Question

The correct order of bond angles (smallest first) 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

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

Let's, try to determine the geometry and structure of each compound.
$H_{2}S\Rightarrow ep=\frac{V+L+A-C}{2}=\frac{6+2}{2}=4\Rightarrow Tetrahedral$

$N{H}_3\Rightarrow ep=\frac{V+L+A-C}{2}=\frac{5+3}{2}=4\Rightarrow Tetrahedral$

$Si{H}_4\Rightarrow ep=\frac{4+4}{2}=4\Rightarrow Tetrahedral$

$B{F}_3\Rightarrow ep=\frac{3+3}{2}=3\Rightarrow TrigonalPlanar$

$H_{2}S\Rightarrow$

$N{H}_3\Rightarrow$

$Si{H}_4\Rightarrow$

$B{F}_3\Rightarrow$

In trigonal Planar, the bond angle is ${120}^o$. The angles for the other compounds correspond to:
$N{H}_3\Rightarrow Around{107}^o$
$H_{2}S\Rightarrow Around{104.5}^o$
$Si{H}_4\Rightarrow Around{109.5}^o$
Therefore, the order will be
$H_{2}S<N{H}_3<Si{H}_4<B{F}_3$