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Heterolytic Cleavage

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Question

The correct order of normal boiling points of $O_{2}, N_{2}, N H_{3}$ and $C H_{4}$ for whom the values of van der Waal's constant $^{'} a^{'}$ are $1.360, 1.390, 4.170 and 2.253 L^{2} {atm mol}^{- 2}$ respectively is :

A

O_{2} < N_{2} < N H_{3} < C H_{4}

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B

O_{2} < N_{2} < C H_{4} < N H_{3}

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C

N H_{3} < C H_{4} < N_{2} < O_{2}

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D

N H_{3} < C H_{4} < O_{2} < N_{2}

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Solution

The correct option is B $O_{2} < N_{2} < C H_{4} < N H_{3}$
The correct order of normal boiling points of $O_{2}, N_{2}, N H_{3}$ and $C H_{4}$ , (for whom the values of van der Waal's constant $^{'} a^{'}$ are $1.360, 1.390, 4.170 and 2.253 L^{2} . {atm mol}^{- 2}$ respectively) is $O_{2} < N_{2} < C H_{4} < N H_{3}$ .
Higher is the value of van der Waal's constant $a$ , higher will be the boiling point as $a$ is an indirect measure of the magnitude of attractive forces between the molecules.

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