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

1 Answer
Nov 9, 2015

Answer:

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.

Explanation:

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.