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

1 Answer

Answer:

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)#