Find the Oxidation Numbers Al(NO_3)_3

$A l {(N O_{3})}_{3}$

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 $O$ is in column $6$ of the periodic table, it will share $6$ electrons and use an oxidation state of $6$ .

$O = 33 (- 2) = - 66$

Step 3

To find the oxidation state of $N$ , set up an equation of each oxidation state found earlier and set it equal to $0$ .

$3 - 66 + 1 x = 0$

Step 4

Solve the equation for $x$ .

Simplify $3 + - 66 + 1 x$ .

Multiply $x$ by $1$ .

$3 + - 66 + x = 0$

Add $3$ and $- 66$ .

$- 63 + x = 0$

Add $63$ to both sides of the equation.

$x = 63$

Step 5

$N$ is using an oxidation state of $63$ .

$63$

Step 6

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

$A l = 3, O = - 66, N = 63$