Find the Oxidation Numbers Ba_3(PO_4)_2

$B a_{3} {(P O_{4})}_{2}$

Step 1

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

$B a = 3 (2) = 6$

Step 2

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

$O = 42 (- 2) = - 84$

Step 3

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

$6 - 84 + 1 x = 0$

Step 4

Solve the equation for $x$ .

Simplify $6 + - 84 + 1 x$ .

Multiply $x$ by $1$ .

$6 + - 84 + x = 0$

Add $6$ and $- 84$ .

$- 78 + x = 0$

Add $78$ to both sides of the equation.

$x = 78$

Step 5

$P$ is using an oxidation state of $78$ .

$78$

Step 6

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

$B a = 6, O = - 84, P = 78$