Question

Question #bbb25

Answer 1

The oxidation state of nitrogen in compound `"Y"` is +3.

Explanation:

So, you know that one mole of hydrazine, `N_2H_4`, loses 10 electrons to form a new compouind, `"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 * ON_(N) + 4 * (+1) = 0`

`ON_(N) = -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_2H_4 -> Y + 10e^(-)`

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"))`

`ON_("N in Y") = -2 - (-5) = color(green)(+3)`