Chemical Bonding & Molecular Structure

The correct order is,

$ \mathrm{H}-\mathrm{H}<\mathrm{O}-\mathrm{H}<\mathrm{C}-\mathrm{H}<\mathrm{C}-\mathrm{C} $

$\mathrm{H}-\mathrm{H}$ is smallest due to small size of H atom.

$\mathrm{C}-\mathrm{C}$ is larger than both H and O thus this single bond is largest among the given options.

$\mathrm{O}_2^{2+}=14$ electron,

B.O. $\left(\mathrm{O}_2^{2-}\right)=\frac{1}{2}(10-8)=1$;

B.O. $\left(\mathrm{O}_2^{+}\right)=\frac{1}{2}(10-5)=2.5$

B.O. $\left(\mathrm{O}_2^{-}\right)=\frac{1}{2}(10-7)=1.5$;

B.O. $\left(\mathrm{O}_2\right)=\frac{1}{2}(10-6)=2$

Unpaired electron

$ \mathrm{O}_2^{2+}, \mathrm{O}_2^{2-}=\mathrm{O} ; \mathrm{O}_2^{+}, \mathrm{O}_2^{-}=\mathrm{I} ; \mathrm{O}_2=2 $

Sum of B.O. and unpaired electron

$ 3+1+2.5+1.5+2=10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ 0+0+1+1+2=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

The correctly matched geometry and hybridisation is given in, I and II. While III and IV are incorrect. Their correct form is :

$\mathrm{SF}_4=s p^3 d-$ Trigonal bipyramidal

$\mathrm{PbCl}_2 \rightarrow s p^2-\mathrm{V}$ shaped geometry.

The correct order of dipole moment is

$ \mathrm{B}<\mathrm{C}<\mathrm{A} $

$\mathrm{H}_2 \mathrm{O}$ has bent shape with two lone pairs on oxygen is highly electronegative, creating strong $\mathrm{O}-\mathrm{H}$ bond dipoles followed by $\mathrm{NH}_3$ and $\mathrm{CHCl}_3$.

Let's match each molecule with its approximate dipole moment:

1. HCl (Hydrogen Chloride): HCl is a diatomic molecule with a polar covalent bond. Chlorine is significantly more electronegative than hydrogen. This creates a net dipole moment. The experimentally observed dipole moment for HCl is approximately 1.08 D.

   • Matching with List-II, (II) 1.07 D is the closest and most accurate value. So, A - II.

2. NH₃ (Ammonia): NH₃ has a trigonal pyramidal geometry due to the presence of a lone pair on the nitrogen atom and three N-H bonds. Nitrogen is more electronegative than hydrogen, so the N-H bond dipoles point towards nitrogen. The lone pair also contributes significantly to the dipole moment, pointing away from the base of the pyramid. All these contributions add up constructively, resulting in a substantial net dipole moment. The experimentally observed dipole moment for NH₃ is approximately 1.47 D.

   • Matching with List-II, (IV) 1.47 D is a perfect match. So, B - IV.

3. H₂O (Water): H₂O has a bent (V-shaped) geometry due to two lone pairs on the oxygen atom and two O-H bonds. Oxygen is much more electronegative than hydrogen. The O-H bond dipoles point towards oxygen, and these dipoles, along with the lone pair contributions, add up constructively because of the bent shape, resulting in a very large net dipole moment. The experimentally observed dipole moment for H₂O is approximately 1.85 D.

   • Matching with List-II, (I) 1.85 D is a perfect match. So, C - I.

4. NF₃ (Nitrogen Trifluoride): NF₃ also has a trigonal pyramidal geometry, similar to NH₃. However, in NF₃, fluorine is more electronegative than nitrogen. Therefore, the N-F bond dipoles point towards the fluorine atoms, i.e., away from the nitrogen atom. The lone pair on nitrogen contributes a dipole moment pointing away from the base, similar to NH₃. However, the N-F bond dipoles oppose the dipole moment from the lone pair. This leads to a partial cancellation of dipoles, resulting in a much smaller net dipole moment compared to NH₃. The experimentally observed dipole moment for NF₃ is approximately 0.23 D.

   • Matching with List-II, (III) 0.23 D is a perfect match. So, D - III.

