Question

Which of the following molecules has the highest bond angle?

• A. $N{H}_3$
• B. $P{H}_3$
• C. $H_{2}O$
• D. $C{H}_4$

Answer 1

The correct option is D $C{H}_4$

$P{H}_3$ is a drago compound i.e. no hybridisation. Therefore, its bond angle is nearly ${90}^{\circ }$.
$N{H}_3$ and $H_{2}O$ and $C{H}_4$ have $1$, $2$ and $0$ lone pairs respectively with the hybridization of $s{p}^3$ and tetrahedral geometry. Their respective shapes are pyramidal, bent and tetrahedral. So, in $N{H}_3$ and $H_{2}O$, the bond angles are less than ${109}^{\circ }{28}^{\prime }$ due to lone pair-bond pair repulsion.
In $C{H}_4$, the bond angle is ${109}^{\circ }{28}^{\prime }$ as it is regular tetrahedral.

The structure of $N{H}_3$:

The structure of $P{H}_3$:

The structure of $H_{2}O$:

The structure of $C{H}_4$: