Question

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

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

Answer 1

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

The correct order = increasing order of Bond angle.

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

$H_{2}S,N{H}_3,Si{H}_4$ are $s{p}^3$ hybridized with 2lp, 1lp, 0lp respectively.

Greater the number of lone pairs, smaller is the bond angle due to lp-bp repulsions.