Combining the matches:

• (A) HCl - (II) 1.07 D

• (B) NH₃ - (IV) 1.47 D

• (C) H₂O - (I) 1.85 D

• (D) NF₃ - (III) 0.23 D

Now, let's check the given options:

Option A: A-II, B-IV, C-I, D-III. This matches our derived connections.

The final answer is $\boxed{\text{A}}$

Only $\mathrm{PCl}_3$ and $\mathrm{ClF}_3$ are correctly matched. While $\mathrm{SO}_2$ and $\mathrm{SF}_4$ are incorrect. Their correct match is

$ \begin{aligned} & \mathrm{SO}_2=1 \text { lone pair }=s p^2 \\ & \mathrm{SF}_4=1 \text { lone pair }=s p^3 d \end{aligned} $

The number of molecules having lone pair of electrons on central atoms are 5 (five).

They are

$\mathrm{SF}_4, \mathrm{NCl}_3, \mathrm{XeF}_6, \mathrm{XeF}_4, \mathrm{SnCl}_2$

Whereas $\mathrm{BF}_3, \mathrm{SiCl}_4, \mathrm{HgCl}_2$ and $\mathrm{PCl}_5=0$ lone pair

There are 4 Lewis bases in the given molecules.

They are $\mathrm{NH}_3, \mathrm{OH}, \mathrm{CO}, \mathrm{C}_2 \mathrm{H}_4$

Lewis bases are molecules or ions that can donate an electron pair.

$\mathrm{ClF}_3$ and $\mathrm{SF}_4$ has hybridisation as that of xenon (II) flouride. Both has $s p^3 d$ hybridisation.

$\mathrm{N}_2, \mathrm{CO}, \mathrm{NO}^{+}$are isoelectronic species.

Isoelectronic species are those species that have same number of electrons. $\mathrm{N}_2$, $\mathrm{Co}, \mathrm{No}^{+}, \mathrm{Au}$ has 14 electrons.

A molecule with $T$-shape has total-five electrons pairs in the valence shell of its central atom. This consists of three bond pairs and two lone pairs. This is in accordance with VSEPR theory.

Bond order is given by

B.O. $=\frac{1}{2}$ [Number of bonding electrons- number of antibonding electrons]

B.O. of $\mathrm{C}_2=2$

B.O. of $\mathrm{O}_2^{2+}=2.5, x=2+3=5$

B.O. $(a+b+c)=4.5$

$ a=\mathrm{B}_2, b=\mathrm{N}_2, C=\mathrm{F}_2 $

B.O. $=1+3+1=5$

Three molecules has two lone pairs of electrons on central atoms. These are $\mathrm{ClF}_3, \mathrm{XeF}_4$ and $\mathrm{H}_2 \mathrm{O}$ while $\mathrm{SF}_6, \mathrm{BF}_3, \mathrm{PCl}_5$ has 0 lone pair and $\mathrm{BrF}_5, \mathrm{SF}_4$ has 1 lone pair.

The pair of molecules that have the same bond order are $\mathrm{O}_2$ and $\mathrm{C}_2$.

Bond order shows how many chemical bonds are between two atoms.

To find bond order, use this formula:

$ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} $

For $\mathrm{O}_2$:

Number of bonding electrons = 10,
Number of antibonding electrons = 6

$ \text{Bond Order} = \frac{10 - 6}{2} = 2 $

For $\mathrm{C}_2$:

Number of bonding electrons = 8,
Number of antibonding electrons = 4

$ \text{Bond Order} = \frac{8 - 4}{2} = 2 $

So, both $\mathrm{O}_2$ and $\mathrm{C}_2$ have a bond order of 2.

Among the given options, only two molecules/ions has trigonal planar structure.

$\left[\mathrm{BO}_3\right]^{3-}, \mathrm{BCl}_3$ : Trigonal planar $\mathrm{NH}_3, \mathrm{PCl}_3, \mathrm{XeO}_3$ : Trigonal pyramidal $\mathrm{ClF}_3$ : T shape

Both $(\mathrm{A})$ and $(\mathrm{R})$ are correct and $(R)$ is the correct explanation of $(A)$.

