent character and is soluble in organic solvents such as ethanol.

The partial covalent character in ionic compounds can be explained on the basis of a phenomenon called polarisation. We know that in an ionic compound, there is an electrostatic attractive force between the cation and anion. The positively charged cation attracts the valence electrons of anion while repelling the nucleus.

This causes a distortion in the electron cloud of the anion and its electron density shift towards the cation, which results in some sharing of the valence electrons between these ions. Thus, a partial covalent character is developed between them. This phenomenon is called polarisation.

The ability of a cation to polarise an anion is called its polarising ability and the tendency of an anion to get polarised is called its polarisability.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 55.
Describe Fajan’s rule.
Answer:
1. The ability of a cation to polarise an anion is called its polarising ability and the tendency of an anion to get polarised is called its polarisibility. The extent of polarisation in an ionic compound is given by the Fajans rule.

2. To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the Cation greater will be the attraction on the electron cloud of the anion.

Similarly higher the magnitude of negative charge on anion, greater is its polansability. For example, Na+ < Mg2+ < Al3+, the covalent character also follows the order – NaCI < MgCl2 < AICI3

3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation. e.g., LiCl is more covalent than NaCI.

4. Cation having ns2np6nd10 configuration shows greater polarising power than the cations with ns2np6 configuration. e.g., CuCI is more covalent than NaCl.

11th Chemistry Guide Chemical Bonding Additional Questions and Answers

I. Choose the best answer:

Question 1.
Which among the following elements has the tendency to form covalent compounds?
a) Ba
b) Be
c) Mg
d) Ca
Answer:
b) Be

Question 2.
Which one of the following forms only covalent bonds?
(a) Alkali metals
(b) Metals
(c) Nonmetals
(d) Metalloids
Answer:
(c) Non-metals

Question 3.
Two elements X and Y have following electronic configurations
X = 1s2 2s2 2p6 3s2 3p6 4s2
Y = 1s2 2s2 2p6 3s2 3p5
The expressed compound formed by the combination of X and Y will be expressed as
a) X2
b) X5Y2
c) X2Y5
d) XY5
Answer:
a) X2

Question 4.
The lattice energy of an ionic compound depends upon:
a) Charge on the ions only
b) Size of the ions only
c) Packing of the ions only
d) Charge and size of the ion
Answer:
d) Charge and size of the ion

Question 5.
Which one of the following elements forms only one bond?
(a) Carbon
(b) Oxygen
(c) Fluorine
(d) Nitrogen
Answer:
(c) Fluorine

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 6.
The formal charge of the O – atoms in the ion Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 32 is
a) -2
b) + 1
c) – 1
d) 0
Answer:
d) 0

Question 7.
A compound with the maximum ionic character is formed from
a) Na and F
b) Cs and F
c) Cs and I
d) Na and Cl
Answer:
b) Cs and F

Question 8.
Which of the following has the highest ionic character?
a) MgCl2
b) CaCl2
c) BaCl2
d) BeCl2
Answer:
c) BaCl2

Question 9.
Which one of the following molecules has a complete octet?
(a) BF3
(b) BeCl2
(c) BCl3
(d) CCI4
Answer:
(d) CCI4

Question 10.
Among the following, the maximum covalent character is shown by the compound
a) FeCl2
b) SnCl2
c) AlCl3
d) MgCl2
Answer:
c) AlCl3

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 11.
Polarization is the distortion of the shape of an anion by an adjacently placed cation. Which of the following statements is correct?
a) maximum polarization is brought about by a cation of high charge
b) Minimum polarization is brought about by a cation of low radius
c) A large cation is likely to bring about a large degree of polarization
d) A small anion is likely to undergo a large degree of polarization
Answer:
a) maximum polarization is brought about by a cation of high charge

Question 12.
Which of the following Lewis structure does not contribute in resonance?

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 33
a) I
b) II
c) III
d) IV
Answer:
b) II

Question 13.
Which one of the following has a coordinated covalent bond?
(a) CaF2
(b) MgO
(c) [Fe(CN)6]4-
(d) KCI
Answer:
(c) [Fe(CN)6]4-

Question 14.
A diatomic molecule has a dipole moment of 1.2 D. If the bond distance is 1 Å. What percentage of electronic charge exists on each atom?
a) 12 % of e
b) 19 % of e
c) 25 % of e
d) 29 % of e
Answer:
c) 25 % of e

Question 15.
The electronegativity difference between two atoms A and B is 2, the percentage of covalent character in the molecule is
a) 54 %
b) 46 %
c) 23 %
d) 72 %
Answer:
a) 54 %

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 16.
The electronegativity of H and Cl are 2.1 and 3.0 respectively. The correct statement (s) about the nature of HCl is/are:
a) 17 % ionic
b) 83 % ionic
c) 50 % ionic
d) 100 % ionic
Answer:
a) 17 % ionic

Question 17.
The value of carbon-carbon double bond length is …………..
(a) 1.43Å
(b) 1.20Å
(c) 1.54A
(d) 1.33A
Answer:
(d) l.33A

Question 18.
Pick out the molecule which has zero dipole moment
a) NH3
b) H2O
c) BCl3
d) SO2
Answer:
c) BCl3

Question 19.
The dipole moment of HBr is 1.6 × 10-30 cm and interatomic spacing is 1 Å. The % ionic character of HBr is
a) 7
b) 10
c) 15
d) 27
Answer:
b) 10

Question 20.
Which one of the flowing has bond order as 2?
(a) N2
(b) C2 – H4
(c) CH4
(d) HCN
Answer:
(b) C2H4

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 21.
Of the following molecules, the one, which has permanent dipole moment is:
a) SiF4
b) BF3
c) PF3
d) PF5
Answer:
c) PF3

Question 22.
Increasing order of dipole moment is:
a) CF4 < NH3 < NF3 < H2O
b) CF4 < NH3 < H2O < NF3
c) CF4 < NF3 < H2O < NH3
d) CF4 < NF3 < NH3 < H2O
Answer:
d) CF4 < NF3 < NH3 < H2O

Question 23.
The correct sequence of dipole moments among the chlorides of methane is
a) CHCl3 < CH2Cl2 > CH3Cl > CCl4
b) CH2Cl2 > CH3Cl > CHCl3 > CCl4
c) CH3Cl > CH2Cl2 > CHCl3 > CCl4
d) CH2Cl2 > CHCl3 > CH3Cl > CCl4
Answer:
c) CH3Cl > CH2Cl2 > CHCl3 > CCl4

Question 24.
Which of the following has been arranged in order of decreasing dipole moment?
a) CH3Cl > CH3F > CH3Br > CH3I
b) CH3F > CH3Cl > CH3Br > CH3I
c) CH3Cl > CH3Br > CH3I > CH3F
d) CH3F > CH3Cl > CH3I > CH3Br
Answer:
a) CH3Cl > CH3F > CH3Br > CH3I

Question 25.
Which of the following has to see-saw shape?
(a) PCl5
(b) IO2F2
(c) SOF4
(d) ClO33
Answer:
(b) IO2F2

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 26.
Among the following compounds, the one that is polar and has central atom with sp2 hybridization is
a) H2CO3
b) SiF4
c) BF3
d) HClO2
Answer:
a) H2CO3

Question 27.
Which of the following will provide the most efficient overlap?
a) s – s
b) s – p
c) sp2 – sp2
d) sp – sp
Answer:
d) sp – sp

Question 28.
The number and type of bonds between two carbon atoms in CaC2 are:
a) one sigma (σ) and one pi (π) bonds
b) one sigma (σ) and two pi (π) bonds
c) one sigma (σ) and one half pi (π) bonds
d) one sigma (σ) bond
Answer:
b) one sigma (σ) and two pi (π) bonds

Question 29.
Which of the following has linear shape?
(a) PCI5
(b) SnBr2
(c) BeCl2
(d) CCl2F2
Answer:
(c) BeCl2

Question 30.
The hybridization of carbon atoms in C – C single bond of H – C ≡ C = CH = CH2 is
a) sp3 – sp3
b) sp2 – sp
c) sp – sp2
d) sp3 – sp
Answer:
b) sp2 – sp

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 31.
The bond in the formation of fluorine molecule will be
a) Due to s – s overlapping
b) Due to s – p overlapping
c) Due to p – p overlapping
d) Due to hybridization
Answer:
c) Due to p – p overlapping

Question 32.
Which of the following overlaps gives a bond with x as an internuclear axis?
a) Pz and Pz
b) s and pz
c) s and Px
d) dx2 – y2 and dx2 – y2
Answer:
c) s and Px

Question 33.
The strength of bonds by overlapping of atomic orbitals is in the order
a) s – s > s – p > p – p
b) s – s < p – p < s – p
c) s – p < s – s < p – p
d) p – p < s – s < s – p
Answer:
a) s – s > s – p > p – p

Question 34.
Which cannot be explained by VBT?
a) Overlapping
b) Bond formation
c) Paramagnetic nature of oxygen
d) Shapes of molecules
Answer:
c) Paramagnetic nature of oxygen

Question 35.
The structure of IF7 is
a) square pyramidal
b) trigonal bipyramidal
c) octahedral
d) pentagonal bipyramidal
Answer:
d) pentagonal bipyramidal

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 36.
The structure of XeOF4 is
a) tetrahedral
b) square pyramidal
c) square planner
d) octahedral
Answer:
b) square pyramidal

Question 37.
The molecule that has linear structure is :
a) CO2
b) NO2
c) SO2
d) SiO2
Answer:
a) CO2

Question 38.
The molecule which has pyramidal shape is:
a) PCl5
b) SO3
c) CO32-
d) NO3
Answer:
a) PCl5

Question 39.
The type of hybrid orbitals used by the chlorine atom in ClO2
a) sp3
b) sp2
c) sp
d) none of these
Answer:
a) sp3

Question 40.
Which one of the following molecule is planar?
a) NF3
b) NCl3
c) PH3
d) BF3
Answer:
d) BF3

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 41.
Which one of the following compounds has sp2 hybridization?
a) CO2
b) SO2
c) NO2+
d) CO
Answer:
b) SO2

Question 42.
Which of the following molecules has trigonal planar geometry?
a) IF3
b) PCl3
c) NH3
d) BF3
Answer:
d) BF3

Question 43.
The percentage s – character of the hybrid orbitals in methane, ethane, and ethyne are respectively
a) 25, 33, 50
b) 25, 50, 75
c) 50, 75, 100
d) 10, 20, 40
Answer:
a) 25, 33, 50

Question 44.
Number of lone pair (s) in XeOF4 is / are
a) 0
b) 1
c) 2
d) 3
Answer:
b) 1

Question 45.
Which of the following molecule contains one pair of non – bonding electrons?
a) CH4
b) NH3
c) H2O
d) HF
Answer:
b) NH3

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 46.
If XeF2, XeF4 and XeF6, the number of lone pair of electrons on Xe are respectively.
a) 2, 3, 1
b) 1, 2, 3
c) 4, 1, 2
d) 3, 2, 1
Answer:
d) 3, 2, 1

Question 47.
The shape of XeO2 F2 molecule is
a) Trigonal bipyramidal
b) square planar
c) tetrahedral
d) see – saw
Answer:
d) see – saw

Question 48.
According to MO theory,
a) O2+ is paramagnetic and bond order is greater than O2
b) O2+ is paramagnetic and bond order is less than O2
c) O2+ is diamagnetic and bond order is less than O2
d) O2+ is diamagnetic and bond order is more than O2
Answer:
a) O2+ is paramagnetic and bond order is greater than O2

Question 49.
Bond order of O2, O2+, O2 and O22- is in order
a) O2< O22-  < O2 < O2+
b) O22 < O2 < O2, < O2+
c) O2+ < O2 < O2 < O22-
d) O2, < O2+ < O2 < O22-
Answer:
b) O22 < O2 < O2, < O2+

Question 50.
Which of the following pairs have Identical values of bond order?
a) N2+ and O2+
b) F2 and Ne2
c) O2 and B2
d) C2 and N2
Answer:
a) N2+ and O2+

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 51.
Which of the following is paramagnetic?
a) O2
b) CN
c) CO
d) NO+
Answer:
a) O2

Question 52.
Which of the following compounds is paramagnetic?
a) CO
b) NO
c) O22-
d) O3
Answer:
b) NO

Question 53.
The number of antibonding electron pairs O22- molecular ion on the basis of molecular orbital theory is
a) 4
b) 3
c) 2
d) 5
Answer:
a) 4

Question 54.
The bond length of the species O2, O2+ and O2 if are in the order of
a) O2+ > O2 > O
b) O2+ > O2 > O2
c) O2 > O2+ > O2
d) O2 > O2 > O2+
Answer:
a) O2+ > O2 > O

Question 55.
Which of the following is a zero overlap which leads to non – bonding?
a) Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 34
b) Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 35
c) Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 36
d) All
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 34

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 56.
Identify the least stable ion amongst the following:
a) Li
b) Be
c) B
d) C
Answer:
b) Be

Question 57.
Which of the following molecular species has unpaired electrons(s)?
a) N2
b) F2
c) O2
d) O22-
Answer:
c) O2

Question 58.
Among the following species which has minimum bond length?
a) B2
b) C2
c) F2
d) O2
Answer:
c) F2

Question 59.
The correct order of bond strength is :
a) O2 < O2 < O2+ < O22-
b) O22- < O2 < O2 < O2+
c) O2 < O22- < O2 < O2+
d) O2+ < O2 < O2 < O22-
Answer:
b) O22- < O2 < O2 < O2+

Question 60.
The species having bond order different from that in CO is
a) NO
b) NO+
c) CN
d) N2
Answer:
a) NO

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 61.
Which one of the following species is diamagnetic in nature?
a) He2+
b) H2
c) H2+
d) H2
Answer:
b) H2

Question 62.
Which of the following species exhibits the diamagnetic behavior?
a) O22-
b) O2+
c) O2
d) NO
Answer:
a) O22-

Question 63.
Which one of the following pairs of species have the same bond order?
a) CN and NO+
b) CN and CN+
c) O2 and CN
d) NO+ and CN+
Answer:
b) CN and CN+

Question 64.
Which one is the electron-deficient compound?
a) ICl
b) NH3
c) BCl3
d) PCl3
Answer:
c) BCl3

Question 65.
The number of electrons shared by each outermost shell of N2 is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 66.
Which of the following has zero dipole moment?
a) CH2Cl2
b) CH4
c) NH3
d) PH3
Answer:
b) CH4

Question 67.
Which of the following statement is not correct?
a) Hybridization is the mixing of atomic orbitals prior to their combining into molecular orbitals
b) sp2 hybrid orbitals are formed from two p atomic orbitals and one s atomic orbital.
c) d2sp3 hybrid orbitals are directed towards the corners of a regular octahedron
d) dsp3 hybrid orbitals are all at 90° to one another
Answer:
d) dsp3 hybrid orbitals are all at 90° to one another

Question 68.
Shape of XeF4 molecule is
a) linear
b) pyramidal
c) tetrahedral
d) square planar
Answer:
d) square planar

Question 69.
Which of the following compounds, the one having a linear structure is
a) NH2
b) CH4
c) C2H2
d) H2O
Answer:
c) C2H2

Question 70.
The isoelectronic pair is
a) Cl2, ICl2
b) ICl2, ClO2
c) IF2+, I3
d) ClO2, ClF2+
Answer:
d) ClO2, ClF2+

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 71.
The bond order is maximum in
a) O2
b) O2
c) O2+
d) O22-
Answer:
c) O2+

Question 72.
Which of the following does not exist on the basis of molecular orbital theory?
a) H2+
b) He2+
c) He2
d) Li2
Answer:
c) He2

Question 73.
Which of the following species have maximum number of unpaired electrons?
a) O2
b) O2+
c) O2
d) O22-
Answer:
a) O2

Question 74.
In a polar molecule, the ionic charge is 4.8 × 10-10 esu. If the interionic distance is 1 Å unit, then the dipole moment is
a) 41.8 debye
b) 4.18 debye
c) 4.8 debye
d) 0.48 debye
Answer:
c) 4.8 debye

Question 75.
The molecule of CO2 has a 180° bond angle. It can be explained on the basis of
a) sp3 hybridisation
b) sp2 hybridisation
c) sp hybridisation
d) d2sp3 hybridisation
Answer:
c) sp hybridisation

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 76.
Which of the following have both polar and non – polar bonds?
a) C2H6
b) NH4Cl
c) HCl
d) AlCl3
Answer:
b) NH4Cl

Question 77.
Blue vitriol has
a) Ionic bond
b) Coordinate bond
c) Hydrogen bond
d) All the above
Answer:
d) All the above

Question 78.
The number of ionic, covalent, and coordinate bond NH4Cl are respectively
a) 1, 3 and 1
b) 1, 3 and 2
c) 1, 2 and 3
d) 1, 1 and 3
Answer:
b) 1, 3 and 2

Question 79.
Which of the following does not contain a coordinate bond?
a) H3O+
b) BF4
c) HF2
d) NH4+
Answer:
c) HF2

Question 80.
Maximum covalent character is associated with the compound
a) NaI
b) MgI2
c) AlCl3
d) AlI3
Answer:
d) AlI3

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 81.
Amongst LiCl, RbCl, BeCl3, and MgCl2 the compounds with the greatest and the least ionic character, respectively, are
a) LiCl and RbCl
b) RbCl and BeCl2
c) RbCl and MgCl2
d) MgCl2 and BeCl2
Answer:
b) RbCl and BeCl2

Question 82.
LiF is least soluble among the fluorides of alkali metals, because
a) smaller size Li+ impart significant covalent character in LiF
b) the hydration energies of Li+ and F are quite higher
c) lattice energy of LiF is quite higher due to the smaller size of Li+ and F
d) LiF has strong polymeric network in solid
Answer:
c) lattice energy of LiF is quite higher due to the smaller size of Li+ and F

Question 83.
The molecule which has zero dipole moment is
a) CH2Cl2
b) BF3
c) NF3
d) ClO2
Answer:
b) BF3

Question 84.
Carbon tetrachloride has no net dipole moment because of
a) Its planar structure
b) Its regular tetrahedral structure
c) similar sizes of carbon and chlorine atoms
d) Similar electron affinities of carbon and chlorine
Answer:
b) Its regular tetrahedral structure

Question 85.
Of the following compounds, which will have a zero dipole moment?
a) 1, 1 dichloroethylene
b) cis – 1,2 dichloroethylene
c) trans – 1,2 dichloroethylene
d) none of these
Answer:
c) trans – 1,2 dichloroethylene

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 86.
Which contains both polar and non – polar bonds?
a) NH 4Cl
b) HCN
c) H2O2
d) CH4
Answer:
c) H2O2

Question 87.
In which of the following species, is the underlined carbon has sp3 hybridisation:
a) CH3COOH
b) CH3CH2OH
c) CH3COCH3
d) CH2 = CH – CH3
Answer:
b) CH3CH2OH

Question 88.
Ration of a and bonds is maximum in
a) naphthalene
b) tetracyano methane
c) enolic form of urea
d) equal
Answer:
c) enolic form of urea

Question 89.
HCN and HNC molecules have equal number of
a) lone pair of σ bonds
b) σ bonds and π bonds
c) π bonds and lone pairs
d) lone pairs, σ bonds, and π bonds
Answer:
d) lone pairs, σ bonds, and π bonds

Question 90.
Allyl cyanide has
a) 9 sigma bonds and 4 pi bonds
b) 9 sigma bonds, 3 pi bonds, and 1 lone pair
c) 8 sigma bonds and 5 pi bonds
d) 8 sigma bonds, 3 pi bonds
Answer:
b) 9 sigma bonds, 3 pi bonds, and 1 lone pair

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 91.
Effective overlapping will be shown by:
a) \(\oplus \Theta+\oplus \Theta\)

b) \(\left(\frac{\oplus}{\Theta}\right)+\left(\frac{\Theta}{\oplus}\right)\)

c) \(\oplus \Theta+\Theta \oplus\)
d) All the above
Answer:
c) \(\oplus \Theta+\Theta \oplus\)

Question 92.
In which of the following pairs, the two species are not Isostructural?
a) CO32- and NO3
b) PCl4+; and SiCl4
c) PF5 and BrF5
d) AlF63- and SF6
Answer:
c) PF5 and BrF5

Question 93.
The hybridization of orbitals of N atom in NO3, NO3+ and NH4+ are respectively.
a) sp, sp2, sp3
b) sp2, sp, sp3
c) sp, sp3, sp2
d) sp2, sp3, sp
Answer:
b) sp2, sp, sp3

Question 94.
On hybridization of one s and one p -orbital we get:
a) two mutually perpendicular orbitals
b) two orbitals at 180°
c) four orbitals directed tetrahedrally
d) three orbitais in a plane
Answer:
b) two orbitals at 180°

Question 95.
A molecule XY2 contains two σ bonds, two π bonds, and one lone pair of electrons in the valence shell of X. The arrangement of lone pair, as well as bond pairs, is
a) Square pyramidal
b) Linear
c) Trigonal planar
d) Unpredictable
Answer:
c) Trigonal planar

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 96.
The maximum number of 90° angles between bond pair of electron is observed in :
a) sp3d2 hybridisation
b) sp3d hybridisation
c) dsp2 hybridisation
d) dsp3 hybridisation
Answer:
a) sp3 d2 hybridisation

Question 97.
Which of the following has a 3 centred 2 electron bond?
a) BF3
b) NH3
c) CO2
d) B2H6
Answer:
d) B2H6

Question 98.
Which has regular tetrahedral geometry?
a) SF4
b) BF4
c) XeF4
d) [Ni(CN)4]2-
Answer:
b) BF4

Question 99.
Which of the following statement is incorrect?
a) During N2+ formation, one electron is removed from the bonding molecular orbital of N2.
b) During O2+ formation, one electron is removed from the anti-bonding molecular orbital of O2.
c) During O2 formation, one electron is added to the bonding molecular orbital of O2
d) During CN formation, one electron is added to the bonding molecular orbital of CN
Answer:
c) During O2 formation, one electron is added to the bonding molecular orbital of O2

Question 100.
Which concept best explains that O – nitrophenol is more volatile than p – nitrophenol?
a) Resonance
b) Hyper conjugation
c) Hydrogen bonding
d) Steric hindrance
Answer:
c) Hydrogen bonding

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

II. Very short question and answers (2 Marks):

Question 1.
State Octet rule.
Answer:
“The atoms transfer or share electrons so that all atoms involved in chemical bonding obtain 8 electrons in their outer shell (valence shell)”.

