Question

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

Answer 1

Answer: 2

$2N_2{H}_4\to X+16e^{-}$

2 moles 16 moles

or 1 mole 8 moles

Or 1 mole of N loses 4 mole of $e^{-}$

$N^{2-}\to N^a+4e^{-}$

(O.N of $N$ in $\stackrel{x}{N_2}\stackrel{+1}{H_4}$ is $2x+4=0$

$\Rightarrow x=-2$

$\therefore a-(-2)=4$

$\Rightarrow a=+2$