$ \text { The reaction } $

$ \mathrm{O}_2^{+} \underset{\mathrm{I}}{\stackrel{ }{\longleftarrow}} \mathrm{O}_2 \underset{\mathrm{II}}{\longrightarrow} \mathrm{O}_2^{-} $

$ \begin{aligned} & \mathrm{O}_2: 16 \text { electrons: Paramagnetic } \\ & \mathrm{O}_2^{+}: 15 \text { electrons (1 removed); } \\ & \text { Paramagnetic } \\ & \mathrm{O}_2^{-}: 17 \text { electrons (1 added) : Paramagnetic } \\ & \mathrm{O}_2^{+}=\sigma \mathrm{ls}^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \\ & \qquad\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^1=\pi^* 2 p_y^0\right) \\ & \qquad \text { B.O. }=\frac{10-5}{2}=2.5 \\ & \mathrm{O}_2^{+}=\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma * 2 s^2 \sigma 2 p_z^2 \\ & \qquad\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^1=\pi^* 2 p_y^1\right) \\ & \text { B.O. }=\frac{10-6}{2}=2 \\ & \mathrm{O}_2^{-}=\sigma l s^2 \sigma * 1 s^2 \sigma 2 s^2 \sigma * 2 s^2 \sigma 2 p_z^2 \\ & \qquad\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^2=\pi * 2 p_y^1\right) \end{aligned} $

$ \begin{aligned} &\text { B.O. }=\frac{10-7}{2}=1.5\\ &\text { Hence, statement } A \text { and } C \text { are correct. } \end{aligned} $

The correct order increasing of lone pair of electron on central atoms are

$ \mathrm{SiH}_4<\mathrm{SF}_4<\mathrm{ClF}_3<\mathrm{XeF}_2 $

zero lone pair one lone pairs two lone pairs 3 lone pair

$ \text { IV }<\text { III }<\text { I }<\text { II } $

$\mathrm{CH}_4, \mathrm{BF}_3, \mathrm{CO}_2$ all has zero dipole moments.

Dipole moments is given by $\mu=\sqrt{n(n+2)}$ B.M. where, $n$ is the number of unpaired electrons on central atom, in $\mathrm{CH}_4$, C has 0 unpaired electrons,

Thus $\mu=\sqrt{0(0+2)}=0$

Same is true for $\mathrm{BF}_3$ and $\mathrm{CO}_2$.

The bond order (BO) is a calculation that indicates the strength and stability of a bond. It is determined using the formula:

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

For calculating the bond order of different diatomic molecules:

For $\text{C}_2$:

The electronic configuration is:

$ \left[\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 \pi 2p_y^2\right] $

Number of bonding electrons ($N_b$) = 8

Number of antibonding electrons ($N_a$) = 4

Using the formula:

$ \text{BO} = \frac{1}{2}[8 - 4] = 2 $

For $\text{O}_2$:

The electronic configuration is:

$ \left[\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^* 2p_x^1 \pi^* 2p_y^1\right] $

Number of bonding electrons ($N_b$) = 10

Number of antibonding electrons ($N_a$) = 6

Using the formula:

$ \text{BO} = \frac{1}{2}[10 - 6] = 2 $

Both $\text{C}_2$ and $\text{O}_2$ have the same bond order of 2, indicating that both molecules have similar bond strength and stability.

Here’s how the three hydrides classify:

Electron‐deficient hydrides

– Need multicenter (3c–2e) bonding because there aren’t enough electrons for simple 2-center bonds.

– Example: $\mathrm{B_2H_6}$

Electron‐precise hydrides

– Have exactly the right number of electrons for one 2-center-2-electron bond per A–H link.

– Examples: $\mathrm{CH_4},\;\mathrm{SiH_4}\,. $

Electron‐rich hydrides

– Carry extra lone‐pair electrons on the central atom beyond those used in A–H bonds.

– Examples: $\mathrm{NH_3},\;\mathrm{H_2O},\;\mathrm{PH_3},\ldots$

Checking your list:

• (i) $\mathrm{B_2H_6}$ – electron‐deficient ✓

• (ii) $\mathrm{NH_3}$ – actually electron‐rich (it has a lone pair), so “electron‐precise” is wrong

• (iii) $\mathrm{H_2O}$ – electron‐rich ✓

Thus only (i) and (iii) are correctly matched ⇒ Option A.