Question 2.
What is an electrovalent bond?
Answer:
The complete transfer of electron leads to the formation of a cation and an anion. Both these ions are held together by the electrostatic attractive force which is known as ionic bond.

Question 3.
What is covalent bond?
Answer:
This type of mutual sharing of one or more pairs of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond

Question 4.
What is Co-ordinate bond?
Answer:
One of the combining atoms donates a pair of electrons i.e., two electrons which are necessary for the covalent bond formation, and these electrons are shared by both the combining atoms. These type of bonds are called coordinate covalent bond or coordinate bond.

Question 5.
Draw the lewis dot structure of the following.
(i) SO3
(ii) NH3
(iii) N2O5
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 37

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 6.
What is bond length?
Answer:
The distance between the nuclei of the two covalently bonded atoms is called bond length.
Example:
Carbon-carbon single bond length (1.54 Å).

Question 7.
What is bond angle?
Answer:
Covalent bonds are directional in nature and are oriented in specific directions in space. This directional nature creates a fixed angle between two covalent bonds in a molecule and this angle is termed as bond angle.
Example :
Molecule = CH4;
Atoms defining the angle = H-C-H
Bond angle is 109° 28′
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 38

Question 8.
What is resonance?
Answer:
When we write Lewis structures for a molecule, more than one valid Lewis structures are possible in certain cases.
They only differ in the position of bonding and lone pair of electrons. Such structures are called structures (canonical structures) and this phenomenon is called resonance.

Question 9.
Draw the resonance structure of CO32- ion?
Answer:

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 39

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

III. Short question and answers (3 Marks):

Question 1.
What is dipole moment of a covalent bond in a polar molecule?
Answer:
Dipole moment of a covalent bond in a polar molecule is defined as the product of the magnitude of the charge present on either of the two atoms and the distance by which the two atoms are separated in the molecule. It is a vector quantity and works in the direction of the line joining the positively charged end to the negatively charged end.
The unit of dipole moment is Debye.
Thus 1D = 1 × 10-18 e.s.u.cm.

Question 2.
Distinguish between σ – molecular orbital & π – molecular orbital.
Answer:

σ molecular orbital π molecular orbital
1. It is formed by head on overlapping of atomic orbitals. It is formed by the sidewise overlapping of atomic orbitals.
2. The overlapping of atomic orbital is maximum. The overlapping of atomic orbital is less.
3. The orbital is symmetrical to rotation about the intemuclear axis. The orbital is not symmetrical to rotation about the internuclear axis.
4. The resulting covalent bond is strong. The resulting covalent bond is weak.

Question 3.
Explain formation of H2 molecule by MO theory.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 40

Molecular orbital diagram of hydrogen molecule (H2)
Electronic configuration of H atom 1s1
Electronic configuration of H2 molecule: σ1s2
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{2-0}{2}\) = 1
Molecule has no impaired electrons hence it is dimagnetic.

Question 4.
Explain the formation of Li2 molecule by MOT.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 41

Molecular orbital diagram of Lithium molecule (Li2)
Electronic configuration of Li atom 1s2 2s1
Electronic configuration of Li2 molecule: σ1s2 σ1s*2 σ2s2
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{4-2}{2}\) = 1
Molecule has no impaired electrons hence it is dimagnetic.

Question 5.
Explain the formation of B2 molecule by MOT.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 42

Molecular orbital diagram of Boron molecule (B2)
Electronic configuration of B atom 1s2 2s2 2P1
Electronic configuration of B2 molecule: σ1s2 σ1s*2 σ2s2 σ2s*2 π2py1 π2pz1
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{6-4}{2}\) = 1
Molecule has two unpaired electrons hence it is paramagnetic.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 6.
Explain the formation of C2 molecule by MOT.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 43

Molecular orbital diagram of Carbon molecule (B2)
Electronic configuration of C atom 1s2 2s2 2p1
Electronic configuration of C2 molecule: σ1s2 σ1s*2 σ2s2 σ2s*2 π2py1 π2pz1
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{8-4}{2}\) = 2
Molecule has two unpaired electrons hence it is diamagnetic.

Question 7.
How is σ and π – bond formed?
Answer:
When two atomic orbitals overlap linearly along the axis, the resultant bond is called a sigma (σ) bond. This overlap is also called ‘Tread – on overlap” or “axial overlap”. Overlap involves an orbital (s- s and s – p overlaps) will always result in a sigma bond as the s orbital is spherical. Overlap between two p orbitals along the molecular axis will also result in sigma bond formation. When we consider x-axis as molecular axis, the px – px overlap will result in σ – bond.

When two atomic orbitals overlaps sideways, the resultant covalent bond is called a pi bond. When we consider x – axis as molecular axis, py – py and pz – pz overlap will result in the formation of a π – bond.

Question 8.
Explain Salient features of VB theory?
Answer:
Salient features :

  1. When half filled orbitais of two atoms overlap, a covalent bond will be formed between them.
  2. The resultant overlapping orbital is occupied by the two electrons with opposite spins. For example, when H2 is formed, the two is electrons of two hydrogen atoms get paired up and occupy the overlapped orbital.
  3. The strength of a covalent bond depends upon the extent of overlap of atomic orbitals. Greater the overlap, larger is the energy released and stronger will be the bond formed.
  4. Each atomic orbital has a specific direction (except s-orbital which is spherical) and hence orbital overlap takes place in the direction that maximizes overlap.
  5. Let us explain the covalent bond formation in hydrogen, fluorine, and hydrogen fluoride using VB theory.

Question 9.
How is HF & F2 molecule formed by using VB theory?
Answer:
Formation of HF molecule:
1. Electronic configuration of hydrogen atom is 1s1
2. Valence shell electronic configuration of fluorine atom: 2s2, 2px2, 2py2, 2pz1
3. When half filled 1s orbital of hydrogen linearly overlaps with a half filled 2pz orbital of fluorine, a σ – covalent bond is formed between hydrogen and fluorine.
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 44

Formation of fluorine molecule (F2):
1. Valence shell electronic configuration of fluorine atom: 2s2, 2px2, 2Py2 2pz1
2. When the half filled pz orbitals of two fluorine overlaps along the z-axis, a σ – covalent bond is formed between them.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

IV. Long question and answers (5 Marks):

Question 1.
What is an ionic bond?
Explain the formation of ionic bond with a suitable example.
Answer:
1. The complete transfer of electrons leads to the fomiation of a cation and an anion. Both these ions are held together by electrostatic attractive forces which is known as ionic bond.

2. KCl: Potassium chloride
Electronic configuration of K [Ar] 4s
Eleçtronic configuration of Cl = [Ne] 3s2 3p5

3. Potassium has 1 electron in its valence shell and chlorine has 7 electrons in its valence shell.

4. By losing one electron potassium attains the nearest inert gas configuration of Argon and becomes a unipositive cation (K) and chlorine accepts this electron to become uninegative chloride ion (CI) to attain the stable configuration of nearest noble gas, Argon.

5. These two ions combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

6. During the formation of one mole of potassium chloride crystal from its constituent ions, 718 kJ of energy is released. This favours the formation of KCl and its stabilisation.

Question 2.
Explain the qualitative treatment of VB theory for the formation of H2 molecule.
Answer:
A simple qualitative treatment of VB theory for the formation of hydrogen molecule is discussed below.
Consider a situation where in two hydrogen atoms (Ha and Hb) are separated by infinite distance. At this stage there is no interaction between these two atoms and the potential energy of this system is arbitrarily taken as zero. As these two atoms approach each other, in addition to the electrostatic attractive force between the nucleus and its own electron (purple arrows), the following new forces begins to operate.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 45
VB theory for the formation of hydrogen molecule

The new attractive forces (green arrows) arise between
(i) nucleus of Ha and valence electron of Hb
(ii) nucleus of Hb and the valence electron of Ha.

The new repulsive forces (red arrows) arise between
(i) nucleus of Ha and Hb
(ii) valence electrons of Ha and Hb

The attractive forces tend to bring Ha and Hb together whereas the repulsive forces tends to push them apart. At the initial stage, as the two hydrogen atoms approach each other, the attractive forces are stronger than the repulsive forces and the potential energy decreases. A stage is reached where the net attractive forces are exactly balanced by repulsive forces and the potential energy of the system acquires a minimum energy.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 46

At this stage, there is a maximum overlap between the atomic orbitals of Ha and Hb, and the atoms Ha and Hb are now said to be bonded together by a covalent bond. The internuclear distance at this stage gives the H – H bond length and is equal to 74 pm.

The liberated energy is 436 kJ mol-1 and is known as bond energy. Since the energy is released during the bond formation, the resultant molecule is more stable. If the distance between the two atoms is decreased further, the repulsive forces dominate the attractive forces and the potential energy of the system sharply increases.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 3.
Explain hybridization and geometry of BeCl2 molecule.
Answer:
Let us consider the bond formation in beryllium chloride. The valence shell electronic configuration of beryllium in the ground state is shown in the figure.
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 47
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 48

In BeCl2 both the Be-Cl bonds are equivalent and it was observed that the molecule is linear. VB theory explain this observed behaviour by sp hybridisation. One of the paired electrons in the 2s orbital gets excited to 2p orbital and the electronic configuration at the excited state is shown.

Now, the 2s and 2p orbitals hybridise and produce two equivalent sp hybridised orbitals which have 50% s-character and 50% p-character. These sp hybridised orbitals are oriented in opposite direction as shown in the figure.

Overlap with orbital of chlorine:
Each of the sp hybridized orbitals linearly overlap with pz orbital of the chlorine to form a covalent bond between Be and Cl as shown in the Figure
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 49
sp hybridization of BeCl2

Question 4.
Distinguish between σ – bond and π – bond.
Answer:

Sigma (σ) Bond Pi (π) Bond
1. Sigma (σ) bond is formed by axial overlap of atomic orbitals. Pi (π) bond is formed by the sidewise overlap of atomic orbitals.
2. This bond can be formed by overlap of s-s, s-p or p-p orbitals. It involves of overlap of p-p orbitals only.
3. The bond is strong because, overlapping can take place to a large extent. The bond is weak because the overlapping occurs to a small extent.
4. The electron cloud formed by axial overlap is symmetrical about the inter nuclear axis and consists of single charged cloud. The electron cloud of Pi bond is discontinuous and consists of two changed clouds above and below the plane of atoms.
5. There can be a free rotation of atoms about the σ bond. Free rotation of atoms around π bond is not possible because it involves breaking of π -bond.
6. The bond may be present between the two atoms either alone or along with a π -bond. The bond is always present between the two  atoms along with the sigma (σ) bond.
7. The shape of molecule is determined by the sigma framework around the central atom. The π bonds do not contribute to the shape.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 5.
Explain hybridisation & geometry of methane molecule?
Answer:
sp3 hybridisation can be explained by considering methane as an example. In methane molecule the central carbon atom bound to four hydrogen atoms. The ground state valence shell electronic configuration of carbon is [He]2s2 2px1 2py1 2Pz0.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 50

In order to form four covalent bonds with the four hydrogen atoms, one of the paired electrons in the 2s orbital of
carbon is promoted to its 2pz orbital in the excite state. The one 2s orbital and the three 2p orbitals of carbon mixes to give four equivalent sp3 hybridised orbitals. The angle between any two sp3 hybridised orbitals is 109° 28′.
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 51

Overlap with 1s orbitals of hydrogen:
The 1s orbitals of the four hydrogen atoms overlap linearly with the four sp3 hybridised orbitals of carbon to form four C-H σ-bonds in the methane molecule, as shown below.
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 52
sp3 hybridization of CH4

Question 6.
Explain hybridization & geometry of PCl5 molecule.
Answer:
In the molecules such as PCl5, the central atom phosphorus is covalently bound to five chlorine atoms. Here the atomic orbitals of phosphorous undergoes sp3d hybridization which involves its one 3s orbital, three 3p orbitals and one vacant 3d orbital (dz2). The ground state electronic configuration of phosphorous is [Ne] 3s2 3px2 3py1 3pz1 as shown below.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 53
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 54

One of the paired electrons in the 3s orbital of phosphorous is promoted to one of its vacant 3d orbital (dz2) in the excite state. One 3s orbital, three 3p orbitals, and one 3dz2. orbital of phosphorus atom mixes to give five equivalent sp3d hybridised orbitals. The orbitals geometry of sp3d hyrbridised orbitals is trigonal bi – pyramidal.

Overlap with 3pz orbitals of chlorine :
The 3pz orbitals of the five chlorine atoms linearly overlap along the axis with the five sp3d hybridized orbitals of phosphorous to form the five P – Cl σ – bonds, as shown below.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 55

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 7.
Explain hybridization & geometry of SF6 molecule.
Answer:
In sulphur hexafluoride (SF6) the central atom sulphur extend its octet to undergo sp3d2 hybridization to generate six sp3d2 hybridized orbitals which accounts for six equivalent S – F bonds. The ground state electronic configuration of sulphur is [Ne] 3s2 3px2, 3py2, 3pz1

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 56
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 57

One electron each from 3s orbital and 3p orbital of sulphur is promoted to its two vacant 3d orbitals (dz2 and dx2 – y2) in the excite state. A total of six valence orbitals from sulphur (one 3s orbital, three 3p orbitals, and two 3d orbitals) mixes to give six equivalent sp3d2 hybridized orbitals. The orbital geometry is octahedral as shown in the figure.

Overlap with 2pz orbitals of fluorine:
The six sp3d2 hybridized orbitals of sulphur overlaps linearly with 2pz orbitals of six fluorine atoms to form the six S – F bonds in the sulphur hexafluoride molecule.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 58

Question 8.
Explain metallic bonding.
Answer:
1. The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is known as metallic bond.

2. According to Drude and Lorentz, metallic crystal is an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to ionisation of the valence electrons of the atoms of the metal.

3. As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is referred to as electronic bonding.

4. The electrostatic attraction between the metal ions and the free electrons yield a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.

5. As the close packed structure contains many slip planes along which movement can occur during mechanical loading, metal acquires ductility.

6. As metal ion is surrounded by electron cloud in all directions, the metallic bonding has no and thermal conductivity. The metallic lustre is due to the reflection of light by the electron cloud.As the metallic bond is strong enough, the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.

7. High thermal conductivity of metals is due. to thermal excitation of many electrons from the valence bond to the conduction band.

Question 9.
Distinguish between molecular orbital & anti bonding molecular orbitals.
Answer:

Bonding molecular orbital Anti-bonding molecular orbital
1. A bonding molecular orbital is formed when the electron waves of combining atoms are in phase, i.e the lobes of atomic orbitals have same sign. An anti-bonding molecular orbital is formed when the electron waves of the combining atoms are not in phase ie the lobes of atomic orbital have opposite sign.
2. For bonding molecular orbital wave function are summed up. For anti-bonding molecular orbital, the wave functions are subtracted.
3. The electron density is centered between the nuclei of the combining atoms. The probability of finding the electron between the nuclei of the combining atom is negligible.
4. The energy of bonding molecular orbital is less than that of the atomic orbitals of the combining atoms. The energy of an antibonding molecular orbital is higher than the atomic orbitals of combining atoms.
5. Electron present in bonding molecular orbital lead to attraction between the atoms and stabilize the molecule. Electrons present in anti-bonding molecular orbitals lead to repulsion between the atoms and destabilize the molecule.

Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding

Question 10.
Explain the formation of NO by MOT?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 10 Chemical Bonding 59
Molecular orbital diagram of nitrogen molecule (NO)

Electronic configuration of N atom 1s2 2s2 2p3
Electronic configuration of NO molecule
σ1s2 σ1s*2 σ2s2 σ2s*2 π2py2 π2pz2 σ2px2 π2py*1
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{10-5}{2}\) = 2.5
Molecule has one unpaired electrons hence it is paramagnetic.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 6 Respiration Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

11th Bio Zoology Guide Respiration Text Book Back Questions and Answers

Part I

I. Choose The Best Options

Question 1.
Breathing is controlled by
a) cerebrum
b) medulla oblongata
c) cerebellum
d) pons
Answer:
b) medulla oblongata

Question 2.
Intercostal muscles are found between the
a) vertebral column
b) sternum
c) ribs
d) glottis
Answer:
c) ribs

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 3.
The respiratory structures of insects are
a) tracheal tubes
b) gills
c) green glands
d) lungs
Answer:
a) tracheal tubes

Question 4.
Asthma is caused due to
a) inflammation of bronchus and bronchioles
b) inflammation of branchiole
c) damage of diaphragm
d) infection of lungs
Answer:
a) inflammation of bronchus and bronchioles

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 5.
The Oxygen Dissociation Curve is
a) sigmoid
b) straight line
c) curved
d) rectangular hyperbola
Answer:
a) sigmoid

Question 6.
The Tidal Volume of a normal person is
a) 800 mL
b) 1000-1200 mL
c) 500 mL
d) 1100-1200 mL
Answer:
c) 500 mL

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 7.
During inspiration, the diaphragm
a) expands
b) unchanged
c) relaxes to become domed-shaped
d) contracts and flattens
Answer:
d) contracts and flattens

Question 8.
CO2 is transported through blood to lungs as
a) carbonic acid
b) oxyhaemoglobin
c) carbamino haemoglobin
d) carboxy haemoglobin
Answer:
c) carbamino haemoglobin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 9.
When 1500 mL air is in the lungs, it is called
a) vital capacity
b) tidal volume
c) residual volume
d) inspiratory reserve volume
Answer:
c) residual volume

Question 10.
Vital capacity is
a) TV + IRV
b) TV + ERV
c) RV + ERV
d) TV + IRV + ERV
Answer:
d) TV + IRV + ERV

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 11.
After a long deep breath, we do not respire for some seconds due to
a) more CO2 in the blood
b) more O2 in the blood
c) less CO2 in the blood
d) less O2 in the blood
Answer:
b) more O2 in the blood

Question 12.
Which of the following substances in tobacco smoke damage the gas exchange system?
a) carbon monoxide and carcinogens
b) carbon monoxide and nicotine
c) carcinogens and tar
d) nicotine and tar
Answer:
b) carbon monoxide and nicotine

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 13.
Column I represents diseases and column II represents their symptoms Choose the correctly paired option.

Column I Column II
P. Asthma i) Recurring of bronchitis
Q. Emphysema ii) Accumulation of W.B.C in alveolus
R. Pneumonia iii) Allergy

a) P – iii, Q – ii, R – i
b) P – iii, Q – i, R – ii
c) P – ii, Q – iii, R – i
d) P – ii, Q – i, R -iii
Answer:
a) P – iii, Q – ii, R – i

Question 14.
Which of the following best describes the process of gas exchange in the lungs?
a) Air moves in and out of the alveoli during breathing.
b) Carbon dioxide diffuses from deoxygenated blood in capillaries into the alveolar air
c) Oxygen and carbon dioxide diffuse down their concentration gradients between blood and alveolar air
d) Oxygen diffuses from alveolar air into deoxygenated blood.
Answer:
c/d

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 15.
Make the correct pairs.

Column I Column II
PIC i) maximum volume of air breathe in after forced
QEC ii) Volume of air present after expiration in lungs
RVC iii) Volume of air inhaled after expiration
SFRC iv) Volume of air exhaled after inspiration

a) P – i, Q – ii, R – iii, S – iv
b) P – ii, Q – iii R – iv, S – i
c) P – ii, Q – iii, R – i, S – iv
d) P – iii, Q – iv, R – i, S – ii
Answer:
d) P – iii, Q – iv, R – i, S – ii

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 16.
Make the correct pairs.

Column I Column II
P. Tidal volume i) 1000 to 1100 ml
Q. Residual volume ii) 500 ml
R. Expiratory reserve volume iii) 2500 to 3000 ml
S. Inspiratory reserve volume iv) 1100 to 1200 ml

a) P – ii, Q – iv, R – i, S – iii
b) P – iii, Q – ii R – iv, S – i
c) P – ii, Q – iv, R – iii, S – i
d) P – iii, Q – iv, R – i, S – ii
Answer:
a) P – ii, Q – iv, R – i, S – iii

Question 17.
Name the respiratory organs of flatworm earthworm, fish, prawn, cockroach, and cat.
Answer:
Flatworm – Body surface
Earthworm – Moist skin
Fish – Gills
Prawn – Gills
Cockroach – Trachea
Cat – Lungs

Question 18.
Name the enzyme that catalyses the bicarbonate formation in RBCs.
Answer:
Carbonic anhydrase

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 19.
Air moving from the nose to the trachea passes through a number of structures. List in order of the structures.
Answer:
External nostrils, Nasal cavity, pharynx Larynx, trachea, the bronchi bronchioles, and the lungs (alveoli)

Question 20.
Which structures seal the Larynx when we swallow?
Answer:
Epiglottis.

Question 21.
Resistance in the airways is typically low why? Give two reasons.
Answer:
The airway resistance is low because:

  • The diameter of most airways is relatively large.
  • For smaller airways there are many in parallel, making their combined diameter large.
  • Air has a low viscosity.

Question 22.
How the body makes long-term adjustments when living in high altitude?
Answer:

  • When a person travels from sea level to elevations where the atmospheric pressure and partial pressure of O2 lowered there is a poor binding of O2 with haemoglobin leads to acute mountain sickness.
  • When the person lives there for a long time the kidney synthesizes the erythropoietin which stimulates the bone marrow to produce more RBCs

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 23.
Why is pneumonia considered a dangerous disease?
Answer:
Inflammation of the lungs due to infection caused by bacteria or virus is called pneumonia. The symptoms are sputum production, nasal congestion, shortness of breath, sore throat etc. The alveoli get filled with fluid or pus, making is difficult to breathe (lung abscesses).

