Question

The correct order of magnitude of bond angles among the compounds $C{H}_4$, $N{H}_3$ and $H_{2}O$ is:

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

Answer 1

Answer: (B)

$H_{2}O<N{H}_3<C{H}_4$

The correct order of magnitude of bond angles is $H_{2}O<N{H}_3<C{H}_4$.
In methane, the bond angle is tetrahedral $\left({109}^o{28}^{\prime }\right)$.
Due to the presence of lone pairs of electrons on $N$ (in ammonia) and $O$ (in water), the bond angle decreases due to repulsion between lone pair and bond pair of electrons. $N$ has only one lone pair of electrons whereas $O$ as two lone pairs of electrons. Hence, the decrease in the bond angle in water is greater than the decrease in bond angle in ammonia. The bond angles in ammonia and water are ${107}^o$ and ${105}^o$ respectively.