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

May 29, 2018

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 $\text{NH"_4^"+}$ ions and $\text{HPO"_4^"2-}$ 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 $\text{H}$ is usually +1.
2. The oxidation number of $\text{O}$ 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 $\text{NH"_4^"+}$

Per Rule 1, the oxidation number of $\text{H}$ is +1.

Write the oxidation number above the $\text{H}$ in the formula:

$\text{N"stackrelcolor(blue)("+1")("H")_4^"+}$

The four $\text{H}$ atoms together have a total oxidation number of +4.

Write this number below the $\text{H}$ atom:

$\text{N"stackrelcolor(blue)("+1")("H")_4^"+}$
color(white)(m)stackrelcolor(blue)("+4")

Per Rule 3, the sum of all the oxidation numbers in $\text{NH"_4^"+}$ must be +1.

Let $x =$ the oxidation number of $\text{N}$. Then

$x + 4 = + 1$

$x = \text{1 - 4 = -3}$

There is only one $\text{N}$ atom, so its oxidation number must be -3.

Write this number above and below the $\text{N}$.

stackrelcolor(blue)("-3")("N")stackrelcolor(blue)("+1")("H")_4^"+"
stackrelcolor(blue)( "-3")""color(white)(l)stackrelcolor(blue)("+4")

(b) The oxidation numbers in $\text{HPO"_4^"-2}$

Per Rule 1, the oxidation number of $\text{H}$ is +1.

Write the oxidation number above and below the $\text{H}$ in the formula:

stackrelcolor(blue)("+1")("H")"P""O"_4^"-2"
stackrelcolor(blue)("+1")

Per Rule 2, the oxidation number of $\text{O}$ is -2.

Write this number above the $\text{O}$.

stackrelcolor(blue)("+1")("H")"P"stackrelcolor(blue)("-2")("O")""_4^"-2"
stackrelcolor(blue)("+1")

There are four $\text{O}$ atoms, so their total oxidation number must be -8.

Write this number below the $\text{O}$.

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 $\text{P}$. Then

$1 + y - 8 = - 2$

$y = \text{-2 + 7 = +5}$

There is only one $\text{P}$ atom, so its oxidation number must be +5.

Write the oxidation number above and below the $\text{P}$.

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 $\text{P}$ 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