Class 11 | Chapter 4 (Chemical Bonding and Molecular Structure) NCERT Chemistry Solutions

Question 1
Explain the formation of a chemical bond.

The attractive force which holds together the constituent particles (atoms, ions or molecules) in a chemical species is known as chemical bond.

Atoms either share or gain or lose electrons to attain stable electronic configuration.

Due to this, a state of minimum energy is obtained and chemical bond is formed. This results in maximum stability.

When two atoms share electrons, covalent bond is formed and when atoms lose or gain electrons, ionic bond is formed.

Question 2
Write Lewis dot symbols for atoms of the following elements:
Mg, Na, B, O, N, Br

Lewis Structures of Mg, Na, B, O, N, Br

Question 3
Write Lewis symbols for the following atoms and ions:
S and S2-
Al and Al3+
H and H

Lewis Dot Structures of S and S2-
Lewis Dot Structures of Al and Al3+
Lewis Dot Structures of H and H-

Question 4
Draw the Lewis structures for the following molecules and ions

Lewis Dot Structure of H2S
Lewis Dot Structure of SiCl4
Lewis Dot Structure of BeF2
Lewis Dot Structure of CO2 3-
Lewis Dot Structure of HCOOH

Question 5
Define octet rule.
Write its significance and limitations.

Octet Rule
The octet rule states that atoms combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons to have an octet i.e., eight electrons in their valence shells.

The octet rule helps to explain the chemical bonding in a chemical compound.

The octet rule illustrates the types of chemical bonding like covalent, ionic, and coordinate bonds.

Incomplete octet – In certain molecules such as BeH2, and BeCl2, BF3, the central atom has less than 8 electrons in its outermost shell, yet these molecules are stable.

Expanded octet – In certain molecules such as PF5, and SF6 the central atom has more than 8 electrons in its outermost shell, yet these molecules are stable.

For Hydrogen (H) to attain stability, it needs to share, gain or lose 1 electron. It does not need to complete an octet to attain stability.  Helium (He) has only 2 electrons and is stable.

Question 6
Write the favourable factors for the formation of ionic bond.

Low ionisation energy and high electron affinity.
When an atom have higher ionisation energy then it can easily loose the electron and as other atom have higher electron affinity therefore its quite easy for it to accept the electron thus forming an ionic bond between atoms.

Question 7
Discuss the shape of the following molecules using the VSEPR model

MoleculeNumber of electron pairs around central atomMolecular GeometryBond Angles
BCl33Trigonal Planar120°
AsF55Trigonal Bi-pyramidalthree 120°
two 90°
H2S6Non linear (bent)92°
PH35Trigonal Pyramidal93.5°

Question 8
Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia.

The central atom N in NH3 has one lone pair and there are three bond pairs. In H2O there are two lone pairs and two bond pairs. The two lone pairs present in the oxygen atom of H2O molecule repels the two bond pairs.

This repulsion is stronger than the repulsion between the lone pair and the three bond pairs on the nitrogen atom. Since the repulsions on the bond pairs in H2O molecule are greater than that in NH3 the bond angle in water is less than that of ammonia.

That’s why bond angle in NH3 is 107.5° and it is 104.5° in H2O

Question 9
How do you express the bond strength in terms of bond order?

The bond order shows the number of chemical bonds present between a pair of atoms. For example, the bond order of diatomic oxygen O = O is 2 and bond order between the carbon atoms in C ≡ C is three. The bond order describes the stability of the bond. Therefore more the bond order, higher is the strength of bond.

Higher the bond order, the stronger pull between atoms in the bond and the shorter the bond length.

Question 10
Define the bond length

Bond length is defined as the average distance between the centres of nuclei of the two bonded atoms in a molecule.