$\mathrm{SnF}_4$ is ionic in nature based on Fajan's rule. There is large differerce in size as well as electronegativity of the atoms, which make the bond ionic between Sn and F .

$\mathrm{PbF}_4$ is also ionic in nature based on Fajan's rule. According to Fajan's rule, an ionic bond is formed by a compound with low positive charge, a large cation and a small anion.

$\mathrm{GeF}_4$ is covalent.

$ \mathrm{BrF}_5 \longrightarrow s p^2 d^2 $

The decomposition of ammonium nitrate is,

$ \mathrm{NH}_4 \mathrm{NO}_3 \longrightarrow \mathrm{~N}_2 \mathrm{O}+2 \mathrm{H}_2 \mathrm{O} $

$\mathrm{N}_2 \mathrm{O}$ is colourless neutral gas.

The complete bond cleavage reaction is,

$ \begin{aligned} &\text { (Negative charge is more on } \mathrm{CH}_3 \mathrm{CH}_2 \text {.) }\\ &\therefore A=\mathrm{CH}_3 \mathrm{CH}_2^{+}, B=\mathrm{I}^{-}, \mathrm{C}=\mathrm{CH}_3 \mathrm{CH}_2^{-}, D=\mathrm{Cu}^{+} \end{aligned} $

Basicity of an acid is the number of replaceable hydrogen in the acid.

For the given reaction below, the hybridisaion of central atom does not change.

$ \begin{aligned} & \mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+} \\ & s p^3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s p^3 \\ & \underset{s p^2}{\mathrm{BF}_3}+\mathrm{F} \longrightarrow \underset{s p^3}{\mathrm{BF}_4^{-}} \\ & \mathrm{PCl}_3+\mathrm{Cl}_2 \longrightarrow \mathrm{PCl}_5 \\ & s p^3 \quad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,s p^3 d \\ & \mathrm{ClF}_3+\mathrm{F}_2 \longrightarrow \mathrm{ClF}_5 \\ & s p^3 d \quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, s p^3 d^2 \end{aligned} $

Explanation of Bond Order Calculation

Bond Order (B.O.) is calculated using the formula:

$ \text{Bond Order} = \frac{1}{2} (\text{number of electrons in bonding orbitals} - \text{number of electrons in antibonding orbitals}) $

Let's analyze each set:

Set (a): $\mathrm{B}_2, \mathrm{CN}^-, \mathrm{O}_2^{2-}$

$\mathrm{B}_2$:

$ \text{Number of bonding electrons} = 4, \text{Number of antibonding electrons} = 2 $

$ \text{B.O.} = \frac{1}{2}(4-2) = 1 $

$\mathrm{CN}^-$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 4 $

$ \text{B.O.} = \frac{1}{2}(10-4) = 3 $

$\mathrm{O}_2^{2-}$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 8 $

$ \text{B.O.} = \frac{1}{2}(10-8) = 1 $

Total Bond Order for Set (a):

$ 1 + 3 + 1 = 5 $

Set (b): $\mathrm{O}_2, \mathrm{F}_2, \mathrm{O}_2^{2+}$

$\mathrm{O}_2$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 6 $

$ \text{B.O.} = \frac{1}{2}(10-6) = 2 $

$\mathrm{F}_2$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 8 $

$ \text{B.O.} = \frac{1}{2}(10-8) = 1 $

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

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 4 $

$ \text{B.O.} = \frac{1}{2}(10-4) = 3 $

Total Bond Order for Set (b):

$ 2 + 1 + 3 = 6 $

Set (c): $\mathrm{O}_2^-, \mathrm{N}_2, \mathrm{O}_2^+$

$\mathrm{O}_2^-$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 7 $

$ \text{B.O.} = \frac{1}{2}(10-7) = 1.5 $

$\mathrm{N}_2$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 4 $

$ \text{B.O.} = \frac{1}{2}(10-4) = 3 $

$\mathrm{O}_2^+$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 5 $

$ \text{B.O.} = \frac{1}{2}(10-5) = 2.5 $

Total Bond Order for Set (c):

