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

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

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Answer 1 · 2018-05-29T19:59:21

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

Explanation:

The oxidation numbers are ${(-3 N {+1 H}_{4})}_{2} +1 H +5 P {-2 O}_{4}$ $(stackrelcolor(blue)("-3")("N")stackrelcolor(blue)("+1")("H")_4)_2stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("P")stackrelcolor(blue)("-2")("O")_4$ .

${({NH}_{4})}_{2} {HPO}_{4}$ $("NH"_4)_2"HPO"_4$ is an ionic compound.

It consists of ${NH}_{4}^{+}$ $"NH"_4^"+"$ ions and ${HPO}_{4}^{2-}$ $"HPO"_4^"2-"$ 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 $H$ $"H"$ is usually +1.
The oxidation number of $O$ $"O"$ 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 ${NH}_{4}^{+}$ $"NH"_4^"+""$

Per Rule 1, the oxidation number of $H$ $"H"$ is +1.

Write the oxidation number above the $H$ $"H"$ in the formula:

$N {+1 H}_{4}^{+}$ $"N"stackrelcolor(blue)("+1")("H")_4^"+"$

The four $H$ $"H"$ atoms together have a total oxidation number of +4.

Write this number below the $H$ $"H"$ atom:

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

Per Rule 3, the sum of all the oxidation numbers in ${NH}_{4}^{+}$ $"NH"_4^"+"$ must be +1.

Let $x =$ $x =$ the oxidation number of $N$ $"N"$ . Then

$x + 4 = + 1$ $x+4 = +1$

$x = 1 - 4 = -3$ $x = "1 - 4 = -3"$

There is only one $N$ $"N"$ atom, so its oxidation number must be -3.

Write this number above and below the $N$ $"N"$ .

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

(b) The oxidation numbers in ${HPO}_{4}^{-2}$ $"HPO"_4^"-2"$

Per Rule 1, the oxidation number of $H$ $"H"$ is +1.

Write the oxidation number above and below the $H$ $"H"$ in the formula:

$+1 H P O_{4}^{-2}$ $stackrelcolor(blue)("+1")("H")"P""O"_4^"-2"$
$+1$ $stackrelcolor(blue)("+1")$

Per Rule 2, the oxidation number of $O$ $"O"$ is -2.

Write this number above the $O$ $"O"$ .

$+1 H P -2 O_{4}^{-2}$ $stackrelcolor(blue)("+1")("H")"P"stackrelcolor(blue)("-2")("O")""_4^"-2"$
$+1$ $stackrelcolor(blue)("+1")$

There are four $O$ $"O"$ atoms, so their total oxidation number must be -8.

Write this number below the $O$ $"O"$ .

$+1 H P -2 O_{4}^{-2}$ $stackrelcolor(blue)("+1")("H")"P"stackrelcolor(blue)("-2")("O")""_4^"-2"$
$+1 m -8$ $stackrelcolor(blue)("+1")""color(white)(m)stackrelcolor(blue)("-8")$

Per Rule 3, the sum of all the oxidation numbers equals -2.

Let $y =$ $y =$ the oxidation number of $P$ $"P"$ . Then

$1 + y - 8 = - 2$ $1 + y - 8 = -2$

$y = -2 + 7 = +5$ $y = "-2 + 7 = +5"$

There is only one $P$ $"P"$ atom, so its oxidation number must be +5.

Write the oxidation number above and below the $P$ $"P"$ .

$+1 H +5 P -2 O_{4}^{-2}$ $stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("P")stackrelcolor(blue)("-2")("O")""_4^"-2"$
$+1 +5 -8$ $stackrelcolor(blue)("+1")""stackrelcolor(blue)("+5")""stackrelcolor(blue)("-8")$

The oxidation number of $P$ $"P"$ is +5.

(c) The oxidation numbers in ${({NH}_{4})}_{2} {HPO}_{4}$ $("NH"_4)_2"HPO"_4$

Now we put the ions together and get the oxidation numbers

${(-3 N {+1 H}_{4})}_{2} +1 H +5 P {-2 O}_{4}$ $(stackrelcolor(blue)("-3")("N")stackrelcolor(blue)("+1")("H")_4)_2stackrelcolor(blue)("+1")("H")stackrelcolor(blue)("+5")("P")stackrelcolor(blue)("-2")("O")_4$

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

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