The bond angles in $\mathrm{H}_{3} \mathrm{O}^{+}$ are less than _______ and greater than _______.
SOLUTION The carbon atom in $\mathrm{CH}_{4}$ has no lone pairs; its bond angles are $109.5^{\circ} .$ The oxygen atom
in $\mathrm{H}_{3} \mathrm{O}^{+}$ has one lone pair. A lone pair is more diffuse than a bonding pair, so the $\mathrm{O}-\mathrm{H}$ bonds squeeze together to minimize electron repulsion. However, they do not squeeze as closely together as they do in water $\left(104.5^{\circ}\right),$ where oxygen has two lone pairs. Therefore, the bond angles in $\mathrm{H}_{3} \mathrm{O}^{+}$ are less than $109.5^{\circ}$ and greater than $104.5^{\circ} .$