Question
The correct order of bond angles (smallest first) in $${H_2}S,\,N{H_3},B{F_3}\,{\text{and}}\,Si{H_4}$$ 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 :
$${H_2}S < N{H_3} < Si{H_4} < B{F_3}$$
Solution :
\[\begin{align}
& \text{The order of bond angles} \\
& \begin{matrix}
B{{F}_{3}} \\
{{120}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
> \\
{} \\
\end{matrix}\begin{matrix}
Si{{H}_{4}} \\
{{109}^{\circ }}28' \\
\end{matrix}\begin{matrix}
> \\
{} \\
\end{matrix}\begin{matrix}
N{{H}_{3}} \\
{{107}^{\circ }} \\
\end{matrix}\,\,\begin{matrix}
> \\
{} \\
\end{matrix}\begin{matrix}
{{H}_{2}}S \\
{{92.5}^{\circ }} \\
\end{matrix} \\
\end{align}\]