Published By Vishal Goyal | Last updated: December 29, 2022
How to calculate formal charges of ozone (O3) with lewis structure?
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In covalently bonded molecules, formal charge is the charge assigned to an atom based on the assumption that the bonded electrons are equally shared between concerning atoms, regardless of their electronegativity.
The overall formal charge present on a molecule is a measure of its stability.
The fewer the formal charges present on the bonded atoms in a molecule (close to zero), the greater the stability of its Lewis structure.
In this article, we will calculate the formal charges present on bonded atoms in the different resonance structures of ozone (O 3 ) and also in its best possible Lewis structure. We will also determine the overall charge present on O 3 .
So for all this interesting information, continue reading!
Page Contents show 1 How to calculate the formal charges on O3 atoms? 2 FAQ 3 Summary
How to calculate the formal charges on O 3 atoms?
The formal charges can be calculated using the formula given below:
The formal charge of an atom = [valence electrons of an atom – non-bonding electrons – ½ (bonding electrons)]
- The valence electrons (V.E) of an atom are the total number of electrons present in its valence shell. Valence electrons can be determined by locating the position of the elemental atom in the Periodic Table.
- Non-bonding electrons (N.E) are the number of lone pairs present on the atom. (1 lone pair means 2 nonbonding electrons).
- Bonding electrons (B.E) are the total electrons shared with the atom via covalent chemical bonds. (1 single bond means 2 bonding electrons).
Now let us use this formula and the ozone Lewis structure is given below to determine the formal charges on three bonded oxygen (O) atoms in O 3 .
The above Lewis structure displays a total of 18 valence electrons. An oxygen (O) atom is present at the center. It is bonded to 2 other O-atoms via a single and a double covalent bond, respectively.
The central O-atom contains 1 lone pair of electrons. The single-bonded O-atom contains 3 lone pairs, while 2 lone pairs are present on the double-bonded O-atom.
Let’s find out how we can determine the formal charges present on each atom in the ozone (O 3 ) Lewis structure.
For the central Oxygen atom
- Valence electrons of Oxygen = It is present in Group VI A = 6 valence electrons
- Bonding electrons = 1 double bond + 1 single bond = 1(4) + 1(2) = 6 electrons
- Non-bonding electrons = One lone pair = 2 electrons
- Formal charge on the central Oxygen atom = 6 – 2 – 6/2 = 6 – 2 – 3 = 6 – 5 = +1
∴ The formal charge on the central O-atom in O 3 is +1.
For double-bonded oxygen atom
- Bonding electrons = 1 double bond = 4 electrons
- Non-bonding electrons = 2 lone pairs = 2(2) = 4 electrons
- Formal charge on the double bonded Oxygen atom = 6 – 4 –4/2 = 6 – 4 – 2 = 6 –6 = 0
∴ The formal charge on the double-bonded O-atom in O 3 is 0.
For each single-bonded oxygen atom
- Bonding electrons = 1 single bond = 2 electrons
- Non-bonding electrons = 3 lone pairs = 3(2) = 6 electrons
- Formal charge on the single bonded Oxygen atom = 6 – 6 – 2/2 = 6 – 6 – 1 = 6 –7 = -1
∴ The formal charge on the single-bonded O-atom in O 3 is -1.
This calculation shows that zero formal charges are present on double-bonded O-atom in O 3 . However, a +1 and a -1 formal charge is present on the other two O-atoms.
A +1 formal charge cancels with -1; therefore, the overall charge present on the molecule is zero.
The actual O 3 structure is a hybrid of the following resonance structures. Each resonance structure is equivalent. It is due to the presence of formal charges on the bonded atoms that the double bond keeps shifting from one position to another to give the best possible O 3 Lewis representation, as shown below.
You must keep in mind that a double bond cannot be formed on both sides of the central O-atom at any one time in the O 3 Lewis structure.
This is because oxygen can accommodate only a total of 8 electrons in its valence shell, unlike sulfur or phosphorus atoms that have an expanded octet.
Also, check –
- How to draw O 3 lewis structure?
- Formal charge calculator
- SO 3 formal charge
- CO 2 formal charge
- HCN formal charge
- SO 4 2- formal charge
- PO 4 3- formal charge
- SO 3 2- formal charge
- CN – formal charge
- SO 2 formal charge
- ClO 3 – formal charge
- SCN – formal charge
- POCl 3 formal charge
- NH 3 formal charge
- CO formal charge
- H 2 O formal charge
- NH 4 + formal charge
- H 3 O + formal charge
- OH – formal charge
- HSO 4 – formal charge
- ClO – formal charge
- BH 4 – formal charge
- N 3 – formal charge
- H 2 SO 4 formal charge
- NCO – formal charge
- NO 3 – formal charge
- NO 2 – formal charge
- CH 3 formal charge
- The best possible Lewis structure of a molecule is the one in which the bonded atoms carry formal charges as close to zero as possible.
- The formal charge formula is [ V.E – N.E – B.E/2].
- In O 3 , a +1 formal charge is present on the central O-atom.
- The double-bonded O-atom also has zero formal charges in O 3 .
- The single-bonded O-atoms have a -1 formal charge in O 3 .
- The overall formal charge on O 3 is also zero as +1 cancels with -1.
About the author
Vishal Goyal
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/
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O3 Lewis Structure, Molecular Geometry, Hybridization, and MO Diagram
We have all heard of the ozone layer depletion, haven’t we? Due to vast global warming and the rapid increase of temperature on earth, the ozone layer of the stratosphere has a hole in it.
This causes severe climate change and environmental damage. A pale blue gas with a molar mass of 47.99 g/ml, this molecular compound is often termed the activated oxygen.
It can be useful for killing bacterial growth and also releases a pungent smell.
Formed from the dioxide molecule, this molecule having three oxygen atoms is very crucial from the chemistry point of view. If you want to dive into his molecule, let us fasten our seat belts!
Because I am going to make you travel through all the essential concepts and explanations related to bonding within ozone.
Lewis Structure
To be very precise, Lewis Structure is the name given to the structural representation of a molecule. It is the diagrammatic layout for understanding the nitty-gritty of chemical bonding.
A very essential concept of molecular chemistry, the following steps dictate how you can successfully draw Lewis Structure:
The initial step towards forming this structure is to find out the total number of valence electrons .
‘+’ stands for positive charge i.e giving away(loss) of electrons.
‘-’ stands for the gain of electrons, or in other words, negative charge.
While calculating the valence electrons, we need to work with these two signs.
We now need to determine the central atom . How can we do so? We can easily find the solution to this with one simple trick!
First, point out the least electronegative atom. You can check this out by calculating the valence number. The one with the highest valence usually has the least electronegativity.
This atom will consist of higher sites of bonding compared to the others.
In this step, the task is to visualize the position of single bonds present in the molecule as a whole to the central atom.
This is carried out by sketching the skeleton diagram of the respective molecules as per requirements
Do you know that when atoms contain less than eight electrons in their outermost electron shell, they are still in their reactive state?
Hence, they react accordingly and tend to form more stable molecular compounds. So, the octet rule is based on the fact that every atom should have eight electrons in its valence shell.
The fourth step of Lewis Structure formation is based on achieving this. Starting with the electropositive ones, slowly fulfill the octet of the atoms.
Once, octet fulfillment has been done, we now need to find out if bond formation is left. Accordingly, multiple bond formation can be done.
We are now done with all kinds of bond formation, The last step is focused on the formal charge concept.
The bond formation (single and multiple ones) leads us to the final step of the process, i.e. the sixth step. Here, we will focus on calculating the formal charge.
We need to check whether all the atoms inside the given molecule are maintained at their least formal charge.
Below is the formula for formal charge:
Lewis Structure of O3
Here, we will be dealing with ozone, the molecular formula is O3.
The below discussion, therefore, will be based on finding out the Lewis Structure of O3.
Ozone consists of three oxygen atoms. Oxygen belongs to group VI of the periodic table with an atomic no of 8. It thus has 6 valence electrons.
Thus, the total number of valence electrons in ozone= 3*6 = 18
Just like triiodide ion where all the atoms are iodine, here, all the atoms are oxygen. So, we will just consider one of the three to be the central atom and place the other two on lateral sides.
