Question

A sample of $2.5mol$ of hydrazine $\left(N_2{H}_4\right)$ loses $25mol$ of electrons on being converted to a new compound $X$. Assuming that there is no loss of nitrogen in the formation of the new compound and the oxidation state of $H^{+}$ is $+1$ throughout the reaction, then what is the oxidation number of nitrogen in compound $X$?

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

Answer 1

The correct option is C $+3$

Let the oxidaion number of $N$ in new compound $X$ be $x$
${\stackrel{-2}{N}}_2{H}_4\to 2N^{+x}+n{e}^{-}$
$\text{Let the no. of electrons lost per mole = n}$
number of moles electron = $\text{change in the oxidation state}\times \text{number of atoms per mole}$
$n=|x-(-2)|\times 2$
$\Rightarrow n=2x+4=2(x+2)$
$\therefore$
$\text{Number of electron lost per mole}=2(x+2)$
$\text{Number of electrons lost by 2.5 moles}=2.5\times 2(2+x)$
Now,
$2.5\times 2(2+x)=25$
$5(2+x)=25$
or $2+x=5$
or $x=+3$