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Q.

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

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a

+5

b

-3

c

-1

d

+3

answer is C.

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Detailed Solution

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$N_{2} H_{4} \to Y + 10 e^{-}, Calculation of O . S . of N in N_{2} H_{4} : 2 x + 4 = 0 \Rightarrow x = - 2$

The two nitrogen atoms will balance the charge of 10 e. Hence, oxidation state of N will increase by +5, i.e. from -2 to +3

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