# 22.6: Assigning Oxidation Numbers

Once we move from the element iron to iron compounds, we need to be able to designate clearly the form of the iron ion. An example of this is iron that has been oxidized to form iron oxide during the process of rusting. Although Antoine Lavoisier first began the idea of oxidation as a concept, it was Wendell Latimer (1893 - 1955) who gave us the modern concept of oxidation numbers. His 1938 book The Oxidation States of the Elements and Their Potentials in Aqueous Solution laid out the concept in detail. Latimer was a well-known chemist who later became a member of the National Academy of Sciences. Not bad for a gentleman who started college planning on being a lawyer.

## Assigning Oxidation Numbers

The oxidation number is a positive or negative number that is assigned to an atom to indicate its degree of oxidation or reduction. In oxidation-reduction processes, the driving force for chemical change is in the exchange of electrons between chemical species. A series of rules have been developed to help us.

1. For free elements (uncombined state), each atom has an oxidation number of zero. $$\ce{H_2}$$, $$\ce{Br_2}$$, $$\ce{Na}$$, $$\ce{Be}$$, $$\ce{K}$$, $$\ce{O_2}$$, $$\ce{P_4}$$, all have oxidation number of 0.
2. Monatomic ions have oxidation numbers equal to their charge. $$\ce{Li^+} = +1$$, $$\ce{Ba^{2+}} = +2$$, $$\ce{Fe^{3+}} = +3$$, $$\ce{I^-} = -1$$, $$\ce{O^{2-}} = -2$$, etc. Alkali metal oxidation numbers $$= +1$$. Alkaline earth oxidation numbers $$= +2$$. Aluminum $$= +3$$ in all of its compounds. Oxygen's oxidation number $$= -2$$ except when in hydrogen peroxide $$\left( \ce{H_2O_2} \right)$$, or a peroxide ion $$\left( \ce{O_2^{2-}} \right)$$ where it is $$-1$$.
3. Hydrogen's oxidation number is $$+1$$, except for when bonded to metals as the hydride ion forming binary compounds. In $$\ce{LiH}$$, $$\ce{NaH}$$, and $$\ce{CaH_2}$$, the oxidation number is $$-1$$.
4. Fluorine has an oxidation number of $$-1$$ in all of its compounds.
5. Halogens ($$\ce{Cl}$$, $$\ce{Br}$$, $$\ce{I}$$) have negative oxidation numbers when they form halide compounds. When combined with oxygen, they have positive numbers. In the chlorate ion $$\left( \ce{ClO_3^-} \right)$$, the oxidation number of $$\ce{Cl}$$ is $$+5$$, and the oxidation number of $$\ce{O}$$ is $$-2$$.
6. In a neutral atom or molecule, the sum of the oxidation numbers must be 0. In a polyatomic ion, the sum of he oxidation numbers of all the atoms in the ion must be equal to the charge on the ion.

Example 22.6.1

What is the oxidation number for manganese in the compound potassium permanganate $$\left( \ce{KMnO_4} \right)$$?

Solution:

The oxidation number for $$\ce{K}$$ is $$+1$$ (rule 2)

The oxidation number for $$\ce{O}$$ is $$-2$$ (rule 2)

Since this is a compound (there is no charge indicated on the molecule), the net charge on the molecule is zero (rule 6)

So we have

\begin{align} +1 + \ce{Mn} + 4 \left( -2 \right) &= 0 \\ \ce{Mn} - 7 &= 0 \\ \ce{Mn} &= +7 \end{align}

When dealing with oxidation numbers, we must always include the charge on the atom.

Another way to determine the oxidation number of $$\ce{Mn}$$ in this compound is to recall that the permanganate anion $$\left( \ce{MnO_4^-} \right)$$ has a charge of $$-1$$. In this case:

\begin{align} \ce{Mn} + 4 \left( -2 \right) &= -1 \\ \ce{Mn} - 8 &= -1 \\ \ce{Mn} &= +7 \end{align}

Example 22.6.2

What is the oxidation number for iron in $$\ce{Fe_2O_3}$$?

Solution:

\begin{align} &\ce{O} \: \text{is} \: -2 \: \left( \text{rule 2} \right) \\ &2 \ce{Fe} + 3 \left( -2 \right) = 0 \\ &2 \ce{Fe} = 6 \\ &\ce{Fe} = 3 \end{align}

If we have the compound $$\ce{FeO}$$, then $$\ce{Fe} + \left( -2 \right) = 0$$ and $$\ce{Fe} = 2$$. Iron is one of those materials that can have more than one oxidation number.

The halogens (except for fluorine) can also have more than one number. In the compound $$\ce{NaCl}$$, we know that $$\ce{Na}$$ is $$+1$$, so $$\ce{Cl}$$ must be $$-1$$. But what about $$\ce{NaClO_3}$$?

\begin{align} \ce{Na} &= 1 \\ \ce{O} &= -2 \\ 1 + \ce{Cl} + 3 \left( -2 \right) &= 0 \\ 1 + \ce{Cl} - 6 &= 0 \\ \ce{Cl} - 5 &= 0 \\ \ce{Cl} &= +5 \end{align}

Not quite what we expected, but $$\ce{Cl}$$, $$\ce{Br}$$, and $$\ce{I}$$ will exhibit multiple oxidation numbers in compounds.

## Summary

• Rules for determining oxidation numbers are listed.
• Examples of oxidation number determinations are provided.

## Contributors

• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.