Find the Oxidation Numbers AlH_3

$A l H_{3}$

Step 1

Since $A l$ is in column $3$ of the periodic table, it will share $3$ electrons and use an oxidation state of $3$ .

$A l = 1 (3) = 3$

Step 2

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

$H = 3 (- 7) = - 21$

Step 3

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

$A l = 3, H = - 21$