Chemical Bonding & Molecular Structure

Since, $Z$ is an alkali metal with atomic number $a+2$

$Y$ has atomic number $a+1$, so it must be a noble gas.$X$ has atomic number a, so it must be a halogen.

$X$ is a halogen which is non-metal. Z is an alkali metal, which is a metal. Compound formed between metal and non-metal is ionic.

Set of II and III i.e.

$\mathrm{HgCl}_2, \mathrm{SO}_3, \mathrm{SF}_6$ and $\mathrm{BeCl}_2, \mathrm{BF}_3, \mathrm{PCl}_5$ has no lone pair of electrons present on central atom

The structure of the given molecules with their bond angles are given below.

Hence, the increasing order of bond angle is

$ \underset{(\approx 104.5)}{\mathrm{H}_2 \mathrm{O}}<\underset{(\approx 107)}{\mathrm{NH}_3} \underset{(\approx 116.8)}{<\mathrm{O}_3} \underset{\substack{\approx 1199 \\ \approx \mathrm{SO}_2}}{\stackrel{\mathrm{SO}_2}{\sim}} $

The boiling point order is primarily determined by hydrogen bonding strength and network. $\mathbf{H}_2 \mathbf{O}$ forms the most extensive hydrogen bond network due to two lone pairs.

HF possesses very strong individual hydrogen bonds, but a less extensive network.

$\mathbf{N H}_{\mathbf{3}}$ exhibits weaker hydrogen bonding compared to $\mathrm{H}_2 \mathrm{O}$ and HF .

$\mathbf{P H}_3$ lacks hydrogen bonding, having only weaker dipole-dipole forces.

Hence, the correct order is

$ \mathrm{H}_2 \mathrm{O}>\mathrm{HF}>\mathrm{NH}_3>\mathrm{PH}_3 $

The structure of $\mathrm{S}_8, \mathrm{P}_4, \mathrm{~S}_6$ and $\mathrm{O}_3$ are given below.

So, the increasing order of their bond angles is $\mathrm{P}_4<\mathrm{S}_6<\mathrm{S}_8<\mathrm{O}_3$.

Step 1: Find bond orders using the formula

Bond order = $\frac{1}{2}$(number of bonding electrons - number of antibonding electrons)

Bond order of $\mathrm{O}_2^{-}$ (17 electrons):

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

Bond order of $\mathrm{O}_2^{2-}$ (18 electrons):

$ =\frac{1}{2}(10-8)=1.0 $

Bond order of $\mathrm{O}_2^{2+}$ (14 electrons):

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

Step 2: Add the bond orders for $\mathrm{O}_2^{-}$ and $\mathrm{O}_2^{2-}$

The question says $x$ is their sum.

$ x=1.5+1.0=2.5 $

Step 3: Express the bond order of $\mathrm{O}_2^{2+}$ in terms of $x$

We already found bond order of $\mathrm{O}_2^{2+}$ is 3.0.

We want to write 3.0 as a multiple of $x$:

$ \begin{aligned} 3.0 & =k \cdot x \\ 3.0 & =k \cdot 2.5 \\ k & =\frac{3.0}{2.5}=1.20 \end{aligned} $

So, bond order of $\mathrm{O}_2^{2+}$ is $1.20x$.

The structure of $\mathrm{CH}_2(\mathrm{CN})_2$ is as follows

So, number of $\sigma$ bonds $=6$

Number of $\pi$ bonds $=4$

So, the ratio of $\sigma: \pi$ bonds $=3: 2$

Bond order

$ =\frac{\text {Bonding electrons - Non-bonding electrons }}{2} $

B.O. of $\mathrm{O}_2=\frac{1}{2}[10-6]=2$

$ \begin{aligned} & \mathrm{O}_2^{+}=\frac{1}{2}[10-5]=2.5 \\ & \mathrm{O}_2^{-}=\frac{1}{2}(10-7)=1.5 \\ & \mathrm{O}_2^{2+}=\frac{1}{2}(10-4)=3 \end{aligned} $

Sum of B.O. of $\mathrm{O}_2, \mathrm{O}_2^{+}, \mathrm{O}_2^{-}$and $\mathrm{O}_2^{2+}$

$ =2+2.5+1.5+3=9 $

Statement I is not correct but statement II is correct. The correct form of statement I is,

Both $\mathrm{SF}_6$ and $\mathrm{BrF}_5$ has same hybridisation i.e., $s p^3 d^2$.

