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Question

$$One\ mole$$ of $$N_2H_4$$ loses $$10\ moles$$ of electrons to form a new compound $$Y$$. Assuming that all nitrogen appear in the new compound, what is the oxidation state of nitrogen in compound $$Y$$?
[There is no change in the oxidation state of hydrogen.]


A
1
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B
3
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C
+3
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D
+5
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Solution

The correct option is C $$+3$$
$$\text{One} $$ $$\text{Mole}$$ of $$N_2H_4$$ loses $$10$$ $$\text{moles}$$ of $$e^-\longrightarrow Y$$

The oxidation state of $$H$$ remains unchanged and all the $$\text{Nitrogen"}$$ appears in $$Y.$$

$$\therefore$$ $$e^-$$ must be lost from $$\text{nitrogen}$$,

$$1$$ $$\text{mole}$$ of $$N_2H_4$$ contains $$2$$ $$\text{moles}$$ of $$\text{nitrogen}$$.

$$\Longrightarrow 2$$ $$\text{moles}$$ of $$N$$ loses $$10$$ $$\text{moles}$$ of $$e^-$$.

$$\therefore 1$$ $$\text{mole}$$ loses $$5$$ $$\text{moles}$$ of $$e^-$$

Initial oxidation state of $$N$$ in $$N_2H_4=-2$$

when it loses $$5$$ $$\text{moles}$$ of $$e^-$$

Final oxidation state of each $$N=-2+5=+3.$$

(since it loses $$e^-$$, therefore it must acquire positive charge).

Hence, the correct option is $$C$$

Chemistry

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