Find the Oxidation Numbers Ca_4C_2

$C a_{4} C_{2}$

Step 1

Since $C a$ is in column $2$ of the periodic table, it will share $2$ electrons and use an oxidation state of $2$ .

$C a = 4 (2) = 8$

Step 2

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

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

Step 3

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

$C a = 8, C = - 8$