Although geometries of $NH_3$ and $H_2O$ molecules are distorted tetrahedral,bond angle in water is less than that of ammonia. Discuss.

(N/A) The central atom $N$ in $NH_3$ has $1$ lone pair and $3$ bond pairs,resulting in a bond angle of $107^{\circ}$.
The central atom $O$ in $H_2O$ has $2$ lone pairs and $2$ bond pairs.
According to $VSEPR$ theory,the repulsion between lone pairs is greater than the repulsion between lone pair and bond pair,which is greater than the repulsion between bond pairs $(lp-lp > lp-bp > bp-bp)$.
In $H_2O$,there are $2$ lone pairs on the oxygen atom,which exert strong repulsion on the $2$ bond pairs,pushing them closer together.
In $NH_3$,there is only $1$ lone pair on the nitrogen atom,which exerts less repulsion on the $3$ bond pairs compared to the $2$ lone pairs in $H_2O$.
Therefore,the bond angle in $H_2O$ $(104.5^{\circ})$ is smaller than the bond angle in $NH_3$ $(107^{\circ})$.