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Hydrides of Nitrogen

Of the follow...

Question

Of the following hydrides which one has the lowest boiling point?

A s H_{3}

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S b H_{3}

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P H_{3}

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N H_{3}

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Solution

The correct option is C

P H_{3}

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

From

P H_{3}

B {i H}_{3}

the boiling points increase with increase in molecular weights, due to the increase in Van der Waal's forces with molecular mass.

However, boiling point of

{N H}_{3} > P H_{3}

even though

{N H}_{3}

has lower molecular mass than

P H_{3}

, due to the intermolecular hydrogen bonding in Ammonia that is not present in

P H_{3}

and other hydrides in Group-15.

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}

15

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

H

N H_{3}

S b

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