Now, we will draw the skeletal structure of ozone based on step no. 3. While drawing, let us now place the valence shell electrons( total count=18) for octet fulfillment.
As discussed above, we can place six electrons surrounding each oxygen as per the periodic table knowledge. Have a look at the above diagram now.
You can see that the side oxygen atoms have both achieved octet. They both have eight electrons surrounding them. But, the central atom only has six electrons around itself.
So, to fulfill the octet rule, what we need to do is:
We have to shift two electrons from either one lateral oxygen atom and place it beside the central oxygen atom.
The octet formation is now successful.
We now have one double bond and one single bond concerning the central O atom.
Since we could have drawn, either way, we now have resonance structures. After checking the formal charge, the final Lewis Structure or electron dot structure of O3 has been done.
Hybridization of O3
What do we mean by hybridization? Why is this such a common topic in chemical bonding and why do we need to study this?
Well, hybridization is one of the vast and major topics in molecular chemistry.
It refers to the process of intermixing orbitals to form hybrid orbitals. How and why several atoms tend to combine with one another forms the basis of hybridization.
Hence, the study is important to know more in detail about a molecule and its properties.
To learn about ozone, therefore, we need to have some knowledge about its hybridization.
Now, how can we find out here?
How many electrons does the central oxygen atom have? 8.
2s orbital has two electrons and the rest six are present in 2px and 2py.
Total no of orbitals= 1s and 2p
Hence, the hybridization of the O3 molecule is sp2.
Molecular Geometry of O3
To find out the molecular geometry of ozone we need to check the VSEPR theory model.
How to proceed with this?
At the very beginning, you have to check the terminal atoms.
Terminal atom no.= 2
Now, find out the no of lone electrons or lone pairs
Lone electrons=2 in the central atom
Lone pair=1
After this, we have to match this with the VSEPR model graph
The ozone molecule is found to be bent trigonal planar shape due to the presence of resonance. Repulsion causes the bond angle to come to about 116 degrees.
The polarity concept is based on the distribution of positive and negative charges inside a molecule surrounding the constituent atoms.
The dipole moment is used to measure or calculate the polarity. This only has a net value when there is a charge difference.
In the case of ozone, the usual dipole moment value varies and there is the presence of partial + and + charges inside the molecule.
The ozone atom at the center will carry the partial+ + charge.
Dipole moments then are responsible for moving the ozone molecule in a downward direction The lone electron pair results in a net dipole in O3 and hence the ozone molecule is considered to be polar in nature.
O3 Molecular Orbital Diagram (MO)
The molecular orbital theory is one of the major revolutionary concepts of chemical bonding.
It uses quantum mechanics to give us a detailed almost explanatory diagram of the bonding nature inside a molecule.
Here is a diagrammatic representation of the MO diagram of ozone.
Ozone is a trigonal planar molecule. Hence, as we take one p orbital from each atom of oxygen(O3), we focus on the 4 electron H3- anion.
According to hybrid orbital approximation, we will consider the 2s, 2py, and 2pz orbitals. Then we will perform semi-empirical Molecular Orbital calculations and involve the concept of group orbitals.
Chemical bonding is indeed one of the vast and yet very important chapters for the entire life. If you are here to learn small details about any molecule, you have come to the right place.
Now that you have gone through the major concepts of the ozone molecule in detail, I hope you have got a basic understanding.
We have discussed Lewis structure and Hybridization. We have mentioned the shape and the polarity of the molecule and also the MO Diagram. So, did you have a good time?
I think so. Always keep learning.
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How to calculate formal charge
Table of Contents
How to calculate formal charge Examples
ot all atoms within a neutral molecule need be neutral. An atom can have the following charges: positive , negative , or neutral , depending on the electron distribution. This is often useful for understanding or predicting reactivity. Identifying formal charges helps you keep track of the electrons.
The formal charge is the charge on the atom in the molecule. The term “formal” means that this charge is not necessarily on the presented atom because in some cases, it is also prevalent on other atoms present in the molecule. It is actually spread out through the other atoms and is not only on the one atom. Identifying a formal charge involves:
- Determining the appropriate number of valence electrons for an atom – This can be accomplished by inspecting the periodic table. The group number indicates the appropriate number of valence electrons for each atom
- Determining whether the atom exhibits the appropriate number of electrons – In the Lewis structure, determine whether some of the atoms show an unexpected number of electrons
The formal charge on an atom can be calculated using the following mathematical equation.
Lewis structures also show how atoms in the molecule are bonded. They can be drawn as lines (bonds) or dots (electrons). One line corresponds to two electrons . The nonbonding electrons, on the other hand, are the unshared electrons and these are shown as dots. One dot is equal to one nonbonding electron. The valence electrons are the electrons in the outermost shell of the atom.
CH 4 , methane
A number of non-bonding electrons: 0 for both H and C
[ Formal charge ] H = 1 – (1/2) × 2 – 0 = 0 ⇒ This applies to each hydrogen. These hydrogens are all zero.
[ Formal charge ] C = 4 – (1/2) × 8 – 0 = 0
⇒ This molecule is neutral .
CH 3 + , methyl cation
[ Formal charge ] H = 1 – (1/2) × 2 – 0 = 0 ⇒ This applies to each hydrogen. These hydrogens are all zero. [ Formal charge ] C = 4 – (1/2) × 6 – 0 = 4 – 3 – 0 = +1
⇒ This is a cation .
CH 3 – , methyl cation
A number of non-bonding electrons: 0 for H, 2 for C
[ Formal charge ] C = 4 – (1/2) × 6 – 2 = 4 – 3 – 2 = -1
⇒ This is a anion .
If you have any questions or would like to share your reviews on the How to calculate formal charge , then comment down below. I would love to hear what you have to think.
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This article illustrates the o3 lewis structure , How to draw the lewis structure, hybridization, shapes, charges, and some other properties of the o3 lewis structure.
The highly reactive gas ozone (o3) naturally exists in our atmosphere.
How to draw lewis structure for O3?
Draw the o3 lewis structure using different steps:
Step 1: calculate the total no. of Valence electron in the o3 lewis structure:
Step 2: draw the skeleton of the given molecule and find the central atom:, step 3: assign the valence electron:.
Thus total valence electrons are 18 in the o3 lewis structure molecule, assign them in between the three atoms of o3 with the help of dots. Complete the octet of each atom that is around the central atom.
Step 5: Perform the following octet for the central atom:
O3 lewis structure shape.
Due to the distortion in the o3 lewis structure , the O3 molecule shape is frequently said to be bent.
O3 lewis structure formal charges
The formal charge on O: 6-2-½ (6)= 1, thus the formal charge on o3 lewis structure is +1 on the central oxygen atomSimilarly, two adjacent oxygen atoms carry (-½ ) partial negative charge, and central oxygen carries +1 formal charge as shown in the figure below.
O3 lewis structure lone pairs
O3 hybridization.
Hybridization refers to joining two or more atomic levels with the same or different energies to form a new orbital. Ozone has only one central oxygen atom with eight electrons in its outermost shell.
O3 lewis structure resonance
Except for the placement of the electrons, these Lewis structures are identical.
From a stability standpoint, these structures are identical because each has a positive and negative formal charge on two oxygen atoms.
O3 lewis structure octet rule
The octet rule is applied to molecules when each atom becomes stable by gaining eight electrons in its outermost shell. So, in the case of the o3 lewis structure , the sideways oxygen follows the octet rule but the central atom has only six electrons.
O3 polar or nonpolar
The ozone molecule is thought to be polar because the lone electron pair produces a net dipole. The typical dipole moment value changes in an ozone molecule, and the molecule contains partial positive and negative charges.
O3 lewis structure bond angle
A double bond exists between the central oxygen atom and one of the lateral oxygen atoms in the O3 Lewis structure. In the O3 Lewis structure, a double bond exists between the central oxygen atom and one of the lateral oxygen atoms .
O3 lewis structure electron geometry
But the molecular geometry is bent.
O3 valence electrons
To calculate the total number of valence electrons provided by a given element, multiply the number of electrons in the valance shell by the number of atoms in that element.