Among the given molecules, total five molecules are polar in nature

$ \mathrm{NH}_3, \mathrm{NF}_3, \mathrm{H}_2 \mathrm{~S}, \mathrm{CHCl}_3, \mathrm{H}_2 \mathrm{O} $

Total three, are non-polar molecules

$ \mathrm{BF}_3, \mathrm{CO}_2, \mathrm{CH}_4 $

$ \begin{aligned} &\text { }\\ &\begin{aligned} \mathrm{ClF}_3 & =T \text {-shape } \\ & =s p^3 d \text { hybridisation } \\ \mathrm{XeF}_4 & =\text { square planar }=s p^3 d^2 \text { hybridisation } \\ \mathrm{SF}_4 & =s e e \text {-saw }=s p^3 d \text { hybridisation } \\ \mathrm{BrF}_5 & =\text { square pyramidal }=s p^3 d^2 \end{aligned} \end{aligned} $

$\mathrm{SO}_2$ and $\mathrm{O}_3$ has $s p^2$ hybridisation with bent geometry.

Statement given in III and IV are correct. While the statements given in I and II are incorrect. Their correct forms are

I. When $\mathrm{O}_2$ is converted to $\mathrm{O}_2^{2+}$, the bond order increases from 2 to 3. II. In conversion of $\mathrm{O}_2$ to $\mathrm{O}_2^{2+}$, the magnetic property changes from paramagnetic to diamagnetic.

Statement given in option (d) is incorrect according to MOT, rest all are correct. The correct form is, $ \mathrm{C}_{2} $ molecules consists of $ 2 \pi $ bonds only.

The melting point of 0 -hydroxybenzaldehyde $ (A) $ is lower than that of $ p $-hydroxybenzaldehyde

(B). This is because $ A $ has intramolecular H -bonding and $ B $ has intermolecular H -bonding.

The bond angles $ \mathrm{H}-\mathrm{O}-\mathrm{N} $ and $ \mathrm{O}-\mathrm{N}-\mathrm{O} $ in the planar structure of nitric acid are $ 102^{\circ} $ and $ 130^{\circ} $ respectively.

The correct order of boiling points for hydrogen halides is:

$ \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI} < \mathrm{HF} $

This order is due to the presence of strong hydrogen bonding between HF molecules. Hydrogen fluoride (HF) forms strong intermolecular hydrogen bonds because of the high electronegativity of fluorine (F), leading to an unusually high boiling point for HF compared to other hydrogen halides.

For the other hydrogen halides, the boiling points increase from HCl to HI as the size of the halogen atom increases from chlorine (Cl) to iodine (I). This increase in boiling points is attributed to the progressively stronger dispersion forces as the atomic size grows.

The bond angle in $ \mathrm{C}-\ddot{\mathrm{O}}-\mathrm{H} $ in alcohol is slightly less than tetrahedral angle $ 108.9^{\circ} $. It is due to repulsion between the unshared electron pairs of oxygen. In alcohols, two lone pair of electrons are present.

The $ \mathrm{C}-\mathrm{O}-\mathrm{C} $ bond angle ( $ 111.7^{\circ} $ ) in ether is slightly greater than the tetrahedral angle due to repulsive interaction between the two bulky $ (-A) $ groups.

Hence $ X < 109^{\circ} 28^{\prime} $ and $ Y > 109^{\circ} 28^{\prime} $

Among the given options, order of dipole moment given in option (c) is incorrect.

Dipole moment occurs due to the difference in electronegativity between two chemically bonded atom.

Fluorine is most electronegative hence the electronegativity difference with hydrogen is maximum and hence, the dipole moment will be maximum.

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

According to MOT, the bond order is given by

$ \mathrm{BO}=\frac{N_B-N_A}{2} $

where, $N_B$ and $N_A$ are number of bonding electrons and antibonding electrons respectively.

The difference is greatest in $\mathrm{O}_2^{2-}$ and $\mathrm{O}_2^{2+}$

from given sets of options.

