Find the Oxidation Numbers CH_4

$C H_{4}$

Step 1

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

$H = 4 (1) = 4$

Step 2

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

$C = 1 (- 4) = - 4$

Step 3

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

$H = 4, C = - 4$