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.

Answer:

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) 2p_y and 2p_y (d) 1s and 2s.

Answer: (c) 2p_y and 2p_y 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.

• All C atoms have sp³ hybridization in CH₃-CH₃
• C1 has sp³ hybridization while C2 has sp² hybridization in CH₃-CH=CH₂
• All C atoms have sp³ hybridization in CH₃-CH₂-OH
• C1 and C2 have sp³ hybridization while C3 has sp² hybridization in CH₃-CHO
• C1 has sp³ hybridization while C2 has sp² hybridization in CH₃COOH

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.

Answer:

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 N_A and N_B and electrons are e_A and e_B. 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. N_A-e_A and N_B-e_B
2. Nucleus of one atom and electron of other atoms, i.e. N_A-e_B and N_B-e_A

Repulsive forces arise between:

1. Electrons of two atoms, i.e. e_A-e_B
2. Nuclei of two atoms, i.e. N_A-N_B

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.

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