# Difference between oxidation and formal charge

What is the difference between oxidation number and formal charge? I am confused about the two.

• Those are both assigned values referring to the definitions, but think of it this way: formal charge assigns charge according to the electrons around the nucleus, assuming that electrons in all chemical bonds are shared equally between atoms. Well, the oxidation number doesn't go for the same assumption, making that $\ce{CO2}$ has for the carbon atom a formal charge of $0$ and an oxidation number of $+4$ – L3ul Mar 20 '16 at 14:12

This is best explained in an example. Let’s take methyl isocyanide:

$$\ce{\overset{-}{C}#\overset{+}{N}-CH3}$$

This structural formula already includes the formal charges. How do we arrive at these? First, we draw a Lewis structure with the correct, required connectivity.

In this case, we have to know that a $\ce{CN}$ fragment is connected to the methyl group via the nitrogen. We can also find out that we need to form a total of seven bonds, three of which must be $\ce{C-H}$ bonds in the methyl group. Meaning that from the intermediate of $\ce{C-N-CH3}$ we need to add two more bonds but the carbon on the right-hand side already has a completed octet so these can only be between nitrogen and the left-hand carbon.

We also know that we should have a single lone pair; the only atom that can accomodate this is the left-hand carbon. All atoms now have a completed octet, our new intermediate solution is $\ce{C#N-CH3}$.

To determine formal charges, each bond is split homolytically, meaning that each bonding electron pair is split apart giving each atom one electron. Lone pairs are left alone, since they are, well, lone. The total we get is compared to the number of electrons an atom should have according to its position in the periodic table. This gives us:

$$\begin{array}{lcccc}\hline \text{atom} & n(\ce{e-}) & \text{valence electrons} & \Delta & \text{charge}\\ \hline \ce{C} & 5 & 4 & 1 & -1 \\ \ce{N} & 4 & 5 & -1 & +1 \\ \ce{C(H3)} & 4 & 4 & 0 & \pm0 \\ \hline\end{array}$$

This concludes the determination of formal charges; the left-hand carbon gets a single formal negative charge while the nitrogen gets a single formal positive charge.

For oxidation states, a different method is employed. Here, all bonds are split heterolytically in such a manner that the more electronegative partner gets both electrons. Nitrogen is more electronegative than carbon which in turn is more electronegative than hydrogen. The resulting number is again compared to the number an atom should have; it will likely be different than before. This method will give us the following:

$$\begin{array}{lcccc}\hline \text{atom} & n(\ce{e-}) & \text{valence electrons} & \Delta & \text{oxidation state} \\ \hline \ce{C} & 2 & 4 & -2 & \mathrm{+II} \\ \ce{N} & 8 & 5 & 3 & \mathrm{-III} \\ \ce{C(H3)} & 6 & 4 & 2 & \mathrm{-II} \\ \hline \end{array}$$

If we add the three $\mathrm{+I}$ of the three hydrogens, we arrive at a neutral $\pm0$ again.

It is important to note that both resuts are quite different and the only thing in common is that they both add up to zero for a neutral molecule or the ionic charge of a molecular ion (unless you did it wrong). The formal charge rarely has any non-formal meaning except in ammonium derivatives where it denotes the actual centre of positive charge. The oxidation states corresponds to the actually observed electron density in an atom somewhat.

Both however are formal concepts and you need to know how to cope with them to use them to your advantage.