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Hydrides of Nitrogen

Which is in t...

Question

Which is in the decreasing order of boiling points of Vth group hydrides?

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

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Solution

The correct option is C $S b H_{3} > N H_{3} > A s H_{3} > P H_{3}$
$N H_{3}$ will have the higher boiling point because $N H_{3}$ molecules can form hydrogen bonds with each other. The other molecules will increase in boiling point with size. So, in order of decreasing boiling point:
$S b H_{3} > N H_{3} > A s H_{3} > P H_{3}$

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}

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

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