$N_{2} + H_{2} \to {N H}_{3}$
Equivalent weight of Nitrogen and ${N H}_{3}$ respectively is: (if $M$ is their molar mass)

\frac{M}{3}, \frac{M}{6}

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\frac{M}{6}, \frac{M}{3}

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\frac{M}{3}, \frac{M}{3}

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\frac{M}{6}, \frac{M}{6}

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Solution

The correct option is B $\frac{M}{6}, \frac{M}{3}$
$N_{2} + H_{2} ⟶ N H_{3}$
Oxidation state of $N$ in $N_{2} = 0$
Oxidation state of N in $N H_{3} = + 3$

$∴$ Change in oxidation state of $N = 3$
& Change in oxidation state of $N_{2} = 6$

$∴$ Equivalent weight of $N_{2} = \frac{M}{6}$
& Equivalent weight of $N H_{3} = \frac{M}{3}$

Therefore, answer is $\frac{M}{6}, \frac{M}{3}$

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