State True or False.
$N H_{3}, P H_{3}$ and $A s H_{3}$ have $s p^{3}$ - hybridised central atom and thus each had bond angle of $109^{o} 28^{'}$

True

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False

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Solution

The correct option is B False
$N H_{3} ({107.8}^{o}) > P H_{3} ({99.5}^{o}) > A s H_{3} ({91.8}^{o}) \approx S b H_{3} ({91.3}^{o})$
As we move down the group, bond angle decreases due to decrease in electronegativity.

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Similar questions

N H_{3}

s p^{3}

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

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

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

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

A s H_{3} < P H_{3} < N H_{3}

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

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

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