Chemical Bonding & Molecular Structure

$\mathrm{XeF}_2 \Rightarrow 2$ sigma bonds and 3 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=5, \mathrm{sp}^3 \mathrm{~d}$ hybridisation , geometry will be trigonal bipyramidal.

$\mathrm{P}-5$

$\mathrm{XeF}_4 \Rightarrow 4$ sigma bonds and 2 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=6, \mathrm{sp}^3 \mathrm{~d}^2$ hybridisation , geometry will be octahedral.

Q-3

$\mathrm{XeO}_3 \Rightarrow 3$ sigma bonds and 1 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=4, \mathrm{sp}^3$ hybridisation , geometry will be tetrahedral.

R-2

$\mathrm{XeO}_3 \mathrm{F}_2 \Rightarrow 5$ sigma bonds and 0 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=5, \mathrm{sp}^3 \mathrm{~d}$ hybridisation , geometry will be trigonal bipyramidal.

S-4

The octet rule states that atoms tend to combine in such a way that they each have eight electrons in their valence shell, giving them the same electronic configuration as a noble gas. Let's examine each option to determine which molecules follow the octet rule.

Option A: $\mathrm{CO}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{NO}$ and $\mathrm{HCl}$

1. $\mathrm{CO}_2$: Carbon dioxide follows the octet rule. Carbon forms double bonds with two oxygen atoms, allowing it to have 8 electrons in its valence shell.

2. $\mathrm{C}_2 \mathrm{H}_4$: Ethylene follows the octet rule. Each carbon forms a double bond with the other carbon and single bonds with two hydrogen atoms, resulting in 8 electrons in the valence shell of the carbons.

3. $\mathrm{NO}$: Nitric oxide does not follow the octet rule. It has an odd number of electrons, making it a free radical.

4. $\mathrm{HCl}$: Hydrogen chloride follows the octet rule. Chlorine has 8 electrons in its valence shell when bonded with hydrogen.

Thus, three molecules $\mathrm{CO}_2$, $\mathrm{C}_2 \mathrm{H}_4$, and $\mathrm{HCl}$ follow the octet rule here.

Option B: $\mathrm{NO}_2, \mathrm{O}_3, \mathrm{HCl}$ and $\mathrm{H}_2 \mathrm{SO}_4$

1. $\mathrm{NO}_2$: Nitrogen dioxide does not follow the octet rule. It has an odd number of electrons.

2. $\mathrm{O}_3$: Ozone follows the octet rule. It has a resonance structure that allows each oxygen atom to have 8 electrons in its valence shell.

3. $\mathrm{HCl}$: As mentioned, hydrogen chloride follows the octet rule.

4. $\mathrm{H}_2 \mathrm{SO}_4$: Sulfuric acid follows the octet rule. All atoms achieve stable configurations through bonding.

Thus, three molecules $\mathrm{O}_3$, $\mathrm{HCl}$, and $\mathrm{H}_2 \mathrm{SO}_4$ follow the octet rule in this option.

Option C: $\mathrm{BCl}_3, \mathrm{NO}, \mathrm{NO}_2$ and $\mathrm{H}_2 \mathrm{SO}_4$

1. $\mathrm{BCl}_3$: Boron trichloride does not follow the octet rule. Boron has only 6 electrons in its valence shell.

2. $\mathrm{NO}$: As mentioned, nitric oxide does not follow the octet rule.

3. $\mathrm{NO}_2$: As mentioned, nitrogen dioxide does not follow the octet rule.

4. $\mathrm{H}_2 \mathrm{SO}_4$: Sulfuric acid follows the octet rule.

Thus, only one molecule, $\mathrm{H}_2 \mathrm{SO}_4$, follows the octet rule in this option.

Option D: $\mathrm{CO}_2, \mathrm{BCl}_3, \mathrm{O}_3$ and $\mathrm{C}_2 \mathrm{H}_4$

1. $\mathrm{CO}_2$: As mentioned, carbon dioxide follows the octet rule.

2. $\mathrm{BCl}_3$: As mentioned, boron trichloride does not follow the octet rule.

3. $\mathrm{O}_3$: As mentioned, ozone follows the octet rule.

4. $\mathrm{C}_2 \mathrm{H}_4$: As mentioned, ethylene follows the octet rule.

Thus, three molecules $\mathrm{CO}_2$, $\mathrm{O}_3$, and $\mathrm{C}_2 \mathrm{H}_4$ follow the octet rule in this option.

Conclusion: The correct options are Option A and Option D, both having at least three molecules that follow the octet rule.

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 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.