Why is the $H - O - H$ bond angle in $H_{2} O$ smaller than $H - N - H$ bond angle in $N H_{3}$ ?

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Solution

According to $V S E P R$ theory, $H_{2} O$ has $2$ lone pairs of electrons whereas ${N H}_{3}$ has only one lone pair.
Hence, repulsive forces between lone pair and bond pair is more in $H_{2} O$ . Which results in bond angle of $104^{0}$ degrees in $H_{2} O$ and $102^{0}$ in ${N H}_{3}$ .

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