Question

Consider the molecules ${CH}_4, {NH}_3$ and ${H}_2 {O}$. Which of the given statements is false?

(a) The ${H}-{O}-{H}$ bond angle in ${H}_2 {O}$ is smaller than the ${H}-{N}-{H}$ bond angle in ${NH}_3$.

(b) The $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle in $\mathrm{CH}_4$ is larger than the $\mathrm{H}-\mathrm{N}-\mathrm{H}$ bond angle in $\mathrm{NH}_3$.

(c) The $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle in $\mathrm{CH}_4$, the $\mathrm{H}-\mathrm{N}-\mathrm{H}$ bond angle in $\mathrm{NH}_3$, and the $\mathrm{H}-\mathrm{O}-\mathrm{H}$ bond angle in $\mathrm{H}_2 \mathrm{O}$ are all greater than $90^{\circ}$.

(d) The $\mathrm{H}-\mathrm{O}-\mathrm{H}$ bond angle in $\mathrm{H}_2 \mathrm{O}$ is larger than the $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle in $\mathrm{CH}_4$.

Answer 1

Correct Option : (D)

Explanation —

Two lone pairs of electrons in $\mathrm{H}_2 \mathrm{O}$ cause more repulsion to suppress the $\mathrm{H}-\mathrm{O}-\mathrm{H}$ bond angle more than in $\mathrm{NH}_3$ (one lone pair of electrons). While in $\mathrm{CH}_4$ there is no lone pair of electrons hence, it has normal tetrahedral bond angles.