Question

A mole of $N_2{H}_4$ loses 10 mol of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y ? ( There is no change in the oxidation number of hydrogen. )

• A. --1
• B. -3
• C. 3
• D. 5

Answer 1

The correct option is C 3

From the given information , we have
$N_2{H}_4\to Y+10e^{-}$

Number of N in Y is 2

Let the oxidation number of N be 'x'.

Oxidation number of $H=+1$

In $N_2{O}_4$: $2x+4\times 1=0$

or $x=-2$

$\therefore$ The oxidation state of N in $N_2{H}_4=-2$
The two nitrogen atoms in Y will balance the charge of $10e^{-}$ where each N losses 5 electron. Hence oxidation state of N changes from -2 to $-2+5=+3$.