Class 11 Chemistry

Chemical Bonding

NCERT Solution

Part 3

Question 25: Describe the change in hybridization (if any) of the Al atom in the following reaction:

AlCl3 + Cl- → AlCl4-

Answer: The valence orbital of Al looks like following in ground state.

↑ ↓

The valence orbital of Al looks like following in excite state.


So, it goes sp2 hybridization to form AlCl3 to result in trigonal planar shape. When additional Cl- is added it results in sp3 hybridization and gives tetrahedral shape.

Question 26: Is there any change in the hybridization of B and N atoms as a result of the following reaction?

BF3 + NH3 → F3B.NH3

Answer: Following is the excited state configuration of Boron.


Following is the excited state configuration of nitrogen.


So, there is sp2 hybridization in case of boron before the reaction. After the reaction, boron undergoes sp3 hybridization, while hybridization of nitrogen remains the same, i.e. sp3.

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


Lewis structure of ethylene

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

Answer: We know that sigma bond is formed in case of single bond, and a pi-bond is always present in multiple bonds. Moreover, a triple bond contains both sigma and pi bonds. So, C2H4 contains four sigma bonds and one pi bond. C2H2 contains three sigma bonds and one pi bond.

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 2p (c) 2py and 2py (d) 1s and 2s.

Answer: (c) 2py and 2py because they will undergo lateral or sideways overlapping.

Question 30: Which hybrid orbitals are used by carbon atoms in the following molecules? (a) CH3-CH3 (b) CH3-CH=CH2 (c) CH3-CH2-OH (d) CH3-CHO (e) CH3COOH

Answer: Excited state configuration of C is as follows:


So, in case of single bond there will be sp3 hybridization and in case of double bond there will be sp2 hybridization. As per convention, we need to count the number of C atoms from right as 1, 2, 3 and so on.

Question 31: What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.

Answer: The pair of electron (in central atom) which participates in bond formation is called bond pair, and remaining pair in central atom is called lone pair. The lone pair is localized on the central atom, while bond pair is shared by two atoms. The following Lewis structure shows H2S which has two bond pairs and two lone pairs.

Question 32: Distinguish between a sigma and a pi bond.


Sigma BondPi Bond
Formed by end to end (head-on) overlap of bonding orbitals along internuclear axis.Formed by sideways overlap of bonding orbitals perpendicular to internuclear axis.
It is stronger.It is weaker.
It contains one electron cloud which is symmetrical.It contains two clouds; one above and another below the internuclear axis.
Free rotation about this bond is possible.Rotation is restricted.

Question 33: Explain the formation of H2 molecule on the basis of valence bond theory.

Answer: Let us assume that two hydrogen atoms A and B are approaching each other. Their nuclei are NA and NB and electrons are eA and eB. When two atoms come closer to each other, new attractive and repulsive forces begin to work.

Attractive forces arise between:

  1. Nucleus of one atom and its own electron, i.e. NA-eA and NB-eB
  2. Nucleus of one atom and electron of other atoms, i.e. NA-eB and NB-eA

Repulsive forces arise between:

  1. Electrons of two atoms, i.e. eA-eB
  2. Nuclei of two atoms, i.e. NA-NB

Experiments have shown that magnitude of new attractive force is greater than that of new repulsive force. So, two atoms come closer to each other and potential energy decreases. Finally, the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage two hydrogen atoms are said to be bonded together to form a stable molecule.