$ 1.5 + 3 + 2.5 = 7 $

Set (d): $\mathrm{C}_2, \mathrm{O}_2, \mathrm{He}_2^{2+}$

$\mathrm{C}_2$:

$ \text{Number of bonding electrons} = 8, \text{Number of antibonding electrons} = 4 $

$ \text{B.O.} = \frac{1}{2}(8-4) = 2 $

$\mathrm{O}_2$:

$ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 6 $

$ \text{B.O.} = \frac{1}{2}(10-6) = 2 $

$\mathrm{He}_2^{2+}$:

$ \text{Number of bonding electrons} = 2, \text{Number of antibonding electrons} = 0 $

$ \text{B.O.} = \frac{1}{2}(2-0) = 1 $

Total Bond Order for Set (d):

$ 2 + 2 + 1 = 5 $

Conclusion

Set (c) has the maximum total bond order sum of $7$.

Diborane - $\mathrm{B}_2 \mathrm{H}_6$

Each boron atom is $s p^3$ hybridised.

$ \theta_1=97^{\circ} \text { and } \theta_2=120^{\circ} $

$\mathrm{H}_t$ is terminal hydrogen atom and two boron atoms lie in the same plane.

The correct order of covalent bond character is:

$ \mathrm{LiCl}<\mathrm{BeCl}_2<\mathrm{BCl}_3<\mathrm{CCl}_4 $

According to Fajan's rules, the covalent character of a bond increases with higher charge on the cation. In this sequence, as the charge on the cation increases from LiCl to $\mathrm{CCl}_4$, the covalent character of the bonds also increases.

Dipole moment, mathematically given by, $\mu=\sqrt{n(n+2)}$ BM only occur when there is unpaired electron present in the ion/molecules.

For $\mathrm{CO}_2$ and $\mathrm{BCl}_3$ there is no unpaired electrons.

In $\mathrm{BCl}_3$ and $\mathrm{NF}_3$ pair only $\mathrm{NF}_3$ has dipole' moment due to present of unpaired electron. ( 0.24 D )

In case of $\mathrm{SO}_2$ and $\mathrm{NF}_3$ both have unpaired electrons and hence, both have dipole moments, 1.60 D and 0.24 D respectively.

Statement given in (i), (iii) and (iv) are correct, while (ii) is incorrect. It's correct from is, Covalent character is the measure of the tendency of an atom to attract the shared pair of electron towards itself. LiF has more covalent character than KF because LiF has more electronegativity difference. Electronegativity difference is directly proportional to covalent character.

The explanation involves analyzing the order of various properties among different molecules, and identifying which sequences are correct.

For sets (ii), (iii), and (iv), the sequences provided are correct:

(ii) Dipole Moment: The order $ \mathrm{H}_2\mathrm{O}>\mathrm{NH}_3>\mathrm{H}_2\mathrm{S} $ is correct due to the polarity and shapes of the molecules, with water having the highest dipole moment.

(iii) Bond Enthalpy: The sequence $ \mathrm{N}_2>\mathrm{O}_2>\mathrm{H}_2 $ is accurate because nitrogen gas has a triple bond, making it much stronger compared to the double bond in oxygen and the single bond in hydrogen.

(iv) Bond Order: The order $ \mathrm{NO}^{+}>\mathrm{O}_2>\mathrm{O}_2^{2-} $ correctly reflects the number of bonds between the atoms, with $ \mathrm{NO}^{+} $ having the highest bond order.

However, for set (i), the provided order for bond angles is incorrect. The correct order should be:

(i) Bond Angle: $ \mathrm{H}_2\mathrm{O}<\mathrm{NH}_3<\mathrm{SO}_2 $, with numerical values $ 104.5^\circ<107^\circ<119^\circ $. This is based on the interplay of lone pair repulsion and the geometric configurations of the molecules.

In the reaction $ \mathrm{XeF}_6 + \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{XeOF}_4 + 2 \mathrm{HF} $, the hybridization of the xenon (Xe) atom changes.

For $\mathrm{XeF}_6$:

Calculating hybridization using the formula:

$ \text{Hybridization} = \frac{1}{2}[V + M - C + A] $

where:

$ V $ is the number of valence electrons of the central atom (Xe), which is 8.

