Question

One mole of hydrazine $\left(N_2{H}_4\right)$ loses $10$ moles of electrons in a reaction to form a new compound $X$. Assuming that all the nitrogen atoms in hydrazine appear in the new compound, what is the oxidation state of nitrogen in $X$? (Note There is no change in the oxidation state of hydrogen in the reaction)

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

Answer 1

Answer: (C)

$+3$

Given, $1$ mol of $H_{2}N-N{H}_2$ (hydrazine), it loses $10$ moles of electrons to form a new compound that contains both the $N$-atoms with same oxidation number means:
$N_2{H}_4⟶10e^{-}+X$ (product)
(Oxidation number of $N$-atom in $N_2{H}_4=-2$)
$\because$ Oxidation number of both the $N$-atoms are same.
The oxidation state of N in hydrazine is -2.

1 mole of hydrazine contains 2 moles of N and loses 10 moles of electrons. Hence, 1 N atom will lose 5 electrons. Hence, its oxidation number will increase by 5.

$\therefore$ New oxidation state of $N$ in $X$ $=-2+5+x=0$
'

$x=+3$