Question

A $2.5\, mol$ of hydrazine, $N_{2}H_{4}$ loses $25$ moles of electrons and being converted to a new compound $X$. Assuming that all of the nitrogen appears in the new compound, what is the oxidation state of nitrogen in compound $X$ ?

• A. $-1$
• B. $-2$
• C. $+3$
• D. $+4$

Answer 1

Answer: (C)

$\overset{-2}{N_{2}}H_{4} \to \overset{x}{N}+ne^{-}$
We are considering the change in oxidation state of N only
Suppose number of electrons lost per mole $= n$
$\therefore 2 (-2) =2x+n(-1)$
$n=2x+4$ or $n=2(x+2)$
i.e., number of electrons lost per mole, $n = 2 (x + 2)$
Number of electrons lost by $2.5\, mol =2.5 \times 2 (x+2)$
$=5 (x+2)$
But $5 (x+2)=25$ (given)
$x=\frac{25-10}{5}=+3$
i.e., oxidation state of $N$ in new compound = $+ 3$