Question

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

Answer 1

The overall charge of `"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 `"H"` in a compound is usually +1.
3. The oxidation number of `"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 `"Na, H"`, and `"O"` are +1, +1, and -2 (Rules 1, 2, and 3):

`stackrelcolor(blue)("+1")("Na")stackrelcolor(blue)("+1")("H")_2"P"stackrelcolor(blue)("-2")("O")_4`

We can use Rule 4 to calculate the oxidation number of `"P"`.

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

`stackrelcolor(blue)("+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 = "-5"`.

Thus, the oxidation number of `"P"` must be `"+5"`.

`stackrelcolor(blue)("+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)`