# Question #81b1f

Jan 11, 2015

Fractional oxidation numbers are sometimes used to describe the average oxidation state of several atoms in a compound. It is important to understand that not all atoms of a particular element must have the same oxidation number in a compound.

Let's take, for example, propane. Propane's molecular formula is ${C}_{3} {H}_{8}$. If you take into account the fact that $H$ has an oxidation number of +1, the resulting oxidation number (ON) for $C$ would be

$3 \cdot$ $\text{ON"_("carbon") + 8*(+1) =0 => "ON"_("carbon") = "-8/3}$

This represents a fractional oxidation number. However, if you take into account the fact that propane's structure is actually

$C {H}_{3} C {H}_{2} C {H}_{3}$

you will see that, in this case, the first $C$ (starting from the left of the molecule) has an ON of -3, the center $C$ has an ON of -2, and the third $C$ has an ON of -3. The average oxidation number for $C$ in this structure is

$\frac{- 3 + \left(- 2\right) + \left(- 3\right)}{3} =$ $\text{-8/3}$

It is very important to try and avoid fractional oxidation numbers, since oxidation numbers are theoretical values used to do a sort of electron bookkeeping; oxidation numbers allow you to compare how many electrons an atom "owns" in a molecule or an ion, as opposed to how many valence electrons it has by itself.

Another example of fractional oxidation states in a compound is $F {e}_{3} {O}_{4}$. Since $O$ has an ON of -2, the ON of $F e$ turns out to be

$3 \cdot \text{ON"_(Fe) + 4*(-2) = 0=> "ON"_(Fe) = "+8/3}$

This indeed represents an average oxidation state for the $F e$ atoms in this structure, since $F {e}_{3} {O}_{4}$ has 2/3 of the iron ions as $F {e}^{3 +}$ and 1/3 as $F {e}^{2 +}$; the actual formula for $F {e}_{3} {O}_{4}$ is actually $F e O \cdot F {e}_{2} {O}_{3}$.

So, if you have an ionic compound, you can keep the different oxidation states of an atom separated by fragmenting the ionic compound into its respective ions.

Organic compounds like propane can be a little tricky at first, but writing their structural molecular formulas will allow you to determine the oxidation number of, for example, each $C$ atom.