Question

The incorrect order of bond angle is :

• A. $\text{ClO}_2 > \text{OCl}_2$
• B. $\text{NO}_2^- < \text{NO}_2$
• C. $\dot{\text{C}}\text{H}_3 > \dot{\text{C}}\text{F}_3$
• D. $\text{O(SiH}_3)_2 < \text{ClO}_2$

Answer 1

Answer: (D)

$\text{O(SiH}_3)_2 < \text{ClO}_2$

Answer 2

1. $\text{ClO}_2$: $sp^2$ hybridized, 2 bonding pairs, 1 lone pair, 1 unpaired electron. Bond angle ~111°.
2. $\text{OCl}_2$: $sp^3$ hybridized, 2 bonding pairs, 2 lone pairs. Bond angle < 109.5°, ~110.7°. Thus, $\text{ClO}_2 > \text{OCl}_2$.
3. $\text{NO}_2^-$: $sp^2$ hybridized, 2 bonding pairs, 1 lone pair. Bond angle ~115°.
4. $\text{NO}_2$: $sp^2$ hybridized, 2 bonding pairs, 1 lone pair, 1 unpaired electron. Unpaired electron is less repulsive than a lone pair. Bond angle ~134°. Thus, $\text{NO}_2^- < \text{NO}_2$.
5. $\dot{\text{C}}\text{H}_3$: $sp^2$ hybridized, 3 bonding pairs, 1 unpaired electron. Bond angle ~119°.
6. $\dot{\text{C}}\text{F}_3$: $sp^2$ hybridized, 3 bonding pairs, 1 unpaired electron. Electronegativity of F and back-bonding reduce bond angle ~108-110°. Thus, $\dot{\text{C}}\text{H}_3 > \dot{\text{C}}\text{F}_3$.
7. $\text{O(SiH}_3)_2$: $sp^3$ hybridized, 2 bonding pairs, 2 lone pairs. Back-bonding with Si increases bond angle to ~140-150°.
8. Comparing $\text{O(SiH}_3)_2$ (~140-150°) and $\text{ClO}_2$ (~111°), the order $\text{O(SiH}_3)_2 < \text{ClO}_2$ is incorrect.