Chemical Bonding & Molecular Structure

The Lewis dot structures of the given compounds are as follows:

Thus, there are 4 compounds in which central atom has one lone pair of electron. i.e. $\mathrm{PbCl}_2, \mathrm{PH}_3, \mathrm{SF}_4$ and $\mathrm{SnCl}_2$.

Thus, $\mathrm{SF}_4$ molecules has same number of lone pair of electrons on central atom and $d$-orbitals involved in the hybridisation of central atom i.e. one.

Species containing same number of electrons will have same bond order.

I. $\mathrm{CN}^{-}-14$ electrons ( 6 from carbon, 7 from nitrogen add 1 for negative charge).

II. $\mathrm{N}_2^{+}-13$ electrons ( 7 from each N -atom, subtract 1 for positive charge).

III. $\mathrm{O}_2^{2-}-18$ electrons ( 8 from each O -atom, add 2 for ( -2 ) negative charge).

IV. $\mathrm{NO}^{+}-14$ electrons ( 8 from O -atom, 7 from N -atom, subtract 1 for positive charge).

Hence, I and IV will have same bond order.

The correct match is A-III, B-IV, C-I, D-III

Lone pair-lone pair (lp-lp) repulsion is the repulsion between various lone pair present in molecule. More the number of lone pair of electrons in molecule, greater will be the lone pair-lone pair repulsion.

1. 1.  $\mathrm{ClF}_3$ - Central atom Cl has 3 bond pairs and 2 lone pairs.

2. 2.  $\mathrm{IF}_5-\mathrm{I}$ has 5 bond pairs and 1 lone pair

3. 3.  $\mathrm{SF}_4-\mathrm{S}$ has 4 bond pairs and 1 lone pair

4. 4.  $\mathrm{XeF}_2-\mathrm{Xe}$ has 2 bond pairs and 3 lone pairs.

Since, lone pairs are maximum in XeF 2. Hence, it has maximum lone pair-lone pair repulsion.

The correct order of dipole moment is $\mathrm{BF}_3<\mathrm{NF}_3<\mathrm{NH}_3$

$\mathrm{NH}_3$ and $\mathrm{NF}_3$ have pyramidal geometry where central nitrogen has 3 bond pairs and 1 lone pair. Hence, both are $s p^3$-hybridised.

$\mathrm{NH}_3$, dipole vector of bond and lone pair are in same direction while in $\mathrm{NF}_3$, they are in opposite direction. Hence, dipole moment of $\mathrm{NH}_3$ is larger than $\mathrm{NF}_3$.

$\mathrm{BF}_3$ is asymmetrical molecule where dipole vectors are arranged at $120^{\circ}$. Therefore, the resultant of two dipole vectors cancel out with the third dipole vector. Hence it's dipole moment will be zero.

Bond Order Calculation

The bond order (BO) of a molecule is calculated using the formula:

$ \text{BO} = \frac{1}{2} [\text{Number of bonding electrons} - \text{Number of antibonding electrons}] $

Oxygen Molecule Configurations and Calculations

For $\mathrm{O}_2^{+}$:

Electronic configuration:

$ \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \approx \pi 2p_y^2, \pi^* 2p_x^1 \approx \pi^* 2p_y^0 $

Bond order calculation:

$ \mathrm{BO} = \frac{1}{2} [10 - 5] = \frac{5}{2} $

Given $\mathrm{BO}$ of $\mathrm{O}_2^+$ is $x$. Therefore:

$ \frac{5}{2} = x \quad \text{...(i)} $

For $\mathrm{O}_2^{-}$:

Electronic configuration:

$ \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \approx \pi 2p_y^2, \pi^* 2p_x^2 \approx \pi^* 2p_y^1 $

Bond order calculation:

$ \mathrm{BO} = \frac{1}{2} [10 - 7] = \frac{3}{2} $

Using equation (i), the bond order of $\mathrm{O}_2^-$ is expressed as:

$ \mathrm{BO} = \frac{3x}{5} $

For $\mathrm{O}_2^{2+}$:

Electronic configuration:

$ \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 \approx \pi 2p_y^2 $

Bond order calculation:

$ \mathrm{BO} = \frac{1}{2} [10 - 4] = 3 $

From equation (i), substituting for $x$, we can express it as related fractions of bond order in terms of $x$.