$ M $ is the number of monovalent atoms attached, here 6 fluorine atoms.

$ C $ is the charge on the cation.

$ A $ is the charge on the anion.

$ = \frac{1}{2}[8 + 6 - 0 + 0] = 7 $

The hybridization for $\mathrm{XeF}_6$ is $ sp^3d^3 $.

For $\mathrm{XeOF}_4$:

The hybridization number is calculated to be 6, resulting in a hybridization of $ sp^3d^2 $.

Thus, the xenon atom goes through a change in hybridization from $ sp^3d^3 $ in $\mathrm{XeF}_6$ to $ sp^3d^2 $ in $\mathrm{XeOF}_4$.

Isostructural species have the same shape or structure. Here's how to determine the hybridization and shape of each species:

For $\mathrm{XeO}_3$:

Hybridization formula: $\frac{1}{2}[V + M - C + A]$

$V =$ Valence electrons of the central atom (Xe), which is 8

$M =$ Number of monovalent atoms attached, which is 0

$C =$ Positive charge, which is 0

$A =$ Negative charge, which is 0

$ = \frac{1}{2}[8 + 0 - 0 + 0] = 4 = \text{sp}^3 $

Xe has one lone pair, resulting in a pyramidal shape.

For $\mathrm{H}_3\mathrm{O}^+$:

Apply the same hybridization formula:

$ = \frac{1}{2}[6 + 3 - 1 + 0] = 4 = \text{sp}^3 $

The structure is pyramidal.

For $\mathrm{SO}_3$:

$ = \frac{1}{2}[6 + 0 - 0 + 0] = 3 = \text{sp}^2 $

Results in a trigonal planar shape.

For $\mathrm{CO}_3^{2-}$:

$ = \frac{1}{2}[4 + 0 - 0 + 2] = 3 = \text{sp}^2 $

Also results in a trigonal planar shape.

Conclusion:

$\mathrm{XeO}_3$ and $\mathrm{H}_3\mathrm{O}^+$ both have a pyramidal shape.

$\mathrm{SO}_3$ and $\mathrm{CO}_3^{2-}$ both have a trigonal planar shape.

These alignments help identify which species are isostructural pairs based on their shapes.

$ \begin{aligned} &\text { The correct order of bond angle is }\\ &\mathrm{HgCl}_2>\mathrm{SO}_3>\mathrm{SiCl}_4>\mathrm{NH}_3 \end{aligned} $

The formula of formal charge is

$ \mathrm{FC}=V-N-\frac{B}{2} $

$V=$ Number of valence electrons of neutral atom.

$N=$ Number of non-bonding electrons on this atom.

$B=$ Number of electrons shared in bond with other atom.

For $1 \mathrm{st}, \mathrm{FC}=6-4-\frac{1}{2} \times 4=0$

For 2 nd, FC $=5-2-\frac{1}{2} \times 6=0$

For 3 rd, FC $=6-6-\frac{1}{2} \times 2=-1$

Hence, the formal charges are $0,0,-1$.

$\mathrm{Al}_2 \mathrm{Cl}_6$ (Dimer) is a $s p^3$ hybridised and tetrahedral at aluminum.

$ b_1=118^{\circ}, b_2=79^{\circ}, b_3=101^{\circ} $

Hence, $\mathrm{XeF}_6$ and $\mathrm{IF}_7$ have same hybridisation of their central atoms.

The correct order is $\mathrm{SO}_3<\mathrm{SnCl}_2<\mathrm{ClF}_3<\mathrm{XeF}_2$.

Bond length/distance is defined as average distance between niaclei of two bonded atom in a molecules.

Bond length $=r_1+r_2$

Element $(x)=143$ pm means it is close to bond length of fluorine $\left(\mathrm{F}_2\right)$.

Atomic number $=9$

Element $(Y)=110 \mathrm{pm}$ is equal to bond length of nitrogen $\left(N_2\right)$.

Atomic number $=7$

Element $(Z)=121$ pm is equal to bond length of axygen $\left(\mathrm{O}_2\right)$.

Atomic number $=8$

A formal charge is the charge assigned to an atom in a molecule in the covalent view of bonding assuming that electrons in all chemical bonds are shared equally between atoms regardless of relative electronegativity.

