Bonding, Structure, and Resonance
A Key Skill: How to Calculate Formal Charge
Last updated: December 13th, 2022 |
How To Calculate Formal Charge
To calculate the formal charge of an atom, we start by:
- evaluating the number of valence electrons (VE) the neutral atom has (e.g. 3 for boron, 4 for carbon, 5 for nitrogen, and so on). (note: this is also equivalent to the effective nuclear charge Zeff , the number of protons that an electron in the valence orbital “sees” due to screening by inner-shell electrons.)
- counting the number of non-bonded valence electrons (NBE) on the atom. Each lone pair counts as 2, and each unpaired electron counts as 1.
- counting the number of bonds (B) to the atom, or alternatively, counting the number of bonding electrons and dividing this by 2.
The formal charge FC is then calculated by subtracting NBE and B from VE.
FC = VE – (NBE + B)
which is equivalent to
FC = VE – NBE – B
The calculation is pretty straightforward if all the information is given to you. However, for brevity’s sake, there are many times when lone pairs and C-H bonds are not explicitly drawn out.
So part of the trick for you will be to calculate the formal charge in situations where you have to take account of implicit lone pairs and C-H bonds.
In the article below, we’ll address many of these situations. We’ll also warn you of the situations where the calculated formal charge of an atom is not necessarily a good clue as to its reactivity, which is extremely important going forward.
Table of Contents
- Formal Charge
- Simple Examples For First-Row Elements
- Formal Charge Calculations When You Aren’t Given All The Details
- Some Classic Formal Charge Problems
- Formal Charges and Curved Arrows
- Quiz Yourself!
- (Advanced) References and Further Reading
Formal charge is a book-keeping formalism for assigning a charge to a specific atom.
To obtain the formal charge of an atom, we start by counting the number of valence electrons [Note 1] for the neutral atom, and then subtract from it the number of electrons that it “owns” (i.e. electrons in lone pairs, or singly-occupied orbitals) and half of the electrons that it shares (half the number of bonding electrons, which is equivalent to the number of bonds)
The simplest way to write the formula for formal charge (FC) is:
FC = VE – NBE – B
- VE corresponds to the number of electrons around the neutral atom (3 for boron, 4 for carbon, 5 for nitrogen, 6 for oxygen, 7 for fluorine)
- NBE corresponds to the number of non-bonded electrons around the atom (2 for a lone pair, 1 for a singly-occupied orbital, 0 for an empty orbital)
- B is the number of bonds around the atom (equivalent to half the number of bonding electrons)
It’s called “formal” charge because it assumes that all bonding electrons are shared equally. It doesn’t account for electronegativity differences (i.e. dipoles).
When all the lone pairs are drawn out for you, calculating formal charge is fairly straightforward.
Let’s work through the first example in the quiz below.
- In the hydronium ion (H3O) the central atom is oxygen, which has 6 valence electrons in the neutral atom
- The central atom has 2 unpaired electrons and 3 bonds
- The formal charge on oxygen is [6 – 2 – 3 = +1 ] giving us H3O+
See if you can fill in the rest for the examples below.
If that went well, you could try filling in the formal charges for all of the examples in this table.
It will take some getting used to formal charge, but after a period of time it will be assumed that you understand how to calculate formal charge, and that you can recognize structures where atoms will have a formal charge.
Let’s deal with some slightly trickier cases.
When we draw a stick figure of a person and don’t draw in their fingers, it doesn’t mean we’re drawing someone who had a bad day working with a table saw. We just assume that you could fill in the fingers if you really needed to, but you’re skipping it just to save time.
Chemical line drawings are like stick figures. They omit a lot of detail but still assume you know that certain things are there.
- With carbon, we often omit drawing hydrogens. You’re still supposed to know that they are there, and add as many hydrogens as necessary to give a full octet (or sextet, if it’s a carbocation).
- If there is a lone pair or unpaired electron on a carbon, it’s always drawn in.
One note. If we draw a stick figure, and we do draw the fingers, and took the time to only draw in only 3, then we can safely assume that the person really does only have 3 fingers. So in the last two examples on that quiz we had to draw in the hydrogens in order for you to know that it was a carbocation, otherwise you would have to assume that it had a full octet!
Oxygen and nitrogen (and the halogens) are dealt with slightly differently.
- Bonds to hydrogen are always drawn in.
- The lone pairs that are often omitted.
- Nitrogen and oxygen will always have full octets. Always. [Note 2 – OK, two exceptions]
So even when the lone pairs aren’t drawn in, assume that enough are present to make a full octet. And when bonds from these atoms to hydrogen are missing, that means exactly what it seems to be: there really isn’t any hydrogen!
Try these examples:
Now see if you can put these examples together!
(Note that some of these are not stable molecules, but instead represent are resonance forms that you will encounter at various points during the course!)
