Question

The correct order of bond angle in $H_{2}S$, $N{H}_3$, $B{F}_3$ and $Si{H}_4$ is :

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

Answer 1

The correct option is A $H_{2}S<N{H}_3<Si{H}_4<B{F}_3$

A) $B{F}_3$ has trigonal planar geometry with 3 bond pairs. The bond angle is ${120}^o$.

B) $Si{H}_4$ is tetrahedral in shape with 4 bond pairs. The bond angle is ${109.28}^o$.

C) $N{H}_3$ is pyramidal in shape with 3 bond pairs and one lone pair. The bond angle is ${107.5}^o$.

D) $H_{2}S$ is bent shaped with 2 bond pairs and two lone pairs. The bond angle is ${92.3}^o$.

$\underset{{92.3}^{\circ }}{H_{2}S}<\underset{{107.5}^{\circ }}{N{H}_3}<\underset{{109}^{\circ }{28}^{\prime }}{Si{H}_4}<\underset{{120}^{\circ }}{B{F}_3}$

Hence, option A is correct.