How does formal charge relate to the number of valence electrons?

1 Answer
Nov 9, 2015

Oxidation/reduction reactions are conceived to occur in an atom's valence shell. Formal charge, therefore, relates to the presence or absence of extra valence electrons.


By way of example, oxygen has atomic number (#Z#) #=#8. The oxygen atom has 8 protons in its nucleus (this is what makes it an oxygen atom). The neutral atom therefore should have 8 electrons, but 2 of these are conceived to be inner core, and most of the time we depict oxygen in a molecule as #R-O-R#, with 2 lone pairs on the oxygen. These 2 lone pairs have a charge that is deemed to be associated entirely with the #O# atom; meanwhile the #O# centre is conceived to have half a share of the 2 electrons that form the #C-O# bond, i.e. 1 electron. So, for the assignment of formal charge, the oxygen atom has 8 protons in its nucleus, 8 positive charges, balanced by 2 inner shell electrons, plus the 2 electrons from the covalent bonds (this is 1 electron from each #O-C# bond, plus the 4 electrons from the 2 lone pairs, whose charge devolves to the #O# atom solely.

Now let's take this example further. Oxygen in alcohol, #R-O-H#, "owns" 6 valence electrons, and is neutral. Oxygen in alkoxide (or hydroxide), "owns" 7 valence electrons, and is therefore formally represented as #R-O^-#, with the negative charge localized on the oxygen atom (3 lone pairs + 1 electron from the #C-O# bond). The oxide dianion, #O^(2-)#, that occurs in many salts but is unstable in water (why?), has formally 8 valence electrons. If I have been unclear or hard to follow please state your objections, and I'll have another go.