Bond Length explained using diagram

Below is a table showing some sample bond lengths

BondBond Length
C – C154 pm
C = C134 pm
C ≡ C120 pm
C – N147 pm
C = N127 pm
C ≡ N116 pm

Question 11
Explain the important aspects of resonance with reference to the 𝐶𝑂32- ion

The pi bond electrons are delocalised to transfer from one to another bond to make different structures. According to experimental findings, all carbon to oxygen bonds in 𝐶𝑂32- are equivalent.

Hence, it is inadequate to represent 𝐶𝑂32- ion by a single Lewis structure having two single bonds and one double bond.

Below is resonance diagram of 𝐶𝑂32- ion

Carbonate Anion Resonance Structures

Question 12
H3PO3 can be represented by structures (i) and (ii) shown below.

Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3?

If not, give reasons for the same.

These two structures of H3PO3 can be called canonical forms because positions of atoms in both of these are different, but in canonical forms position of atoms should be same and only electrons position should be different.

Question 13
Write the resonance structures for

Resonance Structures of SO3

Resonance Structures of SO3

Resonance Structures of NO2

Resonance Structure of NO2

Resonance Structures of NO3

Question 14
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :
(a) K and S
(b) Ca and O
(c) Al and N

(a) K and S

Lewis Dot Structure showing formation of K2S from K and S

(b) Ca and O

(c) Al and N

Reaction of Al and N to form AlN

Question 15
Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.

CO2​ molecule with zero dipole moment is linear as two C=O bond dipoles cancel each other. The net dipole moment of the water molecule is non-zero which indicates that the water molecule is non-linear. Two O−H bonds form angular shape.

Question 16
Write the significance/applications of dipole moment.

Dipole moment is useful in finding the polar nature of the chemical bond. As the magnitude of dipole moment increases, the polar nature of the bond increases.

Molecules with zero dipole moment are non-polar in nature. Molecules having dipole moments are said to be polar.

Dipole moment is useful in finding the structure/shape of the molecule.

Molecules having dipole moment are bent or angular in shape and have unsymmetrical structure. Molecules having zero dipole moment have symmetrical shape.

Dipole moment is useful in distinguishing between the cis- and trans- isomers.

Isomers having high dipole moment are trans-isomers. Isomers having low dipole moment are cis-isomers.

Dipole moment is useful in distinguishing between the ortho, meta and para- isomers.

The isomer having zero dipole moment is the para-isomer. The isomer having moderate dipole moment is the meta-isomer. The isomer having the highest dipole moment is the ortho-isomer.

Dipole moment is useful in finding the percent ionic character.

Question 17
Define electronegativity. How does it differ from electron gain enthalpy?

Electronegativity of an atom in a molecule is defined as it’s ability to attract shared electron pair towards itself. It is measured using Pauling Scale.

On the other hand, Electron Gain Enthalpy is defined as energy required or released when a gaseous atom accepts an electron. It is measured in terms of kJ/mol.

Below are some key differences between Electronegativity and Electron Gain Enthalpy

ElectronegativityElectron Gain Enthalpy
An atom in a molecule tends to attract the shared pair of electrons towards itself known as electronegativityElectron gain enthalpy is defined as the amount of energy released when an electron is added to an isolated gaseous atom
It is the property of a bonded atomIt is the property of an isolated atom
Its unitlessIts unit is kJ/mol
It cannot be determined experimentallyIt can be measured experimentally
It is always positiveIt can be negative or positive depending upon whether energy is being releases or required

If energy is release then it’s negative
If energy is required then it’s positive

Question 18
Explain with the help of suitable example polar covalent bond

Polar covalent bond is a type of chemical bonding in which an electron pair is shared unequally between the two atoms, which means electron pair is attracted more towards one atom as compared to another in the bond.

For example, in hydrogen fluoride (H – F) shared electron pair is attracted towards fluorine atom because it have higher electronegativity as compared to hydrogen atom.

This unequal attraction of electrons towards one atom as compared to another in a covalent bond lead to generation of slightly positive charge on one atom and slightly negative charge on another, thereby forming positive/negative poles in the molecule.

