Chemical Bonding and Molecular Structure

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

Ans: (a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding.

Ans: The intramolecular hydrogen bonding is shown by compound (I) and the intermolecular hydrogen bonding is shown by compound (II) . In compound (I) the NO₂ and OH group are close together in comparison to that in compound (II) . So, that is why compound (I) shows intramolecular hydrogen bonding. The intermolecular hydrogen bonding in compound (II) is shown as:

(b) The melting point of a compound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point.

Ans: Due to the formation of intermolecular hydrogen bonds, compound (II) will have a higher melting point. Thus, due to hydrogen bond formation more molecules are joined together.

(c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with water easily and be more soluble in it?

Ans: Compound (I) will be less soluble in water due to intramolecular hydrogen bonding; it will not be able to form hydrogen bonds with water. Whereas, compound (II) can form hydrogen bonds with water more easily and is thus more soluble in water.

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

Ans: The electronic configurations of O₂⁺ and O₂ ^- according to molecular orbital theory is:

O₂⁺: σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p¹_x

O₂^- : σ1s² σ∗1s² σ2s² σ∗2s² s2p_z², π2p²_y , π2p²_x, π*2p²_y π*2p¹_x

The bond order of O₂⁺

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

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

The bond order of O₂^-

$\frac{1}{2}$ (10-7)= $\frac{3}{2}$ = 1.5

As the bond order of O₂⁺ is higher, it is more stable than O₂ ^-, because higher the bond order more stable is the bond. Both the molecular species have the presence of unpaired electrons. So, they both are paramagnetic in nature.

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

Ans: In the Lewis structure of PCl₃, the phosphorus atom is surrounded by three bond pairs (chlorine atoms) and one lone pair. These four electron pairs are arranged in a tetrahedral geometry around the central phosphorus atom. Due to the presence of lone pair of electron on the phosphorus atom, PCl₃ will have a distorted tetrahedral geometry. Thus, it will form a pyramidal shape and is non-linear in structure.

In H₂S, the central sulphur atom is surrounded by two bond pairs and two lone pairs of electrons. It can be said that these four electron pairs are arranged in a tetrahedral geometry. Due to the presence of two lone pairs on the central sulphur atom, the lone pair-bond pair repulsion happens and due to this the molecule H₂S has a V-shaped geometry and is non-linear in structure.

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.