The correct sequence of decrease in bond angles of 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 C $N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$
The bond angle decreases from $N$ to $S b$ because of the decrease in the inter electronic repulsion due to which the bonded pairs adjust themselves.

The bond angle of hydrides is decreased from top to bottom in the group.

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

Since the repulsion between lone pairs is greater than that of bond pairs, the above phenomenon takes place.

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}

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}

Out of NH₃, PH₃, AsH₃ and SbH₃ the most basic hydride is:

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