Find the Oxidation Numbers NaPO_4

$N a P O_{4}$

Step 1

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

$N a = 1 (1) = 1$

Step 2

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

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

Step 3

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

$1 - 8 + 1 x = 0$

Step 4

Solve the equation for $x$ .

Simplify $1 + - 8 + 1 x$ .

Multiply $x$ by $1$ .

$1 + - 8 + x = 0$

Add $1$ and $- 8$ .

$- 7 + x = 0$

Add $7$ to both sides of the equation.

$x = 7$

Step 5

$P$ is using an oxidation state of $7$ .

$7$

Step 6

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

$N a = 1, O = - 8, P = 7$