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Bond Length

Arrange NH3...

Question

Arrange $N H_{3}, P H_{3}, A s H_{3}, S b H_{3}$ in increasing bond angle.

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

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

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

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

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Solution

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

Suggest Corrections

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}

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}

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

S i F_{4} < S i C l_{4} < S i B r_{4} < S i I_{4}

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

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

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

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