It is the difference between an atoms number of valence electrons in its neutral state and the number allocated to that atom in a Lewis structure.

$ Q_f=v-(V+B) $

where, $V$ is sumber of valence electron in free atom.

$U=$ Number of unshared electron in atom.

$B=$ Number of bond around the atom.

The correct match is

A-II, B-III, C-I, D-IV

Bond enthalpy also known as bond dissociation energy describes the amount of energy stored in a bond between atoms in a molecules.

Specifically it is the energy required to be added for cleavage of bond in a gas phase.

$ \begin{aligned} & \mathrm{Si}-\mathrm{Si} \longrightarrow 297 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{C}-\mathrm{C} \longrightarrow 348 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{Sn}-\mathrm{Sn} \longrightarrow 248 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{Ge}-\mathrm{Ge} \longrightarrow 260 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Down the group, bond enthalpy decreases.

$\begin{aligned} & \text { FC }(\text { oxygen } 1)=6-6-\frac{1}{2} \times 2=-1 \\ & \text { FC }(\text { oxygen } 2)=6-4-\frac{1}{2} \times 4=6-6=0 \\ & \text { FC }(\text { oxygen } 3)=6-6-\frac{1}{2} \times 2=-1 \end{aligned}$

Bond order $(\mathrm{BO})=\frac{1}{2}$ (number of electron in bonding MO's $-$ number of electron in anti-bonding MO's)

$\mathrm{C}_2^{2-}$ : Total number of electrons $=6+6+2=14$

Electronic configuration

$\begin{aligned} & \text { Electronic configuration } \\ & \sigma\left(1 s^2\right), \sigma\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^2\left(2 s^2\right), \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2 \\ & \mathrm{BO}=\frac{1}{2} \times(10-4)=\frac{1}{2} \times 6=3 \end{aligned}$

$\mathbf{N}_{\mathbf{2}}$ : Total number of electrons $=7+7=14$

$\begin{aligned} & \sigma\left(1 s^2\right), \sigma^{\circ}\left(\left(s^2\right), \sigma\left(2 s^2\right), \sigma^*\left(2 s^2\right), \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\right. \\ & \mathrm{BO}=\frac{1}{2}(10-4)=\frac{6}{2}=3 \end{aligned}$

$\begin{aligned} & \begin{array}{l} \mathrm{O}_2^{2-}: \text { Total number of electron }=8+8+2=18 \\ \sigma\left(l s^2\right), \sigma^{\prime}\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^{\prime}\left(2 s^2\right), \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2, \\ \pi^2 2 p_x^2=\pi^{\prime} 2 p_y^2 \end{array} \\ & \begin{aligned} \mathrm{BO} & =\frac{1}{2}(10-8) \\ & =\frac{1}{2} \times 2=1 \end{aligned} \end{aligned}$

Option (a) is incorrect.

$\begin{aligned} & \mathrm{O}_2^{+} \text {Total number of electrons }=8+8-1=15 \\ & \sigma\left(s^2\right), \sigma^*\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^*\left(2 s^2\right), \sigma\left(2 p_z^2\right), \pi\left(2 p_x^2\right) \\ & =\pi\left(2 p_y^2\right), \pi^{\cdot} 2 p_x^1 \\ \end{aligned}$

$\mathrm{BO}=\frac{1}{2}(10-5)=\frac{1}{2} \times 5=2.5$

$\mathrm{O}_2^{-}$ Total number of electrons $=8+8+1=17$

$\sigma\left(1 s^2\right), \sigma^*\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^2\left(2 s^2\right), \sigma\left(2 p_z^2\right), \pi\left(2 p_x^2\right)$

$\begin{aligned} & \sigma\left(1 s^2\right), \sigma\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma\left(2 s^2\right), \sigma\left(2 p_z^2, \pi\left(22_x^2\right)\right. \\ & =\pi\left(2 p_y^2\right), \pi^*\left(2 p_x^2\right)=\pi^*\left(2 p_y^1\right) \\ & \mathrm{BO}=\frac{1}{2}(10-7)=\frac{1}{2} \times 3=1.5 \end{aligned}$

$\mathbf{N}_2^{+}$ Total number of electron $=7+7-1=13$

$\begin{aligned} & \mathbf{N}_2 \text { Total number } \\ & \sigma\left(1 s^2\right), \sigma\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma\left(2 s^2\right), \pi\left(2 p_x^2\right)=\pi\left(2 p_y^2\right), \sigma\left(2 p_z^1\right) \\ & \mathrm{BO}=\frac{1}{2}(9-4)=\frac{1}{2} \times 5=2.5 \end{aligned}$

Option (b) is correct.

