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Bond Angle

Which is the ...

Question

Which is the incorrect order for compounds?
Compound : $N H_{3}, P H_{3}, A s H_{3}, S b H_{3}$

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

(Bond angle)

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

(Bond energy )

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

(Boiling point)

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

(Reducing power)

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Solution

The correct option is B $N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$ (Boiling point)
$N H_{3}$ will have the highest boiling point because $N H_{3}$ molecules can form hydrogen bonds with each other. The other molecules will increase in boiling point with size.

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

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