# Question 76fa4

May 3, 2016

Nitrogen will be in a $+ 3$ oxidation state.

#### Explanation:

The idea here is that you can use the number of moles of electrons lost by one mole of hydrazine, ${\text{N"_2"H}}_{4}$, to determine how many moles of electrons were lost by one mole of nitrogen.

Since hydrogen's oxidation state is said to remain unchanged, you can say for a fact that all the moles of electrons lost by hydrazine were actually lost by nitrogen.

Now, look at hydrazine's molecular formula, which contains

• two atoms of nitrogen, $2 \times \text{N}$
• four atoms of hydrogen, $4 \times \text{H}$

Notice that one mole of hydrazine contains two moles of nitrogen.

This of course means that when $1$ mole of hydrazine loses $10$ electrons, these electrons are actually coming from $2$ moles of nitrogen.

As a result, you can say that every mole of nitrogen present in one mole of hydrazine will lose $5$ moles of electrons.

If you take this down to the level of a single atom, you can say that one atom of nitrogen will lose $5$ electrons.

In hydrazine, hydrogen has a $\textcolor{b l u e}{+ 1}$ oxidation state, which means that nitrogen has an oxidation state equal to

$2 \times O {N}_{\text{N" + 4 xx ON_"H}} = 0$

$2 \times O {N}_{\text{N}} = 0 - 4 \cdot \left(\textcolor{b l u e}{+ 1}\right)$

${\text{ON}}_{N} = \frac{- 4}{2} = - 2$

Now, when each nitrogen atom loses $5$ electrons, its oxidation state increases by $5$, i.e. it's being oxidized. You will have

color(red)(|bar(ul(color(white)(a/a)color(black)(stackrel(color(blue)(-2))("N")_2"H"_4 -> 2stackrel(color(blue)(+3))("N") + 10"e"^(-))color(white)(a/a)|)))#

Here one atom of nitrogen loses $5$ electrons, so two atoms of nitrogen will lose $10$ electrons.

The oxidation state of nitrogen in this new compound $\text{Y}$ will thus be equal to $\textcolor{b l u e}{+ 3}$.

May 4, 2016

+3

#### Explanation:

Let ON of N in ${N}_{2} {H}_{4}$ be x. H ,being more electropositive than N ,it will have +1ON and species being a molecule total ON wil be zero.
So $2 \cdot x + 4 \cdot \left(+ 1\right) = 0 \implies x = - 2$
By the problem
the oxidation process may be written as
${N}_{2} {H}_{4} \to Y \left(\text{*}\right)$
$\text{*molecule contains 2 N-atoms}$

One mole of ${N}_{2} {H}_{4}$ loses 10 moles of electrons to form a new compound Y
So One molecule of ${N}_{2} {H}_{4}$ loses 10 electrons to form one molecule of new compound Y which contains 2 N-atoms As there is no change of ON of H-atom , 2N-atoms of ${N}_{2} {H}_{4}$ loses 10 electrons causing total 10 unit increase in ON of 2 N-atoms .

Now if the ON of 2 N-atoms of Y be y then total increase in ON of N-atoms$= \left(2 y - 2 \cdot \left(- 2\right) = 2 y + 4\right)$
By the given condition of the problem , we have
$2 y + 4 = 10 \implies y = + 3$

Hence the oxidation state of N in the compound Y is +3