An oxygen atom has six electrons in its valence shell. There are three oxygen atoms in an O3 molecule. As a result, the ozone molecule has a total of 18 valence electrons.
Conclusion:
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What is the Charge of O3 (Ozone)? And Why? if(typeof ez_ad_units!='undefined'){ez_ad_units.push([[468,60],'knordslearning_com-box-3','ezslot_5',132,'0','0'])};__ez_fad_position('div-gpt-ad-knordslearning_com-box-3-0');
So let’s calculate the formal charge of O3 (Ozone).
Calculating the formal charge of O3 using lewis structure
Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2
Formal charge on left Oxygen = Valence electrons – Nonbonding electrons – (Bonding electrons)/2 = 6 – 4 – (4/2) = 0
So the formal charge on left oxygen atom is 0.
So according to the formula of formal charge, you will get;
Formal charge on right Oxygen = Valence electrons – Nonbonding electrons – (Bonding electrons)/2 = 6 – 6 – (2/2) = 1-
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What are the formal charges in O 3 (ozone)?
Formal charge: the formal charge of an atom in a molecule is the charge which might exist on the atom if all bonding electrons were evenly shared. a formal charge value is equal to an atom's valence electrons deducting the number of electrons given to it. f . c . = [ total no . of valence e – in free state ] – [ total no . of non - bonding pair e – ( lone pair ) ] – 1 2 [ total no . of bonding e – ] structure of ozone: ozone has a dipole moment of 0 . 53 d and thus a polar molecule. the molecule can be described as a resonance hybrid with significant contributing structures, one with a single bond on one side and the other with a double bond. both sides have an overall bond order of 1 . 5 in this arrangement. formal charge in o 3 ( ozone): in an o 3 molecule, the formal charge on the middle oxygen atom( 2 ) is + 1 . f . c = 6 – 2 – 1 2 ( 6 ) f . c = 6 – 5 f . c = 1 in an o 3 molecule, the formal charge on the left oxygen atom( 3 ) is - 1 . f . c = 6 – 6 – 1 2 ( 2 ) f . c = 6 - 7 f . c = - 1 in an o 3 molecule, the formal charge on the right oxygen atom( 1 ) is 0 . f . c = 6 – 4 – 1 2 ( 4 ) f . c = 6 – 6 f . c = 0.
The formal charge on oxygen which is single bonded in ozone is:
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- Formal Charge
What is Formal Charge?
A formal charge (F.C. or q) is the charge assigned to an atom in a molecule in the covalent view of bonding, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity.
The formal charge is the difference between an atom's number of valence electrons in its neutral free state and the number allocated to that atom in a Lewis structure.
When choosing the optimum Lewis structure (or predominant resonance structure) for a molecule, it is important to keep the formal charge on each of the atoms as low as feasible.
The following equation can be used to compute the formal charge of an atom in a molecule:
F = V - L - \[\frac{B}{2}\]
F = Formal Charge
V = Valence Electron of the neutral atom in isolation
L = Number of non-bonding valence electrons on this atom in the molecule
B = Total number of electrons shared in bonds with other atoms in the molecule
Formula, Calculation, Importance, and Example
The formula for computing a formal charge is:
(Number of valency electrons in neutral atom)-(electrons in lone pairs + 1/2 the number of bonding electrons)
The number of bonding electrons divided by two equals the number of bonds that surround the atom, hence this expression can be reduced to:
Formal Charge = (number of valence electrons in neutral atom)-(non-bonded electrons + number of bonds)
Take the compound BH 4 or tetrahydrdoborate.
Boron (B) possesses three valence electrons, zero non-bonded electrons, and four bonds around it.
This changes the formula to 3-(0+4), yielding a result of -1.
Let us now examine the hydrogen atoms in BH4. One valence electron, zero non-bonded electrons, and one bond make up hydrogen.
In BH4, the formal charge of hydrogen is 1-(0+1), resulting in a formal charge of 0.
Calculate the formal charge on the following:
O atoms of O3
Cl atom in HClO4- ion
S atom in HSO4- ion
Ans: We are showing how to find a formal charge of the species mentioned.
Formal charge on O1: 6 – 6/2 – 2 = +1
Formal charge on O2: 6 – 4/2 – 4 = 0
Formal charge on O3: 6 – 2/2 – 6 = -1
Formal charge on Cl atom of HClO4 ion: 7 – 8/2 – 0 = 3
Formal charge on S atom of HSO4- ion: 6 – 8/2 – 0 = 2
Significance
Molecular Structure
An atom in a molecule should have a formal charge of zero to have the lowest energy and hence the most stable state. If there are numerous alternatives for a molecule's structure, this gives us a hint: the one with the least/lowest formal charges is the ideal structure.
While formal charge can indicate a molecule's preferred structure, the problem becomes more complicated when numerous equally preferred structures exist. This condition could point to resonance structures, especially if the structures have the same atom arrangement but different types of arrangements of bonds.
The formal charge of a molecule can indicate how it will behave during a process. A negative formal charge indicates that an atom is more likely to be the source of electrons in a reaction (a nucleophile). If it has a positive one, on the other hand, it is more likely to take electrons (an electrophile), and that atom is more likely to be the reaction's site.
It's also worth noting that an atom's formal charge differs from its actual charge. Formal charge ignores electronegativity and assumes that electrons in a bond are uniformly distributed.
It's only a courtesy that's utilized to make molecular structures and reaction mechanisms more understandable. The actual charge, on the other hand, is based on the electronegativities of the atoms and the polarity of the bonds and looks at the actual electron density.
Importance Of Formal Charge
Now that we know what is the formal charge and we are familiar with the process for calculating a formal charge, we will learn about its importance.
The formal charge is a theoretical concept, useful when studying the molecule minutely. It does not indicate any real charge separation in the molecule. This concept and the knowledge of ‘what is formal charge' is vital.
The formal charge is crucial in deciding the lowest energy configuration among several possible Lewis structures for the given molecule. Therefore, calculating formal charges becomes essential.
Knowing the lowest energy structure is critical in pointing out the primary product of a reaction. This knowledge is also useful in describing several phenomena.
The structure of least energy is usually the one with minimal formal charge and most distributed real charge.
Besides knowing what is a formal charge, we now also know its significance.
Fun Facts On Formal Charge
In organic chemistry, convention governs that formal charge is essential for depicting a complete and correct Lewis-Kekulé structure. However, the same does not apply to inorganic chemistry.
The structure variation of a molecule having the least amount of charge is the most superior.
The differences between formal charge and oxidation state led to the now widely followed and much more accurate valence bond theory of Slater and the molecular orbital theory of Mulliken.
Home / A Key Skill: How to Calculate Formal Charge
Bonding, Structure, and Resonance
By James Ashenhurst
- 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 Z eff , 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:
- 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 (H 3 O) 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 H 3 O +
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 , H 3 O + .
Here’s a quiz. See if you can draw the curved arrow going from the hydroxide ion to H 3 O+.
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? I f it doesn’t react on oxygen, where is it supposed to react ?”
On the hydrogens! H 3 O+ 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.
6. Halogens
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.
7. Conclusion
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 H 3 O+ 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.
Related Articles
- Partial Charges Give Clues About Electron Flow
- How To Use Electronegativity To Determine Electron Density (and why NOT to trust formal charge)
- How to apply electronegativity and resonance to understand reactivity
- Maybe they should call them, “Formal Wins” ?
- Common Mistakes: Formal Charges Can Mislead
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 , H 3 O+, 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.
1. Valence, Oxidation Number, and Formal Charge : Three Related but Fundamentally Different Concepts Gerard Parkin Journal of Chemical Education 2006 83 (5), 791 DOI : 10.1021/ed083p791
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 DOI : 10.1021/ed068p456
00 General Chemistry Review
- Lewis Structures
- Ionic and Covalent Bonding
- Chemical Kinetics
- Chemical Equilibria
- Valence Electrons of the First Row Elements
- How Concepts Build Up In Org 1 ("The Pyramid")
01 Bonding, Structure, and Resonance
- How Do We Know Methane (CH4) Is Tetrahedral?
