Question

In which of the following pairs, first compound has larger bond angle than another?

• A. $H_{2}O$ and $N{H}_3$
• B. $S{F}_2$ and $Be{F}_2$
• C. $B{F}_3$ and $B{F}_4^{-}$
• D. $N{H}_3$ and $N{F}_3$

Answer 1

Answer: (C), (D)

$B{F}_3$ and $B{F}_4^{-}$
$N{H}_3$ and $N{F}_3$

Concept involved:
As the lone pairs on central atom increase bond angle decreases.
A) $H_{2}O$ and $N{H}_3$ both are possessing $s{p}^3$ hybridization (109.5 ) but $H_{2}O$ have two lone pairs on central atom and $N{H}_3$ have one lone pair on central atom. As the number of lone pairs are more in $H_{2}O$, $H_{2}O$ posses less bond angle and $N{H}_3$ posses more bond angle. Hence $N{H}_3$ > $H_{2}O$.
B) $S{F}_2$ and $Be{F}_2$ possessing $s{p}^3$ (109.5 ) and sp (180) respectively, hence $Be{F}_2$ has more bond angle. Hence $Be{F}_2$ > $S{F}_2$.
C) $B{F}_3$ and $B{F}^{4-}$ possessing $s{p}^2$ (120) and $s{p}^3$ (109.5) hybridization respectively, hence $B{F}_3$ is having more bond angle. Hence $B{F}_3$ > $B{F}^{4-}$.
D) $N{H}_3$ and $N{F}_3$ both are possessing $s{p}^3$ hybridization (105.9), but due to more repulsions are present between florine atoms of $N{F}_3$ (as flourine is more electronegative), $N{F}_3$ is having less bond angle, hence $N{H}_3$ is having more bond angle. Hence $N{H}_3$ > $N{F}_3$.