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
B
3
C
+3
D
+5

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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