Although the geometries of $NH_{3}$ and $H_{2}O$ molecules are based on a distorted tetrahedral arrangement,the bond angle in water is less than that of ammonia. Discuss.

(N/A) In $NH_{3}$,the central $N$ atom has one lone pair and three $N-H$ bonding pairs,making it an $AB_{3}E$ type molecule.
In $H_{2}O$,the central $O$ atom has two lone pairs and two $O-H$ bonding pairs,making it an $AB_{2}E_{2}$ type molecule.
According to the $VSEPR$ theory,the order of repulsion is: $lp-lp > lp-bp > bp-bp$.
$H_{2}O$ has two lone pairs on the oxygen atom,which exert strong $lp-lp$ repulsion,pushing the $O-H$ bonds closer together,resulting in a bond angle of $104.5^{\circ}$.
$NH_{3}$ has only one lone pair on the nitrogen atom,which exerts less repulsion compared to the two lone pairs in water,resulting in a bond angle of $107^{\circ}$.