Question 24.
Diffusion of gases occurs in the alveolar region and only not in any other part of the respiratory system discuss.
Answer:

  • The other parts of the respiratory system do the work of passing the air into the lungs only.
  • Real respiration takes place between alveoli and blood capillaries.

The diffusion membrane of the alveolus is made up of three layers.

  1. The thin squamous epithelial cells.
  2. The endothelium of the alveolar capillaries.
  3. The basement substance found in between them.

The thin squamous epithelial cells of the alveoli provide space for gaseous exchange

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 25.
Sketch a flow chart to show the pathway of airflow during respiration.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration 9

Question 26.
Explain the conditions which create problems in oxygen transport?
Answer:
When a person travels quickly from sea level to elevations above 8000 ft, where the atmospheric pressure and partial pressure of oxygen are lowered, the individual responds with symptoms of acute mountain sickness (AMS)- headache, shortness of breath, nausea, and dizziness due to poor binding of O2 with hemoglobin. When the person moves on a long-term basis to mountains from sea level his body begins to make respiratory and hematopoietic adjustments.

To overcome this situation kidneys accelerate the production of the hormone erythropoietin, which stimulates the bone marrow to produce more RBCs. When a person descends deep into the sea, the pressure in the surrounding water increases which causes the lungs to decrease in volume.

This decrease in volume increases the partial pressure of the gases within the lungs. This effect can be beneficial, because it tends to drive additional oxygen into the circulation, but this benefit also has a risk, the increased pressure can also drive nitrogen gas into the circulation.

This increase in blood nitrogen content can lead to a condition called nitrogen narcosis. When the diver ascends to the surface too quickly a condition called ‘bends’ or decompression sickness occurs and nitrogen comes out of solution while still in the blood-forming bubbles. Small bubbles in the blood are not harmful, but large bubbles can lodge in small capillaries, blocking blood flow or can press on nerve endings.

Decompression sickness is associated with pain in joints and muscles and neurological problems including stroke. The risk of nitrogen narcosis and bends is common in scuba divers. During carbon dioxide poisoning, the demand for oxygen increases. As the 02 level in the blood decreases it leads to suffocation and the skin turns bluish-black.

Part II

11th Bio Zoology Guide Respiration Additional Important Questions and Answers

(1 Mark)

I. Choose The Best Options

Question 1.
What are the respiratory organs of the Limulus?
a) Trachea
b) Gills
c) Bookgills
d) Green glands
Answer:
c) Bookgills

Question 2.
Hamburger’s phenomenon is also known as ……….. [CPMT1988,1991, AMU2001, JLPMER 2002]
(a) HCO3 shift
(b) Na+ shift
(c) H+ shift
(d) Chloride shift
Answer:
(d) Chloride shift

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 3.
These are mucous-secreting cells.
a) Oxynctic cells
b) Chief cells
c) Goblet cells
d) Parietal cells
Answer:
c) Goblet cells

Question 4.
These are the respiratory surfaces.
a) brunchioles
b) Terminal bronchiole
c) alveoli
d) small bronchi
Answer:
c) alveoli

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 5.
At higher CO2 concentration, oxygen dissociation curve of hemoglobin will ……….. [CPMTX 990]
(a) Move to left
(b) Move to right
(c) Become irregular
(d) Move upwardly
Answer:
(b) Move to right

Question 6.
Match:
1) Residual volume i) 6000 ml
2) Expiratory reserve volume ii) 2500 – 3000 ml
3) Inspiratory reserve volume iii) 1000 -1100 ml
4) Total lung capacity iv) 1100 -1200 ml
1) i – iv; 2 – i; 3 – ii; 4 – iii
b) i – i; 2 – ii; 3 – iii; 4 – iv
c) i – iii; 2 – i; 3 – iv; 4 – ii
d) i – iv; 2 – iii; 3 – ii; 4 – i
Answer:
d) i – iv; 2 – iii; 3 – ii; 4 – i

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 7.
1) Atmospheric air i) Partial pressure of O2 -104
2) Alveoli ii) Partial pressure of O2 – 40
3) Tissues iii) Partial pressure of O2 – 95
4) Oxygenated iv) Partial pressure of O2 – blood 159
a) i – i; 2 – ii; 3 – iii; 4 – iv
b) i – ii; 2 – iii; 3 – i; 4 – iv
c) i – iv; 2 – iii; 3 – ii; 4 – i
d) i – iv; 2 – i; 3 – ii; 4 – iii
Answer:
d) i – iv; 2 – i; 3 – ii; 4 – iii

Question 8.
Find out the correct statement and assertion:
Assertion: The high partial pressure of CO2 provides essential space for the dissociation of O2 from oxyhemoglobin
Reason: Haemoglobin takes a maximum of 4 molecules of CO2
a) Assertion and reason are correct
b) Assertion wrong reason wrong
c) Assertion correct reason correct. The reason explains the assertion.
d) Assertion correct Reason wrong
Answer:
d) Assertion correct Reason wrong

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 9.
Assertion: The CO2 when enters into the blood combines with water to form carbonic acid
Reason: Carbonic anhydrase enzyme acts as a catalyst for this reaction
a) Assertion correct Reason wrong
b) Assertion wrong Reason correct
c) Assertion and reason are wrong
d) Assertion and reason are correct
Answer:
d) Assertion and reason are correct

Question 10.
The volume of air remaining in lungs after maximum respiratory’ effort is ……….. [JKCMEE 1992, Har. PMT 2003]
(a) Vital capacity
(b) Residual volume
(c) Total lung capacity
(d) Tidal volume
Answer:
(b) Residual volume

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 11.
What is the cause of pleurisy?
a) embolism
b) constriction of airways
c) alveolei is damaged
d) Pleura becomes inflammed
Answer:
d) Pleura becomes inflammed

Question 12.
People workes in the sand grinding mills may have this disease?
a) asbestosis
b) Fibrosis
c) Silicosis
d) Nephrosis
Answer:
c) Silicosis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 13.
What is the cause of the disease bend?
a) Fibrosis
b) Necrosis
c) Narcosis
d) Silicosis
Answer:
c) Narcosis

Question 14.
Find the wrong pair.

a) Larynx Epiglottis prevents the food from entering into the larynx
b) C-shaped cartilage Ensures that the air passage does not collapse or burst.
c) The rigidity of bronchioles Prevent them from collapsing
d) Fine respiratory brochioler Terminate into air sacs

Question 15.
Read the following statement and find whether they are correct or wrong.
1) The thin squamous epithelial cells of the alveoli are composed of Type-I and the gases can diffuse rapidly through them
2) Type-II cells are thin. The gaseous exchange takes place through diffusion.
3) The spirometer is used to find the volume of air
4) A healthy man respires 10-15 times per minute
a) 1 – True; 2 – False; 3 – True; 4 – False
b) 1 – True; 2 – True; 3 – False; 4 – True
c) 1 – False; 2 – True; 3 – False; 4 – True
d) 1 – True; 2 – False; 3 – True; 4 – False
Answer:
a) 1 – True; 2 – False; 3 – True; 4 – False

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 16.
The volume of air breathed in and out during effortless respiration is ………..
(a) residual volume
(b) vital volume
(c) tidal volume
(d) normal volume
Answer:
(c) Tidal volume

Question 17.
A normal human adult can inspire or expire approximately.
a) 5000-8000 ml
b) 6000 – 7000 ml
c) 6000-8000 ml
d) 5000 – 9000 ml
Answer:
c) 6000-8000 ml

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 18.
Lungs have a number of alveoli for ………..[MPPMT 1995]
(a) Having spongy texture and proper shape
(b) More surface area for diffusion of gases
(c) More space for the increasing volume of inspired air
(d) More nerve supply
Answer:
(b) More surface area for diffusion of gases

Question 19.
Match the following.
i) Inspiratory reserve volume i) 6000 ml
ii) Expiratory reserve volume ii) 1100-1200 ml
iii) Residual volume iii) 1000-31100 ml
iv) Total lung capacity iv) 2500 – 43000 ml
a) i – 2; ii – 3; iii – 4; iv – 2
b) i – 1; ii – 2; iii – 3 ; iv – 4
c) i – 4; ii – 2; iii – 2; iv -1
d) i – 3; ii – 2; iii -1; iv – 4
Answer:
c) i – 4; ii – 2; iii – 2; iv -1

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 20.
The amount of air that moves into the respiratory passage in a minute.
a) 7000ml
b) 8000 ml
c) 9000 ml
d) 6000 ml
Answer:

Question 21.
The amount of air that is not involved in gaseous exchange.
a) 200 ml
b) 150 ml
c) 300 ml
d) 250 ml
Answer:
d) 250 ml

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 22.
Total lung capacity is ………….
a) VC+RC
b) TV+IRV
c) ERV+TV+IRV
d) TV+ERV
Answer:
a) VC+RC

Question 23.
Maximum amount 70-75% of carbon dioxide transport occurs. [RPMT1996, 1998, MPPMT1998, CPMT 1998, BV 2002]
(a) Dissolved in plasma
(b) Carbaminohaemoglobin complex
(c) Bicarbonate
(d) None of the above
Answer:
(c) Bicarbonate

Question 24.
Vital Capacity is:
a) TV+IRV+ERV
b) RV+ERV
c) TV+IRV
d) TV+ERV
Answer:
a) TV+IRV+ERV

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 25.
In what form oxygen is transported in blood.
a) HbO4
b) HbO6
c) HbO2
d) HbO3
Answer:
c) HbO2

Question 26.
How many molecules of oxygen are accepted by haemoglobin?
a) 4
b) 3
c) 2
d) 1
Answer:
a) 4

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 27.
Which one has the lowest value?
(a) Tidal volume
(b) Vital capacity
(c) Inspiratory reserve volume
(d) Expiratory reserve volume
Answer:
(b) Vital capacity

Question 28.
Find out the wrong pair.

a) Dissolved CO2 in blood Plasma 7- 10%
b) The transport of O2 in blood in the dissolved state 7%
c) The dissolved CO2 in haemoglobin 20-25%
d) Emphysema Smoking

Answer:
b) The transport of O2 in blood in the dissolved state – 7%

Question 29.
Where is the respiratory regulatory center present in the brain?
a) Pons Varoli
b) Pons
c) Medulla oblongata
d) cerebellum
Answer:
a) Pons Varoli

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 30.
In which altitude the symptoms of a cute mountain sickness appear?
a) 1000 feet
b) 9000 feet
c) 8000 feet
d) 7000 feet
Answer:
c) 8000 feet

Question 31.
Find out the wrong pair

a) Erythropoietin Increases the red blood cell synthesis
b) Nitrogen narcosis decompression sickness
c) Carbonic anhydrase Synthesis of carbonic acid
d) Normal ferrous Methaemoglobin

Answer:
d) Normal ferrous – Methaemoglobin

Question 32.
Find the wrong pair.

a) Level of O2 in the blood is low Skin turns bluish-black
b) Sigmoid curve Percentage Saturation of haemoglobin
c) Haemoglobin HbO4
d) Emphysema Smoking

Answer:
c) Haemoglobin – HbO4

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 33.
Exchange of gases in lung alveoli occurs through ……….. [AFMC 2002]
(a) Active transport
(b) Osmosis
(c) Simple diffusion
(d) Passive transport
Answer:
(c) Simple diffusion

Question 34.
Confirm:
Assertion (A): Workers working in grinding industries wear protective masks
Reason (B): People working in grinding industries suffers from silicosis
a) A – True, B – True
b) A-False, B-True
c) The assertion A is wrong
d) assertion A is wrong. The reason B is True
Answer:
a) A – True, B – True

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 35.
Confirm:
Part – A: The tar present in the nicotine damages the gaseous exchange.
Part – B: The blood vessels get narrower and the blood pressure increases due to smoking Ans:
a) Part A – False, Part B – Ture
b) Part A- True, Part B-True
c) Part A – True, Part B – is not correct explanation
d) Part A – It is a correct statement. Part B is not a correct statement
Answer:
b) Part A- True, Part B-True

Question 36.
Match and find the correct sequence
i) Pleurisy A) Constriction of alveoli
ii) Atelectasis B) Widening of alveoli
iii) Emphysema c) Accumulation of fluid in the air spaces
iv) Pulmonary edema D) Pleura becomes inflamed
a) I – D, II-A, III – B, IV – C
b) I – A, II – B, III – C, IV – D
c) I – D, II – C, III – B, IV – A
d) I – A, II – C, III-A, IV – B
Answer:
a) I – D, II-A, III – B, IV – C

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 37.
Match and find the correct sequence
i) Tuberculosis A) Alveolei will be affected
ii) Pneumonia B) Inflamation of bronchioles
iii) Asthma Q Mycobacterium
iv) Bronchitis D) Mucous secretion
a) I – A, II – B, III – C, IV – D
b) I – C, II-A, III – D, IV – D
c) I – A, II – C, III – B, IV – D
d) I – A, II – B, III – D, IV – C
Answer:
b) I – C, II-A, III – D, IV – D

Question 38.
The world tuberculosis day
a) March 20th
b) March 21st
c) March 23rd
d) March 24th
Answer:
d) March 24th

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 39.
Respiratory centre of brain is stimulated by ……….. [AllMS 2000]
(a) Carbon dioxide content in venous blood
(b) Carbon dioxide content in arterial blood
(c) Oxygen content in venous blood
(d) Oxygen content in arterial blood
Answer:
(d) Oxygen content in arterial blood

Question 40.
What is the speed of sneeze?
a) 165 km/hr
b) 200 km/hr
c) 250 km / hr
d) 225 km/ hr
Answer:
a) 165 km/hr

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 41.
The adult respires ………………….. time and newborn child respires …………………. times.
a) 12-16;30-60
b) 12-14;30-50
c) 12-20; 30-70
d) 12-30;30-70
Answer:
a) 12-16;30-60

(2 marks)

II. Very Short Questions

Question 1.
What is excretion?
Answer:
The exchange of oxygen and carbon dioxide between the environment and cells of our body, where organic nutrients are broken down oxygenatically to release energy.

Question 2.
How much air can be respired by a normal human adult?
Answer:
A normal adult can respire approximately 6000 to 8000 ml of air per minute. During vigorous exercise, the tidal volume is about 4-10 times higher.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 3.
The rate of breathing in aquatic animals is faster than the of terrestrial animals. Give reason.
Answer:
The amount of dissolved oxygen is very low in water compared to the amount of oxygen in the air. Hence the rate of breathing in aquatic animals is faster than the terrestrial animals.

Question 4.
What is residual volume?
Answer:

  • The volume of air remaining in the lungs after a forceful expiration.
  • Ex.: 1100-1200ml.

Question 5.
What is the function of epiglottis?
Answer:
Epiglottis is a thin elastic flap at the junction of the nasopharynx and larynx. It prevents the food from entering into the larynx and avoids choking on food.

Question 6.
What is meant by inspiratory capacity?
Answer:
The total volume of air a person can inhale after normal expiration. It includes tidal volume and inspiratory reserve volume.
IC = TV + IRV

Question 7.
What is expiratory capacity?
Answer:
The total volume of air a person can exhale after a normal inspiration. It includes tidal volume and expiratory reserve volume.
EC = TV + ERV

Question 8.
How are lungs protected?
Answer:
The lungs are light spongy tissues enclosed in the thoracic cavity surrounded by an air-tight space. It is bound dorsally by the vertebral column and ventrally by the sternum, laterally by the ribs, and on the lower side by the dome-shaped diaphragm.

Question 9.
What is meant by minute respiratory volume?
Answer:

  • The amount of air that moves into the respiratory passage per minute is called minute respiratory volume.
    Normal TV = 500 ml
  • Normal respiratory rate = 12 times / minute Therefore the minute respiratory volume – 6 litre / minute

Question 10.
What are the characteristic features of the respiratory surface?
Answer:
The surface area of the respiratory surface is large and richly supplied with blood vessels.

  • It is extremely thin and kept moist.
  • It is in direct contact with the environment.
  • It is permeable to respiratory gases.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 11.
Give short notes on a ‘C’ shaped cartilage of bronchi?
Answer:
Bronchi have ‘c’ shaped cartilage plates to ensure that the air passage does not collapse or burst as the air pressure changes during breathing.

Question 12.
What should be the characteristic features of the respiratory surface?
Answer:

  • The surface area must be very large and richly supplied with blood vessels.
  • Should be extremely thin and kept moist.
  • Should be in direct contact with the environment.
  • Should be permeable to the respiratory gases.

Question 13.
What is meant by breathing?
Answer:
The movement of air between the atmosphere and the lungs is known as breathing.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 14.
Name the muscle that helps in respiration?
Answer:

  • Diaphragm
  • Intercostal muscle
  • External and internal intercostal muscle.

Question 15.
What is meant by expiratory reserve volume?
Answer:

  • The additional volume of air a person can forcefully exhale by forcefully expiration is called expiratory reserve volume.
  • The normal value is 1000-1100 ml.

Question 16.
What is the cause for the reduction in the elasticity of the lungs?
Answer:

  • Healthy lungs contain large amounts of elastic connective tissue around the alveoli containing elastin which makes the lung tissue elastic.
  • People with emphysema and bronchitis have difficulty in exhaling because the enzyme elastase destroys the elastin around the alveoli and reduces the elasticity of the lungs.

Question 17.
Give notes on Asthma.
Answer:

  • Allergy is caused by allergens, it may be due to dust, pollens some seafood.
  • Allergens provoke an inflammatory response. The allergens affect our respiratory tracts and we immediately start sneezing and coughing.

Question 18.
Why do some people snore?
Answer:
Breathing with a hoarse sound during sleep is caused by the vibration of the soft palate.
Snoring is caused by a partially closed upper airway (nose and throat) which becomes too narrow for enough air to travel through the lungs. This makes the surrounding tissues vibrate and produces the snoring sound.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 19.
Why we should not laugh loudly during eating.
Answer:

  • The oesophagus and trachea lies in the pharynx During swallowing a thin elastic flap called epiglottis prevent the food from entering in to the larynx.
  • If we talk or laugh during swallowing the closing of trachea becomes disturbed and hence the food may enter in to trachea.

Question 20.
Breathing through the nose is healthy than through the mouth? why?
Answer:

  • There are more dust and microbes in the air. If we breathe through the mouth there is a possibility of entering these microbes and dust in to the stomach through oesophagus.
  • When we breathe through the nose the dust will be filtered by the bristles. The dust particular is trapped by the mucous membrane of the nasal cavity.

Question 21.
Write the structure of the alveoli.
Answer:

  • The diffusion membrane of the alveolus is made up of three layers. The thin squamous epithelial cells.
  • The endothelium of the alveolar capillaries
  • The basement substance found in between them. The thin requamous epithelial cells of alveoli are composed of Type-I and Type-II cells.
  • The Type-I cells are very thin so that gases can diffuse rapidly through them. Type-II cells are thicker synthesize and secrete a substance called surfactant.

Question 22.
Give the passage of breathing.
Answer:
External nostrils → Nasal cavity → Pharynx → Larynx → bronchi → bronchioles → Alveolei → Lungs.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

(3 marks)

III. Short Questions

Question 1.
What is meant by dead space?
Answer:
Some of the inspired air never reaches the gas exchange areas but fills the respiratory passages where exchange of gases does not occur. This air is called dead space. Dead space is not involved in gaseous exchange. It amounts to approximately 150mL.

Question 2.
Give an account of the structures of haemoglobin?
Answer:

  • Hemoglobin belongs to the class of conjugated protein.
  • The iron-containing pigment portion haem constitutes only 4% and the rest colourless protein of the histone class globin.
  • The molecular weight of Hb is 68000
  • These four Iron atoms can combine with a molecule of oxygen.

Question 3.
What is meant by methaemoglobin?
Answer:
If the iron component of the haem is in ferric state in stead of normal ferrous state it is called methaemoglobin. Methaeglobin does not bind with O2.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 4.
What Are Surfactants?
Answer:
They are the thin non-cellular films made of protein and phospholipids covering the alveolar membrane.

Question 5.
What are the significances of surfactants?
Answer:
The surfactant lowers the surface tension in the alveoli and prevents the lungs from collapsing.
It also prevents pulmonary oedema.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 6.
What is new born respiratory distress syndrome (NRDS)?
Answer:
Premature Babies have low levels of surfactant in the alveoli may develop the new bom respiratory distress syndrome (NRDS) because the synthesis of surfactants begins only afer the 25th week of gestation.

Question 7.
What is the reason for yawning?
Answer:
When there is a shortage of O2, it is sensed by our brain and sends a message to CNS to correct the imbalance for O2 demand and trigger us to yawn. Yawning helps us to breath more oxygen to the lungs.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 8.
Why are hiccups occured?
Answer:
Hiccups are due to eating too fast or having occasional spasms of the diaphragm.

Question 9.
What is the need of respiration?
Answer:
For all the activities of our body energy is needed. This we receive from the food. Oxygen is utilized by the organisms to break down the biomolecules the glucose and to derive energy. Hence Respiration is necessary.

Question 10.
Why the rate of respiration in aquatic animals is high?
Answer:
The amount of dissolved oxygen is very low in water compared to the amount of oxygen in the air. So the rate of breathing in aquatic organisms is much faster than land animals.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 11.
What is the importance of mucus in the respiratory tract?
Answer:
The goblet cells present in the mucus membrane secrete mucus, a slimy material rich in glycoprotein. Microorganisms and dust particles attach to the mucus films and are carried upwards to pass down the gullet during swallowing.

Question 12.
What is dead space?
Answer:
Some of the inspired air never reaches the gas exchange areas but fills the respiratory passages where the exchange of gases does not occur. This air is called dead space. Dead space is not involved in gaseous exchange. It amounts to approximately 150mL.

