# What is the oxidation number of "NaH"_2"PO"_4?

##### 1 Answer
Jul 25, 2017

The overall charge of ${\text{NaH"_2"PO}}_{4}$ is zero.

#### Explanation:

However, only each atom in the compound has its own oxidation number.

Here are the rules you use to calculate them:

1. The oxidation number of a Group 1 metal in a compound is +1.
2. The oxidation number of $\text{H}$ in a compound is usually +1.
3. The oxidation number of $\text{O}$ in a compound is usually –2.
4. The sum of all oxidation numbers in a neutral compound is zero.

We can use these rules to calculate the oxidation numbers of the atoms in the compound.

Thus, the oxidation numbers of $\text{Na, H}$, and $\text{O}$ are +1, +1, and -2 (Rules 1, 2, and 3):

$\stackrel{\textcolor{b l u e}{\text{+1")("Na")stackrelcolor(blue)("+1")("H")_2"P"stackrelcolor(blue)("-2")("O}}}{_} 4$

We can use Rule 4 to calculate the oxidation number of $\text{P}$.

I will write the total oxidation numbers of each atom below the formula.

$\stackrel{\textcolor{b l u e}{\text{+1")("Na")stackrelcolor(blue)("+1")("H")_2"P"stackrelcolor(blue)("-2")("O}}}{_} 4$
stackrelcolor(blue)("+1")color(white)("Na")stackrelcolor(blue)("+2")color(white)("H"_2)color(white)("P")stackrelcolor(blue)("-8")color(white)("O"_4)

We know from Rule 4 that the sum of all the oxidation numbers must equal zero.

The sum of all the oxidation numbers showing is $+ 1 + 2 - 8 = \text{-5}$.

Thus, the oxidation number of $\text{P}$ must be $\text{+5}$.

$\stackrel{\textcolor{b l u e}{\text{+1")("Na")stackrelcolor(blue)("+1")("H")_2stackrelcolor(red)("+5")("P")stackrelcolor(blue)("-2")("O}}}{_} 4$
stackrelcolor(blue)("+1")color(white)("Na")stackrelcolor(blue)("+2")color(white)("H"_2)stackrelcolor(blue)("+5")color(white)("P")stackrelcolor(blue)("-8")color(white)("O"_4)