- Hybrid Orbitals and Hybridization
- How To Determine Hybridization: A Shortcut
- Orbital Hybridization And Bond Strengths
- Sigma bonds come in six varieties: Pi bonds come in one
- The Four Intermolecular Forces and How They Affect Boiling Points
- 3 Trends That Affect Boiling Points
- Introduction to Resonance
- How To Use Curved Arrows To Interchange Resonance Forms
- Evaluating Resonance Forms (1) - The Rule of Least Charges
- How To Find The Best Resonance Structure By Applying Electronegativity
- Evaluating Resonance Structures With Negative Charges
- Evaluating Resonance Structures With Positive Charge
- Exploring Resonance: Pi-Donation
- Exploring Resonance: Pi-acceptors
- In Summary: Evaluating Resonance Structures
- Drawing Resonance Structures: 3 Common Mistakes To Avoid
- Bond Hybridization Practice
- Structure and Bonding Practice Quizzes
- Resonance Structures Practice
02 Acid Base Reactions
- Introduction to Acid-Base Reactions
- Acid Base Reactions In Organic Chemistry
- The Stronger The Acid, The Weaker The Conjugate Base
- Walkthrough of Acid-Base Reactions (3) - Acidity Trends
- Five Key Factors That Influence Acidity
- Acid-Base Reactions: Introducing Ka and pKa
- How to Use a pKa Table
- The pKa Table Is Your Friend
- A Handy Rule of Thumb for Acid-Base Reactions
- Acid Base Reactions Are Fast
- pKa Values Span 60 Orders Of Magnitude
- How Protonation and Deprotonation Affect Reactivity
- Acid Base Practice Problems
03 Alkanes and Nomenclature
- Meet the (Most Important) Functional Groups
- Condensed Formulas: Deciphering What the Brackets Mean
- Hidden Hydrogens, Hidden Lone Pairs, Hidden Counterions
- Don't Be Futyl, Learn The Butyls
- Primary, Secondary, Tertiary, Quaternary In Organic Chemistry
- Branching, and Its Affect On Melting and Boiling Points
- The Many, Many Ways of Drawing Butane
- Wedge And Dash Convention For Tetrahedral Carbon
- Common Mistakes in Organic Chemistry: Pentavalent Carbon
- Table of Functional Group Priorities for Nomenclature
- Summary Sheet - Alkane Nomenclature
- Organic Chemistry IUPAC Nomenclature Demystified With A Simple Puzzle Piece Approach
- Boiling Point Quizzes
- Organic Chemistry Nomenclature Quizzes
04 Conformations and Cycloalkanes
- Staggered vs Eclipsed Conformations of Ethane
- Conformational Isomers of Propane
- Newman Projection of Butane (and Gauche Conformation)
- Introduction to Cycloalkanes (1)
- Geometric Isomers In Small Rings: Cis And Trans Cycloalkanes
- Calculation of Ring Strain In Cycloalkanes
- Cycloalkanes - Ring Strain In Cyclopropane And Cyclobutane
- Cyclohexane Conformations
- Cyclohexane Chair Conformation: An Aerial Tour
- How To Draw The Cyclohexane Chair Conformation
- The Cyclohexane Chair Flip
- The Cyclohexane Chair Flip - Energy Diagram
- Substituted Cyclohexanes - Axial vs Equatorial
- Ranking The Bulkiness Of Substituents On Cyclohexanes: "A-Values"
- The Ups and Downs of Cyclohexanes
- Cyclohexane Chair Conformation Stability: Which One Is Lower Energy?
- Fused Rings - Cis-Decalin and Trans-Decalin
- Naming Bicyclic Compounds - Fused, Bridged, and Spiro
- Bredt's Rule (And Summary of Cycloalkanes)
- Newman Projection Practice
- Cycloalkanes Practice Problems
05 A Primer On Organic Reactions
- The Most Important Question To Ask When Learning a New Reaction
- The 4 Major Classes of Reactions in Org 1
- Learning New Reactions: How Do The Electrons Move?
- How (and why) electrons flow
- The Third Most Important Question to Ask When Learning A New Reaction
- 7 Factors that stabilize negative charge in organic chemistry
- 7 Factors That Stabilize Positive Charge in Organic Chemistry
- Nucleophiles and Electrophiles
- Curved Arrows (for reactions)
- Curved Arrows (2): Initial Tails and Final Heads
- Nucleophilicity vs. Basicity
- The Three Classes of Nucleophiles
- What Makes A Good Nucleophile?
- What makes a good leaving group?
- 3 Factors That Stabilize Carbocations
- Three Factors that Destabilize Carbocations
- What's a Transition State?
- Hammond's Postulate
- Grossman's Rule
- Draw The Ugly Version First
- Learning Organic Chemistry Reactions: A Checklist (PDF)
- Introduction to Addition Reactions
- Introduction to Elimination Reactions
- Introduction to Free Radical Substitution Reactions
- Introduction to Oxidative Cleavage Reactions
06 Free Radical Reactions
- Bond Dissociation Energies = Homolytic Cleavage
- Free Radical Reactions
- 3 Factors That Stabilize Free Radicals
- What Factors Destabilize Free Radicals?
- Bond Strengths And Radical Stability
- Free Radical Initiation: Why Is "Light" Or "Heat" Required?
- Initiation, Propagation, Termination
- Monochlorination Products Of Propane, Pentane, And Other Alkanes
- Selectivity In Free Radical Reactions
- Selectivity in Free Radical Reactions: Bromination vs. Chlorination
- Halogenation At Tiffany's
- Allylic Bromination
- Bonus Topic: Allylic Rearrangements
- In Summary: Free Radicals
- Synthesis (2) - Reactions of Alkanes
- Free Radicals Practice Quizzes
07 Stereochemistry and Chirality
- Types of Isomers: Constitutional Isomers, Stereoisomers, Enantiomers, and Diastereomers
- The Single Swap Rule
- Introduction to Assigning (R) and (S): The Cahn-Ingold-Prelog Rules
- Assigning Cahn-Ingold-Prelog (CIP) Priorities (2) - The Method of Dots
- Enantiomers vs Diastereomers vs The Same? Two Methods For Solving Problems
- Assigning R/S To Newman Projections (And Converting Newman To Line Diagrams)
- How To Determine R and S Configurations On A Fischer Projection
- The Meso Trap
- Optical Rotation, Optical Activity, and Specific Rotation
- Optical Purity and Enantiomeric Excess
- What's a Racemic Mixture?
