$H - N - H$ bond angle in $N H_{3}$ is than/to $H - P - H$ bond angle in $P H_{3}$ .

greater

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less

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equal

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Solution

The correct option is A greater
$H - N - H$ bond angle in $N H_{3}$ is $107^{o}$ .
$H - P - H$ bond angle in $P H_{3}$ is $90^{o}$ .

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