Bond angles of $N H_{3}, P H_{3}, A s H_{3} a n d S b H_{3}$ are in the order

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$S b H_{3} > A s H_{3} > N H_{3} > P H_{3}$

$P H_{3} > A s H_{3} > S b H_{3} > N H_{3}$

$N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$

$S b H_{3} > A s H_{3} > P H_{3} > N H_{3}$

answer is D.

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Detailed Solution

Bond angle
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On moving down the group, the electro negativity of the central atom decreases. N being more electronegative will attract the shared pair of electrons towards itself thus bp-bp repulsion increases. Hence bond angle increases.

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