- Chiral Allenes And Chiral Axes
- On Cats, Part 4: Enantiocats
- On Cats, Part 6: Stereocenters
- Stereochemistry Practice Problems and Quizzes
08 Substitution Reactions
- Introduction to Nucleophilic Substitution Reactions
- Walkthrough of Substitution Reactions (1) - Introduction
- Two Types of Nucleophilic Substitution Reactions
- The SN2 Mechanism
- Why the SN2 Reaction Is Powerful
- The SN1 Mechanism
- The Conjugate Acid Is A Better Leaving Group
- Comparing the SN1 and SN2 Reactions
- Polar Protic? Polar Aprotic? Nonpolar? All About Solvents
- Steric Hindrance is Like a Fat Goalie
- Common Blind Spot: Intramolecular Reactions
- The Conjugate Base is Always a Stronger Nucleophile
- Substitution Practice - SN1
- Substitution Practice - SN2
09 Elimination Reactions
- Elimination Reactions (1): Introduction And The Key Pattern
- Elimination Reactions (2): The Zaitsev Rule
- Elimination Reactions Are Favored By Heat
- Two Elimination Reaction Patterns
- The E1 Reaction
- The E2 Mechanism
- E1 vs E2: Comparing the E1 and E2 Reactions
- Antiperiplanar Relationships: The E2 Reaction and Cyclohexane Rings
- Bulky Bases in Elimination Reactions
- Comparing the E1 vs SN1 Reactions
- Elimination (E1) Reactions With Rearrangements
- E1cB - Elimination (Unimolecular) Conjugate Base
- Elimination (E1) Practice Problems And Solutions
- Elimination (E2) Practice Problems and Solutions
10 Rearrangements
- Introduction to Rearrangement Reactions
- Rearrangement Reactions (1) - Hydride Shifts
- Carbocation Rearrangement Reactions (2) - Alkyl Shifts
- Pinacol Rearrangement
- The SN1, E1, and Alkene Addition Reactions All Pass Through A Carbocation Intermediate
11 SN1/SN2/E1/E2 Decision
- Identifying Where Substitution and Elimination Reactions Happen
- Deciding SN1/SN2/E1/E2 (1) - The Substrate
- Deciding SN1/SN2/E1/E2 (2) - The Nucleophile/Base
- Deciding SN1/SN2/E1/E2 (3) - The Solvent
- Deciding SN1/SN2/E1/E2 (4) - The Temperature
- Wrapup: The Quick N' Dirty Guide To SN1/SN2/E1/E2
- Alkyl Halide Reaction Map And Summary
- SN1 SN2 E1 E2 Practice Problems
12 Alkene Reactions
- E and Z Notation For Alkenes (+ Cis/Trans)
- Alkene Stability
- Addition Reactions: Elimination's Opposite
- Selective vs. Specific
- Regioselectivity In Alkene Addition Reactions
- Stereoselectivity In Alkene Addition Reactions: Syn vs Anti Addition
- Markovnikov Addition Of HCl To Alkenes
- Alkene Hydrohalogenation Mechanism And How It Explains Markovnikov's Rule
- Arrow Pushing and Alkene Addition Reactions
- Addition Pattern #1: The "Carbocation Pathway"
- Rearrangements in Alkene Addition Reactions
- Bromination of Alkenes
- Bromination of Alkenes: The Mechanism
- Alkene Addition Pattern #2: The "Three-Membered Ring" Pathway
- Hydroboration - Oxidation of Alkenes
- Hydroboration Oxidation of Alkenes Mechanism
- Alkene Addition Pattern #3: The "Concerted" Pathway
- Bromonium Ion Formation: A (Minor) Arrow-Pushing Dilemma
- A Fourth Alkene Addition Pattern - Free Radical Addition
- Alkene Reactions: Ozonolysis
- Summary: Three Key Families Of Alkene Reaction Mechanisms
- Palladium on Carbon (Pd/C) for Catalytic Hydrogenation
- OsO4 (Osmium Tetroxide) for Dihydroxylation of Alkenes
- m-CPBA (meta-chloroperoxybenzoic acid)
- Synthesis (4) - Alkene Reaction Map, Including Alkyl Halide Reactions
- Alkene Reactions Practice Problems
13 Alkyne Reactions
- Acetylides from Alkynes, And Substitution Reactions of Acetylides
- Partial Reduction of Alkynes With Lindlar's Catalyst or Na/NH3 To Obtain Cis or Trans Alkenes
- Hydroboration and Oxymercuration of Alkynes
- Alkyne Reaction Patterns - Hydrohalogenation - Carbocation Pathway
- Alkyne Halogenation: Bromination, Chlorination, and Iodination of Alkynes
- Alkyne Reactions - The "Concerted" Pathway
- Alkenes To Alkynes Via Halogenation And Elimination Reactions
- Alkynes Are A Blank Canvas
- Synthesis (5) - Reactions of Alkynes
- Alkyne Reactions Practice Problems With Answers
14 Alcohols, Epoxides and Ethers
- Alcohols - Nomenclature and Properties
- Alcohols Can Act As Acids Or Bases (And Why It Matters)
- Alcohols - Acidity and Basicity
- The Williamson Ether Synthesis
- Williamson Ether Synthesis: Planning
- Ethers From Alkenes, Tertiary Alkyl Halides and Alkoxymercuration
- Alcohols To Ethers via Acid Catalysis
- Cleavage Of Ethers With Acid
- Epoxides - The Outlier Of The Ether Family
- Opening of Epoxides With Acid
- Epoxide Ring Opening With Base
- Making Alkyl Halides From Alcohols
- Tosylates And Mesylates
- PBr3 and SOCl2
- Elimination Reactions of Alcohols
- Elimination of Alcohols To Alkenes With POCl3
- Alcohol Oxidation: "Strong" and "Weak" Oxidants
- Demystifying The Mechanisms of Alcohol Oxidations
- Intramolecular Reactions of Alcohols and Ethers
- Protecting Groups For Alcohols
- Thiols And Thioethers
- Calculating the oxidation state of a carbon
- Oxidation and Reduction in Organic Chemistry
- Oxidation Ladders
- SOCl2 Mechanism For Alcohols To Alkyl Halides: SN2 versus SNi
- Alcohol Reactions Roadmap (PDF)
- Alcohol Reaction Practice Problems
- Epoxide Reaction Quizzes
- Oxidation and Reduction Practice Quizzes
15 Organometallics
- What's An Organometallic?
- Formation of Grignard and Organolithium Reagents
- Organometallics Are Strong Bases
- Reactions of Grignard Reagents
- Protecting Groups In Grignard Reactions
- Grignard Practice Problems: Synthesis (1)
- Grignard Reactions And Synthesis (2)
- Organocuprates (Gilman Reagents): How They're Made
- Gilman Reagents (Organocuprates): What They're Used For
- The Heck, Suzuki, and Olefin Metathesis Reactions (And Why They Don't Belong In Most Introductory Organic Chemistry Courses)
- Reaction Map: Reactions of Organometallics
- Grignard Practice Problems
16 Spectroscopy
- Degrees of Unsaturation (or IHD, Index of Hydrogen Deficiency)
- Conjugation And Color (+ How Bleach Works)
- Introduction To UV-Vis Spectroscopy
- UV-Vis Spectroscopy: Absorbance of Carbonyls
- UV-Vis Spectroscopy: Practice Questions
- Bond Vibrations, Infrared Spectroscopy, and the "Ball and Spring" Model
- Infrared Spectroscopy: A Quick Primer On Interpreting Spectra
- IR Spectroscopy: 4 Practice Problems
- 1H NMR: How Many Signals?
- Homotopic, Enantiotopic, Diastereotopic
- Diastereotopic Protons in 1H NMR Spectroscopy: Examples
- C13 NMR - How Many Signals
- Liquid Gold: Pheromones In Doe Urine
- Natural Product Isolation (1) - Extraction
- Natural Product Isolation (2) - Purification Techniques, An Overview
- Structure Determination Case Study: Deer Tarsal Gland Pheromone
17 Dienes and MO Theory
- What To Expect In Organic Chemistry 2
- Are these molecules conjugated?
- Conjugation And Resonance In Organic Chemistry
- Bonding And Antibonding Pi Orbitals
- Molecular Orbitals of The Allyl Cation, Allyl Radical, and Allyl Anion
- Pi Molecular Orbitals of Butadiene
- Reactions of Dienes: 1,2 and 1,4 Addition
- Thermodynamic and Kinetic Products
- More On 1,2 and 1,4 Additions To Dienes
- s-cis and s-trans
- The Diels-Alder Reaction
- Cyclic Dienes and Dienophiles in the Diels-Alder Reaction
- Stereochemistry of the Diels-Alder Reaction
- Exo vs Endo Products In The Diels Alder: How To Tell Them Apart
- HOMO and LUMO In the Diels Alder Reaction
- Why Are Endo vs Exo Products Favored in the Diels-Alder Reaction?
- Diels-Alder Reaction: Kinetic and Thermodynamic Control
- The Retro Diels-Alder Reaction
- The Intramolecular Diels Alder Reaction
- Regiochemistry In The Diels-Alder Reaction
- The Cope and Claisen Rearrangements
- Electrocyclic Reactions
- Electrocyclic Ring Opening And Closure (2) - Six (or Eight) Pi Electrons
- Diels Alder Practice Problems
- Molecular Orbital Theory Practice
18 Aromaticity
- Introduction To Aromaticity
- Rules For Aromaticity
- Huckel's Rule: What Does 4n+2 Mean?
