The correct sequence of decreasing 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 $N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$
Bond angle of group 15 element hydrides follow the order
$N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$
Here, the central atom is surrounded by one lone pair and three bond pairs. When you go down in a group, the size of atoms increases and the electronegativity decreases. As a result, the repulsions between bond pairs decreases and therefore, the bond angle also decreases.

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

Why does bond angle in hydrides decrease on going down the group?

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