# What is the oxidation number of phosphorus in "PO"_4^(3-)?

Oct 23, 2015

$+ 5$

#### Explanation:

You're dealing with an ion, the phosphate anion to be precise, which means that the oxidation states of each atom must add up to give the overall charge of the ion.

So, start by writing down what you know. You know that

• the ion has a $\text{(3-)}$ overall charge
• the ion is made up of one phosphous atom and four oxygen atoms
• oxygen's oxidation state in most compounds is $- 2$.

This means that you can write

overbrace(1 xx ON_"P")^(color(blue)("one phosphorus atom")) + underbrace(4 xx ON_"oxygen")_(color(red)("four oxygen atoms")) = -3

Solve the equation for $O {N}_{\text{P}}$ to get

$O {N}_{\text{P}} + 4 \times \left(- 2\right) = - 3$

$O {N}_{\text{P}} = - 3 + 8 = \textcolor{g r e e n}{+ 5}$

So, phosphorus has a $+ 5$ oxidation state in the phosphate anion, ${\text{PO}}_{4}^{3 -}$.