# Why are electrons shared equally when calculating formal charge, but unequally when calculating oxidation state?

• When calculating formal charge - electrons are shared equally between the atoms in the bond.
• When calculating oxidation state - electrons are both given to the most electronegative atom.

Why is this different?

I am assuming it is something to do with the intended use of calculating formal charge and oxidation state, which I believe are:

• Formal charge can be used to find the most stable structure.
• Oxidation numbers can be used to determine what will become oxidised in a redox reaction.

However, even based on their uses, I am unsure why they are calculated differently.

*Images from Wikipedia

• Are talking about a complex, for instance like $\ce{FeCl4^−}$ ? – MaxW Feb 19 '17 at 22:47
• I am unsure what a the implications of something being a complex are (I just googled what a complex itself referred to). But my question was for a compound such as $\ce{CO2}$. I have added some images to clarify. Thank you. – K-Feldspar Feb 19 '17 at 22:53

The oxidation state is a form to distribute the charge of atoms in molecules considering to have only ionic bonds. In contrast, the formal charge is the distribution of the electrons considering to have only 100% electron-pair covalent bonds.

Actually, the formal charge is not very realistic and is more or less a bookkeeping tool. Now there exists theoretical and computational methods to calculate more reasonably charges on atoms from electronic structure calculations like Hartree-Fock, post-HF, semi-empirical and density-functional methods.

One example, taken from Anslyn's Modern Physical Organic Chemistry, is the case of the tetramethylammonium. One would place a positive formal charge on the nitrogen atom, but this is more electronegative than carbon, so there is a contradiction. The calculations mentioned before, indicate that the nitrogen is essentially neutral and the positive charge resides on the methyl groups. So the formal charge is more adequate for indicate the charge on the molecule and not on the atoms.

• Nice explanation to cut through Gordian knot. I'd start off second paragraph with "No bond is purely ionic or purely covalent. Thus neither model is absolutely realistic and both are more or less ..." – MaxW Feb 20 '17 at 0:21
• @MaxW you're right. I was going to write it but I forgot. – Verktaj Feb 20 '17 at 0:30
• Thanks for the answer Andres. I am a bit confused when you say formal charge is for charge on the molecule. We still calculate the formal charge for each atom right (and then maybe sum them up to find the charge on the molecule). But in calculating the charge of each atom, why is 1 electron distributed to each atom in the covalent bond for formal charge, whereas both electrons are taken by the more electronegative atom for oxidation state? Why isn't the same process used for both formal charge and oxidation state? Thank you. – K-Feldspar Feb 20 '17 at 5:05
• @K-Feldspar the formal charge is only a convenient form to visualize an electron distribution in a Lewis structure. In the basics, the formal charge is what you know, I mean, the sum of the atoms in a neutral molecule is zero and so on. The thing is that charge is actually not what you "see" in, for example, electronic structure calculations. – Verktaj Feb 20 '17 at 14:16
• @K-Feldspar Think as you have two types of bonds, the ionic bond and the covalent bond. In covalent bonding, two atoms share one electron pair per bond. In ionic bonding, the electronegative atom take the electron from the electropositive. In formal charge, you only account for 100% covalent bonding, i.e. the electronegativity stuff is fully neglected. In oxidation states, you only account for asymmetrical electron distribution, associated with a 100% ionic bond, for example in Na$^{+}$Cl$^{-}$. By definition, formal charge and oxidation state accounts for different types of bonds. – Verktaj Feb 20 '17 at 14:20