Question

The correct order of bond angles is :

• A. $H_{2}S<N{H}_3<B{F}_3<Si{H}_4$
• B. $N{H}_3<H_{2}S<Si{H}_4<B{F}_3$
• C. $H_{2}S<N{H}_3<Si{H}_4<B{F}_3$
• D. $H_{2}S<Si{H}_4<N{H}_3<B{F}_3$

Answer 1

Answer: (C)

$H_{2}S<N{H}_3<Si{H}_4<B{F}_3$

The bond angles are as shown below.
$H_{2}SN{H}_{3}Si{H}_{4}B{F}_3$
${92.6}^{\circ }{107}^{\circ }{109}^{\circ }28{120}^{\circ }$
Thus, the correct order of the bond angles is $H_{2}S<N{H}_3<Si{H}_4<B{F}_3$.
$B{F}_3$ undergoes $s{p}^2$ hybridization which results in bond angle of 120${}^0$. $N{H}_3$ and $Si{H}_4$ undergoes $s{p}^3$ hybridization which results in bond angle around ${109}^0$ . For ammonia, due to lone pair bond pair repulsion, the bond angle decreases to ${107}^0$.
In case of $H_{2}S$, the hybridization is $s{p}^3$ but due to lone pair - lone pair repulsion bond angle is ${92.6}^0$.