- Aromatic, Non-Aromatic, or Antiaromatic? Some Practice Problems
- Antiaromatic Compounds and Antiaromaticity
- The Pi Molecular Orbitals of Benzene
- The Pi Molecular Orbitals of Cyclobutadiene
- Frost Circles
- Aromaticity Practice Quizzes
19 Reactions of Aromatic Molecules
- Electrophilic Aromatic Substitution: Introduction
- Activating and Deactivating Groups In Electrophilic Aromatic Substitution
- Electrophilic Aromatic Substitution - The Mechanism
- Ortho-, Para- and Meta- Directors in Electrophilic Aromatic Substitution
- Understanding Ortho, Para, and Meta Directors
- Why are halogens ortho- para- directors?
- Disubstituted Benzenes: The Strongest Electron-Donor "Wins"
- Electrophilic Aromatic Substitutions (1) - Halogenation of Benzene
- Electrophilic Aromatic Substitutions (2) - Nitration and Sulfonation
- EAS Reactions (3) - Friedel-Crafts Acylation and Friedel-Crafts Alkylation
- Intramolecular Friedel-Crafts Reactions
- Nucleophilic Aromatic Substitution (NAS)
- Nucleophilic Aromatic Substitution (2) - The Benzyne Mechanism
- Reactions on the "Benzylic" Carbon: Bromination And Oxidation
- The Wolff-Kishner, Clemmensen, And Other Carbonyl Reductions
- More Reactions on the Aromatic Sidechain: Reduction of Nitro Groups and the Baeyer Villiger
- Aromatic Synthesis (1) - "Order Of Operations"
- Synthesis of Benzene Derivatives (2) - Polarity Reversal
- Aromatic Synthesis (3) - Sulfonyl Blocking Groups
- Birch Reduction
- Synthesis (7): Reaction Map of Benzene and Related Aromatic Compounds
- Aromatic Reactions and Synthesis Practice
- Electrophilic Aromatic Substitution Practice Problems
20 Aldehydes and Ketones
- What's The Alpha Carbon In Carbonyl Compounds?
- Nucleophilic Addition To Carbonyls
- Aldehydes and Ketones: 14 Reactions With The Same Mechanism
- Sodium Borohydride (NaBH4) Reduction of Aldehydes and Ketones
- Grignard Reagents For Addition To Aldehydes and Ketones
- Wittig Reaction
- Hydrates, Hemiacetals, and Acetals
- Imines - Properties, Formation, Reactions, and Mechanisms
- All About Enamines
- Breaking Down Carbonyl Reaction Mechanisms: Reactions of Anionic Nucleophiles (Part 2)
- Aldehydes Ketones Reaction Practice
21 Carboxylic Acid Derivatives
- Nucleophilic Acyl Substitution (With Negatively Charged Nucleophiles)
- Addition-Elimination Mechanisms With Neutral Nucleophiles (Including Acid Catalysis)
- Basic Hydrolysis of Esters - Saponification
- Transesterification
- Proton Transfer
- Fischer Esterification - Carboxylic Acid to Ester Under Acidic Conditions
- Lithium Aluminum Hydride (LiAlH4) For Reduction of Carboxylic Acid Derivatives
- LiAlH[Ot-Bu]3 For The Reduction of Acid Halides To Aldehydes
- Di-isobutyl Aluminum Hydride (DIBAL) For The Partial Reduction of Esters and Nitriles
- Amide Hydrolysis
- Thionyl Chloride (SOCl2)
- Diazomethane (CH2N2)
- Carbonyl Chemistry: Learn Six Mechanisms For the Price Of One
- Making Music With Mechanisms (PADPED)
- Carboxylic Acid Derivatives Practice Questions
22 Enols and Enolates
- Keto-Enol Tautomerism
- Enolates - Formation, Stability, and Simple Reactions
- Kinetic Versus Thermodynamic Enolates
- Aldol Addition and Condensation Reactions
- Reactions of Enols - Acid-Catalyzed Aldol, Halogenation, and Mannich Reactions
- Claisen Condensation and Dieckmann Condensation
- Decarboxylation
- The Malonic Ester and Acetoacetic Ester Synthesis
- The Robinson Annulation
- Haloform Reaction
- The Hell–Volhard–Zelinsky Reaction
- Enols and Enolates Practice Quizzes
- The Amide Functional Group: Properties, Synthesis, and Nomenclature
- Basicity of Amines And pKaH
- 5 Key Basicity Trends of Amines
- The Mesomeric Effect And Aromatic Amines
- Nucleophilicity of Amines
- Alkylation of Amines (Sucks!)
- Reductive Amination
- The Gabriel Synthesis
- Some Reactions of Azides
- The Hofmann Elimination
- The Hofmann and Curtius Rearrangements
- The Cope Elimination
- Protecting Groups for Amines - Carbamates
- The Strecker Synthesis of Amino Acids
- Introduction to Peptide Synthesis
- Reactions of Diazonium Salts: Sandmeyer and Related Reactions
- Amine Practice Questions
24 Carbohydrates
- D and L Notation For Sugars
- Pyranoses and Furanoses: Ring-Chain Tautomerism In Sugars
- What is Mutarotation?
- Reducing Sugars
- The Big Damn Post Of Carbohydrate-Related Chemistry Definitions
- The Haworth Projection
- Converting a Fischer Projection To A Haworth (And Vice Versa)
- Reactions of Sugars: Glycosylation and Protection
- The Ruff Degradation and Kiliani-Fischer Synthesis
- Isoelectric Points of Amino Acids (and How To Calculate Them)
- Carbohydrates Practice
- Amino Acid Quizzes
25 Fun and Miscellaneous
- Organic Chemistry GIFS - Resonance Forms
- Organic Chemistry and the New MCAT
- A Gallery of Some Interesting Molecules From Nature
- The Organic Chemistry Behind "The Pill"
- Maybe they should call them, "Formal Wins" ?
- Planning Organic Synthesis With "Reaction Maps"
- Organic Chemistry Is Shit
- The 8 Types of Arrows In Organic Chemistry, Explained
- The Most Annoying Exceptions in Org 1 (Part 1)
- The Most Annoying Exceptions in Org 1 (Part 2)
- Reproducibility In Organic Chemistry
- Screw Organic Chemistry, I'm Just Going To Write About Cats
- On Cats, Part 1: Conformations and Configurations
- On Cats, Part 2: Cat Line Diagrams
- The Marriage May Be Bad, But the Divorce Still Costs Money
- Why Do Organic Chemists Use Kilocalories?
- What Holds The Nucleus Together?
- 9 Nomenclature Conventions To Know
- How Reactions Are Like Music
Comment section
57 thoughts on “ a key skill: how to calculate formal charge ”.
sir 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
nice, concise explanation
Very good explanation.I finally understood how to calculate the formal charge,was having some trouble with it.Thanks:)
Glad you found it helpful.
thank you for excellent explanation
Glad you found it useful Peter!
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 collaboration of formal charge
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.
Great!i can use this for my exam!thanks!
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.
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?
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.
http://chemwiki.ucdavis.edu/Physical_Chemistry/Acids_and_Bases/Aqueous_Solutions/The_hydronium_Ion
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?
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!
Thank you very very more for the simple explanation! Unbelievably easy and saves so much time!!!!!!
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!
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.
what does it means if we determine a molecule with zero charge ?
It’s neutral!
I am beryllium and i got offended!!!!!!……..LOL Just kidding…….BTW, I found this article very useful.Thanks!!!!!!!!!!
AM REALLY LOST NOW ON THAT EXAMPLE OF CH3 CARBON # 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)] FORMAL CHARGE=4+[(1/2*3)+1] =1.5
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 .
Thank u very much my exam is today and i wouldn’t pass without this information
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!
But when I used this formula it works. Thus #valence electrons_#lone pair__#1/2.bond pairs
Thanks for the easy approach.
This was so helpful n the best explanation about the topic…
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 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.
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.
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.
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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.
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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.
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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.
Great teaching , can I know where did u studied ??
Nice simple explanation
Thank you so much sir. Finally i understood how to calculate the formal charge
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!
