How do oxidation numbers relate to electron configuration?

How do oxidation numbers relate to electron configuration

Oxidation numbers are not so much related to electron configuration as they are to the polarity of the bonds to the atoms.

Consider a carbon atom, for example. It could either accept or give away four electrons in order to gain a noble gas configuration. However, because of the way in which we count electrons for oxidation number purposes, carbon can have any of the oxidation numbers -4, -3, -2, -1, 0, +1, +2, +3, or +4. The atom still has eight valence electrons surrounding it, but we count the electrons according to certain arbitrary rules. We should therefore review the rules for calculating oxidation numbers.

1. Lone pair electrons belong entirely to the atom on which they are located.
2. Electrons that are shared between identical atoms are shared equally (each atom gets one electron).
3. Electrons that are shared between nonidentical atoms belong entirely to the more electronegative atom.

Let’s apply these rules to the carbon atoms in acetic acid.

The left hand C is attached to three H atoms and the other C atom. Since C is more electronegative than H, it gets all six electrons from the C-H bonds. Since it shares its electrons equally with the other C atom, it gets only one of these electrons. This gives carbon 7 valence electrons, which is 3 more than it normally has. The oxidation number of this C atom is -3.

The right hand C atom is attached to two (more electronegative) O atoms, so it loses all of these shared electrons to the O atoms. However, it retains the one electron from the shared bond with the other C. Since it now has only 1 valence electron, it has officially lost 3 electrons, and its oxidation number is +3.

The basic rule is:
The more electronegative atom gets ALL of the shared electrons.

Hope this helps.