In the reacti...

Question

In the reaction,
$3 C u O + 2 {N H}_{3} ⟶ N_{2} + 3 H_{2} O + 3 C u$
the change of ${N H}_{3}$ to $N_{2}$ involves:

Loss of 6 electrons per mole of

N_{2}

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Loss of 3 electrons per mole of

N_{2}

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Gain of 6 electrons per mole of

N_{2}

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Gain of 3 electrons per mole of

N_{2}

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Solution

The correct option is B Loss of 6 electrons per mole of $N_{2}$
$3 C u O + 2 {N H}_{3} ⟶ N_{2} + 3 H_{2} O + 3 C u$

Oxidation state of N in $N H_{3} = - 3$

Oxidation state of N in $N_{2} = 0$

The change in oxidation state of $N$ is loss of $3$ electrons.

There are two atoms of nitrogen, hence the change involves loss of $6 e^{-}$ for $1$ mol of $N_{2}$ .

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