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