Question 19
Arrange the bonds in order of increasing ionic character in the molecules – LiF, K2O, N2, SO2 and ClF3

Ionic character of a molecule depends upon electronegativity difference of atoms in the molecule.
More the difference in electronegativity of atoms involved, higher is the ionic character of molecule.

Therefore order of ionic character is N2 < SO2 < ClF3 < K2O < LiF

Question 20
The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.

Incorrect structure of CH3COOH

In the above diagram, we can clearly see that one of hydrogen atom on left side is making double bond which is impossible as hydrogen just have valency of 1.

Correct structure is given in the below diagram.

Correct Structure of CH3COOH

Question 21
Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar?

The square planar geometry has a bond angle of 90°. This is lesser than the tetrahedral bond angle which is 109.5°
Due to this, in the square planar geometry, the repulsive forces are more and hence, square planar geometry is less stable.

Question 22
Explain why BeH2 molecule has a zero dipole moment although the Be — H bonds are polar?

BeH2 is linear, although Be – H individual bonds are polar and have dipole moment. Total dipole moment of molecule is zero because both dipoles are have opposite directions with respect to each other.

BeH2 Structure showing opposite dipole moments which total to zero

Question 23
Which out of NH3 and NF3 has higher dipole moment and why?

The dipole moment of ammonia NH3(1.47D) is higher than the dipole moment of NF3​ (0.24D). The molecular geometry is pyramidal for both the molecules. In each molecule, N atom has one lone pair.

F is more electronegative than H and N−F bond is more polar than N−H bond. Hence, NF3​ is expected to have much larger dipole moment than NH3​.

However reverse is true as in case of ammonia, the direction of the lone pair dipole moment and the bond pair dipole moment is same whereas in case of NF3​ it is opposite.

Thus, in ammonia molecule, individual dipole moment vectors add whereas in NF3​, they cancel each other.

Question 24
What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.

Hybridisation is defined as the concept of mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals.

Salient Features of Hybridisation
The number of hybrid orbitals is equal to the number of the atomic orbitals that undergo hybridisation.
The hybrid orbitals are always equivalent in energy and shape.
The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
These hybrid orbitals are directed in space in some preferred directions to have minimum repulsion between electron pairs and thus have a stable arrangement.
Therefore, the type of hybridisation indicates the geometry of the molecules.
Some Important Conditions for Hybridisation
Only those orbitals that are present in the valence shell of the atom undergo hybridisation.
The orbitals undergoing hybridisation should have almost equal energy.
Promotion of electron is not essential condition prior to hybridisation.
It is not necessary that only half filled orbitals participate in hybridisation.
In some cases, even filled orbitals of valence shell take part in hybridisation.

sp hybridisation 
When one s-and one p-orbital, intermix then it is called sp-hybridisation. For example, in BeF2, Be atom undergoes sp-hybridisation. It has linear shape. Bond angle is 180°.

One s-and two p-orbitals get hybridised to form three equivalent hybrid orbitals. The three hybrid orbitals directed towards three corners of an equilateral triangle. It is, therefore, known as trigonal hybridisation.

sp3 hybridisation 
One s-and three p-orbitals get hybridised to form four equivalent hybrid orbitals. These orbitals are directed towards the four corners of a regular tetrahedron.

Question 25
Describe the change in hybridisation (if any) of the Al atom in the following reaction.
AlCl3 + Cl → AlCl4

The hybridisation of Al atom changes from sp2 in AlCl3​ to sp3 in AlCl4​

In AlCl3​​, Al contains 3 bond pairs of electrons and zero lone pair of electrons. Hence, it is sp2 hybridised.

In AlCl4, Al atom contains 4 bond pairs of electrons and zero lone pairs of electrons. Hence, it is sp3 hybridised.