Question 13.
Why should we avoid breathing with our mouths?
Answer:
Breathing through mouth results in bladder shrinkage and creates an urge to urinate in the middle of the night.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

IV. Competitive Exam Corner

Question 1.
Sarojini’s father has congestion of the lungs. His doctor advised him to take bed rest and prescribed him an inhaler. What disease is he suffering from? List the symptoms of the disease.
Answer:
He is suffering from pneumonia.
Symptoms of pneumonia:

  1. Sputum Production,
  2. Nasal congestion,
  3. Shortness of breath,
  4. Sore throat

Question 2.
A villager who came to the city was affected by severe respiratory illness due to the inhalation of particulate pollutants. Suggest the reason for his illness and how do particulate pollutants affect him.
Answer:
He is suffering from a dust allergy. As he entered in a polluted area he started sneezing and coughing. The allergens in that place affecting his respiratory tracts and provoked inflammatory response prolonged allergy leads to Asthma.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 3.
Kumar’s mother works in a stone grinding factory. Suddenly she faints and taken to the hospital. The doctor notices fibers in the lungs. What kind of disease is she affected with? How can it be rectified?
Answer:
Long exposure to sand particles can give rise to inflammation leading to fibrosis. She must be hospitalized and have to give medication like anti-coagulation the imatinib. that fight against the disease.

(5 marks)

V. Essay Questions

Question 1.
List the primary functions of the respiratory system?
Answer:

  • It helps in the exchange of O2 and CO2 between the atmosphere and the blood.
  • It maintains homeostatic regulation of body pH.
  • It protects us from inhaled pathogens and pollutants.
  • It maintains the vocal cords for normal communication.
  • It removes the heat produced during cellular respiration through breathing.

Question 2.
Describe the structure of trachea with a diagram.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration 1

  • The trachea is semiflexible tube supported by cartilaginous rings.
  • It starts from the pharynx and ends in the lungs there it divides into right and left primary bronchi.
  • With in the lungs the bronchi divided repeatedly into secondary and tertiary bronchi.
  • That further divides into terminal bronchioles and respiratory bronchioles.
  • Bronchi have ‘c’ shaped curved cartilage plates.
  • This plate helps in preventing collapsing as the air pressure changes during breathing.
  • There is no cartilaginous stingray the brarichioles,
  • The rigidity of the bronchioles prevents them from collapsing.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 3.
Describe the process of inspiration & expiration with a diagram?
Answer:
Inspiration occurs if the pressure inside the lungs is less than the atmospheric pressure.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration 2

Inspiration:

  • There is a contraction of diaphragm muscles and external intercostal muscles which pulls the ribs and sternum upwards and downwards and increases the volume of the thoracic chamber in the dorsoventral axis.
  • Hence the pulmonary pressure is less than the atmospheric pressure.
  • This forces the fresh air from outside to enter the air passages into the lungs to equalize the pressure.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration 3

Expiration:

  • Expiration takes place when the pressure within the lungs is higher than the atmospheric pressure.
  • Relaxation of the diaphragm leads to its original dome-shaped nature.
  • The internal intercostal muscles contract pulling the ribs downward reducing the thoracic volume and pulmonary volume.
  • Thin results in an increase in the intrapulmonary pressure slightly above the atmospheric pressure causing the expulsion of air from the lungs.

Question 4.
Describe the structure of lung with a diagram.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration 4

  • The lungs are light spongy tissue.
  • It is enclosed inthe thoracic cavity surrounded by an air-tight space.
  • The thoracic cavity is bound dorsally by the ventral column and ventrally by the sternum. laterally by the ribs and on the lower side by the dome-shaped diaphragm.
  • The lungs are covered by a double walled pleural membrane and the plural cavity is filled with pleural fluid which reduces friction.
  • The trachea is a semi-flexible tube supported by cartilaginous rings which extends upto the 5th thoracic vertebra.
  • It divides into right and left bronchi and enters in to the lungs. There it divides further many times and ends in alveoli.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 5.
Explain the transport of oxygen In blood.
Answer:
Molecular oxygen is carried in blood in two ways: bound to haemoglobin within the red blood cells and dissolved in plasma. Oxygen is poorly soluble in water, so only 3% of the oxygen is transported in the dissolved form. 97% of oxygen binds with haemoglobin in a reversible manner to form oxyhaemoglobin (Hb02). The rate at which haemoglobin binds with O2 is regulated by the partial pressure of O2.

Each haemoglobin carries maximum of four molecules of oxygen. In the alveoli high pO2, low pCO2, low temperature and less H+ concentration, favours the formation of oxyhaemoglobin, whereas in the tissues low p02, high pCO2, high H+ and high temperature favours the dissociation of oxygen from oxyhaemoglobin.

A sigmoid curve (S-shaped) is obtained when percentage saturation of haemoglobin with oxygen is plotted against pO2. This curve is called oxygen haemoglobin dissociation curve. This S-shaped curve has a steep slope for pO2 values between 10 and 50 mm Hg and then flattens between 70 and 100 mm Hg. Under normal physiological conditions, every 100 mL of oxygenated blood can deliver about 5 mL of O2 to the tissues.

Question 6.
Give the tabulated column of partial pressure of O2 and CO2 in comparison to the gases in the atmosphere.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration 5

Question 7.
Describe the process of CO2 transport.
Answer:

  • 7-10% of CO2 is transported in a dissolved form in the plasma.
  • 20-25% of dissolved CO2 is bound and carried in the RBCs as carbamino haemoglobin.
    CO2 + H6  ⇌ H6 CO2
  • About 70% of CO2 is transported as bicarbonate ions.
  • At the tisoues the PCO2 is high due to catabolism and diffuses in the blood to form HCO3 and H+
  • When CO2 diffuses into RBCs it combines with water forming carbonic acid catalyzed by carbonic anhydrase.
  • Carbonic acid is dissociated into hydrogen and bicarbonate.
    Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration 6
  • Every 100 ml of deoxygenated blood delivers 4ml CO2 to the alveoli for elimination.

Question 8.
Describe the process of regulation of respriation.
Answer:

  • Medulla oblongata is a repiratory regulation centre.
  • The pneumotaxic centre present in the pons varoli is the respiratory rhythm centre.
  • The chemosensitive area found close to the rhythm centre is highly sensitive to CO2 and H+
  • H+ are eliminated out by respiratory process.
  • Receptiors associated with the aortic arch and carotid artery send signals to the rhythm centre for remedial action.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 9.
Write an essay on respiratory disorders.
Answer:
The respiratory system is affected by environmental occupational personal and social factors.
Following are some of the respiratory disorders:

Asthma: It is characterized by narrowing and inflammation of bronchi and bronchioles and difficulty in breathing.

Causes: Allergens like dust drugs pollen grains, certain food items like fish.

Emphysema: It is chronic breathlessness. It is caused by gradual breakdown of the thin wall of the alveoli decreasing the total surface area of a gaseous exchange.

Causes: The widening of the alveoli is called Emphysema.
Cigarette smoking reduces the respiratory surface of the alveolar walls.

Bronchitis: It is the inflammation of the bronchi.

Causes: Pollution smoke ciagratte smoking.

Symptoms: Cough shortness of breath sputum in the lungs.

Pneumonia: It is the inflammation of the lungs.

Causes: Bacteria and virus

Symptoms

  • Sputum production Nasal congestion, Shortness of breath sore throat.
  • Tuberculosis

Causes

  • Tuberculosis is caused by mycobacterium tubercular.
  • Infection mainly occurs in the lungs and bones.

Symptoms
Collection of fluid between the lungs and the chest wall is the main complication of this

Question 10.
List the ill effects of smoking.
Answer:

  • It increases the heartbeat rate.
  • It narrows the blood vessels results in raised blood pressure and leads to coronary heart diseases.
  • Smoking can cause lung diseases by damaging the airways and alveoli and results in emphysema and chronic bronchitis.

Question 11.
Tabulate the organism. respiratary organs and the
Answer:

ORGANISMS RESPIRATORY ORGANS
1. Sponges, Coelenterates Body surface
2. Earth worm The moist skin
3. Insects Trachea
4. Aquatic Arthropods mollusca Gills
5. Fishes Gills
6. Amphibians, Reptiles Aves mammals Lungs
7. Frog Lungs, Moist skin

Question 12.
What are the steps involved in the respiratory process?
Answer:
Steps involved in respiration are

  1. The exchange of air between the atmosphere and the lungs.
  2. The exchange of O2 and CO2 between the lungs and the blood.
  3. Transport of O2 and CO2 by the blood.
  4. Exchange of gases between the blood and the cells.
  5. Intake of O2 by the cells for various activities and release of CO2

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 13.
Tabulate the disorders of respiratory system.
Answer:

Disorders Symptoms
1. Pulmonary Embolism Blood clot occurs in the lung
2. Bronchitis Inflammation of the lining of your bronchial tubes
3. Asthma Swelling and narrowing of air ways and there is excess secretion of mucus.
4. Lung cancer Smoking causes cancer
5. Pneumonia Inflammation of lungs affecting alveoli
6. Pulmonary edema fluid accumulation of the tissue and air spaces of lung.
7. Emphysema Shortness of breath due to widening of alveoli
8. Atelectasis Alveoliand lungs get deflated
9. Tuberculosis It affects lungs and bones and effasion (fluid accumulation in the lungs)
10. Pleurisy Pleura becomes inflammed

Question 14.
List the problems in oxygen transport.
Answer:

  • When a person travels from sea level to elevations above 8000 ft there is a poor binding of O2 with haemoglobin.
  • There is a symptom of headache shortness of breath nausea and dizziness develop. (Acute mountain sickness)
  • To overcome this situation kidneys accelerate the production of the hormone erythropoietin which stimulates the synthesis of RBCs.

II. Inthedeepsea

  • When a person descends deep in to the sea the pressure in the water increases which causes the lungs to decrease in volume.
  • There is an increased nitrogen level in the blood lead to nitrogen narcosis.
  • When the diver ascends to the surface a condition called decompression sickness occurs. As nitrogen comes out of solution while still in the blood-forming bubbles.
  • The large bubbles can block the blood flowor can press on the nerve ending. This also causes pain in joints, muscles and causes neurological problems.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 15.
List the toxic substances present in tobacco. What are the ill-effects of smoking.
Answer:
a) Toxic substances present in tobacco.
Nicotine tar, carbon monoxide ammonia, arsenic and sulphur dioxide.

b) ill effects

  • Carbon monoxide and Nicotine damaged the cardie vascular system.
  • The tar damages the gaseous exchange system.
  • Nicotine stimulate the heart to beat faster and narrowing the blood vessels results in raised blood pressure and coronary heart diseases.
  • Carbon monoxide reduces O2 Supply.
  • Smoking causes lung, stomach, and pancreases and bladder cancer.
  • It lowers sperm count in men.

Question 16.
What is meant by chronic obstructive pulmonary disease?
Answer:

  • Smoking can cause lung diseases by damaging the airways and alveoli and results in emphysema and chronic bronchitis.
  • These two diseases along with asthma are referred to as a chronic obstructive pulmonary disease.
  • When a preson smokes 85% of the smoke released is inhaled by the smoker himself and others in the vicinity called passive smokers are also affected indirectly.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 17.
List the events in inspiration and expiration.
Answer:

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration 7

Question 18.
Describe the relationship between partial pressure of O2 and the nature of O2 dissolving the haemoglobin.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration 8

  • In the alveoli high PO2 low PCO3 Low temperature and less H+ Concentration favours the formation of oxyhemoglobin wjiere as in the tissues low PO2 high PCO2 high H and high-temperature favoures the dissocation of O2 from oxyhemoglobin.
  • A sigmoid curve is obtained when the percentage saturation of haemoglobin with O2 is plotted against PO2.
  • This S, Shaped curve has a steep slope for PO2 valuer between 10 and 50 mm Hg and then flattens between 70 and 100 mm Hg.
  • Under normal physiological conditions, every 100 ml of oxygenated blood can deliver about 5 ml of O2 to the tissues.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 6 Respiration

Question 19.
List the PO2 and PCO2 during inspiration expiration and in lungs and blood vessels.
Answer:

Location Partial Pressures of Oxygen PO2 The partial pressure of CO2 PCO2
Inspiration 159 mm. Hg 0.3 mm. Hg
Expiration 120 mm. Hg 127 mm. Hg
Alveoli 104 mm. Hg 40 mm. Hg
Pulmonary artery 40 mm. Hg 45 mm.Hg
Pulmonary vein 95 mm. Hg 40 mm. Hg
Oxygenated blood 95 mm. Hg 40 mm. Hg
Deoxygenated blood 40 mm. Hg 45 mm. Hg

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 1.
Find the expected value for the random variable of an unbiased die
Solution:
When a un based die is thrown , any one of the number 1, 2, 3, 4, 5, 6, can turn up that x denote the random variable taking the values from 1 to 6
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 1

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 2.
Let X be a random variable defining number of students getting A grade. Find the expected value of X from the given table
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 2
Solution:
Expected value of X, E(X) = \(\sum_{x} x P_{x}(x)\)
E(X) = (0 × 0.2) + (1 × 0.1) + (2 × 0.4) + (3 × 0.3)
= 0 + 0. 1 + 0.8 + 0.9
= 1.8

Question 3.
The following table is describing about the probability mass function of the random variable X
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 3
Find the standard deviation of x.
Solution:
Let x be the random variable taking the values 3, 4, 5
E(x) = Σpixi
= (0.1 × 3) + (0.1 × 4) + (0.2 × 5)
0.3 + 0.4 + 1.0
E(x) = 1.7
E(x²) = Σpixi²
= (0.1 × 3²)+ (0.1 × 4²) + (0.2 × 5²)
= (0.1 × 9) + (0.1 × 16) + (0.2 × 25)
E(x²) = 7.5
Var(x) = E(x²) – (E(x)]²
= 7.5 – (1.7)²
= 7.5 – 2.89
Var(x) = 4.61
Standard deviation(S.D) = \(\sqrt { var (x)}\)
= \(\sqrt { 4.61}\)
σ = 2.15

Question 4.
Let X be a continuous random variable with probability density function
fx (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected value of X.
Solution:
Let x be a continuous random variable. In the probability density function, Expected
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 4

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 5.
Let X be a continuous random variable with probability density function
fx (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the mean and variance of X.
Solution:
Let x be a continuous random variable. In the probability density function,
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 5
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 6

Question 6.
In investment, a man can make a profit I of ₹ 5,000 with a probability of 0.62 or a loss of ₹ 8,000 with a probability of 0.38. Find the expected gain.
Solution:

X 5000 -8000
P(x = x) 0.62 0.38

Let x be the random variable of getting gain in an Investment
E(x) be the random variable of getting gain in an Investment
E(x) = ΣPixi
= (0.62 × 5000) + [0.38 × (-8000)]
= 3100 – 3040
E(x) = 60
∴ Expected gain = ₹ 60

Question 7.
What are the properties of mathematical expectation?
Solution:
The properties of Mathematical expectation are as follows:

  1. E(a) = a, where ‘a’ is a constant
  2. Addition theorem: For two r.v’s X and Y, E(X + Y) = E(X) + E(Y)
  3. Multiplication theorem: E(XY) = E(X) E(Y)
  4. E(aX) = aE(X), where ‘a’ is a constant
  5. For constants a and b, E(aX + b) = a E(X) + b

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 8.
What do you understand by mathematical expectation?
Solution:
The average value of a random Phenomenon is termed as mathematical expectation or expected value.
The expected value is a weighted average of the values of a random variable may assume

Question 9.
How do you define variance in terms of Mathematical expectation?
Solution:
Let X be a random variable. Let E(X) denote the expectation of X.
Then the variance is defined in terms of the mathematical expectation as follows.
(a) X is discrete r.v with p.m.f p(x). Then Var(X) = \(\sum[x-\mathrm{E}(\mathrm{X})]^{2} p(x)\)
(b) X is continuous r.v with p.d.f fx(x). Then Var(X) = \(\int_{-\infty}^{\infty}[X-E(X)]^{2} f_{X}(x) d x\)

Question 10.
Define mathematical expectation in tears of a discrete random variable?
Solution:
Let X be a discrete random variable with probability mass function (p.m.f.) p(x). Then, its expected value is defined by
E(X) = \(\sum_{ x }\) x p(x) ……(1)

Question 11.
State the definition of mathematical expectation using continuous random variables.
Solution:
Let X be a continuous random variable with probability density function f(x). Then the expected value of X is
\(\mathrm{E}(\mathrm{X})=\int_{-\infty}^{\infty} x f(x) d x\)
If the integral exists, E(X) is the mean of the values, otherwise, we say that the mean does not exist.

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 12.
In a business venture, a man can make a profit of ₹ 2,000 with a probability of 0.4 or have a loss of ₹ 1,000 with a probability of 0.6. What are his expected, variance and standard deviation of profit?
Solution:
Let X be the random variable of getting profit in a business

X 2000 -1000
P(x = x) 0.4 0.6

E(x) = Σxxpx(x)
= (0.4 × 2000) +[0.6 × (-1000)]
= 800 – 600
E(X) = 200
∴ Expected value of profit = ₹ 200
E(X²) = Σx² Px(x)
= [(2000)² × 0.4] + [(-1000)² × 0.6]
= (4000000 × 0.4) + (1000000 × 0.6)
E(X²) = 2200000
Var(X) = E(X²) – [E(X)]²
= 22000000 – (200)²
= 2200000 – 40000
Var(X) = 21,60,000
Variance of his profit = ₹ 21,60,000
Standard deviation(S.D) = \(\sqrt { var (x)}\)
σ = \(\sqrt { 2160000}\)
σ = 1469.69
Standard deviation of his profit is ₹ 1,469.69

Question 13.
The number of miles an automobile tire lasts before it reaches a critical point in tread wear can be represented by a p.d.f.
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 7
Find the expected number of miles (in thousands) a tire would last until it reaches the critical tread wear point.
Solution:
We know that,
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 8

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 14.
A person tosses a coin and is to receive ₹ 4 for a head and is to pay ₹ 2 for a tail. Find the expectation and variance of his gains.
Solution:
Let X denote the amount the person receives in a game
Then X takes values 4,-2 and
So P(X = 4) = P (of getting a head)
= \(\frac { 1 }{2}\)
P(X = – 2) = P (of getting a tail)
= \(\frac { 1 }{2}\)
Hence the Probability distribution is

X 4 -2
P(x = x) 1/2 1/2

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 9
E(x²) = 10
Var(x) = E(x²) – E(x²)
= 10 – (1)²
Var(x) = 9
∴ His expected gain = ₹ 1
His variance of gain = ₹ 9

Question 15.
Let X be a random variable and Y = 2X + 1. What is the variance of Y if the variance of X is 5?
Solution:
Given X is a random variable and Y = 2X + 1 and Var(X ) = 5
Var (Y) = Var (2X + 1) = (2)2 = 4
Var X = 4(5) = 20

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 7 Body Fluids and Circulation Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

11th Bio Zoology Guide Body Fluids and Circulation Text Book Back Questions and Answers

Part I

I. Choose The Best Options

Question 1.
What is the function of lymph?
a. Transport of O2 into brain
b. Transport of CO2 into lungs
c. Bring interstitial fluid in blood
d. Bring RBC and WBC in lymph node
Answer:
c. Bring interstitial fluid in blood

Question 2.
Which one of the following plasma proteins is involved in the coagulation of blood?
a. Globulin
b. Fibrinogen
c. Albumin
d. Serum amylase
Answer:
b. Fibrinogen

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 3.
Which of the following WBCs are found in more numbers?
a. Eosinophil
b. Neutrophil
c. Basophil
d. Monocyte
Answer:
b. Neutrophil

Question 4.
Which of the following is not involved in blood clotting?
a. Fibrin
b. Calcium
c. Platelets
d. Bilirubin
Answer:
d. Bilirubin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 5.
Lymph is colourless because
a. WBC are absent
b. WBC are present
c. Haemoglobin is absent
d. RBC are absent
Answer:
d. RBC are absent

Question 6.
Blood group is due to the presence or absence of surface
a. Antigens on the surface of WBC
b. Antibodies on the surface of RBC
c. Antigens on the surface of RBC
d. Antibodies on the surface of WBC
Answer:
c. Antigens on the surface of RBC

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 7.
A person having both antigen A and antigen B on the surface of RBCs belongs to blood group
a. A
b. B
c. AB
d. O
Answer:
c. AB

Question 8.
Erythroblastosis foetalis is due to the destruction of
a. Foetal RBCs
b. Foetus suffers from atherosclerosis
c. Foetal WBCs
d. Foetus suffers from mianmata
Answer:
a. Foetal RBCs

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 9.
Dub sound of heart is caused by
a. Closure of atrio-ventricular valves
b. Opening of semi-lunar valves
c. Closure of semi-lunar values
d. Opening of atrio-ventricular valves.
Answer:

Question 10.
Why is the velocity of blood flow the lowest in the capillaries?
a. The systemic capillaries are supplied by the left ventricle, which has a lower cardiac output than the right ventricle.
b. Capillaries are far from the heart, and blood flow slows as distance from the heart increases.
c. The total surface area of the capillaries is larger than the total surface area of the arterioles.
d. The capillary walls are not thin enough to allow oxygen to exchange with the cells.
e. The diastolic blood pressure is too low to deliver blood to the capillaries at a high flow rate.
Answer:
c. The total surface area of the capillaries is larger than the total surface area of the arterioles.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 11.
An unconscious patient is rushed into the emergency room and needs a fast blood transfusion. Because there is no time to check her medical history or determine her blood type, which type of blood should you as her doctor, give her?
a. A
b. AB
c.  Q+ve
d. O ve
Answer:
d. O ve

Question 12.
Which of these functions could or could not be carried out by a red blood cell?
a. Protein synthesis
b. Cell division
c. Lipid synthesis
d. Active transport
a. Protein synthesis – Red blood cells have no DNA.
b. Cell division – Red blood cells have no nucleus. So do not undergo any mitosis and mesiosis.
c. Lipid synthesis – Red blood contains none of the cellular orgakelles.
d. Active transport – Yes, RBC transport O2 and CO2 with in the cell.
Answer:
d. Active transport – Yes, RBC transport O2 and CO2 with in the cell.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 13.
At the venous end of the capillary bed, the osmotic pressure is
a. Greater than the hydrostatic pressure
b. Result in net outflow of fluids
c. Results in net absorption of fluids
d. No change occurs.
Answer:
a. Greater than the hydrostatic pressure

Question 14.
A patient’s chart reveals that he has a cardiac output of 75G0mL per minute and a stroke volume of 50 mL. What is his pulse rate (in beats/min)
a. 50
b. 100
c. 150
d.400
Answer:
c. 150

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 15.
At any given time there is more blood in the venous system than that of the arterial system. Which of the following features of the veins allows this?
a. relative lack of smooth muscles
b. presence of valves
c. proximity of the veins to lymphatic’s
d. thin endothelial lining
Answer:
b. presence of valves

Question 16.
Distinguish between arteries and veins.
Answer:

Arteries Veins
1. They carry blood away from the heart They carry blood from the parts of the body towards the heart.
2. They lie deep inside the body. They lie on the surface beneath the skin.
3. The walls are thick and non-collapsible They have thinner walls.
4. There are no valves. They have valves.
5. Except for the pulmonary artery all the arteries carry oxygenated blood. Except for the pulmonary vein, all the veins carry deoxygenated blood.
6. Blood pressure is high in the arteries Blood pressure is low in the veins.
7. A small sphincter lies at the junction between the arterioles and capillaries to regulate the blood supply. There is no sphincter muscles

Question 17.
Distinguish between open and closed circulation
Answer:

Open circulation Closed circulation
1. The blood pumps from the heart pass into the coelom through blood vessels. The body cavity is known as haemocoel. The blood pumps from the heart push into the blood vessels.
2. (Eg.) Arthropoda Mollusca (Eg.) Annelida vertebrates

Question 18.
Distinguish between the mitral valve and semilunar valve.
Answer:

Mitral valve Semilunar valve
1. It guards the opening between the left atrium and left ventricle It guards the opening of the pulmonary artery and aorta.
2. It allows the blood to flow from auricle to ventricle It allows the blood to flow from the ventricle to the pulmonary artery and aorta.
3. It prevents the backflow of blood It prevents the backflow of blood.
4. The mitral valve closes during contraction of heart and produces a sound lub. The dub sound is produced due to the closure of lunar valves on dialation of heart
5. The chordae tendinae is not connected to this valve. The chordae tendinae is connected to this valve.

