Assigned oxidation States to atoms underlined in the compound (NH4)2HPO4?

1 Answer
May 29, 2018

Answer:

Warning! Long Answer. Here's what I get.

Explanation:

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:

  1. The oxidation number of #"H"# is usually +1.
  2. The oxidation number of #"O"# is usually -2.
  3. 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#