Question

Consider the molecules $C{H}_4,N{H}_3,\text{and}{H}_{2}O$. Which of the given statements is false?

• A. The $H-C-H$ bond angle in $C{H}_4$, the $H-N-H$ bond angle in $N{H}_3$, and the $H-O-H$ bond angle in $H_{2}O$ are all greater than ${90}^o$
• B. The $H-O-H$ bond angle in $H_{2}O$ is larger than the $H-C-H$ bond angle in $C{H}_4$
• C. The $H-O-H$ bond angle in $H_{2}O$ is smaller than the $H-N-H$ bond angle in $N{H}_3$
• D. The $H-C-H$ bond angle in $C{H}_4$ is larger than the $H-N-H$ bond angle in $N{H}_3$

Answer 1

The correct option is B The $H-O-H$ bond angle in $H_{2}O$ is larger than the $H-C-H$ bond angle in $C{H}_4$

All the three molecules are $s{p}^3$ hybridised but the bond angles are different due to the presence of lone pairs. The bond angle of $H_{2}O$ is ${104.5}^o$ due to lone pair-lone pair and bond pair-lone pair repulsion and the bond angle of $C{H}_4$ is ${109}^o{28}^{\prime }$.
Bond angle of $H_{2}O$ is smaller than $N{H}_3$ due to presence of two lone pair of electron on central atom(O) of $H_{2}O$.

Hence, option (B) is the correct answer.