Here are some classic formal charge problems.
Note that although the structures might look weird, the formal charge formula remains the same.
The formal charge formula can even be applied to some fairly exotic reactive intermediates we’ll meet later in the semester.
Don’t get spooked out. Just count the electrons and the bonds, and that will lead you to the right answer.
We use curved arrows to show the movement of electron pairs in reactions and in resonance structures. (See post: Curved Arrows For Reactions)
For example, here is a curved arrow that shows the reaction of the hydroxide ion HO(-) with a proton (H+).
The arrow shows movement of two electrons from oxygen to form a new O–H bond.
Curved arrows are also useful for keeping track of changes in formal charge. Note that the formal charge at the initial tail of the curved arrow (the oxygen) becomes more positive (from -1 to 0) and the formal charge at the final tail (the H+) becomes more negative (from +1 to 0).
When acid is added to water, we form the hydronium ion, H3O+.
Here’s a quiz. See if you can draw the curved arrow going from the hydroxide ion to H3O+.
If you did it successfully – congratulations!
But I’m willing to bet that at least a small percentage of you drew the arrow going to the positively charged oxygen.
What’s wrong with that?
There isn’t an empty orbital on oxygen that can accept the lone pair. If you follow the logic of curved arrows, that would result in a new O–O bond, and 10 electrons on the oxygen, breaking the octet rule.
Hold on a minute, you might say. “I thought oxygen was positively charged? If it doesn’t react on oxygen, where is it supposed to react?”
On the hydrogens! H3O+ is Brønsted acid, after all. Right?
This is a great illustration of the reason why it’s called “formal charge”, and how formal charge not the same as electrostatic charge (a.ka. “partial charges” or “electron density”).
Formal charge is ultimately a book-keeping formalism, a little bit like assigning the “win” to one of the 5 pitchers in a baseball game. [Note 3] It doesn’t take into account the fact that the electrons in the oxygen-hydrogen bond are unequally shared, with a substantial dipole.
So although we draw a “formal” charge on oxygen, the partial positive charges are all on hydrogen. Despite bearing a positive formal charge bears a partially negative electrostatic charge.
This is why bases such as HO(-) react at the H, not the oxygen.
Just to reiterate:
- Positive charges on oxygen and nitrogen do not represent an empty orbital. Assume that oxygen and nitrogen have full octets! [Note 2]
- In contrast, positive charges on carbon do represent empty orbitals.
Positive formal charges on halogens fall into two main categories.
We’ll often be found drawing halonium ions Cl+ , Br+, and I+ as species with six valence electrons and an empty orbital (but never F+ – it’s a ravenous beast)
It’s OK to think of these species as bearing an empty orbital since they are large and relatively polarizable. They can distribute the positive charge over their relatively large volume.
These species can accept a lone pair of electrons from a Lewis base, resulting in a full octet.
Cl, Br, and I can also bear positive formal charges as a result of being bonded to two atoms.
It’s important to realize in these cases that the halogen bears a full octet and not an empty orbital. They will therefore not directly accept a pair of electrons from Lewis bases; it’s often the case that the atom adjacent to the halogen accepts the electrons.
If you have reached the end and did all the quizzes, you should be well prepared for all the examples of formal charge you see in the rest of the course.
- Formal charge can be calculated using the formula FC = VE – NBE – B
- Line drawings often omit lone pairs and C-H bonds. Be alert for these situations when calculating formal charges.
- Positively charged carbon has an empty orbital, but assume that positively charged nitrogen and oxygen have full octets.
- The example of the hydronium ion H3O+ shows the perils of relying on formal charge to understand reactivity. Pay close attention to the differences in electronegativity between atoms and draw out the dipoles to get a true sense of their reactivity.
Note 1. Using “valence electrons” gets you the right answer. But if you think about it, it doesn’t quite make sense. Where do positive charges come from? From the positively charged protons in the nucleus, of course!
So the “valence electrons” part of this equation is more properly thought of as a proxy for valence protons – which is another way of saying the “effective nuclear charge”; the charge felt by each valence electron from the nucleus, not counting the filled inner shells.
Note 3. In baseball, every game results in a win or a loss for the team. Back in the days of Old Hoss Radborn, where complete games were the norm, a logical extension of this was to assign the win to the individual pitcher. In today’s era, with multiple relief pitchers, there are rules for determining which pitcher gets credited with the win. It’s very possible for a pitcher to get completely shelled on the mound and yet, through fortuitous circumstance, still be credited for the win. See post: Maybe They Should Call Them, “Formal Wins” ?
In the same way, oxygen is given individual credit for the charge of +1 on the hydronium ion, H3O+, even though the actual positive electrostatic charge is distributed among the hydrogens.
Note 4. This image from a previous incarnation of this post demonstates some relationships for the geometry of various compounds of first-row elements.