Correct order of decreasing thermal stability is as:

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

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

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A s H_{3} > P H_{3} > N 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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Solution

The correct option is A $N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$
The stability of hydrides decreases from ammonia to bismuthine. This is because the central atom E increases in size down the group. With the increase in the size of the central atom, the E – H bond becomes weaker. where E is nitrogen, phosphorus, arsenic, antimony, bismuth.

With the increase in size, the overlapping between the orbitals of E and hydrogen will not be effective. This accounts for the decrease in the stability of hydrides as we move down the group.

The correct order is : $N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$

Hence, option $A$ is correct.

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}

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}

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}

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