By understanding these calculations, you can determine the bond orders of $\mathrm{O}_2^-$ and $\mathrm{O}_2^{2+}$ based on the given $x$ value and their respective electronic configurations.

The correct order of bond angles is

$ \mathrm{P}_4<\mathrm{S}_6<\mathrm{S}_8<\mathrm{O}_3 $

In $\mathrm{P}_4$, the four $s p^3$ hybridised P -atom lie at corner of regular tetrahedron with $\angle \mathrm{PPP}=60^{\circ}$.

In $\mathrm{S}_8$, the eight S -atoms arranged in crown shaped structure with $\angle S S S=107^{\circ}$.

In $\mathrm{S}_6$, the six S -atoms arranged in chair form with

$ \angle S S S=102.2^{\circ} $

In $\mathrm{O}_3$, hybridisation of oxygen atom is $s p^2$. So, angle between 0 -atom will be $120^{\circ}$ but due to presence of one lone pair, the bond angle decreases from $120^{\circ}$ to $116.8^{\circ}$.

The set of molecules in option (d), consisting of SF₄, XeF₄, and CH₄, has central atoms with different types of hybridization.

SF₄:

Calculation: $ 4 + \frac{1}{2} \times (6 - 4) = 4 + 1 $

Hybridization: $ \text{sp}^3d $

XeF₄:

Calculation: $ 4 + \frac{1}{2} \times (8 - 4) = 4 + 2 $

Hybridization: $ \text{sp}^3d^2 $

(Square planar shape)

CH₄:

Calculation: $ 4 + \frac{1}{2} \times (4 - 4) = 4 $

Hybridization: $ \text{sp}^3 $

This demonstrates that the hybridizations are $ \text{sp}^3d $, $ \text{sp}^3d^2 $, and $ \text{sp}^3 $ for SF₄, XeF₄, and CH₄, respectively, indicating distinct hybridization states among the central atoms.

Thus, statements II and IV are incorrect while statement I is correct. Also; from the structure of compound $Z$ i.e., $\mathrm{XeO}_3$ it is clear that statement III is also correct

$3 \pi$-bonds and 1 lone pair of electrons.

On finding number of bond pairs in each

$ \begin{aligned} (A)\,& \mathrm{H}_2 \mathrm{CO}_3 \frac{2 \times 1+4+3 \times 6}{8}=\frac{24}{8}=3 \\ & \Rightarrow s p^2 \text { hybridised } \end{aligned} $

(B) $\mathrm{SiF}_4 \frac{4+4 \times 7}{8}=\frac{32}{8}=4 \Rightarrow s p^3$

hybridised

(C) $\mathrm{BF}_3 \frac{3+3 \times 7}{8}=\frac{24}{8}=3 \Rightarrow s p^2$

hybridised

(D) $\mathrm{HClO}_2 \frac{1+7+2 \times 6}{8}=\frac{20}{8}$ ( 2 lone

pair +4 bond pair $s p^3$ hybridised

A and C only are having $s p^2$ hybridised central atom.

Hybridisation $=\frac{1}{2}$ (Number of valence electron + number of monovalent atoms other than central atom + negative charge - positive charge)

$ =\frac{1}{2}(7+2+1) $

$ \begin{aligned} &=\frac{10}{2}\\ &=5 \cong s p^3 d . \end{aligned} $

$ =5 \cong s p^3 d . $

Also it can be determined by sum of number of lone pairs and number of $\sigma$-bonds central metal atom will form. As I is further linked to two iodine atoms and number of lone pair on each will be 3 .

$ \therefore 3+2=5 \cong s p^3 d \text { hybridised. } $

So, shape is linear and hybridisation is $s p^3 d$.

Dipole moment of the given compounds along with their structures are given below.

Isostructural compounds are those that have same structure as well as same hybridisation.

(a) $\mathrm{PF}_6^{-}\left(6\right.$ bond pairs) and $\mathrm{SF}_6$ ( 6 bond pairs) are isostructural.

(b) $\mathrm{IO}_3^{-}(4$ pairs, i.e. 3 bond pair +1 lone pair) and $\mathrm{XeO}_3$ (4 pairs, i.e. 3 bond pairs +1 lone pair) are isostructural.

(c) $\mathrm{BH}_4^{-}$(4 bond pairs) and $\mathrm{NH}_4^{+}$( 4 bond pairs) are isostructural.