Question 19.
The right ventricular wall is thinner than the left ventricular wall. Why?
Answer:
The right ventricle pumps deoxygenated blood, to the lungs through pulmonary artery. The left ventricle pumps the oxygenated blood to all parts of the body through the aorta. Hence, left ventricle has to exert more pressure. Hence right ventricular wall is thinner but the left ventricular walls is thicker.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 20.
What might be the effect on a person whose diet has less iron content?
Answer:

  • The number of red blood cells decreases.
  • Due to the depletion of haemoglobin he finds it difficult to breath.
  • Due to the deficiency of iron anaemia may develop.
  • Due to the deficiency of iron, the oxygen-carrying capacity of haemoglobin reduces.

Question 21.
Describe the mechanism by which the human heartbeat is initiated and controlled.
Answer:

  • The human heart is myogenic in nature.
  • The heartbeat is originated from a pacemaker. The total rate of heartbeat is decided by this node.
  • This pacemaker is situated in the right sinuatrial node.
  • On the left side of the right atrium is a node called auricula ventricular node.
  • Two special cardiac muscle fibres originate from the auriculo ventricular node and are called the bundle of His which runs down into the interventricular septum and the fibres spread into the ventricle called Purkinje fibres.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 1

Origin of heartbeat:

  • Pacemaker cells produce excitation through depolarisation of their cell membrane. The j excitation is spread in to the auricle. Then this is passed on to bundle of His through auriculo ventricular node.
  • The purkinje fibres cause ventricular contraction.

Regulation:

  • The pacemaker cells produce excitation through depolarisation of their cell membrane. Each polarisation is slow taken place by sodium influx and reduction in potassium efflux.
  • Minimal potential is required to activate voltage gated calcium (Ca+) channels that causes rapid depolarisation which results in action potential.
  • The pacemaker cells repolarise slowly via K+ efflux.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 22.
What is lymph? Write its function.
Answer:
About 90% of fluid that leaks from capillaries eventually seeps back into the capillaries and the remaining 10% is collected and returned to blood system by me of a series of tubules known as lymph vessels or lymphatics.

The fluid inside the lymphatics is called lymph. The lymphatic system consists of a complex network of thin walled ducts (lymphatic vessels), filtering bodies (lymph nodes) and a large number of lymphocytic cell concentrations in various lymphoid organism.

The lymphatic vessels have smooth walls that run parallel to the blood vessels, in the skin, along the respiratory and digestive tracts. These vessels serve as return ducts for the fluids that are continually diffusing out of the blood capillaries into the body tissues.

Lymph fluid must pass through the lymph nodes before it is returned to the blood. The lymph nodes that filter the fluid from the lymphatic vessels of the skin are highly concentrated in the neck, inguinal, axillaries, respiratory and digestive tracts.

The lymph fluid flowing out of the lymph nodes flow into large collecting duct which finally drains into larger veins that runs beneath the collar bone, the subclavian vein and is emptied into the bloodstream. The narrow passages in the lymph nodes are the sinusoids that are lined with macrophages.

The lymph nodes successfully prevent the invading microorganisms from reaching the bloodstream. Cells found in the lymphatics are lymphocytes. Lymphocytes collected in the lymphatic fluid are carried via the arterial blood and are recycled back to the lymph. Fats are absorbed through lymph in the lacteals present in the villi of the intestinal wall.

Question 23.
What are the heart sounds? When and how are these sounds produced?
Answer:
Rhythmic contraction and expansion of the heart are called heartbeat. The contraction of the heart is called systole and the relaxation of the heart is called diastole. The heart normally beats 70-72 times per minute in a human adult. During each cardiac cycle, two sounds are produced that can be heard through a stethoscope.

The first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves whereas the Second heart sound (dub) is associated with the closure of the semilunar valves. These sounds are of clinical diagnostic significance. An increased heart rate is called tachycardia and decreased heart rate is called bradycardia.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 24.
Select the correct biological term.
Lymphocytes, red cells, leukocytes, plasma, erythrocytes, white cells, hemoglobin, phagocyte, platelets, blood clot.
Answer:

a. Disc-shaped cells which are concave on both sides. Red cells
b. Most of these have a large bilobed nucleus Leucocytes
c. Enable red cells to transport blood. Haemoglobin
d. The liquid part of the blood Plasma
e. Most of them move and change shape like an amoeba Leucocytes
f. Consists of water and important dissolved substances. Plasma
g. Destroyed in the liver and spleen after circulating in the blood for four months. Red cells
h. The substances which give red cells their colour Haemoglobin
i. Another name for red cells Erythrocytes
j. Blood that has been changing to jelly Blood clot
k. A word that means cell eater Phagocytes
l. Cells without nucleus Red cells
m. White cells made in the lymphatic tissue Lymphocytes
n. Blocks wound and prevent excessive bleeding. Blood clot
o. Fragment of cells which are made in the bone marrows Platelets
p. Another name for white blood cells. Leucocytes
q. Slowly releases oxygen to blood cells. Haemoglobin
r. Their function is to help blood clots in wounds. Platelets

Question 25.
Select the correct biological term. Cardiac muscle, atria, tricuspid systole, auricles, arteries, diastole, ventricles, bicuspid valve, pulmonary artery, cardiac cycle, semilunar valve, veins, pulmonary vein, capillaries, vena cava, aorta.
Answer:

a. The main artery’ of the blood. Dorsal aorta
b. Valves between the left atrium and ventricle. Bicuspid valve
c. Technical name for relaxation of the heart. Diastole
d. Another name for atria. Auricle
e. The main vein. Vena cava
f. Vessels which carry blood away from the heart. Aorta
g.Two names for the upper chambers of the heart. Auricle
h. Thick-walled chambers of the heart. Ventricle
I. Carries blood from the heart to the lungs. Pulmonary artery
j. Takes about 0.8 sec to complete. Cardiac cycle
k. Valves situated at the point where blood flows out of the heart. Semilunar valve
l. Vessels which carry blood towards the heart. Vein
m. Carries blood from the lungs to the heart. Pulmonary vein
n. The two lower chambers of the heart. Ventricle
o. Prevent blood from re-entering the ventricles after entering the aorta. Semilunar valve
P. Technical name for one heartbeat. Cardiac cycle
q. Valves between right atrium and ventricles. Tricuspid valve
r. The technical name for the contraction of the heart. Systole
s. Very narrow blood vessels. Capillaries

Question 26.
Name and Label the given diagrams to show A, B, C, D, E, F, and G.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 2

Part II

11th Bio Zoology Guide Body Fluids and Circulation Additional Important Questions and Answers

I. Choose The Best Options

Question 1.
Which of the following is not the function of circulatory system?
(a) Transport of respiratory gases
(b) Carrying of digested food materials
(c) Transport of hormones to target organism
(d) Removal of nitrogenous wastes from the body
Answer:
(d) Removal of nitrogenous wastes from the body

Question 2.
Find out the wrong statement.
a. The density of protein in tissue fluid is lesser than plasma
b. Blood is a fluid connective tissue
c. The amount of blood present in the 70 kg man is 5 litre
d. The plasma protein albumin provided immunity.
Answer:
d. The plasma protein albumin provided immunity.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 3.
What is the function of albumin?
(a) Transport of hormones
(b) Blood clothing
(c) Maintenance of osmotic pressure
(d) Immunity
Answer:
(c) Maintenance of osmotic pressure

Question 4.
Which of the following is rich in urea?
a. Hepatic vein
b. Splenic vein
c. Pancreatic vein
d. Pulmonary vein
Answer:
a. Hepatic vein

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 5.
The red colour of the RBC is due to the presence of a respiratory pigment ……………
(a) Haemoerythrin
(b) Haemoglobin
(c) Haemocyanin
(d) Chlorocronin
Answer:
(b) Haemoglobin

Question 6.
Which is a heterophil?
a. Neutrophils
b. Eosinophil
c. Basophil
d. Monocytes
Answer:
a. Neutrophils

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 7.
What is hematocrit/packed cells volume?
(a) The ratio of WBCs to blood plasma
(b) The ratio of RBCs to blood plasma
(c) The ratio of platelets to blood plasma
(d) The ratio of plasma and blood cells
Answer:
(b) The ratio of RBCs to blood plasma

Question 8.
Name the blood cell secreted by megakaryocytes?
a. Red blood cells
b. White blood cells
c. Platelets
d. None of the above
a. I – d, II – c, III – b, IV – a
b. I – a, II – d, III – b, IV – c
c. I – a, II – d, III – b, IV – c
d. I – d, II – c, III – a, IV – b
Answer:
c. I – a, II – d, III – b, IV – c

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 10.
Name the protein which is similar to the protein present in the red blood cells of Rhesus Monkey.
a. D – antigen
b. A – antigen
c. B – antigen
d. A, B – antigen
Answer:
a. D – antigen

Question 11.
Confirm A: Red blood cell contains more hemoglobin
Reason B: There are no cell organelles in the red blood cells.
a. A – True, B – False
b. A – True, B – True
c. A – False, B – True
d. A – False, B – False
Answer:
b. A – True, B – True

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 12.
‘A’ blood group has ………………….. antigen and …………………. antibody
(a) A, anti B
(b) AB, no antibodies
(c) No antigen, anti A, Anti B
(d) B, Anti A
Answer:
(d) B, Anti A

Question 13.
Name the white blood cells present in lymph?
a. Lymphocytes
b. Monocytes
c. Neutrophils
d. Basophils
Answer:
a. Lymphocytes

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 14.
Name the layer which is not seen in capillaries
a. Tunica intima
b. Tunica media
c. Tunica external
d. Tunica internal
Answer:
b. Tunica media

Question 15.
How much time is taken for a single cardiac cycle?
a. 0.7 secs
b. 0.8 secs
c. 0.6 secs
d. 10.9 secs
Answer:
b. 0.8 secs

Question 16.
Pulmonary veins carry ………………. blood from lungs to ……………….
(a) Oxygenated, right auricle
(b) Deoxygenated, right auricle
(c) Deoxygenated, left auricle
(d) Oxygenated, left auricle
Answer:
(c) Deoxygenated, left auricle

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 17.
The slow excitation of the ‘QRS’ wave indicates ………………………….. defect.
a. Inflammation of ventricle
b. Defects in bicuspid valve
c. Block in the coronary artery
d. Defects in the atrioventricular node
Answer:
a. Inflammation of ventricle

Question 18.
Who has explained first about blood circulation?
a. Raymond deviessens
b. William Harvey
c. Robert William
d. James Elam
Answer:
b. William Harvey

Question 19.
What is the normal blood pressure of a man?
a. 120/90 mm Hg
b. 120/80 mm Hg
c. 120 /80 mm Hg
d. 130/80 mm Hg
Answer:
c. 120 /80 mm Hg

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 20.
Which of the following is the cause of a stroke?
a. Rupture of blood vessels in the brain
b. A clot in the blood vessels of brain
c. Deposition in the blood vessels of the brain
d. All the above
Answer:
d. All the above

Question 21.
Does ischemic heart disease indicate?
a. Myocardial infarction
b. Rheumatoid heart disease
c. Angino pectoralis
d. Stroke
Answer:
a. Myocardial infarction

Question 22.
Which hormone increases the heartbeat?
(a) Acetylcholine
(b) Gastrin
(d) Epinephrine
(d) Oxytocin
Answer:
(c) Epinephrine

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 23.
Match the following and find the correct answer?
I. Erythropoietin – a. Agglutinization
II. Haematocrit – b. It takes an important role in the inflammation of body tissues
III. Heparin – c. Finding the ratio between blood plasma and red blood cells.
IV. Antigen – d. It stimulates the synthesis of RBCs in the bone marrow.
a. I-d, II-c, III – b, IV-a
b. I – a, II – b, III – c, IV – d
c. I – a, II – c, III – d, IV – b
d. I – d, II – c, III – b, IV – a
Answer:
a. I-d, II-c, III – b, IV-a

Question 24.
Note the given diagram and find out the correct answers.
a. T wave represents the repolarisation of auricle
b. The ‘P’ wave represents the functions of auricle
c. Te ‘Q’ wave represents the depolarization of ventricular septum
d. ‘R’ and ‘S’ wave represents depolarization of auricle
Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 3
Answer:
b. The ‘P’ wave represents the functions of the auricle

(2 marks)

II. Very Short Questions

Question 1.
What are the types of body fluids?
Answer:
The intra-cellular fluid present inside the cells and the extracellular fluid present outside the cells are the two types of body fluids.

Question 2.
What is meant by interstitial fluid or tissue fluid?
Answer:
A fluid that surrounds the cell is known as interstitial fluid.
(Eg.) Plasma, Lymph

Question 3.
Give short notes on blood?
Answer:

  • Blood is the most common body fluid that transports substances from one part of the body to the other. It is known as fluid connective; tissue.
    • The plasma constitutes 55% of total blood volume.
  • The average blood volume is about 5000 mZ (51) is an adult weighing 70 kg.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 4.
What are the components of blood?
Answer:

  1. Red blood cells
  2. White blood cells.
  3. Platelets

Question 5.
Why is the spleen considered a graveyard of red blood cells?
Answer:

  • The average life span of red cells is 120 days.
  • After 120 days the red cells are destroyed in the spleen. Hence the spleen said to be a graveyard of RBCs.

Question 6.
What is hematocrit?
Answer:
The ratio of red blood cells to blood plasma is expressed as hematocrit (packed cell volume).

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 7.
What are the types of lymphocytes? What are its uses?
Answer:

  • B – lymphocytes – Produces antibodies
  • T – lymphocytes – Involves in cell-mediated immunity.

Question 8.
Classify the monocytes based on its location?
Answer:

Location Name
1. Central nervous system Microglia
2. Liver sinuses Kupffer cells
3. Lungs Macrophages of alveolei

Question 9.
What are the types of ‘ABO’ blood groups?
Answer:

  • A -blood group
  • B – blood group
  • AB – blood group
  • O- blood group

Question 10.
What is meant by alleleic genes?
Answer:
The gene that regulates the synthesis of A, B and O blood groups in ‘ABO’ type.

Question 11.
What are agglutinogens? What is it’s composition?
Answer:
Antigens present on the surface of RBCs are called agglutinogens.

Composition:
Sucrose, D – galactose, N – acetyl glucosamine, 11 terminal amino acids

Question 12.
What are the steps to be taken to prevent erythroblastosis f oetalis?
Answer:
After the first delivery if the first child is the Rh+ve, we should give anti D – antibodies -Rhocum to the
Rh-ve mother.

Question 13.
What is serum?
Answer:
Plasma without fibrinogen is called serum.

Question 14.
What is an anticoagulant substance? Where is it synthesized?
Answer:
Heparin is an anti-coagulant substance. It is synthesized in the mast cells.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 15.
What are the layers of blood vessels
Answer:

  1. Tunica externa – Outer layer
  2. Tunica media – Mid layer
  3. Tunica indima – Inner layer

Question 16.
Give notes on capillaries.
Answer:

  • There is no tunica media in the capillaries
  • It is the site for the exchange of materials between blood and tissues.
  • The blood volume is high but the flow of blood is low.
  • The walls of the capillaries are guarded by semilunar valves.
  • The oxygenated and deoxygenated blood is present in the capillaries.

Question 17.
What is meant by single circulation?
Answer:

  • Single circulation is seen in fishes. There is only one auricle and ventricle in the heart of fishes.
  • The blood flows from heart to gills there it gets oxygenated and supplies to the organ and then returns to the heart.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 18.
What is meant by incomplete double circulation?
Answer:
Reptiles have an incompletely divided ventricle. The oxygenated and deoxygenated blood is mixed here. Hence it is known as incomplete double circulation.

Question 19.
What is meant by complete double circulation?
Answer:

  • There are well divided 2 auricle and 2 ventricles in the heart of birds, crocodiles and mammals.
  • The oxygenated and deoxygenated blood is completely separated.
  • The pulmonary and systemic circulation is well defined.

Question 20.
Differentiate the tachycardia and bradycardia?
Answer:

Tachycardia
1. The rate of heartbeat increases
Bradycardia
The rate of heartbeat decreases.

Question 21.
What is meant by cardiac output?
Answer:
The amount of blood pumped out by each ventricle per minute is called the cardiac output.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 22.
What is meant by pulse or pulse rate?
Answer:
The rate of heartbeat per minute is called a pulse.

Question 23.
What is meant by pulse pressure?
Answer:
Pulse pressure: Systolic pressure -diastolic pressure.

Question 24.
What is meant by stroke volume?
Answer:
Stroke volume is the volume of blood pumped out by one ventricle with each beat.
CO = HR x SV

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 25.
When will the stroke volume double?
Answer:

  • During vigorous exercise, SV may double as a result of venous return.
  • The amount of blood pumps out of the ventricle is also increased.

Question 26.
What is meant by mean arterial pressure?
Answer:
Mean arterial pressure is a function of cardiac output and resistance in the arterioles.

Question 27.
What is the baroreceptor reflex?
Answer:
The primary reflex pathway for homeostatic control of mean arterial pressure is the baroreceptor reflex.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 28.
What is meant by orthostatic hypotension?
Answer:
When we are lying flat the gravitational force is evenly distributed. When you stand up gravity causes blood to pool in the lower extremities. The decrease in blood pressure upon standing is known as orthostatic hypotension.

Question 29.
What is myogenic heart?
Answer:
The heartbeat of man is originated from the cardiac muscles. Hence human heart is a myogenic heart.

Question 30.
Tabulate the cardiac diseases?
Answer:

Diseases Defects
1. Coronary heart disease When the coronary arteries are blocked the amount of blood goes to heart muscles decreases leads to oxygen and nutrient deficiency.
2. Vascular diseases Infection in the arteries veins and lymphatic glands
3. Aorta disease The wall of the aorta weakened and bulges to form a balloon-like sac aneurysm.
4. Pericarditis Inflammation in the layers of the pericardium
5. Cardio myopathy An abnormally thick heart muscle causing the heart to pump weaker than normal and leads to heart failure.
6. Heart valve disease One or more of the heart valves does not work.
7. Heart failure The heart cannot pump as powerfully as it need to in order to supply like body with 0, and nutrients carrying heart muscles to overwork and weaker.
8. Arrhythmia The heart beats irregularly

Question 31.
What is edema?
Answer:
The concentration of proteins in the blood becomes much lower than usual leads to the accumulation of fluid.

Question 32.
The walls of arteries nearer to the heart are more elastic than the arteries away from the heart? Why?
Answer:

  • When the heart contracts the blood is pushed into the artery hence the pressure in the arteries increases.
  • To withstand the pressure the artery walls nearer to the heart are more elastic and hence they relax and reduce the pressure.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 33.
How is the blood needed for the skeletal muscle during exercise compensated?
Answer:

  • During exercise, more blood is needed for skeletal muscle.
  • Hence the blood is diverted from the digestive system to skeletal muscle.

Question 34.
Define Laplace law? What do you infer from this?
Answer:

  • It states that the tension in the walls of the blood vessel is proportional to the blood pressure and vessel radius.
  • This law is used to understand the structure and function of blood vessels and the heart.

