Chemistry

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Ans: (i) Due to the electrostatic forces between the two opposite charges, ionic bonds are non-directional. The bonding direction does not matter as the electrostatic field of an ion is non-directional. Whereas, the formation of covalent bonds happens with the overlap of the atomic orbitals. The direction of the bonds is given by the direction of overlapping.

(ii) The hybridization of the oxygen atom in water molecules is 3 sp due to the presence of two lone pairs of electrons on the oxygen atom. Tetrahedral geometry is acquired by these four 3 sp hybridized orbitals with two bonds of hydrogen atoms and two lone pairs of electrons. Due to the presence of lone pairs of electron the bond angle gets reduced as there is a repulsion between the LP-LP and thus the molecule acquires a bent shape. The hybridization of carbon atoms in the carbon dioxide molecule is sp. The two sp are arranged in such a way that it makes an angle of 180° . That is why carbon dioxide has a linear shape.

 

 

(iii) Both the carbon atoms in ethyne molecules are sp hybridized. Ethyne has two unhybridized 2p_x 2p_y orbitals. The two sp hybrid orbitals of both the carbon atoms are arranged such that they form an angle of 180° with each other. That is why ethyne molecules are linear in shape.

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

Ans: (i) N₂ → N₂⁺ + e^-

The electronic configuration of N₂ is:

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

So, its bond order will be 3

The electronic configuration of N₂⁺ is:

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

Its bond order will be 2.5

Hence the bond order decreases in this reaction N₂ → N₂⁺ + e^-

(ii) O₂→ O₂⁺ + e-

The electronic configuration of O₂ is:

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

Its bond order will be 2

The electronic configuration of O₂⁺ will be:

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

Its bond order will be 2.5 .

Hence, the bond order in this reaction increases O₂→ O₂⁺ + e-

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

Ans: The general sequence of the energy level of the molecular orbital is σ1s < σ*1s < σ2s< σ*2s < π2px = π2py < σ2pz

N₂    =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p²_z

N₂⁺  =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y σ2p¹_z

N₂^-   =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x= π2p²_y σ2p²_z σ2p²_x

N₂²⁺ =   σ1s² σ*1s² σ2s² σ*2s² π2p²_x = π2p²_y

The formula for finding the bond order (B. O) for any molecular species is:

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

Hence, the bond orders of the given molecular species are:

N₂= $\frac{1}{2}$ ( 10-4)= 3

N₂⁺= $\frac{1}{2}$ ( 9-4)= 2.5

N₂^-= $\frac{1}{2}$ ( 10-5)= 2.5

N₂²⁺= $\frac{1}{2}$ (8-4)= 2

The order for the stability of the given molecular species thus will be:

N_2.> N₂^-> N₂⁺> N₂²⁺

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

Ans: Although the hybridisation of the central atom oxygen in both the molecules is 3 sp but dimethyl ether will have a higher bond angle than water molecule.

Due to the presence of two bulky methyl groups in dimethyl ether, the repulsive forces will be greater in them than the two hydrogens in water molecules. In dimethyl ether the -CH₃ is a group attached to three hydrogen atom through s bonds. Thus, the C- H bond pairs increase the electron density on the carbon atom which results in lone pair-bond pair repulsions. Due to this lone pair-bond pair repulsions, the bond angle of dimethyl ether is greater than that of water molecules.

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.