Structures of AlCl3 And AlCl4-

Question 26
Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF3 + NH3 → F3B.NH3

In BF3 hybridisation of B is sp2
In NH3 hybridisation of N is sp3

After the reaction, hybridisation of B changes to sp3 but that of N still remains sp3

Structure of BF3
Structure of NH3
Structure of BF3 NH3

Question 27
Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules

Question 28
What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2
(b) C2H4

(a) C2H2

Structure of C2H2

From the above diagram its clear that
Number of sigma bonds = 3
Number of pi bonds = 2

(b) C2H4

Structure of C2H4

From the above diagram its clear that
Number of sigma bonds = 5
Number of pi bonds = 1

Question 29
Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
(a) 1s and 1s
(b) 1s and 2px
(c) 2py and 2py
(d) 1s and 2s

(c) 2py and 2py

The sigma bonds are formed if the atomic orbitals overlap head to head along the same nuclear axis and pi bonds are formed if they overlap sideways or lateral in a direction perpendicular to the internuclear axis.

Question 30
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH3CH3
(b) CH3CH = CH2
(c) CH3CH2OH
(d) CH3CHO

(a) sp3, sp3

(b) sp3, sp2, sp2

(c) sp3, sp3

(d) sp3, sp2

(e) sp3, sp2

Question 31
What do you understand by bond pairs and lone pairs of electrons?

Illustrate by giving one example of each type.

In the formation of a covalent bond between two atoms, each atom will share one electron. The electrons present in the covalent bond are known as the bond pair of electrons.

The pair of electrons left in the outermost valence shell without forming covalent bonds are known as lone pairs of electrons.

Lone and Bond Pairs

Question 32
Distinguish between sigma and a pi bond

Difference Between Sigma and Pi bond
Sigma Bond
It is denoted by the symbol σ.
Pi Bond
It is usually denoted by the symbol π.
During sigma bond formation overlapping orbitals can either be one hybrid orbital and a single pure orbital, or two pure orbitals and two hybrid orbitals.During pi bond formation overlapping orbitals must be two unhybridized orbitals.
Sigma bonds are known to exist independently and allow free rotation.Pi-bond must always exist along with sigma bond and the rotation is restricted.
Sigma bonds are stronger bonds.Pi bonds are usually weaker compared to sigma bonds.
Sigma bonds are formed first when atoms interact.Pi bonds between two atoms are formed after sigma bonds are formed between them.
During the bonding between two given atoms,Only one sigma bond is formed.Here two pi bonds can exist between two atoms.
Sigma bonds are known to have cylindrical charge symmetry around the axis of the bond.No symmetry exists in pi bonds.
Atoms with sigma bonds are less reactive.Atoms with pi bonds are highly reactive when compared to those with only sigma bonds.
Sigma bond can be used to determine the shapes of molecules.Pi bond cannot be used for the same purpose.

Question 33
Explain the formation of H2 molecule on the basis of Valence Bond Theory

Let us consider the combination between atoms of hydrogen HA and HB and eA and eB be their respective electrons.

As they tend to come closer, two different forces operate between the nucleus and the electron of the other and vice versa. The nuclei of the atoms as well as their electrons repel each other.

Energy is needed to overcome the force of repulsion. Although the number of new attractive and repulsive forces is the same, but the magnitude of the attractive forces is more.

Thus, when two hydrogen atoms approach each other, the overall potential energy of the system decreases. Thus, a stable molecule of hydrogen is formed.

Question 34
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals

The atomic orbital with one electron (completely half filled) can take part in combination or overlapping.

During the time of overlap, the combining orbitals must have opposite spins of electrons.

The combining atomic orbitals must have the same or nearly the same energy. This means that in a homonuclear molecule, the 1s-atomic orbital of an atom can combine with the 1s-atomic orbital of another atom, and not with the 2s-orbital.

The combining atomic orbitals must have proper orientations to ensure that the overlap is maximum.

The extent of overlapping should be large.