Question 35.
When blood volume drops down abruptly? What happens to the stroke volume?
Answer:
When there is a drop in blood volume the blood flows to the heart decreases hence the blood coming out of the heart during contraction decreases. (Stroke volume decreases)

(3 marks)

III. Short Questions

Question 1.
List the characteristics of the circulatory system?
Answer:

  • Oxygen and carbon dioxide are exchanged in the luiìgs and tissues.
  • Nutrients are taken from the digestive system and are carried to the liver and through blood taken to all parts of the body.
  • Wastes from the tissues are carried by the blood and finally removed by the kidneys.
  • The hormones are transported to their target organs.
  • Circulatory system helps to maintain the homeostasis of the body fluids and body temperature.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 4

Question 2.
Explain the role of the Rh factor?
Answer:
Rh factor is a protein (D antigen) present on the surface of the red blood cells in the majority (80%) of hum. This protein is similar to the protein present in Rhesus monkey, hence the term Rh. Individuals who carry the antigen D on the surface of the red blood cells are Rh+ (Rh-positive) and the individuals who do not carry antigen D, are Rh (Rh-negative). Rh factor compatibility is also checked before blood transfusion.

When a pregnant woman is Rh+ and the foetus is Rh+ incompatibility (mismatch) is observed. During the first pregnancy, the Rh antigens of the foetus do not get exposed to the mother’s blood as both their blood are separated by the placenta. However, small amount of the foetal antigen becomes exposed to the mother’s blood during the birth of the first child.

The mother’s blood starts to synthesize D antibodies. But during subsequent pregnancies, the Rh antibodies from the mother (Rh) enters the foetal circulation and destroys the foetal RBCs. This becomes fatal to the foetus because the child suffers from anaemia and jaundice. This condition is called erythroblastosis foetalis. This condition can be avoided by administration of anti D antibodies (Rhocum) to the mother immediately after the first childbirth.

Question 3.
Describe red blood cells?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 5

  1. Red blood cells are abundant than other blood cells. There are about 5 – 5.5 million RBC mm of blood in a healthy man and 4.5 – 5 million RBC mm” in healthy women.
  2. The red colour of the RBC is due to the respiratory pigment haemoglobin and it involves in the transport of respiratory gases.
  3. The biconcave-shaped RBC s increase the surface area.
  4. The RBCs are devoid of nucleus mitochondria ribosomes and endoplasmic reticulum.
  5. The average life span of RBC is about 120 days after which they are destroyed in the spleen.
  6. RBCs are synthesized in the bone marrow.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 4.
Give notes on platelets?
Answer:

  • Platelets are known as thrombocytes.
  • They are synthesized by the megakaryocytes of bone marrow.
  • They are devoid of a nucleus.
  • Blood normally contains 150000 – 350000 platelets mm3 of blood.
  • They are involved in blood coagulation.
  • The reduction in platelet number can lead to clotting disorders leads to excessive loss of blood from the body.

Question 5.
Arrange the blood groups, their antigens and antibodies and tabulate them.
Answer:

Blood group Agglutiongens (antigens) on the RBC Agglutinin antibodies in the plasma
A A Anti B
B B Anti A
AB AB No antibodies
O No antigens Anti A and Anti B

Question 6.
What are anastomoses?
Answer:

  • These are connections of one blood vessel with another blood vessel.
  • They provide an alternate route of blood flow if the original blood vessel is blocked.
  • Arteries in the joints contain numerous anastomoses. This allows blood to flow freely even if one of the arteries closes during bending of the joints.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 7.
Write notes on coronary blood vessels.
Answer:

  • Blood vessels that supply blood to the cardiac muscles with all nutrients and remove wastes are the coronary artries and veins.
  • Heart muscle is supplied by two arteries namely the right and left coronary arteries.
  • These arteries are the first branch of the aorta.
  • These arteries usually surround the heart in the manner of a crown hence called the coronary artery.
  • Right ventricle and posterior portion of the left ventricle are supplied by the right coronary artery.
  • Anterior and lateral part of the left ventricle is supplied by the left coronary arteries.

Question 8.
Give notes on the heartbeat.
Answer:

  1. Rhythmic contraction and expansion of heart is called heartbeat. The contraction of the heart is called systole and the relaxation of the heart is called diastole.
  2. The heart normally beats 70 – 72 times per min in a normal adult. Lub and dub sound is produced. These are heart sounds.
  3. The sound lub is associated with the closure of the tricuspid and bicuspid and the dub sound is associated with the closure of the semilunar valves.
  4. The heart sounds can be heard through a stethoscope.
  5. These sounds are of clinical diagnostic significance.
  6. An increased heart rate is called tachycardia and decreased heart rate is called bradycardia.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 9.
What is blood pressure?
Answer:

  • Blood pressure is the pressure exerted on the surface of blood vessels by the blood.
  • This pressure circulates the blood through arteries veins and capillaries.
  • There are 2 types of pressure the systolic pressure and diastolic pressure.
  • Systolic pressure is the pressure in the arteries as the chambers of the heart contract.
  • Diastolic pressure is the pressure in the arteries when the heart chambers relax
  • Blood pressure is measured using a sphygmomanometer.
  • Normal blood pressure in man is about 120 / 80 mm Hg.

Question 10.
What is single circulation and what is double circulation?
Answer:
Single circulation:

  • The blood circulates once through heart and supplies blood to all the parts of the body. This is single circulation.
  • There is systemic and pulmonary circulation. (Eg.) The two-chambered heart of fishes.

Double circulation:
There are two types of circulation.
Systemic circulation:

  • The oxygenated blood entering the aorta from the left ventricle is carried by a network of arteries to the tissues.
  • The deoxygenated blood from the tissue is collected and emptied into the right atrium.

Pulmonary circulation:
The blood from right ventricle is taken to the lungs by the pulmonary artery and the oxygenated blood from the lungs is emptied into the left auricle by the pulmonary vein.

Question 11.
Why the pressure in the blood vessels nearer to the alveolei of lung is low and the pressure of arteries nearer to the heart is high?
Answer:

  • The alveoli are very thin. Hence exchange of gases are taking place easily.
  • If the pressure of the blood vessels of alveoli increases the blood vessels will damage and there is collection of tissue fluid

Question 12.
Explain heart failure or myocardial infarction?
Answer:

  • This heart failure is due to a decrease in cardiac muscle contractility.
  • When the blood supply to the heart muscle is remarkably reduced it leads to the death of the muscle fibres.
  • The blood clot or thrombosis blocks the blood supply to the heart and weakens the muscle fibres.
  • It is also called Ischemic heart disease due to a lack of oxygen supply to the heart muscles.
  • If this persists it leads to chest pain or angina.
  • Prolonged angina leads to death of the heart muscle resulting in heart failure.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 13.
What is cardiopulmonary resuscitation?
Answer:
1. Cardiopulmonary resuscitation is a life-saving procedure that is done at the time of emergency conditions such as when a person’s breath on heartbeat has stopped abruptly in case of drowning electric shock or heart attack.

2. CPR includes rescue of breath which is achieved by mouth to mouth breathing to deliver oxygen to the victim’s lungs by external chest compression which helps to circulate blood to the vital organs.

3. CPR must be performed within 4 to 6 minutes. Brief electric shock is given to the heart to recover the function of the heart (defibrillation).

Question 14.
What is meant by varicose veins?
Answer:

  • The veins are so dilated that the valves prevent the backflow of blood.
  • The veins lose their elasticity and become congested.
  • Common sites are legs rectal anal regions, and spermatic cords.

Question 15.
What is embolism?
Answer:

  • It is the obstruction of the blood vessel.
  • It is due to the abnormal mass of materials such as fragments of the blood clot.
  • If embolus occurs in the lungs coronary artery or liver that leads to death.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 16.
Write notes on Rheumatoid heart disease?
Answer:

  • Rheumatic fever is an auto immense disease.
  • It is due to the streptococcal infection in the throat.
  • The fever occurs 2-4 weeks after the infection.
  • The antibodies developed to combat the infection cause damage to the heart.
  • The symptoms include fibrous nodules on the mitral valve.
  • Fibrosis of the connective tissue and accumulation of fluid in the pericardial cavity.

Question 17.
Write notes on stroke and Angina pectoris?
Answer:
Stroke:
Stroke is a condition when the blood vessels in the brain bursts or when there is a block in the artery that supplies the brain. The part of the brain tissue that is supplied by this damaged artery dies due to lack of oxygen, (cerebral infarction)

Angina Pectoris:
If Atheroma may partially block the coronary artery and reduce the blood supply to the heart. As a result, there is tightness or choking with difficulty in breathing.
This leads to angina or chest pain. It lasts for a short duration of time.
reduces the blood flow.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 19.
Give short notes on heart transplantation?
Answer:

  • The first heart transplantation surgery was performed by South African professor – Christian Bernard in the year 1959.
  • He has done heart transplantation operation on December 3rd, 1967 Inkrute shour hospital at Capetown.
  • In India, in 1994 at AIMS hospital on August 3rd Dr. Anangipalli Venu Gopal has performed heart transplantation surgery.

Question 20.
What is an aneurysm?
Answer:

  • The weekened regions of the wall of the artery orvein bulge to forma baloon like sac. This is called aneurysm.
  • Unruptured aneurysm may exert pressure on the adjacent tissues or may burst causing massive hemorrhage.

( 5 marks)

V. Essay Questions

Question 1.
Describe white blood cells.
Answer:

  • White blood cells are colourless amoeboid nucleated cells devoid of hemoglobin and hence colourless.
  • 6000 – 8000 per cubic mm of WBC s are seen in the blood. WBCs are synthesized in the bone marrow these are of two types.
  • Granulocytes and agranolocytes.

I Granulocytes:
a) Neutrophils:

  • They are also called heterophils. Hence the nucleus has 3-4 lobes they are called polymorphonuclear.
  • This constitutes about 60 – 65 % of the total WBC.

b) Eosinophils:

  • They have a bilobed nucleus.
    Eosinophils Basophils Neutrophils
  • It constitutes about 2 – 3 % of total WBCs.
  • Their number increases during allergic reactions.

c) Basophils:

  • They are less numerous than any other type of WBCs constituting 0.5 % -1 % of total WBCs.
  • The nucleus is large and has granules in the cytoplasm.
  • They secrete heparin serotonin and histamines.

II A granulocytes:
a) Lymphocytes:

  • They are secreted in the lymph gland and spleen.
  • Lymphocytes constitute 28% of WBCs. They have large nucleus and small amount of cytoplasm.
  • The two types of lymphocytes are B and T cells.
  • B cells produce antibodies to neutralize the harmful effects of foreign substances. T cells are involved in cell-mediated immunity.

b) Monocytes:

  • They are phagocytic cells. They have kidney-shaped nucleus. They constitute 1 – 3 % of the total WBCs.
  • The macrophages of the central nervous system are the microglia and in the liver they are called “Kupffer cells” and in the pulmonary region they are the alveolar macrophages.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 2.
Explain cardiac output in a man?
Answer:
The amount of blood pumped out by each ventricle per minute is called cardiac output (CO). It is a product of heart rate (HR) and stroke volume (SV). Heart rate or pulse is the number of beats per minute. Pulse pressure = systolic pressure – diastolic pressure. Stroke volume (SV) is the volume of blood pumped out by one ventricle with each beat. SV depends on ventricular contraction.

CO = HR x SV. SV represents the difference between EDV (amount of blood that collects in a ventricle during diastole) and ESV (volume of blood remaining in the ventricle after contraction). SV = EDV – ESV. According to Frank-Starling law of the heart, the critical factor controlling SV is the degree to which the cardiac muscle cells are stretched just before they contract.

The most important factor stretching cardiac muscle is the amount of blood returning to the heart and distending its ventricles, venous return.
During vigorous exercise, SV may double as a result of venous return.

Heart’s pumping action normally maintains a balance between cardiac output and venous return. Because the heart is a double pump, each side can fail independently of the other. If the left side of the heart fails, it results in pulmonary congestion and if the right side fails, it results in peripheral congestion. Frank – Starling effect protects the heart from abnormal increase in blood volume.

Question 3.
What is the coagulation of blood?
Answer:
The mechanism by which excessive blood loss is prevented by the formation of clot is called blood coagulation.

  • The clotting process begins when the endothelium of the blood vessel is damaged and the connective tissue in its wall is exposed to the blood.
  • Platelets adhere to collagen fibres in the connective tissue and release blood clotting factors.
  • The blood clotting factors with platelets form the platelet plug which provides emergency protection against blood loss.
  • Clotting factors released from the clumbed platelet mix with clotting factors in the plasma.
  • The inactivated prothrombin is converted into active thrombin in the presence of calcium and vitamin K.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 6

Thrombin converts soluble fibrinogen into insoluble fibrin in plasma Fibrinogen Thrombin Fibrin
Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 7

The threads of fibers become interlinked into a patch that traps blood cell and seals the injured vessel and prevents blood loss.

Question 4.
Give an account of the composition of lymph and explains its significances?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 8

  • About 90% of fluid that leaks from capillaries seeps back into the capillaries and the remaining 10% is collected and returned to the blood system by means of lymph vessel.
  • The fluid inside the lymphatics is called lymph.

Lymphatic system:

  • The lymphatic system consists of a complex network of thin-walled ducts having group of immune response cells.
  • The lymphatic vessels have smooth walls that run parallel to the blood vessels in the skin along the respiratory and digestive tracts.
  • These vessels serve as return ducts for the fluids that are continually diffusing out of the blood capillaries into the body tissues.
  • The lymphatic nodes arc concentrated in the neck ingunial axillaries respiratory and digestive tracts.
  • The lymph fluid flowing out of the lymph nodes flows into large collecting ducts which finally drains into larger veins that run beneath the collar bone the subclavian vein and is emptied into the bloodstream.
  • The lymph nodes contain macrophage cells and they prevent the invading microorganisms from reaching the bloodstream. Cells found in the lymphatics are lymphocytes.
  • Fats are absorbed through lymph in the lacteals present in the villi of the intestinal wall

Question 5.
Describe the structure of the heart with a diagram?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 9

  • The structure of the human heart was described by Raymond devises in 1706.
  • It is situated in the thoracic cavity and its apex portion is slightly tilted towards left. It weighs about 300 g in a adult. The size of our heart is roughly equal to a closed fist.
  • Heart is divided into four chambers upper two auricles and lower two ventricles.
  • The walls of the ventricles are thicker than the auricle.
  • The heart is covered with pericardium. The pericardial space is filled with pericardial fluid.
  • The heart wall is made up of three layers. The outer epicardium middle myocardium the inner endocardium.
  • The two auricles are separated by inter auricular septum and the two ventricles are separated by interventricular septum.
  • Tricuspid valve is present in between the opening of right auricle and right ventricle and the bicuspid valve is present in between the opening of left auricle and left ventricle.
  • From the right ventricle arises pulmonary artery and from the left ventricle arises the dorsal aorta.
  • The semilunar valve is present at the beginning of these arteries.
  • The deoxygenated blood from all the parts of body reaches left auricle through superior and inferior venacava.
  • The oxygenated blood reaches the left auricle through four pulmonary vein.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation

Question 6.
Describe the functioning of heart with a diagram or Describe about the cardiac cycle?
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 7 Body Fluids and Circulation 10
The events that occur at the beginning of heart beat and lasts until the beginning of next beat is called cardiac cycle. It lasts for 0.8 seconds.

Phase I:
Ventricular diastole – During this phase the blood pressure increases AV valves are opened and the semilunar valves are closed. Blood flows from the auricles into the ventricles passively.

Phase II:
During atrial systole the ventricle is in relaxed position. The contraction of the auricles pushes maximum volume of blood to the ventricles.
The end-diastolic volume is related to the length of the cardiac muscle fibre. More the muscle is stretched greater the EDV and the stroke volume.

Phase III:
Ventricular systole – During this phase the ventricular pressure increases and the AV value closes.
The blood is pumped from the ventricles into the aorta.

Phase IV:
Ventricular systole – During this phase the ventricular pressure increases that forces semi lunar valve to open.
Blood is ejected out of the ventricles without back flow of blood.
This point is the end of systolic volume.

Phase V:
Ventricular diastole – The ventricles begins to relax pressure in the arteries exceeds ventricular pressure resulting in the closure of semilunar valves.
The heart returns to phase I of the cardiac cycle.

Question 7.
Explain Cardio Pulmonary Resuscitation (CPR)?
Answer:
In 1956, James Elam and Peter Safar were the first to use mouth-to-mouth resuscitation. CPR is a life-saving procedure that is done at the time of emergency conditions such as when a person’s breath or heart beat has stopped abruptly in case of drowning, electric shock or heart attack.

CPR includes rescue of breath, which is achieved by mouth to mouth breathing, to deliver oxygen to the victim’s lungs by external chest compressions which helps to circulate blood to the vital organiser.

CPR must be performed within 4 to 6 minutes after cessation of breath to prevent brain damage or death. Along with CPR, defibrillation is also done. Defibrillation me a brief electric shock is given to the heart to recover the function of the heart.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Zoology Guide Pdf Chapter 9 Locomotion and Movement Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

11th Bio Zoology Guide Locomotion and Movement Text Book Back Questions and Answers

Part I

Question 1.
Muscles are derived from
a) ectoderm
b) mesoderm
c) endoderm
d) neuroectoderm
Answer:
b) mesoderm

Question 2.
Muscles are formed by
a) myocytes
b) leucocytes
c) osteocytes
d) lymphocytes
Answer:
a) myocytes

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 3.
The muscles attached to the bones are called
a) skeletal muscle
b) cardiac muscle
c) involuntary muscle
d) smooth muscles
Answer:
a) skeletal muscle

Question 4.
Skeletal muscles are attached to the bones by
a) tendon
b) ligament
c) pectin
d) fibrin
Answer:
a) tendon

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 5.
The bundle of muscle fibers is called
a) Myofibrils
b) fascicle
c) sarcomere
d) sarcoplasm
Answer:
b) fascicle

Question 6.
The pigment present in the muscle fiber to store oxygen is
a) myoglobin
b) troponin
c) myosin
d) actin
Answer:
a) myoglobin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 7.
The functional unit of a muscle fibre is
a) sarcomere
b) sarcoplasm
c) myosin
d) actin
Answer:
a) sarcomere

Question 8.
The protein present in the thick filament is
a) myosin
b) actin
c) pectin
d) leucine
Answer:
a) myosin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 9.
The protein present in the thin filament is
a) myosin
b) actin
c) pectin
d) leucine
Answer:
b) actin

Question 10.
The region between two successive Z-discs is called a
a) sarcomere
b) microtubule
c) myoglobin
d) actin
Answer:
a) sarcomere

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 11.
Each skeletal muscle is covered by
a) epimysium
b) perimysium
c) endomysium
d) hyponychium
Answer:
a) epimysium

Question 12.
The knee joint is an example of
a) saddle joint
b) hinge joint
c) pivot joint
d) gliding joint
Answer:
b) hinge joint

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 13.
The name of the joint present between the atlas and axis is
a) synovial joint
b) pivot joint
c) saddle joint
d) hinge joint
Answer:
b) pivot joint

Question 14.
ATPase enzyme needed for muscle contraction is located in
a) actinin
b) troponin
c) myosin
d) actin
Answer:
c) myosin

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 15.
Synovial fluid is found in
a) Ventricles of the brain
b) Spinal cord
c) immovable joint
d) freely movable joints
Answer:
d) freely movable joints

Question 16.
Inflammation of joints due to accumulation of uric acid crystals is called as
a) Gout
b) myasthenia gravis
c) osteoporosis
d) osteomalacia
Answer:
a) Gout

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 17.
The acetabulum is located in
a) collar bone
b) hip bone
c) shoulder bone
d) thigh bone
Answer:
b) hip bone

Question 18.
Appendicular skeleton is
a) girdles and their limbs
b) vertebrae
c) skull and vertebral column
d) ribs and sternum
Answer:
a) girdles and their limbs

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 19.
The type of movement exhibits by the macrophages are
a) flagellar
b) ciliary
c) muscular
d) amoeboid
Answer:
d) amoeboid

Question 20.
The pointed portion of the elbow is
a) acromion process
b) glenoid cavity
c) olecranon process
d) symphysis
Answer:
c) olecranon process

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 21.
Name the different types of movement
Answer:

  • Amoeboid movement
  • Ciliary movement
  • Flagellar movement
  • Muscular movement

Question 22.
Name the filaments present in the Sarcomere.
Answer:
Thick and thin filaments are the two types of filaments present inside the sarcomere.

Question 23.
Name the contractile proteins present in the skeletal muscle.
Answer:

  • Myosin – thick filament
  • Actin – Thin filament

Question 24.
When describing a skeletal muscle. What does “Striated mean?
Answer:
Each skeletal muscle fibre has a repeated series of dark and light bands. The dark A-bands and light I-bands give a striated appearance to the muscle.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 25.
How does an isotonic contraction take place?
Answer:
In isotonic contraction, the length of the muscle changes but the tension remains constant.
(eg) lifting dumbbells and weight lifting.

Question 26.
How does an isometric contraction take place?
Answer:
In isometric contraction, the length of the muscle does not change but the tension of the muscle changes. The force produced is changed, e.g., pushing against a wall, holding a heavy bag.

Question 27.
Name the bones of the skull.
Answer:
The cranial bones are 8 in number.

  1. Paired parietal
  2. Paired temporal
  3. frontal
  4. Sphenoid
  5. Occipital
  6. Ethmoid

Question 28.
Which is the only jointless bone in the human body?
Answer:
The jointless bone is the hyoid bone in our throat. The hyoid bone (lingual bone) is a horseshoe.

Question 29.
List the three main parts of the axial skeleton.
Answer:

  1. Cranium
  2. Hyoid (Lingual)
  3. Vertebral column
  4. Thoracic cavity.

Question 30.
How is tetany caused?
Answer:
Tetany is caused when rapid muscle spasms occur in the muscles due to deficiency of parathyroid hormone resulting in reduced calcium levels in the body.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 31.
How does rigor mortis happen?
Answer:
After the death of an individual, the membrane of muscle cells becomes more permeable to calcium ions. This happens due to the partial contraction of skeletal muscles. The contracted muscles are unable to relax. This condition is known as rigor mortis.

Question 32.
What are the different types of rib bones that form the rib cage?
Answer:
Thoracic vertebrae ribs and sternum together constitute the ribcage.

