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Polar Covalent Compounds

Amongst NH3...

Question

Amongst $N H_{3}, P H_{3}, A s H_{3}$ and $S b H_{3}$ , the one with highest boiling point is:

A s H_{3}

because of lower molecular weight

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S b H_{3}

because of higher molecular weight

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N H_{3}

because of H-bonding

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P H_{3}

because of lower molecular weight

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Solution

The correct option is B $S b H_{3}$ because of higher molecular weight
Solution
$S b H_{3}$ will have highest boiling point because of higher molecular weight which increase van de waals force of attraction.
The correct option is B

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}

Amongst $H_{2} O$ , $H_{2} S$ , $H_{2} S e$ and $H_{2} T e$ , the one with the highest boiling point is -

H_{2} O, H_{2} S, H_{2} S e

H_{2} T e

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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