Find the Oxidation Numbers H_2N_2O_2

$H_{2} N_{2} O_{2}$

Step 1

Since $H$ is in column $1$ of the periodic table, it will share $1$ electrons and use an oxidation state of $1$ .

$H = 2 (1) = 2$

Step 2

Since $O$ is in column $6$ of the periodic table, it will share $6$ electrons and use an oxidation state of $6$ .

$O = 2 (- 2) = - 4$

Step 3

To find the oxidation state of $N$ , set up an equation of each oxidation state found earlier and set it equal to $0$ .

$2 - 4 + 2 x = 0$

Step 4

Solve the equation for $x$ .

Add $2$ and $- 4$ .

$- 2 + 2 x = 0$

Add $2$ to both sides of the equation.

$2 x = 2$

Divide each term in $2 x = 2$ by $2$ and simplify.

Divide each term in $2 x = 2$ by $2$ .

$\frac{2 x}{2} = \frac{2}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{2}{2}$

Divide $x$ by $1$ .

$x = \frac{2}{2}$

Simplify the right side.

Divide $2$ by $2$ .

$x = 1$

Step 5

$N$ is using an oxidation state of $1$ .

$1$

Step 6

This is the full list of oxidation states for this molecule.

$H = 2, O = - 4, N = 1$