Question 33.
What are the bones that make the pelvic girdle?
Answer:

  • Ilium
  • Ischium
  • Pubis

Question 34.
List the disorders of the muscular system.
Answer:

  1. Myasthenia gravis
  2. Tetany
  3. Muscle fatigue
  4. Atrophy
  5. Muscle pull
  6. Muscular dystrophy

Question 35.
Explain the sliding-filament theory of muscle contraction.
Answer:
Sliding filament theory is an active process. It is proposed by Andrw F. Huxley in 1954 and Rolf Niedergerke.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 1

  • Muscle contraction is initiated by a nerve impulse sents by the central nervous system through a motor neuron
  • When the nerve impulse reaches neuromuscular junction acetylcholine is released and created action potential.
  • This action potential triggers the release of calcium from the sarcoplasmic reticulum
  • The released calcium ions bind to troponin on thin filaments.
  • The active sites are exposed to the heads of myosin to form a cross bridge. Hence actin and myosin form a protein complex called actomyosin.
  • Utilizing the energy released from the hydrolysis of ATP the myosin head rotates until it forms a 90° angle with a long axis of the filament.
  • The power stroke begins after the myosin head and hinges region tilt from a 90° angle to a 45° angle.
  • The cross-bridge transforms into a strong high force bond which allows the myosin head to swells it.
  • When the myosin head swells it pulls the attached actin filament towards the centre of the A – band.
  • The myosin returns back to its relaxed state and releases ADP and phosphate ions. A Newer ATP molecule binds to the head of the myosin and the cross-bridge is broken.
  • At the end of each power stroke each myosin head detaches from actin then swivels back and binds to a new actin molecule to start another contraction cycle.
  • The power stroke repeats many times and the thin filaments move toward the centre of the sarcomere.
  • In this process, there is no change in the lengths of thick or thin filaments.
  • The Z – discs attached to the actin filaments are also pulled inwards from both sides causing the shortening of the sarcomere. This process continues.
  • When motor impulse stops the calcium ions are purnbed back into the sarcoplasmic reticulum results in the masking of the active sites of the actin filament and the myosin head fails to bind with the actin and causes Z – discs back to their original relaxed position.

Question 36.
What are the Benefits of regular exercise?
Answer:

  1. The benefits of regular exercise are:
  2. The muscles used in exercise grow larger and stronger.
  3. The resting heart rate goes down.
  4. More enzymes are synthesized in the muscle fiber.
  5. Ligaments and tendons become stronger.
  6. Joints become more flexible.
  7. Protection from a heart attack.
  8. Influences hormonal activity.
  9. Improves cognitive functions.
  10. Prevents obesity.
  11. Promotes confidence, esteem.
  12. Aesthetically better with a good physique.
  13. Overall well-being with good quality of life.
  14. Prevents depression, stress, and anxiety.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Part II

11th Bio Zoology Guide Locomotion and Movement Additional Important Questions and Answers

I. Choose The Best Options

Question 1.
Which of the following is not related to skeletal muscle?
(a) It is attached to the bone
(b) It is striated
(c) It is an involuntary muscle
(d) It brings about movement of the organ
Answer:
(c) It is an involuntary muscle

Question 2.
Where is a ciliary movement taking place?
Answer:
a) Respiratory tract
b) Stomach
c) oesophagus
d) Reproductive tract
Answer:
d) Reproductive tract

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 3.
The cytoplasm of the muscle fibre is ……………….
(a) Sarcolemma
(b) Sarcoplasm
(c) Ectoplasm
(d) Endoplasm
Answer:
(b) Sarcoplasm

Question 4.
Muscles are made up of ………………….
a) Myocytes
b) Lymphocytes
c) adenocytes
d) leucocytes
Answer:
a) Myocytes

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 5.
The cranial bones are ………………..
(a) 22
(b) 14
(c) 8
(d) 3
Answer:
(c) 8

Question 6.
Name the connective tissue which covers the entire muscle?
a) Epimycium
b) Perimycium
c) Endomyciurn
d) Mesornyscium
Answer:
a) Epimycium

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 7.
The number of vertebrates is ………………..
(a) 8
(b) 12
(c) 5
(d) 33
Answer:
(d) 33

Question 8.
Name the membrane which covers each muscle fiber?
a) Epimycium
b) Perimycium
c) Endomycium
d) Mesomysciurn
Answer:
c) Endomycium

Question 9.
Match and Find the Correct Pair
1. Sarcoplasm – a) Respiratory pigment
2. Myoglobin – b) Glucose giver
3. Glycosome – c) Unit of skeletal muscle
4. Sarcomere – d) cytoplasm
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 2
Answer:
a) I- d, II – a, III – b,IV- c

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 10.
Find the correct and wrong statement and arrange the following statement.
1. The contraction of muscle fibres depends on the actin and myosin protein.
2. The thick muscle fibres depend on Myosin.
3. Each meromyosin molecule will have a globular head with a long arm.
4. The head of the meromyosin bears an actin-binding site and an ATP binding site.
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 3
Answer:
a) True True False True

Question 11.
The joint between carpal and metacarpals is ………………..
(a) Pivot joint
(b) Ball and socket joint
(c) Saddle joint
(d) Hinge joint
Answer:
(c) Saddle joint

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 12.
Who has proposed sliding – filaments hypothesis?
a) Andrew F. Huxley and Rolf Nieder gerke
b) Andrew F. Huxley and Nelson
c) Andrew F. Pluxley and Darwin
d) Andrew F. Huxley and Mendal
Answer:
a) Andrew F. Huxley and Rolf Nieder gerke

Question 13.
Which of the following disorders is related to the endocrine gland?
(a) Myasthenia gravis
(b) Tetany
(c) Atrophy
(d) Muscular dystrophy
Answer:
(b) Tetany

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 14.
Find out the wrong pair
a) Fast – oxidative fibres – have high ATP ase activity
b) Slow – oxidative fibres – low rates of ATP ase activity
c) Oxidative fibres – less number of mitochondria
d) Red muscle fibres – oxidative fibres
Answer:
c) Oxidative fibres – less number of mitochondria

Question 15.
Which of the following is deficiency disorder?
(a) Osteoarthritis
(b) Rheumatoid arthritis
(c) Gouty arthritis
(d) Osteoporosis
Answer:
(d) Osteoporosis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 16.
………………………… a number of bones from the endoskeleton of man?
a) 210
b) 220
c) 206
d) 209
Answer:
c) 206

Question 17.
Hove many bones are there in the axial and appendicular skeleton?
a) 80 and 126
b) 126 and 80
c) 80 and 120
d) 80 and 118
Answer:
a) 80 and 126

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 18.
How many bones are there in the facial and cranial bones?
a) 14 and 9
b) 14 and 8
c) 14 and 10
d) 14 and 12
Answer:
b) 14 and 8

Question 19.
Name the opening of the temporal bone
a) External auditory meatus
b) Nasal opening
c) Optic opening
d) Mouth
Answer:
a) External auditory meatus

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 20.
Name the U-shaped single bone present at the base of the buccal cavity.
a) Palantine bone
b) Hyoid bone
c) Ethmoid bone
d) Sphenoid bone
Answer:
b) Hyoid bone

Question 21.
How many bones from the vertebral column?
a) 33
b) 32
c) 30
d) 36
Answer:
a) 33

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 22.
Match and find the correct pair
1) Cervical vertebrae – a) – 5
2) Thoracic – b)-l
3) Pelvic bone – 0-7
4) Coccyx bone – d) -12
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 4
Answer:
a) I -c, II – d, III – a,IV – b

Question 23.
In which bone ………….. situated?
a) Cervical vertebra
b) thoracic vertebra
c) Pelvic vertebra
d) all the above
Answer:
d) all the above

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 24.
Name the first vertebra?
a) Atlas
b) Maleus
c) Incus
d) Stapes
Answer:
a) Atlas

Question 25.
Name the second vertebra?
a) Atlas
b) Axis
c) Maleus
d) Stapes
Answer:
b) Axis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 26.
How many are the number of ribs?
a) 14 pair
b) 13 pair
c) 12 pair
d) 15 pair
Answer:
c) 12 pair

Question 27.
Match and find the correct answer
1) ribs – a) – 1, 7
2) True ribs – b) -11, 12
3) False ribs – c) – Bicephalic
4) Floating ribs – d) – 8, 10
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 5
Answer:
b) I – c, II – a, III – d,IV -b

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 28.
How many bones are there appendicular skeleton?
a) 130
b) 140
c) 126
d) 122
Answer:
c) 126

Question 29.
Which forms the appendicular skeleton?
a) Upper and hindlimbs
b) Upper limbs and thoracic bones
c) Hind and vertebral column
d) Hind and cranial bones
Answer:
a) Upper and hindlimbs

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 30.
Name the bones which join the axial and appendicular skeleton.
a) Clavicle bone
b) Scapula
c) Acromian process
d) None of the above
Answer:
a) Clavicle bone

Question 31.
Flow many bones are there in the upper arm.
a) 30
b) 32
c) 34
d) 36
Answer:
a) 30

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 32.
Find the wrong pair.
a) Wrist bone – 8
b) Fore arm bones – 30
c) Facial bone – 16
d) Cranial bones – 8
Answer:
c) Facial bone – 16

Question 33.
Find the wrong pair
a) Palm bones – 5
b) Phalanges – 14
c) Thoracic bone – 12
d) Vertebral column – 33
Answer:
d) Vertebral column – 33

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 34.
Match and find the correct pair.
1) Cervical vertebra – a) 12
2) Thoracic vertebra – b) 1
3) Pelvic vertebra – c) 7
4) Coccyx – d) 5
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 6
Answer:
a) I- c, II – a, III – d, IV – b

Question 35.
The oxidative skeletal muscle fibres are termed as
a) Fatty muscle fibres
b) White muscle fibres
c) red muscle fibres
d) yellow muscle
Answer:
c) red muscle fibres

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 36.
Whether the following statement is correct or wrong find out the correct sequence.
1) Pelvic bone Ilium, Ischium, and pubis
2) Acetabulum cavity is present in the sacrum
3) The head of the thigh bone fits in the acetabulum cavity
4) The pubic bones articulate anteriorly at the pubic symphysis
Sequence:
a) 1 – True, 2 – False, 3 – True, 4 – True
b) 1 – False, 2 – False, 3 – True, 4 – True
c) 1 – False, 2 – True, 3 – True, 4 – True
d) 1 – True, 2 – True, 3 – False, 4 – True
Answer:
a) 1 – True, 2 – False, 3 – True, 4 – True

Question 37.
Which is the prominent bone of the pelvic bone?
a) Ilium
b) Ischium
c) Pubis
d) all the above
Answer:
a) Ilium

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 38.
What is the number of hind limbs?
a) 32
b) 37
c) 30
d) 34
Answer:
c) 30

Question 39.
Name the longest bone
a) Femur
b) Humerus bone
c) Tibia
d) all the above
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 7
Answer:
a) I – a, II – d, III – c, IV – b

Question 40.
Match and find the correct pair.
1)Patella – a) 14
2) Tarsus – b) 5
3) Metatarsus – c) 7 bones
4) Phalanges – d) kneecap
Answer:
3) Metatarsus – c) 7 bones

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 41.
Name the membrane which covers the femur bone.
a) Periosteum
b) Endosteum
c) Osteoclast
d) Osteoblast
Answer:
a) Periosteum

Question 42.
Name the joints seen in the cranial region
a) Fibrous joints
b) Cartilagenous joints
c) Diarthroses joints
d) Synovial joints
Answer:
a) Fibrous joints

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 43.
The Synovial fluid is seen in joints.
a) Cartilagenous joints
b) Diarthroses joints
c) Fibrous joints
d) all the above
Answer:
b) Diarthroses joints

Question 44.
Name the diseases due to the deficiency of acetylcholine.
a) Myasthenia gravis
b) Tetany
c) Duchene muscular dystrophy
d) Muscle pull
Answer:
a) Myasthenia gravis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 45.
Name the disease due to the deficiency of ATP
a) Muscle fatigue
b) Muscle pull
c) Muscular dystrophy
d) Tetany
Answer:
a) Muscle fatigue

Question 46.
A tear in the muscle
a) Muscle fatigue
b) Muscle pull
c) Atrophy
d) Muscular dystrophy
Answer:
b) Muscle pull

Question 47.
Name arthritis due to aging.
a) Osteoarthritis
b) Rheumatoid arthritis
c) Gout
d) Osteoporosis
Answer:
a) Osteoarthritis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 48.
Osteoporosis is due to
a) Calcium
b) Sodium
c) Magnesium
d) Potassium
Answer:
a) Calcium

Question 49.
The deposition of urate crystals on the joints is called as
a) Gout
b) Rheumatoid
c) arthritis
d) Osteoporosis
Answer:
a) Gout

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 50.
Assertion A: Acetylcholine is secreted in the Neuromuscular junction.
Reason B: If acetylcholine is not secreted there won’t be any multiple gated channels in the sarcolemma.
a) A true B wrong
b) A true B this explains the action of A
c) A wrong B true
d) A wrong B wrong
Answer:
b) A true B this explains the action of A

Question 51.
Assertion A: The upper limbs are attached to the pectoral girdles.
Reason B: The pectoral girdles are very light and allow the mobility of the hand
a) A wrong B wrong
b) A True B does not explain the A
c) A True B explains the functions of A
d) A True B wrong
Answer:
c) A True B explains the functions of A

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 52.
Assertion A: In the pelvis bone a deep socket is present called acetabulum.
Reason B: The head of the femur bone fits in the acetabulum.
a) A True B explains the functions of A
b) A wrong B explains the functions of A
c) A True B wrong
d) B does not explain the structure of B
Answer:
a) A True B explains the functions of A

Question 53.
Assertion A: The pelvis of the female is shallow wide and flexible in nature
Cause B: This helps during pregnancy.
a) A True B explains the functions of A
b) A wrong B explains the functions of A
c) A True B wrong
d) B does not explain the structure of A
Answer:
a) A True B explains the functions of A

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 54.
Assertion A: The lower arm carries the entire weight of the body and is subjected to exceptional forces when we jump or run.
Reason B: To bear the weight of the body it has 46 bones
a) A True B True
b) A wrong B wrong
c) A True B does not explain the functions of A
d) A True B Wrong
Answer:
c) A True B does not explain the functions of A

Question 55.
Assertion A: The cranium belongs to immovably fixed joints.
Reason B: Structures of the flat skull bones are fibrous joints.
a) A True B Wrong
b) A True B explains the structure of A
c) A True B True
d) B does not explain the structure of A
Answer:
b) A True B explains the structure of A

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 56.
Assertion A: The decreased synthesis of acetylcholine in the neuromuscular junction, causes myasthenia gravis.
Reason B: This leads to muscle fatigue weakness, paralysis.
a) A and B are True
b) A and B are wrong
c) A True B wrong
d) A wrong B True
Answer:
a) A and B are True

Question 57.
Match and arrange the sequence:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 8
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 9
Answer:
a) I – D, II – C, III – A, IV – B

Question 58.
Match and arrange the sequence:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 10
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 11
Answer:
b) I – C, II – D, III – B, IV -A

2 marks

II. Very Short answers

Question 1.
What is amoeboid movement?
Answer:
The movement of cells by streaming movements of the cytoplasm forming pseudo-podia is known as amoeboid movement, e.g., macrophages.

Question 2.
What are the types of muscles?
Answer:

  • Skeletal muscle
  • Visceral muscles
  • Cardiac muscles

Question 3.
What is flagellar movement?
Answer:
The movement due to the lashing of flagella is known as flagellar movement, e.g., sperm cells.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 4.
Name the regulatory proteins in the thin filaments.
Answer:

  • Tropomyosin
  • Troponin

Question 5.
What is a fascicle?
Answer:
Each muscle is made up of bundles of muscle fibres called fascicles.

Question 6.
On the basis of ATP formation, how are muscles classified?
Answer:

  • Oxidative fibres
  • Glycolytic fibres

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 7.
What is epimysium?
Answer:
The connective tissue covering the whole muscle is the epimysium.

Question 8.
What is perimysium?
Answer:
The connective tissue covering around each fascicle is the perimysium.

Question 9.
Differentiate the oxidative fibre from the glycolytic fibre.
Answer:

Oxidative fibre Glycolytic fibre
1. Numerous mitochondria There are few mitochondria
2. Depends on blood flow Not depend on blood flow
3. Myoglobin is present No myoglobin
4. These are known as red muscle fibres These are called muscle fibres as white muscle fibres

Question 10.
What is endomysium?
Answer:
The connective tissue surrounding the muscle fibre is called the endomysium.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 11.
What is a tendon?
Answer:
Skeletal muscle is attached to the bone by a bundle of collagen fibres known as tendons.

Question 12.
What is an endoskeleton?
Answer:
It is found inside the body of vertebrates. It is composed of bones and cartilages, (eg) Man.

Question 13.
What is sarcoplasm?
Answer:
The cytoplasm of the muscle fibre is called the sarcoplasm.

Question 14.
What are Glycosomes?
Answer:
Glycosomes are the granules of stored glycogen that provide glucose during the period of muscle fiber activity.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 15.
What is sarcomere?
Answer:
The functional unit of the skeletal muscle is known as the sarcomere.

Question 16.
Name the ear ossicles?
Answer:

  • Malleus
  • incus
  • stapes

Question 17.
What is meromyosin?
Answer:
The monomer of the myosin molecule is called meromyosin.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 18.
Name the openings of the skull?
Answer:

  • The orbits
  • Nasal cavity
  • Foramen magnum

Question 19.
What is meant by foramen magnum?
Answer:

  • It is a large opening found at the posterior base of the skull.
  • Through this opening, the medulla oblongata of the brain descends down as the spinal cord.

Question 20.
What are oxidative fibres?
Answer:
The muscle fibres that contain numerous mitochondria and have a high capacity for oxidative phosphorylation are classified as oxidative fibres. They are also called red muscle fibres.

Question 21.
Name the first two bones of the vertebral column.
Answer:

  1. Atlas
  2. Axis

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 22.
What are the functions of the vertebral column?
Answer:

  • It protects the spinal cord.
  • Supports the head
  • Serves as the point of attachment for the ribs and musculature of the back.

Question 23.
Give short notes on sternum?
Answer:

  • The sternum is a flat bone on the midventral line of j the thorax.
  • It provides space for the attachment of the thoracic ribs and abdominal muscles.

Question 24.
What is a metaphysis?
Answer:
The bone region where the diaphysis and epiphyses meet is called the metaphysis.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 25.
What are true ribs?
Answer:
The first seven pairs of ribs are called true ribs. Daily they are attached to the thoracic vertebrae and ventrally connected to the sternum.

Question 26.
What is endosteum?
Answer:
Internal bone surfaces are covered with a delicate connective tissue membrane called the endosteum.

Question 27.
What is the sternum?
Answer:
The sternum is a flat bone on the midventral line of the thorax. It provides space for the attachment of the thoracic ribs and abdominal muscles.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 28.
What is meant by an appendicular skeleton?
Answer:
The bones of the upper and lower limbs along with their girdles constitute the appendicular skeleton. It is composed by 126 bones.

Question 29.
White the 3 segments of the lower limb.
Answer:

  1. The thigh
  2. the leg or the shank and
  3. the foot.

Question 30.
What is meant by the acromion process?
Answer:
The scapula has a slightly elevated ridge called the spine which projects as a flat expanded process called the acromion.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 31.
What is meant by the glenoid cavity?
Answer:
Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.

Question 32.
What is meant by the olecranon process?
Answer:
The radius and ultra bones present in the forearm that form the pointed portion of the elbow called the olecranon process.

Question 33.
What is meant by carpal tunnel?
Answer:

  • There are 8 bones in the wrist arranged in two rows of four each.
  • The anterior surface of the wrist has a tunnel-like appearance. This tunnel is termed a carpal tunnel.

Question 34.
Name the bones which form the coxal bones.
Answer:

  • Ilium
  • Ischium
  • Pubis

Question 35.
What is meant by pubic symphysis?
Answer:
Ventrally the two halves of the pelvic girdle meet and form the pubic symphysis containing fibrous cartilage.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 36.
Where calcium ion binds with the muscle fibre? Name the molecules which bind with calcium?
Answer:

  • The calcium released from the sarcoplasm binds with the thin fibre of the muscle.
  • The released calcium binds to troponin thin filaments.

III. Fill Up The Blanks With Suitable Options

1. Scapula – Acromian process
……………. – Bones of the upper arm
2. First 7 pair of rib bones – True ribs
11 and 12th pair of ribs – …………….
3. Cervical vertebrae – 7
……………. – Lumbar bones
4. Skull bones – 22
……………. – Skull bones
5. Thick fibres – Myosin
……………. – Thin fibres
6. Amoeboid movement – Macrophage cells
……………. – Sperm cells
Answer:

  1. Olecranon process
  2. Floating ribs
  3. 5
  4. 8
  5. Actin
  6. Flagellated movement

3 marks

IV. Short answers

Question 1.
Give short notes on skeletal muscle and their covering membranes.
Answer:

  • Each muscle is made up of bundles of muscle fibres called fascicles. Each muscle fibre contains rod-like structures called myofibrils.
  • The connective tissue covering the muscle is the epimysium.
  • The covering around each fascicle is the perimysium.
  • The muscle fibre is surrounded by endomysium.

Question 2.
Give the structure of a skeletel muscle fibre.
Answer:

  • Each muscle fibre is thin and elongated.
  • Most of the taper at one or both ends.
  • Muscle fibres are surrounded by sarcolemma the cytoplasm of the muscle fibre is called the sarcoplasm.
  • It contains glycosomes myoglobin and sarcoplasmic reticulum.
  • Myoglobin is a red-coloured respiratory pigment and glycosomes are reserved glycogen.
  • Muscle fibres contain muscle protein actin and myosin.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 3.
Give notes on slow oxidative fibres.
Answer:

  • These fibers have low rates of myosin ATP hydrolysis but have the ability to make large amounts of ATP.
  • This type of fiber seen in long-distance swimmers and long-distance runners.

