Chemistry NCERT Exemplar Solutions Class 11th Chapter Four

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: Option (ii)

(i) The electronic configuration of O₂ is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p¹_y π*2p¹_x

Bond order = $\frac{1}{2}$ ( N_b- N_a)

Bond order = $\frac{1}{2}$ (10-6)=2.0

The electronic configuration of N₂ :

σ1s² σ*1s² σ2s² σ*2s² π2p²_x, π2p²_y σ2p²_z

Bond order = $\frac{1}{2}$ (10-4)=3.0

So, the bond order of O₂ , N₂ will not be equal.

(ii) The electronic configuration of O₂⁺ is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z²,π2p²_y , π2p²_x, π*2p¹_x

Bond order = $\frac{1}{2}$ (10-5)=2.5

The electronic configuration of N₂^-

σ1s² σ∗1s² σ2s² σ∗2s² ,π2p²_x , π2p²_y, s2p_z² π*2p¹_x

Bond order = $\frac{1}{2}$ (10-5)=2.5

Hence, the bond order of O₂⁺,N₂^- will be equal.

(iii) The electronic configuration of O₂ ^- is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z²,π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

Bond order = $\frac{1}{2}$ (10-7)=1.5

The electronic configuration of N₂⁺ is:

σ1s² σ*1s² σ2s² σ*2s² π2p²_x , π2p²_y σ2p¹_z

Bond order = $\frac{1}{2}$ (9-4)=2.5

Hence, the bond order of O₂^-, N₂⁺ will not be equal.

(iv) The electronic configuration of O₂ ^- is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z²,π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

Bond order = $\frac{1}{2}$ (10-7)=1.5

The electronic configuration of N₂^-

σ1s² σ∗1s² σ2s² σ∗2s² ,π2p²_x , π2p²_y, s2p_z² π*2p¹_x

Bond order = $\frac{1}{2}$ (10-5)=2.5

Hence, the bond order of O₂ ^- , N₂^- will not be equal.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: (i) The electronic configuration of dioxygen is:

σ1s² σ∗1s² σ2s² σ∗2s² s2p_z² π2p²_y π2p²_x π*2p¹_y π*2p¹_x

(ii) As it can be seen from the electronic configuration of oxygen atom π*2p¹_X π*2p¹_Y  the are partially. So, the statement given is incorrect.

(iii) The statement given is correct because there are five bonding molecular orbitals and four antibonding molecular orbital in oxygen molecules. Hence, the bonding and antibonding molecular orbitals are no equal.

(iv) The filled bonding orbitals are not the same as the number of filled antibonding orbitals. The filled bonding orbitals are five and only two filled antibonding orbitals are present in dioxygen molecules.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: (i) Different types of hybridization in a carbon atom are:

(a) sp hybridization: The carbon atoms forming triple bonds with each other determines hybridized carbon. ( Cº C ).

(b) sp² hybridization: The carbon atoms forming double bonds with each other determine sp² hybridized carbon. (C=C). (c) sp³ hybridization: The carbon atoms forming single bonds with each other determine sp³ hybridized carbon. ( C-C).

(ii)The starred C atom is sp² hybridised.

B] The starred C atom is sp³ hybridised.

C] The starred C atom is sp² hybridised.

D] The starred C atom is sp³ hybridised.

E] The starred C atom is sp² hybridised.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Ans: Hybridisation in the case of PCl₅ :

The hybridisation of phosphorus PCl₅ is sp₃d and it has a trigonal bipyramidal geometry. The axial bonds in PCl₅ are slightly longer as compared to the equatorial bonds because axial bonds experience greater repulsive forces from other bonds as compared to the equatorial bonds.

Hybridisation in the case of SF₆:

The hybridisation of sulphur in SF₆ is sp³d² and the molecule has an octahedral geometry. The bond length of axial as well as the equatorial bonds are similar because all the bonds in octahedral geometry experience similar repulsive forces from other bonds.