Find the Oxidation Numbers HCl

$H C l$

Step 1

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

$H = 1 (1) = 1$

Step 2

Since $C l$ is in column $7$ of the periodic table, it will share $7$ electrons and use an oxidation state of $7$ .

$C l = 1 (- 1) = - 1$

Step 3

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

$H = 1, C l = - 1$