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
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- 7.4 Formal Charges and Resonance
- Introduction
- 1.1 Chemistry in Context
- 1.2 Phases and Classification of Matter
- 1.3 Physical and Chemical Properties
- 1.4 Measurements
- 1.5 Measurement Uncertainty, Accuracy, and Precision
- 1.6 Mathematical Treatment of Measurement Results
- Key Equations
- 2.1 Early Ideas in Atomic Theory
- 2.2 Evolution of Atomic Theory
- 2.3 Atomic Structure and Symbolism
- 2.4 Chemical Formulas
- 2.5 The Periodic Table
- 2.6 Ionic and Molecular Compounds
- 2.7 Chemical Nomenclature
- 3.1 Formula Mass and the Mole Concept
- 3.2 Determining Empirical and Molecular Formulas
- 3.3 Molarity
- 3.4 Other Units for Solution Concentrations
- 4.1 Writing and Balancing Chemical Equations
- 4.2 Classifying Chemical Reactions
- 4.3 Reaction Stoichiometry
- 4.4 Reaction Yields
- 4.5 Quantitative Chemical Analysis
- 5.1 Energy Basics
- 5.2 Calorimetry
- 5.3 Enthalpy
- 6.1 Electromagnetic Energy
- 6.2 The Bohr Model
- 6.3 Development of Quantum Theory
- 6.4 Electronic Structure of Atoms (Electron Configurations)
- 6.5 Periodic Variations in Element Properties
- 7.1 Ionic Bonding
- 7.2 Covalent Bonding
- 7.3 Lewis Symbols and Structures
- 7.5 Strengths of Ionic and Covalent Bonds
- 7.6 Molecular Structure and Polarity
- 8.1 Valence Bond Theory
- 8.2 Hybrid Atomic Orbitals
- 8.3 Multiple Bonds
- 8.4 Molecular Orbital Theory
- 9.1 Gas Pressure
- 9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
- 9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
- 9.4 Effusion and Diffusion of Gases
- 9.5 The Kinetic-Molecular Theory
- 9.6 Non-Ideal Gas Behavior
- 10.1 Intermolecular Forces
- 10.2 Properties of Liquids
- 10.3 Phase Transitions
- 10.4 Phase Diagrams
- 10.5 The Solid State of Matter
- 10.6 Lattice Structures in Crystalline Solids
- 11.1 The Dissolution Process
- 11.2 Electrolytes
- 11.3 Solubility
- 11.4 Colligative Properties
- 11.5 Colloids
- 12.1 Chemical Reaction Rates
- 12.2 Factors Affecting Reaction Rates
- 12.3 Rate Laws
- 12.4 Integrated Rate Laws
- 12.5 Collision Theory
- 12.6 Reaction Mechanisms
- 12.7 Catalysis
- 13.1 Chemical Equilibria
- 13.2 Equilibrium Constants
- 13.3 Shifting Equilibria: Le Châtelier’s Principle
- 13.4 Equilibrium Calculations
- 14.1 Brønsted-Lowry Acids and Bases
- 14.2 pH and pOH
- 14.3 Relative Strengths of Acids and Bases
- 14.4 Hydrolysis of Salts
- 14.5 Polyprotic Acids
- 14.6 Buffers
- 14.7 Acid-Base Titrations
- 15.1 Precipitation and Dissolution
- 15.2 Lewis Acids and Bases
- 15.3 Coupled Equilibria
- 16.1 Spontaneity
- 16.2 Entropy
- 16.3 The Second and Third Laws of Thermodynamics
- 16.4 Free Energy
- 17.1 Review of Redox Chemistry
- 17.2 Galvanic Cells
- 17.3 Electrode and Cell Potentials
- 17.4 Potential, Free Energy, and Equilibrium
- 17.5 Batteries and Fuel Cells
- 17.6 Corrosion
- 17.7 Electrolysis
- 18.1 Periodicity
- 18.2 Occurrence and Preparation of the Representative Metals
- 18.3 Structure and General Properties of the Metalloids
- 18.4 Structure and General Properties of the Nonmetals
- 18.5 Occurrence, Preparation, and Compounds of Hydrogen
- 18.6 Occurrence, Preparation, and Properties of Carbonates
- 18.7 Occurrence, Preparation, and Properties of Nitrogen
- 18.8 Occurrence, Preparation, and Properties of Phosphorus
- 18.9 Occurrence, Preparation, and Compounds of Oxygen
- 18.10 Occurrence, Preparation, and Properties of Sulfur
- 18.11 Occurrence, Preparation, and Properties of Halogens
- 18.12 Occurrence, Preparation, and Properties of the Noble Gases
- 19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
- 19.2 Coordination Chemistry of Transition Metals
- 19.3 Spectroscopic and Magnetic Properties of Coordination Compounds
- 20.1 Hydrocarbons
- 20.2 Alcohols and Ethers
- 20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
- 20.4 Amines and Amides
- 21.1 Nuclear Structure and Stability
- 21.2 Nuclear Equations
- 21.3 Radioactive Decay
- 21.4 Transmutation and Nuclear Energy
- 21.5 Uses of Radioisotopes
- 21.6 Biological Effects of Radiation
- A | The Periodic Table
- B | Essential Mathematics
- C | Units and Conversion Factors
- D | Fundamental Physical Constants
- E | Water Properties
- F | Composition of Commercial Acids and Bases
- G | Standard Thermodynamic Properties for Selected Substances
- H | Ionization Constants of Weak Acids
- I | Ionization Constants of Weak Bases
- J | Solubility Products
- K | Formation Constants for Complex Ions
- L | Standard Electrode (Half-Cell) Potentials
- M | Half-Lives for Several Radioactive Isotopes
Learning Objectives
By the end of this section, you will be able to:
- Compute formal charges for atoms in any Lewis structure
- Use formal charges to identify the most reasonable Lewis structure for a given molecule
- Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule
In the previous section, we discussed how to write Lewis structures for molecules and polyatomic ions. As we have seen, however, in some cases, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable.
Calculating Formal Charge
The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.
Thus, we calculate formal charge as follows:
We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure. The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.
We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.
Example 7.6
Calculating formal charge from lewis structures.
- Step 2. We assign lone pairs of electrons to their atoms . Each Cl atom now has seven electrons assigned to it, and the I atom has eight.
- Step 3. Subtract this number from the number of valence electrons for the neutral atom: I: 7 – 8 = –1 Cl: 7 – 7 = 0 The sum of the formal charges of all the atoms equals –1, which is identical to the charge of the ion (–1).
Check Your Learning
Example 7.7
- Step 2. Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
- Step 3. Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge: Br: 7 – 7 = 0 Cl: 7 – 7 = 0 All atoms in BrCl 3 have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.
N: 0; all three Cl atoms: 0
Using Formal Charge to Predict Molecular Structure
The arrangement of atoms in a molecule or ion is called its molecular structure . In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure—different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion:
- A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
- If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
- Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
- When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.
To see how these guidelines apply, let us consider some possible structures for carbon dioxide, CO 2 . We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:
Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).
As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: NCS – , CNS – , or CSN – . The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:
Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).
Example 7.8
Using formal charge to determine molecular structure.
The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:
The number of atoms with formal charges are minimized (Guideline 2), there is no formal charge with a magnitude greater than one (Guideline 2), the negative formal charge is on the more electronegative element (Guideline 4), and the less electronegative atom is in the center position.
Notice that the more likely structure for the nitrite anion in Example 7.8 may actually be drawn in two different ways, distinguished by the locations of the N-O and N=O bonds:
If nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. Experiments show, however, that both N–O bonds in NO 2 − NO 2 − have the same strength and length, and are identical in all other properties.
It is not possible to write a single Lewis structure for NO 2 − NO 2 − in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance : if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in NO 2 − NO 2 − is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms . The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms.
We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).
The carbonate anion, CO 3 2− , CO 3 2− , provides a second example of resonance:
One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three C–O bonds are exactly the same.
Link to Learning
Use this online quiz to practice your skills in drawing resonance structures and estimating formal charges.
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Chemical Bonding and Molecular Geometry
Formal Charges and Resonance
OpenStaxCollege
[latexpage]
Learning Objectives
By the end of this section, you will be able to:
- Compute formal charges for atoms in any Lewis structure
- Use formal charges to identify the most reasonable Lewis structure for a given molecule
- Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule
In the previous section, we discussed how to write Lewis structures for molecules and polyatomic ions. As we have seen, however, in some cases, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable.
