The correct sequence of decrease in the bond angles of the following hydrides is

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

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

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

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

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Solution

The correct option is A $N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$
$N H_{3}$ has $s p^{3}$ hybridisation with bond angle $107^{0}$ where as $P H_{3}, A s H_{3}, S b H_{3}$ has pure orbitals overlapping with bond angle around $90^{0}$

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