Chemical Bonding & Molecular Structure

$\mathrm{XeF}_2$ has $s p^3 d$ hybridisation involving 5 orbitals (one of them $d$ ). The rest are $s p^3$ (no $d$-orbitals).

To find the bond dissociation energy of $ X_2 $, let's assume the bond dissociation energy of $ X_2 $ is $ a \, \text{kJ/mol} $.

Therefore, $ \text{BE}(X_2) = a \, \text{kJ/mol} $.

Given that the bond dissociation energy ratios are $ 1:0.5:1 $, then:

$ \text{BE}(Y_2) = 0.5a \, \text{kJ/mol} $

$ \text{BE}(XY) = a \, \text{kJ/mol} $.

We know that the formation reaction of $ XY $ is:

$ \frac{1}{2} X_2 + \frac{1}{2} Y_2 \rightarrow XY, \quad \Delta H = -200 \, \text{kJ/mol} $

Using the enthalpy change equation:

$ \Delta_r H = \text{BE}(\text{Reactants}) - \text{BE}(\text{Products}) $

Substituting the bond dissociation energies into the equation gives:

$ \Delta_r H = \frac{1}{2} \text{BE}(X_2) + \frac{1}{2} \text{BE}(Y_2) - \text{BE}(XY) $

Plugging in the known values:

$ -200 = \frac{a}{2} + \frac{0.5a}{2} - a $

Simplifying:

$ -200 = \frac{a}{2} + \frac{0.25a}{2} - a = \frac{a}{2} + \frac{a}{4} - a $

Combine the terms:

$ -200 = \frac{a}{2} + \frac{0.25a}{2} - \frac{2a}{2} $

$ -200 = \frac{0.5a + 0.25a - 2a}{2} = \frac{-0.75a}{2} $

Solving for $ a $:

$ -200 = \frac{-0.75a}{2} $

$ -200 \times 2 = -0.75a $

$ -400 = -0.75a $

Dividing both sides by $-0.75$:

$ a = \frac{400}{0.75} = 800 \, \text{kJ/mol} $

Thus, the bond dissociation energy of $ X_2 $ is $ 800 \, \text{kJ/mol} $.

$ I_{3}^{-} $is linear with central iodine atom undergoing $ s p^{3} d $-hybridisation and three lone pairs occupying equatorial position.

Thus, this structure include $ d $-orbital contribution to the hybridisation scheme.

The structure is,

$ \mathrm{PCl}_{3} $ structure is $ s p^{3} $-hybridised with one lone pair. $ \mathrm{NO}_{3}^{-} $is $ s p^{2} $-hybridised and $ \mathrm{H}_{2} \mathrm{Se} $ is $ s p^{3} $ - hybridised with two lone pairs of Se atom.

In $\mathrm{BrF}_3$ molecule, $\mathrm{Br}$ is $s p^3 d$ hybridised, but geometry is T-shaped due to distortion of geometry from trigonal-bipyramidal to T-shaped by the involvement of lone pair-lone pair repulsion. The lone pair occupies equitorial position in order to minimize lone pair-lone pair and lone pair-bond pair repulsion.

Here,

$\begin{array}{l}l p-l p \text { repulsion }=1 \\ l p-b p \text { repulsion }=2 \\ b p-b p \text { repulsion }=2\end{array}$

$\mathrm{I}_3^{-}$ ion is made up of an $\mathrm{I}_2$ molecule with an $\mathrm{I}^{-}$ bonded to it by means of a coordinate bond in which $\mathrm{I}_2$ is lone pair acceptor (Lewis acid) and $I^{-}$ is lone pair donor (Lewis base).

There are two bond pairs and three lone pairs in the outer shell of central atom. To minimise the repulsive forces the three lone pairs occupy the equatorial position.

The ion is therefore, linear in shape with a bond angle of exactly $180^{\circ}$.

$ \underset{\substack{\text { Lewis } \\ \text { acid }}}{\mathrm{I}_2}+\underset{\substack{\text { Lewis } \\ \text { base }}}{\mathrm{I}^{-}} \longrightarrow[\mathrm{I}-\mathrm{I} \leftarrow \mathrm{I}]^{-}$

Similarly, $\mathrm{N}_3^{-}$ ion is also linear in shape. $\mathrm{I}_3^{+}$ does not have linear structure.

This is one unpaired electron in (O$-$F), hence, paramagnetic, thus (a) is correct.

As bond order of O$-$F > F$-$F thus, bond energy of O$-$F > F$-$F.

Hence, (a) and (b) both are correct.

Hydrogen peroxide (H₂O₂) is a non-planar molecule with twisted symmetrical structure. It has non-linear structure with an open book structure.

Larger the bond order, stronger the bond formed.

CH₄ ; bp = 4; sp³-hybridisation

NF₃ ; bp = 3; $l$p = 1; sp³-hybridisation

NH$_4^ + $ ; bp = 4; $lp$ = 0; sp³-hybridisation

H₂O; bp = 2; lp = 2; sp³-hybridisation

Hence, all have the same slope.

o-nitrophenol forms intramolecular H-bonding whereas molecules of p-nitrophenol get associated through intermolecular H-bond.

$\therefore$ o-nitrophenol is the most volatile compound.

The shape of CIF₃ is as follows

It has two lone pair at equatorial position.

Molecular electronic configuration for given species are

(a) ${O_2} = \sigma \,1{s^2}\,\sigma *1{s^2}\,\sigma 2{s^2}\,\sigma *2{s^2}\,\sigma 2p_x^2\,\pi \,2p_y^2\,\pi \,2p_z^2\,\pi *2p_y^1\,\pi *2p_z^1$

It has two unpaired electrons.

(b) $O_2^{2 - }$ : It has no unpaired electron.

(c) $N_2^{2 + }$ : It has one unpaired electron.

(d) B₂ : It has two unpaired electrons.

$\therefore$ $O_2^{2 - }$ is diamagnetic in nature.

Bond order for the various compounds are as follows

for SO₂,

Number of bonds = 4

Number of (SO) groups = 2

$\therefore$ Bond order $ = {4 \over 2} = 2$

For SO₃,

Number of bonds = 6

Number of (SO) groups = 3

$\therefore$ Bond order $ = {6 \over 3} = 2$

For SO$_4^{2 - }$,

Number of bonds = 6

Number of (SO) groups = 4

$\therefore$ Bond order $ = {6 \over 4} = 1.5$

For SO$_3^{2 - }$,

Number of bonds = 4

Number of (SO) groups = 3

$\therefore$ Bond order $ = {4 \over 3} = 1.33$

Hence, correct order is $S{O_2} = S{O_3} > SO_4^{2 - } > SO_3^{2 - }$.

The correct option is (b).

A $\to$ 5, B $\to$ 3, C $\to$ 4, D $\to$ 2, E $\to$ 1.

The shapes of molecules are given below.

Some complex ions of nitrogen show similarities with halide ions. These are called pseudohalide ions e.g., OCN^$-$, NCO^$-$, CN^$-$ etc.

Halogens, among themselves, from complex ions which are called polyhalide ions. e.g., I$_3^ - $, Brl$_2^ {-1} $ etc.

Similarly, interhalogens are the compounds of halogens in which one halogen is cation and other halogen is anion. e.g., IF₇, ICI₅, BrF₅, IF₅ etc.