# Question bbb25

Jun 17, 2015

The oxidation state of nitrogen in compound $\text{Y}$ is +3.

#### Explanation:

So, you know that one mole of hydrazine, ${N}_{2} {H}_{4}$, loses 10 electrons to form a new compouind, $\text{Y}$.

The oxidation state of hydrogen in hydrazine is +1, which means that, in order for the compound to be neutral, the oxidation state of nitrogen must be

$2 \cdot O {N}_{N} + 4 \cdot \left(+ 1\right) = 0$

$O {N}_{N} = - \frac{4}{2} = - 2$

A very important pirce of information given to you is the fact that the oxidation state of hydrogen remains unchanged. Since 1 mole of hydrazine contains 2 moles of nitrogen, this means that the 10 moles of electrons will come from the 2 moles of nitrogen.

As a result, each nitrogen atom will lose 5 electrons.

${N}_{2} {H}_{4} \to Y + 10 {e}^{-}$

If each nitrogen atom loses 5 electrons, its oxidation state will go from -2 to +3

ON_("N in hydrazine") = ON_("N in Y") + underbrace(5e^(-))_(color(blue)("= -5"))#

$O {N}_{\text{N in Y}} = - 2 - \left(- 5\right) = \textcolor{g r e e n}{+ 3}$