Question

$2mole$ of $N_2{H}_4$ loses $16mole$ of electron is being converted to a new compound $X$. Assuming that all the $N$ appears in the new compound. What is the oxidation state of ${}^{\prime }{N}^{\prime }$ in $X$?

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

Answer 1

Answer: (C)

$+2$

$1$ mole $N_2{H}_4$ loses $8$ mole $e^{-}$
$1$ mole $N$ loses $4$ mole of $e^{-}$
$\therefore$ New oxidation state of $N$ is $-2+4\Rightarrow 2$

hence C is correct