# 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

• @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