I am trying to understand formal charge and how it is being calculated. I have copied the image from organic chemistry as a second language below:

How do I calculate formal charge? In the book formal charge is calculated by adding up the electrons where each bond counts as 1 electron, so in the image for the oxygen with a formal charge of -1, we get a total of 7 electrons which is greater than the 6 that oxygen originally has. But for the octet rule I count the bond as 2 electrons (total of 8 electrons), so I am confused as to how the formal charge calculation calculation is being made?

  • 4
    $\begingroup$ Formal charge and octet rule follow a different formalism. This shouldn’t confuse you, you should simply keep the concepts separate. Note that oxidation state follows yet another formalism (all electrons given to the electronegative partner). $\endgroup$
    – Jan
    Commented Feb 23, 2017 at 21:29

2 Answers 2


As Jan hinted in the comments, there are multiple electron bookkeeping methods which serve different uses. We want to keep track of electrons because it gives us an idea of what type of chemical behavior might be attributable to an atom or molecule.

Octet Rule

The Octet Rule (and the related 18 electron rule for some transition metal compounds) serves to give us some idea of whether the structure proposed meets quantum mechanical requirements for valid structures. It also gives us a quick estimate of reactivity.

As an example, second period elements do not exceed the Octet Rule, so you know your structure is invalid if you have 14 electrons on an oxygen atom.

Boron compounds often do not have complete octets on boron (for example $\ce{BH3}$), but that leads to boron compounds being powerful electrophiles and having some weird bonding situations (example: diborane $\ce{B2H6}$).

Both Lewis bond theory and valence bond theory attributes all shared bonding electrons to count toward the total number of electrons for an atom.

Formal Charge

Formal charge is attempting to identify regions of high and low electron density within a structure, which help predict how some compounds react.

For formal charge, we act like each atom in a structure is an ion that has assigned to it all non-bonding electrons and half of all bonding electrons. For your structure, we can learn where the negative charge on the anion is located and where it is not.

Oxidation Number

Oxidation number is one final way to approximate electron distribution, and therefor the propensity to undergo certain kinds of reactions. Oxidation number is calculated by assuming all atoms in the structure are ionic and assigning all bonding electrons to the more electronegative atom in the pair.


So, for your compound, the anionic oxygen would meet the octet rule, have a formal charge of 1-, and an oxidation number of -2. The neutral oxygen would meet the octet rule, have a formal charge of 0, and an oxidation number of -2. If you needed to predict where a cation would be attracted to this compound, you could choose one of the two oxygen atoms appropriately.

The two carbon atoms would both have full octets and zero formal charge, but one has an oxidation number of -4 and the other has an oxidation number of +3. In this case, if you needed to predict where a reducing agent would react with this compound, you would be able to choose one carbon atom over the other.

For monoatomic species, the three counting methods will give you the same number of electrons and the formal charge will equal the oxidation number. As an exmaple: $\ce{O^2-}$ — 8 electrons, formal charge 2-, oxidation number -2.

  • $\begingroup$ So, octet rule gives us a quick estimate of reactivity (that means everything in period 2 reacts the same, O & C have same reactivity? Formal charge says how a reaction might occur? In the case of the cation, it will attack the O with FC -1 1st? I need to read more about oxidation number (haven't read it yet), but that is telling you where the highest distribution of e- are, hence it makes sense Oxygen would be more reactive in my compound. It seems that you are layering on different characterizations of a atom in order to model it's likely behavior in the system (aka the molecule)? $\endgroup$
    – ahat
    Commented Feb 24, 2017 at 6:53
  • $\begingroup$ chemistry.stackexchange.com/questions/68875/… This one helps: basically the bonds are a combination of ionic and covalent bonds, pure covalent would share e- with both atoms while as pure ionic would make 1 atom hog all the electrons. So would it be correct to think of them as a components of a vector and that vector describes the type of bond? Just like how AC current has 2 parts to it magnitude and phase? $\endgroup$
    – ahat
    Commented Feb 24, 2017 at 7:10

The following is a quick check on for the formal charge:

formal charge = Valence Electrons - ½ (bonded electrons) - (non-bonded electrons)

Consider the following Lewis structure for sulfuric acid:

Lewis Structure - Sulfuric Acid

For the sulfur atom there are six valence electrons ($V=6$), eight bonded electrons ($B=8$) and zero non-bonded electrons ($N=0$). Therfore the formal charge of sulfur $q_{\ce{S}}^\mathrm{f}$ is $$q_{\ce{S}}^\mathrm{f} = V - \frac{B}{2} - N = 6 - \frac{8}{2} - 0 = +2$$

Oxygen has 6 valence electrons, and when it is terminal, it has 6 non-bonded electrons and 2 bonded electrons: $$q_{\ce{O}}^\mathrm{f} = V - \frac{B}{2} - N = 6 - \frac{2}{2} - 6 = -1$$

In the hydroxyl group it has 4 non-bonded electrons and 4 bonded electrons: $$q_{\ce{O\color{gray}{(H)}}}^\mathrm{f} = V - \frac{B}{2} - N = 6 - \frac{4}{2} - 4 = 0$$


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