The oxidation numbers are #(stackrelcolor(blue)("-3")("N")stackrelcolor(blue)("+1")("H")_4)_2stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("P")stackrelcolor(blue)("-2")("O")_4#.
#("NH"_4)_2"HPO"_4# is an ionic compound.
It consists of #"NH"_4^"+"# ions and #"HPO"_4^"2-"# ions, so we can calculate the oxidation numbers in each ion separately.
The critical oxidation number rules for this problem are:
- The oxidation number of #"H"# is usually +1.
- The oxidation number of #"O"# is usually -2.
- The sum of the oxidation numbers of all atoms in an ion equals charge on the ion.
(a) The oxidation numbers in #"NH"_4^"+""#
Per Rule 1, the oxidation number of #"H"# is +1.
Write the oxidation number above the #"H"# in the formula:
#"N"stackrelcolor(blue)("+1")("H")_4^"+"#
The four #"H"# atoms together have a total oxidation number of +4.
Write this number below the #"H"# atom:
#"N"stackrelcolor(blue)("+1")("H")_4^"+"#
#color(white)(m)stackrelcolor(blue)("+4")#
Per Rule 3, the sum of all the oxidation numbers in #"NH"_4^"+"# must be +1.
Let #x =# the oxidation number of #"N"#. Then
#x+4 = +1#
#x = "1 - 4 = -3"#
There is only one #"N"# atom, so its oxidation number must be -3.
Write this number above and below the #"N"#.
#stackrelcolor(blue)("-3")("N")stackrelcolor(blue)("+1")("H")_4^"+"#
#stackrelcolor(blue)( "-3")""color(white)(l)stackrelcolor(blue)("+4")#
(b) The oxidation numbers in #"HPO"_4^"-2"#
Per Rule 1, the oxidation number of #"H"# is +1.
Write the oxidation number above and below the #"H"# in the formula:
#stackrelcolor(blue)("+1")("H")"P""O"_4^"-2"#
#stackrelcolor(blue)("+1")#
Per Rule 2, the oxidation number of #"O"# is -2.
Write this number above the #"O"#.
#stackrelcolor(blue)("+1")("H")"P"stackrelcolor(blue)("-2")("O")""_4^"-2"#
#stackrelcolor(blue)("+1")#
There are four #"O"# atoms, so their total oxidation number must be -8.
Write this number below the #"O"#.
#stackrelcolor(blue)("+1")("H")"P"stackrelcolor(blue)("-2")("O")""_4^"-2"#
#stackrelcolor(blue)("+1")""color(white)(m)stackrelcolor(blue)("-8")#
Per Rule 3, the sum of all the oxidation numbers equals -2.
Let #y =# the oxidation number of #"P"#. Then
#1 + y - 8 = -2#
#y = "-2 + 7 = +5"#
There is only one #"P"# atom, so its oxidation number must be +5.
Write the oxidation number above and below the #"P"#.
#stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("P")stackrelcolor(blue)("-2")("O")""_4^"-2"#
#stackrelcolor(blue)("+1")""stackrelcolor(blue)("+5")""stackrelcolor(blue)("-8")#
The oxidation number of #"P"# is +5.
(c) The oxidation numbers in #("NH"_4)_2"HPO"_4#
Now we put the ions together and get the oxidation numbers
#(stackrelcolor(blue)("-3")("N")stackrelcolor(blue)("+1")("H")_4)_2stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("P")stackrelcolor(blue)("-2")("O")_4#
