Question

$Onemole$ of $N_2{H}_4$ loses $10moles$ 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. $+5$

Answer 1

The correct option is C $+3$

$\text{One}$ $\text{Mole}$ of $N_2{H}_4$ loses $10$ $\text{moles}$ of $e^{-}⟶Y$

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 $N_2{H}_4$ contains $2$ $\text{moles}$ of $\text{nitrogen}$.

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

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

Initial oxidation state of $N$ in $N_2{H}_4=-2$

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 $C$