Question

Question #76fa4

Answer 1

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, `"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 xx "N"`
• four atoms of hydrogen, `4 xx "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 `color(blue)(+1)` oxidation state, which means that nitrogen has an oxidation state equal to

`2 xx ON_"N" + 4 xx ON_"H" = 0`

`2 xx ON_"N" = 0 - 4 * (color(blue)(+1))`

`"ON"_N = (-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 `"Y"` will thus be equal to `color(blue)(+3)`.

Answer 2

+3

Explanation:

Let ON of N in `N_2H_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*x+4*(+1)=0=>x=-2`
By the problem
the oxidation process may be written as
`N_2H_4->Y("*")`
`"*molecule contains 2 N-atoms"`

One mole of `N_2H_4` loses 10 moles of electrons to form a new compound Y
So One molecule of `N_2H_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_2H_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`=(2y-2*(-2)=2y+4)`
By the given condition of the problem , we have
`2y+4=10=>y=+3`

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