Question

Among the following molecules which one have smallest bond angle-

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Let's, try to determine the geometry and structure of each compound.
$N{H}_3\Rightarrow ep=\frac{V+L+A-C}{2}=\frac{5+3}{2}=4\Rightarrow Tetrahedral$

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

$H_{@}O\Rightarrow ep=\frac{6+2}{2}=4\Rightarrow Tetrahedral$

$H_{2}Se\Rightarrow ep=\frac{6+2}{2}=4\Rightarrow Tetrahedral$

$N{H}_3\Rightarrow$

$H_{2}O\Rightarrow$

$P{H}_3\Rightarrow$

$H_{2}Se\Rightarrow$

$\to$ Always keep in mind that, one lone pair reduces the angle by around ${2.5}^o.$

Generally, tetrahedral has ${109.5}^o$ angle. But because of lone pair and bond pair repulsions, the bond angle will decreases.
In $N{H}_3\Rightarrow Around{107}^o$

$P{H}_3\Rightarrow Around{107}^o$

$H_{2}O\Rightarrow Around{104.5}^o$[Two lone pairs]

$H_{2}Se\Rightarrow Around{104.5}^o$[Two lone pairs]

However, in practice, the bond angle is the least for $H_{2}Se$

The actual value are:
$N{H}_3\to {107}^o$
$P{H}_3\to {107}^o$
$H_{2}O\Rightarrow {104.5}^o$
$H_{2}Se={91}^o$

There variation in the bond angles between $N{H}_3$ and $P{H}_3$ is because of the differences in electronegativity and atomic size.