Why bond angle in water is less that that of ammonia although their geometries are distorted tetrahedral?

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Solution

Here, $N H_{3}$ has only one lone pair and water has two lone pairs.
The two lone pairs present in the oxygen atom of the water molecule repel the two bond pairs. This repulsion present in the oxygen atom is more than that of $N H_{3}$ where one lone pair and three bond pair are present on the nitrogen atom.
thus, the repulsion on the bond pair in $H_{2} O$ molecule is greater then the $N H_{3}$ , the bond angle in $H_{2} O$ is less than that of $N H_{3}$ .

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