Bonding, Structure, and Resonance
A Key Skill: How to Calculate Formal Charge
Last updated: May 22nd, 2023 |
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
1. Formal Charge
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).
For that reason formal charge isn’t always a good guide to where the electrons actually are in a molecule and can be an unreliable guide to reactivity. We’ll have more to say on that below.
2. Simple Examples For First-Row Elements
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.
3. Formal Charge Calculations When You Aren’t Given All The Details
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!)
4. Some Classic Formal Charge Questions
We can use the exact same formal charge formula, above, along with the rules for implicit lone pairs and hydrogens, to figure out the formal charge of atoms in some pretty exotic-looking molecules.
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.
5. Formal Charges and Curved Arrows
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 2. Nitrenes are an exception. Another exception is when we want to draw bad resonance forms.
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.
(Advanced) References and Further Reading
1. Valence, Oxidation Number, and Formal Charge: Three Related but Fundamentally Different Concepts
Journal of Chemical Education 2006 83 (5), 791
2. Lewis structures, formal charge, and oxidation numbers: A more user-friendly approach
John E. Packer and Sheila D. Woodgate
Journal of Chemical Education 1991 68 (6), 456
57 thoughts on “A Key Skill: How to Calculate Formal Charge”
Your explanations and examples were clear and easy to understand. I appreciate the detailed step-by-step instructions, which made it easy to follow along and understand the concept. Thank you for taking the time to create this helpful resource
I think for Quiz ID: 2310, the formal charge for the carbon in the fourth molecule should be +1 instead of -1.
Fixed. Thanks for the spot!
Thank you so much sir. Finally i understood how to calculate the formal charge
Nice simple explanation
Great teaching , can I know where did u studied ??
Hi I am extremely confused. The two formulas for calculating FC that you provided are not the same and don’t produce the same results when I tried them out.
Formal charge = [# of valence electrons] – [electrons in lone pairs + 1/2 the number of bonding electrons]
Formal Charge = [# of valence electrons on atom] – [non-bonded electrons + number of bonds].
They do not produce the same result…
If I have the formula BH4, and use the first formula provided to find FC of B, I would get:
(3) – (0 + 2) = +1
Using the second formula provided:
(3) – (0+4) = -1
Aren’t these formulas supposed to produce the same results? I am quite confused and I don’t know if I missed something.
Ah. I should have been more clear. The number of bonding electrons in BH4 equals 8, since each bond has two electrons and there are 4 B-H bonds. Half of this number equals 4. This should give you the same answer.
I have updated the post to make this more explicit.
That was the best i have seen but i have a problem with the formula,i think the side where the shared pair electrons came was suppose to be negative but then yours was positive,so am finfding it difficult to understand because the slides we were given by our lecturer shows that it was subtracted not added. i would love it when u explain it to me.
It was a very great explanation! Now I have a good concept about how to find formula charge.
And also i am just a grade nine student so i want to say thank you for this.
YOU ARE THE BEST. I GOT THE HIGHEST MARK IN MY FIRST QUIZ, AND I KNOW THAT THROUGH THIS I WILL GET THE BEST IN MIDTERM AND FINAL. I want you guys to go on youtube and follow the steps. THANK YOU VERY MUCH.
I remember learning that in the cyanide ion, the carbon is nucleophilic because the formal negative charge is on carbon, not nitrogen, despite nitrogen being more electronegative. So I think a different explanation could me more accurate, but I’m not sure how to properly address it. I better keep reading.
In cyanide ion, there are two lone pairs – one on carbon, one on nitrogen. The lone pair on carbon is more nucleophilic because it is less tightly held (the atom is less electronegative than nitrogen).
On all the examples I show that are negatively charged (eg BH4(-) ) there isn’t a lone pair to complicate questions of nucleophilicity.
This really helped for neutral covalent molecules. However, I’m having trouble applying this technique for molecules with an overall charge other than 0. For instance, in (ClO2)- , the formal charge of Cl should be 1. However, with your equation the charge should be 0. With the conventional equation, the charge is indeed 1.
I’d appreciate it if you replied sooner rather than later, as I do have a chemistry midterm on Friday. I’m quite confused with formal charges :)
Thanks for the study guide.
This method is wrong
For CH3 , the valence eloctron is 4 , no : of bonds is 3 and no of non bonded electrons is 1
Then by this equation
F.C= 4-(1+3) = 0 but here it is given as +1
That analysis would be accurate for the methyl radical. However it fails for the methyl carbocation.
That example referred to the carbocation. For the methyl radical, the formal charge is indeed zero.
This was so helpful n the best explanation about the topic…
Thanks for the easy approach.
But when I used this formula it works. Thus
#valence electrons_#lone pair__#1/2.bond pairs
Thanks for the easy approach.
I have a problem in finding the FC on each O atom in ozone. Can you help me with that ASAP?
