Question

One mole of hydrazine $(N_2{H}_4)$ 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)

Given, $1mol$ 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.

New total oxidation number of new compound (X)atoms)

$=4-10+x=0$ (due to 4 H- atoms)

$\therefore x=+6$

$\therefore$ Each $N$ -atom has oxidation state $=+3$

Alternate method

$\underset{(-2)}{N_2{H}_4}⟶X$

Number of electrons lost per $N$ - atom $=5$

$\therefore$ New oxidation state of $N$ in $X$

$=-2+5+x=0$

$x=+3$