The compound that has the largest $\mathrm{H}-\mathrm{M}-\mathrm{H}$ bond angle $(\mathrm{M}=\mathrm{N}, \mathrm{S}, \mathrm{C})$, is :
Correct Option:
$\mathrm{H}_{2} \mathrm{O}-104.5^{\circ}\left(\mathrm{sp}^{3}\right.$ with 2 lone pair at $\left.\mathrm{O}\right)$
$\mathrm{NH}_{3}-107^{\circ}\left(s p^{3}\right.$ with 1 lone pair at $\left.\mathrm{N}\right)$
$\mathrm{CH}_{4}-109.5^{\circ}\left(s p^{3}\right)$
$\mathrm{H}_{2} \mathrm{~S}-92^{\circ}\left(s p^{3}\right.$ with 2 lone pair at $\mathrm{O}$ )
Lone pair-bond pair repulsion in $\mathrm{H}_{2} \mathrm{~S}$ will increase because ' $\mathrm{S}$ ' has lower electronegativity than ' $\mathrm{O}$ '. So there will be lesser electron density on 'S' and thus H-S-H bond angle will be smaller than $\mathrm{H}_{2} \mathrm{O}$.