The FC on central atom would be +1 because [6-(2+3)]
FC on O atom with coordinate bond would be: -1 because [6-(6+1)].
FC on O atom with double bond is: 0 because [6-(4+2)].
Hope I solved your question!
Thank u very much my exam is today and i wouldn’t pass without this information
AM REALLY LOST NOW
ON THAT EXAMPLE OF CH3
# OF VALENCE ELECTRON=4
# OF BONDING=3
# OF UNSHARED=1
SO WHEN I CALCULATE
FORMAL CHARGE=(#OF VALENCE ELEC)+[(1/2#OF BOND)+(#OF UNSHARED)]
PLZ HELP IF AM MAKING MISTAKE
Should be 1/2 [# of bonding ELECTRONS] + # unshared.
This gives you 4 – [3 – 1]
= 0 for ch3 radical.
Should be for CH3(+), not the methyl radical •CH3 .
I am beryllium and i got offended!!!!!!……..LOL Just kidding…….BTW, I found this article very useful.Thanks!!!!!!!!!!
what does it means if we determine a molecule with zero charge
you said that non bonded electrons in carbon is 2, but how ?
because i see it as only 1 because out of the 4 valence electrons in carbon, three are paired with hydrogen so it’s only 1 left
If the charge is -1, there must be an “extra” electron on carbon – this is why there’s a lone pair. If there was only one electron, it would be neutral.
This works! I would take your class with organic chemistry if you are a professor. I am taking chemistry 2 now. Organic is next.
Thank you so much!
Thank you very very more for the simple explanation! Unbelievably easy and saves so much time!!!!!!
Thank you!!! this was awesome, I’m a junior in chemistry and this finally answered all my questions about formal charge :)
Glad it was helpful Haley!
If formal charges bear no resemblance to reality, what are their significance?
I hope the post doesn’t get interpreted as “formal charges have no significance”. If it does I will have to change some of the wording.
What I mean to get across is that formal charges assigned to atoms do not *always* accurately depict electron density on that atom, and one has to be careful.
In other words, formal charge and electron density are two different things and they do not always overlap.
Formal charge is a book-keeping device, where we count electrons and assign a full charge to one or more of the atoms on a molecule or ion.
Electron density, on the other hand, is a measurement of where the electrons actually are (or aren’t) on a species, and those charges can be fractional or partial charges.
First of all, the charge itself is very real. The ions NH4+ , HO-, H3O+ and so on actually do bear a single charge. The thing to remember is that from a charge density perspective, that charge might be distributed over multiple atoms.
Take an ion like H3O+, for example. H3O *does* bear a charge of +1,
However, if one thinks about where the electrons are in H3O+, one realizes that oxygen is more electronegative than hydrogen, and is actually “taking’ electrons from each hydrogen. If you look at an electron density map of H3O+ , one will see that the positive charge is distributed on the three hydrogens, and the oxygen actually bears a slight negative charge. There’s a nice map here.
When we calculate formal charge for H3O+, we assign a charge of +1 to oxygen. This is for book keeping reasons. As a book-keeping device, it would be a royal pain to deal with fractions of charges like this. So that’s why we calculate formal charge and use it.
Sometimes it does accurately depict electron density. For example, in the hydroxide ion, HO- , the negative charge is almost all on the oxygen.
If you have a firm grasp of electronegativity then it becomes less confusing.
Does that help?
There are meny compounds which bears various structure among these which one is more stable or less energetic is it possible to predicu from the formal charge calculation?
Hey great explanation. I have a question though. Why is the FC commonly +/- 1? Could you give me an example when the FC is not +/- 1? Thanks.
Sure, try oxygen with no bonds and a full octet of electrons.
Great!i can use this for my exam!thanks!
Shouldn’t the formal charge of CH3 be -1? I was just wondering because in your example its +1 and in the chart its -1.
In the question.. its mentioned that CH3 without any lone pairs.. which means the valence would be 4 but there will not be any (2electrons) lone pairs left.. Hence it will be (4-)-(0+3)= 1
In CH3 i think FC on C should be -1 as carbon valency is 4 it has already bonded with 3 hydrogen atom one electron is left free on carbon to get bond with or share with one electron H hence, number of non bonded electrons lone pair of electrons is considered as 2. 4-(2+3) = -1.
In your case if we take 0 than valency of c is not satisfied.
thank you for collaboration of formal charge
The answer to the question in the post above is “carbenes” – they have two substitutents, one pair of electrons, and an empty p orbital – so a total of four electrons “to itself”, making it neutral.
thank you for excellent explanation
Glad you found it useful Peter!
Very good explanation.I finally understood how to calculate the formal charge,was having some trouble with it.Thanks:)
Glad you found it helpful.
nice, concise explanation
the sheet posted by u is really very excellent.i m teacher of chemistry in india for pre engineering test.if u send me complete flow chart of chemistry i will great full for u