Find the Oxidation Numbers C_6H_6

$C_{6} H_{6}$

Step 1

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

$H = 6 (1) = 6$

Step 2

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

$C = 6 (- 4) = - 24$

Step 3

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

$H = 6, C = - 24$