Question 4.
Give notes on fast – oxidative fibres.
Answer:

  • These fibres have high myosin ATP as activity and can make large amounts of ATP.
  • They are suited for rapid action.

Question 5.
Give notes on fast glycolytic fibres.
Answer:

  • These fibres have myosin ATP ase activity but cannot make as much ATP as oxidative fibres because their source of ATP in glycolysis.
  • These fibres are best suited for rapid intense actions such as short sprints at maximum speed.

Question 6.
Name the facial bones.
Answer:
There are 14 facial bones.

  1. Pair of maxilla
  2. Pair of Zygomatic
  3. Pair of Palantine
  4. Pair of lacrimal
  5.  Pair of Nasal
  6. Mandible or lower jaw
  7. Vomer

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 7.
Give notes on fibrous joints.
Answer:

  • They are immovably fixed joints in which no movement between the bones is possible.
  • Sutures of the flat skull bones are fibrous joints.

Question 8.
Give notes on cartilaginous joints.
Answer:
They are slightly movable joints in which the joint surface is separated by cartilage and slight movement is only possible.

Question 9.
Give notes on synovial joints.
Answer:
They are freely movable joints the articulating bones are separated by a cavity which is filled with synovial fluid.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 10.
Give notes on myasthenia gravis.
Answer:

  • It is an autoimmune disorder affecting the action of acetylcholine at the neuromuscular junction leading to fatigue.
  • Weakening and paralysis of skeletal muscles.
  • Acetylcholine receptors on the sarcolemma are blocked by antibodies leading to weakness of muscles.
  • When the disease progresses it can make chewing swallowing talking and even breathing difficult.

Question 11.
Give notes on muscle fatigue.
Answer:

  • It is the inability of a muscle to contract after repeated muscle contraction.
  • This is due to lack of ATP and accumulation of lactic acid by anaerobic break down of glucose.

Question 12.
Give notes on the Atrophy of muscles.
Answer:

  • A decrease in the activity of muscles results in the atrophy of muscles.
  • There is a reduction in the size of the muscle and makes the muscle become weak which occurs with lack of usage as in chronic bedridden patients.

Question 13.
What is meant by muscle pull?
Answer:

  • Muscle pull is actually a muscle tear.
    Atraumatic pulling of the fibres produces a tear known as a sprain.
  • This can occur due to the sudden stretching of muscle beyond the point of elasticity.
  • Back pain is a common problem caused by muscle pull due to improper posture with static sitting for long hours.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 14.
What is meant by muscular dystrophy?
Answer:

  • The group of diseases collectively called muscular dystrophy is associated with the progressive degeneration of skeletel muscle fibers weakening the muscles and leading to death from lung or heart failure.
  • (eg) Duchene muscular dystrophy.

Question 15.
What is meant by skeletel muscle glycogen analysis?
Answer:

  • This is used to measure an athlete’s muscle glycogen.
  • Muscle glycogen provides the main source of energy during anaerobic exercise.
  • A single glycogen molecule may contain 5000 glucose molecules.

Question 16.
Give notes on osteoporosis?
Answer:

  • It occurs due to deficiency of vitamin D and hormonal imbalance.
  • It causes rickets in children and osteomalacia in adult females.
  • The bones become soft and fragile.
  • It can be minimized with adequate calcium intake vitamin D intake and regular physical activities.

Question 17.
What is carpal tunnel syndrome?
Answer:

  • The narrow passage bounded by bones and ligaments in the wrist gets narrowed and pinches the median nerve.
  • This syndrome is mostly seen among clerks, software professionals, and people who constantly play or text on mobile phones.

5 Marks

V. Give Detailed Answers 

Question 1.
Explain the structure of skeletal muscle fibre?
Answer:

  1. Each muscle fibre is thin and elongated.
  2. It has multiple oval nuclei beneath sarcolemma.
  3. The cytoplasm of the muscle fibre is called sarcoplasm.
  4. It contains glycosomes, the stored glycogen granules, myoglobin, respiratory pigment, and sarcoplasmic reticulum.
  5. Actin and myosin are muscle proteins present in the muscle fibre.
  6. Each myofibril has a repeated series of dark and light bands called A-bands and I-bands.
  7. Each dark band has a lighter region in its middle called the H-zone.
  8. Each H-zone is bisected vertically by a dark line called the M-line.
  9. Each light I-band has a darker mid-line area called the Z-disc.
  10. The sarcomere is the functional unit of the skeletal muscle. It is a region of a myofibril between two successive Z-discs.
  11. Sarcomere has thick and thin filaments. The thick filaments extend the entire length of the A-band, the thin filaments extend across the I-band and partly into the A-band.
  12. The invagination of the sarcolemma forms transverse tubules (T-tubules) and they penetrate into the junction between the A and I-bands.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 2.
Describe the structure of the sarcomere.
Answer:

  • The unit of the skeletal muscle is the sarcomere A sarcomere is the region of a myofibril between two successive z – discs.
  • It contains an ‘A’ band with a half I band which are perfectly aligned with one another.
  • This type of arrangement gives the cell a striated appearance.
  • Each dark band has a lighter region in its middle called the M – zone.
  • Each H – zone is bisected vertically by a dark line called the M – line.
  • The I bands have a darker mid-line area called the z – disc.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 12

  • Inside the sarcomere, two types of filaments are present namely the thick filaments and thin filaments.
  • The thick filaments extend the entire length of the A band, the thin filaments extend across the I band and partly into the A – band.
  • The invagination of the sarcolemma forms transverse (T- tubules) tubules and they penetrate into the junction between the A and I bands.

Question 3.
Describe the structure of muscle protein.
Answer:
Contraction of the muscle depends on the presence of contractile proteins such as actin and myosin.
Myosin fibre:

  • The thick filaments are composed of the protein myosin.
  • Each myosin molecule is made up of a monomer called meromyosin.
  • The meromyosin have a globular head with a j short arm and a tail.
  • The short arm have heavy meromyosin and the tail portion have light meromyosin.
  • The head bears actin-binding site and an ATP binding site
  • It also contains ATP ase enzyme that split ATP to generate energy for the contraction of muscle.

Actin filament:

  • Actin has polypeptide subunits called globular actin or G – actin and filamentous form F – actin.
  • Each thin filament is made of two F – actins helically wound to each other.
  • Each F – actin is a polymer of monomeric G – actins, It also contains a binding site for myosin.
  • The thin filament contain several regulatory protein like tropomyosin, troponin, which help in regulating the contraction of muscles along with actin and myosin.

Thick filament:
Each thick filament consists of many myosin molecules whose heads produce at opposite ends of the filament Portion of a thick filament

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 13

Thin filament:
A thin filament consists of two strands of actin subunits twisted into a helix plus two types of regulatory proteins (troponin and tropomyosin) Portion of a thin filament.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 14

Question 4.
Give the schematic representation of muscle contraction.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 15

Question 5.
Write the functions of skeletal system?
Answer:
Functions of skeletal system

  1. Support -It forms a rigid framework and supports the weight of the body against gravity.
  2. Shape – It provides and maintains the shape of the body.
  3. Protection – It protects the delicate internal organs of the body.
  4. Acts as reservoir – It stores minerals such as calcium and phosphate. Fat (triglyceride) is stored in yellow bone marrow and represents a source of stored energy for the body.
  5. Locomotion – It acts as lever along with the muscles attached to it.
  6. Strength – It can withstand heavyweight and absorbs mechanical shock.
  7. Asa hemopoietic tissue – Red and white blood cells are produced in the bone marrow of the ribs, spongy bones of vertebrae and extremities of long bones.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 6.
Explain the bones that form the skull?
Answer:
The skull is composed of two sets of bones – cranial and facial bones. It consists of 22 bones of which 8 are cranial bones and 14 are facial, bones. The cranial bones form the hard protective outer covering of the brain and called the brain box. The capacity of the cranium is 1500 cm3.

These bones are joined by sutures which are immovable. They are paired parietal, paired temporal and individual bones such as the frontal, sphenoid, occipital and ethmoid. The large hole in the temporal bone is the external auditory meatus. In the facial bones maxilla, zygomatic, palatine, lacrimal, nasal are paired bones whereas mandible or lower jaw and vomer are unpaired bones. They form the front part of the skull.

A single U-shaped hyoid bone is present at the base of the buccal cavity. It is the only bone without any joint. Each middle ear contains three tiny bones- malleus, incus, and stapes collectively are called ear ossicles. The upper jaw is formed of the maxilla and the lower jaw is formed of the mandible.

The upper jaw is fused with the cranium and is immovable. The lower jaw is connected to the cranium by muscles and is movable. The most prominent openings in the skull are the orbits and the nasal cavity. The foramen magnum is a large opening found at the posterior base of the skull. Through this opening, the medulla oblongata of the brain descends down as the spinal cord.

Question 7.
Write a short note on the vertebral column?
Answer:
The vertebral column is also called the backbone. It consists of 33 serially arranged vertebrae which are interconnected by cartilage known as an intervertebral disc. The vertebral column extends from the base of the skull to the pelvis and forms the main framework of the trunk. The vertebral column has five major regions.

They are the cervical, thoracic, lumbar, sacrum (5 sacral vertebrae found in the infant which are fused to form one bone in the adult), and coccyx (4 coccygeal vertebrae found in the infant which are fused to form one bone in the adult).

Each vertebra has a central hollow portion, the neural canal, through which the spinal cord passes. The first vertebra is called the atlas and the second vertebra is called the axis. Atlas is articulated with the occipital condyles. The vertebral column protects the spinal cord, supports the head, and serves as the point of attachment for the ribs and musculature of the back.

Question 8.
Give an account of the ribcage.
Answer:

  • There are 12 pairs of ribs.
  • Each rib bone is connected dorsally to the vertebral column and ventrally to the sternum.
  • It has two articulation surfaces on its dorsal end called bicephalic.
  • The first 7 pairs of ribs are called true ribsorvertebro – sternal ribs.
  • Dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages.
  • The 8th, 9th, and 10th pairs of ribs do not articulate directly with the sternum but joined with the cartilaginous part of the seventh rib.
  • These are called false ribs or vertebro – chondral ribs.
  • The last 11th and 12th pairs of ribs are not connected ventrally.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 16

  • They are called floating ribs or vertebral ribs.
  • Thoracic vertebrae ribs and sternum from the rib cage.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 9.
Give an account of the pectoral girdle?
Answer:

  • The upper limbs are attached to the pectoral girdles.
  • These are very light and allow the upper limbs a degree of mobility not seen anywhere else in the body.
  • The girdle is formed of two halves.
  • Each pectoral girdle consists of a clavicle or collar bone and a scapula.
  • The scapula is a large triangular bone situated in the dorsal surface of the ribcage between the second and seventh ribs.
  • It has an elevated expanded process called the acromion.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 17

  • The clavicle articulates this process.
  • Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint.
  • Each clavicle is a long slender bone with two curvatures which lie horizontally and connect the axial skeleton with the appendicular skeleton.

Question 10.
Describe the structure of the upper limb.
Answer:

  • The upper limb consists of 30 separate bones and is specialized for mobility.
  • The region between the shoulder and elbow is the humerus.
  • The head of humerus articulates with the glenoid cavity of the scapula and forms the shoulder joint.
  • The distel end of humerus articulates with the two forearm bones the radius and ulna
  • Olecranon process is situated at the upper end of the ulna which forms the pointed portion of the elbow.
  • The hand consists of carpals metacarpals and phalanges.
  • Carpals the wrist bones 8 in number are arranged in two rows of four each and form a tunnel termed as carpal tunnel.
  • Meta carpals the palm bones are 5 in number and phalanges the digit bones are 14 in number.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 18

Question 11.
Give an account of pelvic girdle.
Answer:

  • The pelvic girdle is a heavy structure specialised for weight-bearing.
  • It is composed of two hib bones called coxal bones that secure the lower limbs to the axial skeleton.
  • Together with the sacrum and coccyx the hib bones form the basin-like bony pelvis.
  • Each coxal bone consists of three fused bones ilium, ischium, and pubis.
  • At the point of fusion of these three bones forms a deep hemispherical socket called the acetabulum present on the lateral surface of the pelvis.
  • It receives the head of the femur at hip joint and helps in the articulation of the femur.
  • Ventrally the two halves of the pelvic girdle meet and form the pubic symphysis containing fibrous cartilage.
  • The ilium is the superior flaring portion of the hip bone. Each ilium forms a secure joint with the sacrum posteriorly.
  • The ischium is a curved bar of bone. The ‘V’ shaped pubic bones articulate anteriorly at the pubic symphysis.
  • The pelvis of male is deep and narrow with larger heavier bones and the female is shallow wide and flexible in nature and this helps during pregnancy which is influenced by female hormones.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 19

Question 12.
Give an account of the lower limb.
Answer:

  • The lower limb consists of 30 bones which carries the entire weight of the erect body and is subjected to exceptional forces when we jump or run.
  • The bones of the lower limbs are thicker and stronger than the upper limbs.
  • Each lower limb consists of the thigh, the leg or the shank and the foot.
  • The femur is the strongest and longest bone of the body.
  • The head of femur articulates with the acetabulum of the pelvis to form the hip joint.
  • The tibia and fibula form the skeleton of the shank.
  • A thick triangular patella forms the knee cap which protects the knee joint arteriorly and improves the leverages of thigh muscles acting across the knee.
  • The foot includes the bones of ankle the tarsus (7) the metatarsus (5) and the phalanges or toe ebones. (14)
  • The foot supports our body weight and acts as a lever to propel the body forward while walking and running.
  • The phalanges of the foot are smaller than those of the fingers.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 20

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 13.
Give an account of a structure of a typical long bone.
Answer:

  • The typical long bone has a diaphysis, epiphysis, and membranes.
  • A tubular diaphysis or shaft forms the long axis of the bone and has a central medullary cavity.
  • The epiphyses are the bone ends.
  • Compact bone forms the exterior of epiphyses and their interior contains spongy bone with red marrow.
  • The region where the diaphysis and epiphysis meet is called metaphysics.
  • The external surface of the entire bone except the joint surface is covered by a double-layered membrane called the periosteum.
  • The outer fibrous layer is dense irregular connective tissue.
  • The inner osteogenic layer consists of osteoblasts cell. ( bone-forming cells) and osteoclasts cells (E bone – destroying cells)
  • There are primitive stem cells osteogenic cells that give rise to the osteoblasts.
  • The periosteum is richly supplied with nerve fibres lymphatic vessels and blood vessels.
  • Internal bone surfaces are covered with a delicate connective tissue membrane called the endosteum It also contains osteoblasts and osteoclasts cells.
  • Between the epiphysis and diaphysis growth plate or epiphyseal plate is present.

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 21

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 14.
Explain the disorders of the skeletal system?
Answer:
Arthritis and osteoporosis are the major disorders of the skeletal system.
1. Arthritis: Arthritis is an inflammatory or degenerative disease that damages the joints. There are several types of arthritis.

(I) Osteoarthritis: The bone ends of the knees and other freely movable joints wear away as a person ages. The joints of the knees, hip, fingers, and vertebral column are affected.

(II) Rheumatoid arthritis: The synovial membranes become inflamed and there is an accumulation of fluid in the joints. The joints swell and become extremely painful. It can begin at any age but symptoms usually emerge before the age of fifty.

(III) Gouty arthritis or gout: Inflammation of joints due to accumulation of uric acid crystals or inability to excrete it. It gets deposited in synovial joints.

2. Osteoporosis: It occurs due to deficiency of vitamin D and hormonal imbalance. The bone becomes soft and fragile. It causes rickets in children and osteomalacia in adult females. It can be minimized with adequate calcium intake, vitamin D intake, and regular physical. activities.

Question 15.
Tabulate the differentiate between joints man
Answer:

Pivot joint between atlas and axis
Gliding joint between the carpals
Saddle joint between the carpal and metacarpal
Ball and socket joint between humerus and pectoral gridle
Hinge joint Knee joint
Condyloid or Angular or Ellipsoid Joint between radius joint and carpal

Question 16.
Draw the diagram of different types of fracture and arrange them.
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 22

Question 17.
Bones of the skeletal system. Table: 1 Bones of skeletal system
Answer:
Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement 23
Notes:

  1. The strongest muscle in the human: Massetter in cheeks
  2. The smallest muscle in the human: Middle ear in stapedius
  3. Well moving muscle: Tongue
  4. The largest muscle in the human: Buttock in Glutens Maximus
  5. The longest muscle in the human : Hip to knee (sartorius)
  6. Total number of bones is adults = 206

Samacheer Kalvi 11th Bio Zoology Guide Chapter 9 Locomotion and Movement

Question 18.
Explain the basic categories of exercise and physical activity?
Answer:
Exercise and physical activity fall into four basic categories. Endurance, strength, balance, and flexibility. Endurance or aerobic activities increase the breathing and heart rate. They keep the circulatory system healthy and improve overall fitness.

Strength exercises make the muscles stronger. They help to stay independent and cany out everyday activities such as climbing stairs and carrying bags.

Balance exercises help to prevent falls which is a common problem in older adults. Many strengthening exercises also improve balance.

Flexibility exercises help to stretch body muscles for more freedom of joint movements.

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Miscellaneous Problems

Question 1.
Using three yearly moving averages, Determine the trend values from the following data.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 1
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 2

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 2.
From the following data, calculate the trend values using fourly moving averages.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 3
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 4

Question 3.
Fit a straight line trend by the method of least squares to the following data.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 5
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 6
Therefore, the required equation of the straight line trend is given by
y = a + bx
y = 55.9875 + 0.830 x
⇒ y = 55.9875 + 0.83 (\(\frac { x-1983.5 }{0.5}\))
The trend values can be obtained by
When x = 1980
y = 55.9875 + 0.83 (\(\frac { 1980-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-7)
= 55.9875 – 5.81
= 50.1775
When x = 1981
y = 55.9875 + 0.83 (\(\frac { 1981-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-5)
= 55.9875 – 4.15
= 51.8375
When x = 1982
y = 55.9875 + 0.83 (\(\frac { 1981-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-3)
= 55.9875 – 2.49
= 53.4975
When x = 1983
y = 55.9875 + 0.83 (\(\frac { 1983-1983.5 }{0.5}\))
= 55.9875 + 0.83 (-1)
= 55.9875 – 0.83
= 55.1575
When x = 1984
y = 55.9875 + 0.83 (\(\frac { 1984-1983.5 }{0.5}\))
55.9875 + 0.83 (1)
= 56.8175
when x = 1985
y = 55.9875 + 0.83 (\(\frac { 1985-1983.5 }{0.5}\))
= 55.9875 + 0.83 (3)
= 55.9875 + 2.49
= 58.4775
when x = 1986
y = 55.9875 + 0.83 (\(\frac { 1986-1983.5 }{0.5}\))
= 55.9875 + 0.83 (5)
= 55.9875 + 4.15
= 60.1375
when x = 1987
y = 55.9875 + 0.83 (\(\frac { 1987-1983.5 }{0.5}\))
= 55.9875 + 0.83 (7)
= 55.9875 + 5.81
= 61.7975

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 4.
Fit a straight line trend by the method of least squares to the following data.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 7
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 8
Lasperyre’s price Index number
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 9
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 10
Hence Fisher’s Ideal Index satisfies Time reversal test

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 5.
Using the following data, construct Fisher’s Ideal Index Number and Show that it satisfies Factor Reversal Test and Time Reversal Test?
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 11
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 12
Factor reversal test
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 13
Hence Fisher’s Ideal Index satisfies Factor reversal test.

Question 6.
Compute the consumer price index for 2015 on the basis of 2014 from the following data.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 14
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 15

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 7.
An Enquiry was made into the budgets of the middle class families in a city gave the following information.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 16
What changes in the cost of living have taken place in the middle class families of a city?
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 17
Conclusion:
The cost of living has increased up to 26.10% in 2011 as compared to 2010.

Question 8.
From the following data, calculate the control limits for the mean and range chart.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 18
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 19
UCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 51 + 0.577(6.5)
= 51 + 3.7505
= 54.7505
= 54.75
CL = \(\bar { \bar x}\) = 51
UCL = \(\bar { \bar x}\) – A2\(\bar { R}\)
= 51 – 0.577(6.5)
= 51 – 3.7505
= 47.2495
= 47.25
The control limits for Range chart is
UCL = D4\(\bar { R}\)
= 2.114(6.5)
= 13.741
CL = \(\bar { R}\) = 6.5
LCL = D3\(\bar { R}\) = 0(6.5) = 0

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems

Question 9.
The following data gives the average life(in hours) and range of 12 samples of 5 lamps each. The data are
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 20
Construct control charts for mean and range Comment on the control limits.
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 21
UCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 1367.5 + 0.577(427.5)
= 1367.5 + 246.6675
= 1614.1675
= 1614.17
CL = \(\bar { \bar x}\) = 1367.5
LCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 1367.5 – 0.577(427.5)
= 1367.5 – 246.6675
= 1120.8325
= 1120.83
The control limits for Range chart is
UCL = D4\(\bar { R}\)
= 2.115(427.5)
= 904.1625
= 904.16
CL = \(\bar { R}\) = 427.5
LCL = D3\(\bar { R}\)
= 0(427.5)
= 0

Question 10.
The following are the sample means and I ranges for 10 samples, each of size 5. Calculate ; the control limits for the mean chart and range chart and state whether the process is in control or not.
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 22
Solution:
Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems 23
UCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 4.982 + 0.577(0.36)
= 4.982 + 0.20772
= 5.18972
= 5.19
CL = \(\bar { \bar x}\) = 4.982
LCL = \(\bar { \bar x}\) + A2\(\bar { R}\)
= 4.982 – 0.577(0.36)
= 4.982 – 0.20772
= 4.77428
= 4.774
The control limits for range chart is
UCL = D2\(\bar { R}\) = 2.115(3.6)
= 7.614
CL = \(\bar { R}\) = 3.6
LCL = D3\(\bar { R}\)
= 0(0.36) = 0

Samacheer Kalvi 12th Business Maths Guide Chapter 9 Applied Statistics Miscellaneous Problems