Question

The correct order of bond angles (smallest first) for $H_{2}S,N{H}_3,B{F}_3$ 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 $H_{2}S<N{H}_3<Si{H}_4<B{F}_3$

$B{F}_3$ is $s{p}^2$ hybridised with 3 bond pairs and hence, the bond angle is ${120}^{\circ }$.

$Si{H}_4$ and $N{H}_3$ are $s{p}^3$ hybridised. $N{H}_3$ has one lone pair whereas $Si{H}_4$ has none. So, the bond angle of $N{H}_3$ is smaller than that of $Si{H}_4$ because, the lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion.

If the central atom belongs to the groups 13, 14, 15 or 16 (and is from third or higher period) and is surrounded by atoms having electronegativity less than 2.5, it does not involve hybridization. The lone pair of electrons are present in the pure s-orbital of the atom. Hence, the bond angle is around ${90}^o$ for such compounds. They are called Drago's molecules.

$H_{2}S$ is drago's molecule and does not involve hybridization. Hence, the bond angle is ${92.6}^{\circ }$
Therefore, the correct order of bond angle is:
$H_{2}S<N{H}_3<Si{H}_4<B{F}_3$
${92.6}^{\circ }<{107}^{\circ }<{109}^{\circ }{28}^{\prime }<{120}^{\circ }$