Question

The correct order of bond angle(smallest first) in $H_{2}S,N{H}_3,B{F}_3$ and $Si{H}_4$ 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

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

Correct order bond angle is $H_{2}S<{NH}_3<{SiH}_4<{BF}_3$.

This is because of the repulsion between the lone pair-lone pair and bond pair-lone pair.