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One mole of $N_{2} H_{4}$ loses 10 mol of electrons to form a new compound Y. Assuming that all nitrogen appear in the new compound, what is the oxidation state of $N_{2}$ in Y ? (There is no change in the oxidation state of hydrogen ?

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+ 3

$- 3$

$- 1$

+ 5

answer is A.

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Detailed Solution

$N_{2}^{2 -} \to 2 N^{+ a} + 10 e^{-} ∴ 2 a - [2 \times (- 2)] = 10 ∴ a = + 3 .$

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