Question

A mole of $N_2{H}_4$ loses $10$ mol 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 number of hydrogen)

• A. $-1$
• B. $-3$
• C. $+3$
• D. $+5$

Answer 1

The correct option is C $+3$

As given,
$N_2{H}_4⟶2y+10e^{-}$
Let $x$ be the oxidation state of $N$ in $N_2{H}_4$.
Since, the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$.
$2x+4=0$
$2y=2x$, by replacing $y$, we get
$2x+4=10$
or, $x=3$