Find the Oxidation Numbers C_2O_4

$C_{2} O_{4}$

Step 1

Since $C$ is in column $4$ of the periodic table, it will share $4$ electrons and use an oxidation state of $4$ .

$C = 2 (4) = 8$

Step 2

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

$O = 4 (- 2) = - 8$

Step 3

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

$C = 8, O = - 8$