Similarly, bond order of $\mathrm{O}_2=\frac{1}{2}(10-6)=\frac{1}{2} \times 4=2$

Bond order of $\mathrm{Li}_2=\frac{1}{2}(4-2)=1$

Bond order of $\mathrm{H}_2^{+}=\frac{1}{2}(1-0)=0.5$

Bond order of $C_2=\frac{1}{2}(8-4)=\frac{1}{2} \times 4=2$

Rest other options are incorrect, as they don't have bond order in fraction.

Lewis dot structure of $\mathrm{SF}_6$ is Central atom, S has total 12 electrons. So, it does not obey octet rule.

Lewis dot structure of $\mathrm{PCl}_5$ is central atom, P has total 10 electrons. So, it also does not obey octet rule.

Lewis dot structure of $\mathrm{XeF}_2$ is central atom, Xe has total 10 electrons. So, it does not obey octet rule. Hence, the set of molecules that does not obey octet rule-is $\mathrm{SF}_6, \mathrm{PCl}_5$ and $\mathrm{XeF}_2$.

Formal charge (FC) can be calculated by using formula : $\mathrm{FC}=$ (number of electrons in neutral atom) $-\frac{1}{2}$ (number of electrons in covalent bonds) - (number of electrons in lone pairs)

Formal charge on 1st atom i.e., oxygen is

$\mathrm{FC}=6-\frac{1}{2} \times 4-4=6-2-4=0$

Formal charge on 2 nd atom i.e., nitrogen is

$\mathrm{FC}=5-\frac{1}{2} \times 8-0=5-4=+1$

Formal charge on 3rd atom i.e., oxygen is

$\mathrm{FC}=6-\frac{1}{2} \times 4-4=6-2-4=0$

Hence, formal charge of atoms 1, 2 and 3 in the ion $\mathrm{{[\mathop O\limits_1 = \mathop N\limits_2 = \mathop O\limits_3 ]^ + }}$ is 0, +1 and 0 respectively.

In graphite, each carbon atom combines with three other carbon atoms with three sigma bonds. So, hybridisation is $s p^2$. In diamond, each carbon atom combines with four other carbon atoms with four sigma bonds. So, hybridisation is $s p^3$. In fullerene, hybridisation of carbon atoms is $s p^2$. Hence, the hybridisation of carbon in graphite, diamond and fullerene, $\mathrm{C}_{60}$ are $s p^2, s p^3$ and $s p^2$ respectively.

According to Fajan's rule; covalent character in ionic compound depends on

(i) size and charge of cation

(ii) size and charge of anion

(iii) pseudo-inert gas configuration

(a) Size of iodide $(\mathrm{I}^{-})$ is more than $\mathrm{F}^{-}$, so $\mathrm{I}^{-}$ has more polarisibility. Thus, KI is more covalent than KF . So, option (a) is incorrect.

(b) Charge on Sn in $\mathrm{SnCl}_4$ is +4 and in $\mathrm{SnCl}_2$ is +2 $\mathrm{So}, \mathrm{Sn}$ has more polarising power in $\mathrm{SnCl}_4$. Thus, $\mathrm{SnCl}_4$ is more covalent than $\mathrm{SnCl}_2$. So, option (b) is incorrect.

(c) Size of $\mathrm{Li}^{+}$ is less than $\mathrm{K}^{+}$. So, it has more polarising power. Thus, LiF is more covalent than KF. Hence, option (c) is correct.

(d) Charge on Zn is +2 and on Na is +1 . So, Zn has more polarising power. Thus, $\mathrm{ZnCl}_2$ is more covalent than NaCl . Hence, option (d) is incorrect.