(d) $\mathrm{BrF}_5$ ( 5 bond pairs) and $\mathrm{XeF}_4$ ( 4 bond pairs and 1 lone pair) are not isostructural.

Hydrogen fluoride (HF) has higher boiling point as it possess strongest hydrogen bonding due to high electronegativity of fluorine. Hence, both $(A)$ and $(R)$ are true and $(\mathrm{R})$ is the correct explanation of (A).

Intramolecular hydrogen bonding is present in 2-hydroxybenzoic acid.

Intramolecular hydrogen bonding is not present.

The correct order of bond angle is

$ \mathrm{BeCl}_2>\mathrm{BF}_3>\mathrm{SiCl}_4>\mathrm{SF}_6 $

In $\mathrm{SF}_6$, the molecule geometry is octahedral where bond angle is $90^{\circ}$.

In $\mathrm{SiCl}_4$, the geometry is tetrahedral and the bond angle is $109.5^{\circ}$.

In $\mathrm{BF}_3$, the geometry is trigonal planar where bond angle is $120^{\circ}$.

In $\mathrm{BeCl}_2$, the geometry is linear and bond angle is $180^{\circ}$.

If the bond order of species is zero, then species will not exist.

$ =\frac{\text { Bond order bonding molecular orbital electrons - antibonding molecular orbital electrons }}{2} $

$ \begin{aligned} & \mathrm{H}_2^{+}-\sigma \mathrm{l} s^1 \quad \therefore \text { Bond order }=\frac{1-0}{2}=0.5 \\ & \mathrm{He}_2^{2+}-\sigma l s^2 \quad \therefore \text { Bond order }=\frac{2-0}{2}=1 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \mathrm{Li}_2^{2-}-\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2 \\ & \begin{aligned} \therefore & \text { Bond order }=\frac{4-4}{2} \\ & =0(\text { do not exist }) \end{aligned} \\ & \begin{aligned} \mathrm{Ne}_2: \sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \\ \begin{aligned} & 2 p_y^2, \pi^* 2 p_x^2 \approx \pi^* 2 p_y^2, \sigma^* 2 p_z^2 \\ & \therefore \text { Bond order }=\frac{10-10}{2} \\ &=0 \text { (do not exist) } \\ & \mathrm{Be}_2^{-}-\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^1 \\ & \approx \pi 2 p_y^0 \end{aligned} \\ \therefore \mathrm{Bond} \mathrm{order}^2=\frac{5-4}{2}=0.5 \\ \mathrm{He}_2-\sigma \mathrm{l} s^2, \sigma^* 1 s^2 \\ \therefore \text { Bond order }=\frac{2-2}{2}=0 \text { (do not exist) } \end{aligned} \end{aligned}\\ &\text { Hence, } \mathrm{Li}_2^{2-}, \mathrm{Ne}_2 \text { and } \mathrm{He}_2 \text { will not exist. } \end{aligned} $

Molecular orbital electronic configurations are

$ \begin{aligned} & \mathrm{O}_2 \rightarrow \sigma l s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1 \pi^* 2 p_y^1 \end{aligned} $

Bond order $\rightarrow \frac{6-2}{2}=\frac{4}{2}=2$

paramagnetic

$ \begin{aligned} & \mathrm{O}_2^{+} \rightarrow \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1 \end{aligned} $

Bond order $\rightarrow \frac{10-5}{2}=\frac{5}{2}=2.5$,

paramagnetic

$ \begin{array}{r} \mathrm{O}_2^{2-} \rightarrow \sigma l s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \quad \pi^* 2 p_x^2 \pi^* 2 p_y^2 \end{array} $

Bond order $=\frac{10-8}{2}=\frac{2}{2}=1$,

diamagnetic

$ \begin{aligned} \mathrm{O}_2^{-} \rightarrow \sigma l s^2 \sigma^* 1 s^2 & \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^2 \pi^* 2 p_y^1 \end{aligned} $

$ \text { Bond order }=\frac{10-7}{2}=1.5=\frac{3}{2}=1.5, $

paramagnetic

The species with $(A)$, the highest bond order is $\mathrm{O}_2^{+}$and ( $B$ ), diamagnetic character is $\mathrm{O}_2^{2-}$. Hence, option (b) is correct.