Bond angle order is $N H_{3} > S b H_{3} > A s H_{3} > P H_{3}$ .
If true 1 else 0.
0

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Solution

The correct option is A 0
As we move down the group, radius of elements increases and electronegativity decreases so bond angle decreases.
so bond order is $N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$
107 93 92 91

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Similar questions

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}

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

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

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

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