Bond order of $\mathrm{O}_2^{2-}=\frac{N_B-N_A}{2}$

$ \begin{aligned} N_B & =10, N_A={ } 8 \\\\ \mathrm{BO}\left(\mathrm{O}_2^{2-}\right) & =\frac{10-8}{2}=1 \quad \ldots \text { (i) } \end{aligned} $

Similarly, for $\mathrm{O}_2^{2+}=\frac{N_B-N_A}{2}$

$ \begin{aligned} N_B & =10, N_A=4 \\\\ \mathrm{BO}\left(\mathrm{O}_2^{2+}\right) & =\frac{10-4}{2}=3 \quad \ldots \text { (ii) } \end{aligned} $

Hence, difference is 2 which is greatest among the given species.

Among the given pairs, $\mathrm{SnCl}_2$ and $\mathrm{H}_2 \mathrm{O}$ has same geometry but central atoms in them has different state of hybridisation.

Geometry-trigonal pyramidal

Hybridisation in $\mathrm{SnCl}_2$ in $s p^2$ while in $\mathrm{H}_2 \mathrm{O}$ is $s p^3$.

Correct order are given in $A$ and $C$ while $B$ is incorrect. The correct form of $B$ is,

The order of dipole moment is

$ \underset{(1.85 \mathrm{D})}{\mathrm{H}_2 \mathrm{O}}>\underset{(1.82 \mathrm{D})}{\mathrm{HF}}>\underset{(1.47 \mathrm{D})}{\mathrm{NH}_3} $

The bond angle of $\mathrm{SO}_2$ is more than $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ because $\mathrm{SO}_2$ is $s p^2$-hybridised while $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ are $s p^3$-hybridised. Hence, in $\mathrm{SO}_2$ bond angle will be $120^{\circ}$ while in $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3$ it will be approx. $109.5^{\circ}$.

Out of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{NH}_3, \mathrm{NH}_3$ has more bond angle as in $\mathrm{NH}_3$ there is one lone pair and in $\mathrm{H}_2 \mathrm{O}$ there are two lone pair of electrons. Hence, repulsion is less in $\mathrm{NH}_3$ and bond angle is more.

Thus, the order of bond angles will be

$ \mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\mathrm{SO}_2 $

Bond order $=\frac{1}{2}$ [No. of bonding electrons - No. of anti-bonding electrons]

$ =\frac{1}{2}\left[N_b-N_a\right] $

Molecular orbital configuration

$ \begin{aligned} & C_2=\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2 \approx \pi 2 p_y^2 \\ & \text { Bond order }=\frac{1}{2}[8-4]=2 \end{aligned} $

$ \begin{aligned} &\begin{array}{rl} \mathrm{N}_2=\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma * & 2 s^2, \pi 2 p_x^2 \\ & \approx \pi 2 p_y^2, \sigma 2 p_z^2 \end{array}\\ &\begin{aligned} & \text { Bond order }=\frac{1}{2}[10-4]=3 \\ & \begin{array}{l} \mathrm{He}_2=\sigma \mathrm{l} s^2, \sigma^* 1 s^2=\frac{1}{2}[2-2]=0 \\ \mathrm{~B}_2=\sigma \mathrm{l} s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^1, \pi 2 p_y^1 \\ \quad=\frac{1}{2}[6-4]=1 \end{array} \end{aligned}\\ &\text { Thus, order of bond order will be }\\ &\mathrm{He}_2<\mathrm{B}_2<\mathrm{C}_2<\mathrm{N}_2 \end{aligned} $

Molecular orbital configuration of $\mathrm{C}_2$ is

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

So, $\mathrm{C}_2$ molecule contains only $\pi$-bonds according to MOT.

$\mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+}$

$\mathrm{NH}_3$ is $s p^3$-hybridised and it has 3 -bond pairs and 1-lone pair.

$\mathrm{NH}_4^{+}$is also $s p^3$-hybridised, it has 4 -bond pair and no lone pair.

$ \begin{aligned} \text { Lone pair } & =6 \\ \text { Bond pair } & =3 \\ & 6: 3 \text { or }(2: 1) \end{aligned} $

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.