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 O_A = C = O_B, the formal charge on O_A, C and O_B 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

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

Question 11.
Which of these represents the correct order of their increasing bond order.
a) C₂⁺ < C₂^2- < O₂^2- < O₂
b) C₂^2- < C₂⁺ < O₂ < O₂^2-
c) O₂^2- < O₂ < C₂^2- < C₂⁺
d) O₂^2- < C₂⁺ < O₂ < C₂^2-
Answer:
d) O₂^2- < C₂⁺ < O₂ < C₂^2-

Question 12.
Hybridisation of central atom in PCl₅ involves the mixing of orbitals.
a) s, P_x, P_y, d_x², d_{x² – y²}
b) s, p_x, p_y, p_xy, d_{x² – y²}
c) s, p_x, p_y, p_z, d_{x² – y²}
d) s, p_x, P_y, d_xy, d_{x² – y²}
Answer:
c) s, p_x, p_y, p_z, d_{x² – 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) O₂^2-
c) O₂⁺
d) None of these
Answer:
b) O₂^2-

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

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

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

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

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 p_y – p_y and p_z – p_z overlaps will result in the formation of a π – bond.

Question 33.
In CH₄, NH_3, 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.

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 2p_y orbital in the excited state. In boron, the s orbital and two p orbitals (p_x and p_y) 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 2p_z orbitals of fluorine:
The three sp² hybridised orbitals of boron now overlap with the 2p_z orbitals of fluorine (3 atoms). This overlap takes place along the axis as shown below.

Question 35.
Draw the M.O diagram for oxygen molecule calculate its bond order and show that O₂ is paramagnetic.
Answer:

Electronic configuration of O atom:
1s² 2s² 2p⁴

Electronic configuration of O₂ molecule:
σ_1s², σ_1s^*2, σ_2s², σ_2s^*2, σ_2px², π_2py² π_2pz² π_2py^*1 π_2pz^*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:

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 :
σ_1s², σ_1s^*2, σ_2s², σ_2s^*2, π_2py², π_2pz² σ_2px²

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 ψ_A and ψ_B with comparable energy, combines to form two molecular orbitals. One is bonding molecular orbital(ψ_bonding) and the other is antibonding molecular orbital (ψ_antibonding) The wave functions for these two molecular orbitals can be obtained by the linear combination of the atomic orbitals ψ_A and ψ_B as below,

ψ_bonding = ψ_A + ψ_B;
ψ_antibonding = ψ_A – ψ_B

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,

Destructive interaction:
The two orbitals are out phase

Question 38.
Discuss the formation of N₂ molecule using MO theory.
Answer:

Molecular orbital diagram of nitrogen molecule (N₂)
Electronic configuration of N atom 1s² 2s² 2p³
Electronic configuration of N₂ molecule:
σ_1s², σ_1s^*2, σ_2s², σ_2s^*2, π_2py², π_2pz² σ_2px²
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.

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

Question 41.
Draw the Lewis structures for the following species.
(i) NO₃^–
(ii) SO₄^2-
(iii) HNO₃
(iv) O₃
Answer:
(i) NO₃^–

(ii) SO₄^2-

(iii) HNO₃

(iv) O₃

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

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

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 (σ_1s²). The molecular orbital energy level diagram of the H₂ molecule is given in

Here, N₂ = 2, N_a = 0
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{2-0}{2}\) = 1

He₂: σ_1s² σ_1s^*2
The molecular orbital energy level diagram of He₂ (hypothetical) is given in

Here, N_b = 2 and N_a = 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.

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.

Question 47.
considering x-axis as the molecular axis which out of the following will form a sigma bond.
i) 1s and 2p_y
ii) 2p_x and 2p_y
iii) 2p_x and 2p_z
iv) 1s and 2p_z
Answer:
i) 1s and 2p_y: No sigma bond
ii) 2p_x and 2p_y: sigma bond
iii) 2p_x and 2p_z: No sigma bond
iv) 1s and 2p_z: 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.

(b) Resonance structure of CO₃^2-:

Resonance Hybrid structure of CO₃^2-:
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 2p_x¹ 2p_y¹ 2p_z⁰. To satisfy the valency of carbon promote an electron from 2s orbital to 2p_z orbital in the excited state.

In ethylene both the carbon atoms undergo sp² hybridization involving 2s, 2p_x, and 2p_y, orbitals, resulting in three equivalent sp² hybridised orbitals lying in the XY plane at an angle of 120° to each other. The unhybridized 2p_z 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 2p_z 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.

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 2p_x¹ 2p_y¹ 2p_z⁰. 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 2p_x orbitals, resulting in two equivalent sp hybridized orbitals lying in a straight line along the molecular axis (x-axis). The unhybridized 2p_y and 2p_z orbitals lie perpendicular to the molecular axis.

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 2p_y and 2p_z orbitals of each carbon overlap sideways. This lateral overlap results in the formation of two pi bonds (p_y – p_y and p_z – p_z) between the two carbon atoms as shown in the figure.

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

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

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

Question 52.
CO₂ and H₂O both are triatomic molecule but their dipole moment values are different. Why?
Answer:

Sum of the dipole moment are cancelled.

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
σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py² π_2pz² π_2px²
Bond order = \(\frac{6}{2}\) = 3

(ii) N₂⁺
σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py² π_2pz² π_2px¹
Bond order = \(\frac{5}{2}\) = 2.5

(iii) N₂^–
σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py² π_2pz² π_2px^*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.

