Question

The increasing order of bond angles 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 $H_{2}S<N{H}_3<Si{H}_4<B{F}_3$

The repulsion order is: $lonepair-lonepair>bondpair-lonepair>bondpair-bondpair$


$H_{2}S$ - has 2 lone pair of electrons and has a bent shape.



$N{H}_3$ - $s{p}^3$ hybridised, pyramidal shape


$Si{H}_4$ - $s{p}^3$ hybridised, tetrahedral shape


$B{F}_3$ - $s{p}^2$ hybridised, trigonal planar shape