Find the Oxidation Numbers H_2CO_3

$H_{2} C O_{3}$

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 $6$ .

$O = 3 (- 2) = - 6$

Step 3

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

$2 - 6 + 1 x = 0$

Step 4

Solve the equation for $x$ .

Simplify $2 + - 6 + 1 x$ .

Multiply $x$ by $1$ .

$2 + - 6 + x = 0$

Add $2$ and $- 6$ .

$- 4 + x = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 5

$C$ is using an oxidation state of $4$ .

$4$

Step 6

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

$H = 2, O = - 6, C = 4$