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

Question 6.
The formal charge of the O – atoms in the ion 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

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?

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 %

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

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^–

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

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) P_z and P_z
b) s and p_z
c) s and P_x
d) d_{x² – y²} and d_{x² – y²}
Answer:
c) s and P_x

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

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) CO₃^2-
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

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

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 O₂^2- is in order
a) O₂^– < O₂^2-  < O₂ < O₂⁺
b) O₂² < O₂^– < O_2, < O₂⁺
c) O₂⁺ < O₂ < O₂^– < O₂^2-
d) O_2, < O₂⁺ < O₂^– < O₂^2-
Answer:
b) O₂² < O₂^– < O_2, < 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⁺

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) O₂^2-
d) O₃
Answer:
b) NO

Question 53.
The number of antibonding electron pairs O₂^2- 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)
b)
c)
d) All
Answer:
a)

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) O₂^2-
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⁺ < O₂^2-
b) O₂^2- < O₂^– < O₂ < O₂⁺
c) O₂^– < O₂^2- < O₂ < O₂⁺
d) O₂⁺ < O₂ < O₂^– < O₂^2-
Answer:
b) O₂^2- < 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^–

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) O₂^2-
b) O₂⁺
c) O₂
d) NO
Answer:
a) O₂^2-

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

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⁺

Question 71.
The bond order is maximum in
a) O₂
b) O₂^–
c) O₂⁺
d) O₂^2-
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) O₂^2-
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

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

Question 81.
Amongst LiCl, RbCl, BeCl_3, 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

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

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) CO₃^2- and NO₃^–
b) PCl₄⁺; and SiCl₄
c) PF₅ and BrF₅
d) AlF₆^3- 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

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

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:

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′

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 CO₃^2- ion?
Answer:

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:

Molecular orbital diagram of hydrogen molecule (H₂)
Electronic configuration of H atom 1s¹
Electronic configuration of H₂ molecule: σ_1s²
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:

Molecular orbital diagram of Lithium molecule (Li₂)
Electronic configuration of Li atom 1s² 2s¹
Electronic configuration of Li₂ molecule: σ_1s² σ_1s^*2 σ_2s²
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:

Molecular orbital diagram of Boron molecule (B₂)
Electronic configuration of B atom 1s² 2s² 2P¹
Electronic configuration of B₂ molecule: σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py¹ π_2pz¹
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{6-4}{2}\) = 1
Molecule has two unpaired electrons hence it is paramagnetic.

Question 6.
Explain the formation of C₂ molecule by MOT.
Answer:

Molecular orbital diagram of Carbon molecule (B₂)
Electronic configuration of C atom 1s² 2s² 2p¹
Electronic configuration of C₂ molecule: σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py¹ π_2pz¹
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 p_x – p_x 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, p_y – p_y and p_z – p_z 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, 2p_x², 2p_y², 2p_z¹
3. When half filled 1s orbital of hydrogen linearly overlaps with a half filled 2p_z orbital of fluorine, a σ – covalent bond is formed between hydrogen and fluorine.

Formation of fluorine molecule (F₂):
1. Valence shell electronic configuration of fluorine atom: 2s², 2p_x², 2P_y² 2p_z¹
2. When the half filled p_z orbitals of two fluorine overlaps along the z-axis, a σ – covalent bond is formed between them.

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 (H_a and H_b) 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.

VB theory for the formation of hydrogen molecule

The new attractive forces (green arrows) arise between
(i) nucleus of H_a and valence electron of H_b
(ii) nucleus of H_b and the valence electron of H_a.

The new repulsive forces (red arrows) arise between
(i) nucleus of H_a and H_b
(ii) valence electrons of H_a and H_b

The attractive forces tend to bring H_a and H_b 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.

At this stage, there is a maximum overlap between the atomic orbitals of H_a and H_b, and the atoms H_a and H_b 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.

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.

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 p_z orbital of the chlorine to form a covalent bond between Be and Cl as shown in the Figure

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.

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 2p_x¹ 2p_y¹ 2P_z⁰.

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 2p_z 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′.

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.

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 (d_z²). The ground state electronic configuration of phosphorous is [Ne] 3s² 3p_x² 3p_y¹ 3p_z¹ as shown below.

One of the paired electrons in the 3s orbital of phosphorous is promoted to one of its vacant 3d orbital (d_z²) in the excite state. One 3s orbital, three 3p orbitals, and one 3d_z². 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 3p_z orbitals of chlorine :
The 3p_z 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.

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 3p_x², 3p_y², 3p_z¹

One electron each from 3s orbital and 3p orbital of sulphur is promoted to its two vacant 3d orbitals (d_z² and d_{x² – 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 2p_z orbitals of fluorine:
The six sp³d² hybridized orbitals of sulphur overlaps linearly with 2p_z orbitals of six fluorine atoms to form the six S – F bonds in the sulphur hexafluoride molecule.

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.

Question 10.
Explain the formation of NO by MOT?
Answer:

Molecular orbital diagram of nitrogen molecule (NO)

Electronic configuration of N atom 1s² 2s² 2p³
Electronic configuration of NO molecule
σ_1s² σ_1s^*2 σ_2s² σ_2s^*2 π_2py² π_2pz² σ_2px² π_2py^*1
Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{10-5}{2}\) = 2.5
Molecule has one unpaired electrons hence it is paramagnetic.