Question 35
Use molecular orbital theory to explain why Be2 molecule does not exist

Beryllium has an electron configuration 1s2 2s2

It is clear from the electron configuration that beryllium does not have any atomic orbitals that are only partially filled with electrons.

Because the overlapping cannot take place in the absence of the half-filled orbital, Be2 molecule does not exist.

Question 36
Compare the relative stability of the following species and indicate their magnetic properties;
𝑂2 (superoxide)
𝑂22- (peroxide)

Therefore decreasing order of stability is 𝑂+2 > O2 > 𝑂2 > 𝑂22-

Question 37
Write the significance of a plus and a minus sign shown in representing the orbitals.

The sign of the electron wave is represented by plus and minus sign. Generally plus sign represents the crest of the electron wave and the minus sign represents the troughs of the electron wave.

Question 38
Describe the hybridisation in case of PCl5

Why are the axial bonds longer as compared to equatorial bonds?

PCl5 is sp3d hybridised molecule.

PCl5 has trigonal bi-pyramidal geometry.

In this case the axial bonds are slightly longer than the equatorial bonds.

This is because the axial bonds experience greater repulsion from other bonds than the equatorial bonds.

Question 39
Define hydrogen bond.

Is it weaker or stronger than the van der Waals forces?

Hydrogen bond is defined as interaction involving a hydrogen atom located between a pair of other atoms having high electro negativities.

While on the other hand, Van Der Waals Forces is defined as the relatively weak attractive forces that act on neutral atoms and molecules, it arises because of the electric polarisation induced in each of the particles by the presence of other particles.

Hydrogen bond is stronger than van der waals forces.

Question 40
What is meant by the term bond order?

Calculate the bond order of:

Bond order is defined as the half of the difference between the number of electrons present in bonding and anti-bonding molecular orbitals.

Bond Order = [Number of electrons in bonding molecular orbitals – Number of electrons in anti bonding molecular orbitals]/2

Bond Order of N2

Molecular Orbital Electron Configuration of N2
[σ (1s)]2 [σ* (1s)]2 [σ (2s)]2 [σ* (2s)]2 [𝝅 (2px)]2 [𝝅 (2py)]2 [σ (2pz)]2

Total number of bonding electrons = 10
Total number of anti-bonding electrons = 4
Bond Order = (10 – 4)/2 = 6/2 = 3

Therefore Bond Order of N2 is 3

Bond Order of O2

Molecular Orbital Electron Configuration of O2
[σ (1s)]2 [σ* (1s)]2 [σ (2s)]2 [σ* (2s)]2 [σ (2pz)]2 [𝝅 (2px)]2 [𝝅 (2py)]2 [𝝅* (2px)]1 [𝝅* (2py)]1

Total number of bonding electrons = 8
Total number of anti-bonding electrons = 4
Bond Order = (8 – 4)/2 = 4/2 = 2

Therefor Bond Order of O2 is 2

Bond Order of O2+

Molecular Orbital Electron Configuration of O2+
[σ (1s)]2 [σ* (1s)]2 [σ (2s)]2 [σ* (2s)]2 [σ (2pz)]2 [𝝅 (2px)]2 [𝝅 (2py)]2 [𝝅* (2px)]1

Total number of bonding electrons = 8
Total number of anti-bonding electrons = 3
Bond Order = (8 – 3)/2 = 5/2 = 2.5

Therefor Bond Order of O2+ is 2.5

Bond Order of O2

Molecular Orbital Electron Configuration of O2
[σ (1s)]2 [σ* (1s)]2 [σ (2s)]2 [σ* (2s)]2 [σ (2pz)]2 [𝝅 (2px)]2 [𝝅 (2py)]2 [𝝅* (2px)]2 [𝝅* (2py)]1

Total number of bonding electrons = 8
Total number of anti-bonding electrons = 5
Bond Order = (8 – 5)/2 = 3/2 = 1.5

Therefor Bond Order of O2 is 1.5

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