Calculating Formal Charge
The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.
Thus, we calculate formal charge as follows:
We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure. The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.
We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.
Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen ion \({\text{ICl}}_{4}{}^{\text{−}}.\)
- We assign lone pairs of electrons to their atoms . Each Cl atom now has seven electrons assigned to it, and the I atom has eight.
I: 7 – 8 = –1
Cl: 7 – 7 = 0
Check Your Learning Calculate the formal charge for each atom in the carbon monoxide molecule:
Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen molecule BrCl 3 .
- Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
Br: 7 – 7 = 0
Check Your Learning Determine the formal charge for each atom in NCl 3 .
N: 0; all three Cl atoms: 0
Using Formal Charge to Predict Molecular Structure
The arrangement of atoms in a molecule or ion is called its molecular structure . In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure—different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion:
- A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
- If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
- Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
- When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.
To see how these guidelines apply, let us consider some possible structures for carbon dioxide, CO 2 . We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:
Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).
As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: CNS – , NCS – , or CSN – . The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:
Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).
Using Formal Charge to Determine Molecular Structure Nitrous oxide, N 2 O, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the likely structure for nitrous oxide?
Solution Determining formal charge yields the following:
The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:
The number of atoms with formal charges are minimized (Guideline 2), and there is no formal charge larger than one (Guideline 2). This is again consistent with the preference for having the less electronegative atom in the central position.
Check Your Learning Which is the most likely molecular structure for the nitrite \(\left({\text{NO}}_{2}{}^{\text{−}}\right)\) ion?
You may have noticed that the nitrite anion in [link] can have two possible structures with the atoms in the same positions. The electrons involved in the N–O double bond, however, are in different positions:
If nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. Experiments show, however, that both N–O bonds in \({\text{NO}}_{2}{}^{\text{−}}\) have the same strength and length, and are identical in all other properties.
It is not possible to write a single Lewis structure for \({\text{NO}}_{2}{}^{\text{−}}\) in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance : if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in \({\text{NO}}_{2}{}^{\text{−}}\) is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms . The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms. Thus, the electronic structure of the \({\text{NO}}_{2}{}^{\text{−}}\) ion is shown as:
We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).
The carbonate anion, \({\text{CO}}_{3}{}^{\text{2−}},\) provides a second example of resonance:
One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three C–O bonds are exactly the same.
The online Lewis Structure Make includes many examples to practice drawing resonance structures.
Key Concepts and Summary
In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. These hypothetical formal charges are a guide to determining the most appropriate Lewis structure. A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written. The actual distribution of electrons (the resonance hybrid) is an average of the distribution indicated by the individual Lewis structures (the resonance forms).
Key Equations
- \(\text{formal charge}\phantom{\rule{0.2em}{0ex}}=\text{# valence shell electrons (free atom)}\phantom{\rule{0.1em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\text{# one pair electrons}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.3em}{0ex}}\text{# bonding electrons}\)
Chemistry End of Chapter Exercises
Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
(a) selenium dioxide, OSeO
(b) nitrate ion, \({\text{NO}}_{3}{}^{\text{−}}\)
(c) nitric acid, HNO 3 (N is bonded to an OH group and two O atoms)
(d) benzene, C 6 H 6 :
(e) the formate ion:
(a) sulfur dioxide, SO 2
(b) carbonate ion, \({\text{CO}}_{3}{}^{\text{2−}}\)
(c) hydrogen carbonate ion, \({\text{HCO}}_{3}{}^{\text{−}}\) (C is bonded to an OH group and two O atoms)
(d) pyridine:
(e) the allyl ion:
Write the resonance forms of ozone, O 3 , the component of the upper atmosphere that protects the Earth from ultraviolet radiation.
Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, \({\text{NO}}_{\text{2}}{}^{\text{–}}\text{.}\)
In terms of the bonds present, explain why acetic acid, CH 3 CO 2 H, contains two distinct types of carbon-oxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown:
Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.
Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate.
Determine the formal charge of each element in the following:
(a) H: 0, Cl: 0; (b) C: 0, F: 0; (c) P: 0, Cl 0; (d) P: 0, F: 0
(a) H 3 O +
(b) \({\text{SO}}_{4}{}^{\text{2−}}\)
(d) \({\text{O}}_{2}{}^{\text{2−}}\)
(e) H 2 O 2
Calculate the formal charge of chlorine in the molecules Cl 2 , BeCl 2 , and ClF 5 .
Cl in Cl 2 : 0; Cl in BeCl 2 : 0; Cl in ClF 5 : 0
Calculate the formal charge of each element in the following compounds and ions:
(c) \({\text{BF}}_{4}{}^{\text{−}}\)
(d) \({\text{SnCl}}_{3}{}^{\text{−}}\)
(e) H 2 CCH 2
(h) \({\text{PO}}_{4}{}^{\text{3−}}\)
Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures:
(c) \({\text{NO}}_{2}{}^{\text{−}}\)
(d) \({\text{NO}}_{3}{}^{\text{−}}\)
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH?
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO?
Draw the structure of hydroxylamine, H 3 NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges?
The structure that gives zero formal charges is consistent with the actual structure:
Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule:
Write the Lewis structure and chemical formula of the compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, and determine the formal charge of the atoms in this compound.
Which of the following structures would we expect for nitrous acid? Determine the formal charges:
Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H 2 SO 4 , which has two oxygen atoms and two OH groups bonded to the sulfur.
Formal Charges and Resonance by OpenStaxCollege is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.
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Share 141K views 9 years ago O3 Lewis, Resonance, Shape, Polarity, and more. In order to calculate the formal charges for O3 we'll use the equation Formal charge = [# of valence...
Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge: Br: 7 - 7 = 0. Cl: 7 - 7 = 0. All atoms in BrCl 3 have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule. Exercise 3.3.3.2.
Thus, we calculate formal charge as follows: (4.3.1) formal charge = # valence shell electrons (free atom) − # lone pair electrons − 1 2 # bonding electrons We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure.
The formal charge of the ozone molecule is zero. Its Lewis structures do present charge separation. With simple VSEPR considerations, there are 18 valence electrons to distribute around the 3 oxygen atoms (24 electrons in total; 6 are inner core). Typically, a Lewis structure of O=stackrel(ddot)O^(+)-O^(-), would be depicted. Going from left O to right O and including the 2 inner core ...
Center 6 valence −5 assigned = 1 formal charge Right: 6 valence −7 assigned = −1 formal charge Notice that even though the atoms have varying formal charges, the overall charge of O3 is the sum of the formal charges in the molecule: 0 + 1 + ( −1) = 0. Ions' formal charge sums are ≠ 0. Answer link
Master Organic Chemistry 34.3K subscribers How to calculate the formal charges on the atoms of ozone (O3) MOC members get access to over 1500 quizzes on O3 and many other topics, plus...
The formal charges present on the bonded atoms in O 3 can be calculated using the formula given below: V.E - N.E - B.E/2 Where - ⇒ V.E = valence electrons of an atom ⇒ N.E = non-bonding electrons, i.e., lone pairs ⇒ B.E = bonding electrons What is the formal charge on the central O-atom in the O3 Lewis structure?
Oxygen belongs to group VI of the periodic table with an atomic no of 8. It thus has 6 valence electrons. Thus, the total number of valence electrons in ozone= 3*6 = 18 Just like triiodide ion where all the atoms are iodine, here, all the atoms are oxygen.
An atom can have the following charges: positive, negative, or neutral, depending on the electron distribution. This is often useful for understanding or predicting reactivity. Identifying formal charges helps you keep track of the electrons. The formal charge is the charge on the atom in the molecule. The term "formal" means that this ...
Using the formula charge formula for each atom present, we can calculate the formal charge by observing the Lewis Dot structure of OH. Formal charge = valence electrons - unbonded electrons...
The formal charge on O: 6-2-½ (6)= 1, thus the formal charge on o3 lewis structure is +1 on the central oxygen atomSimilarly, two adjacent oxygen atoms carry (-½ ) partial negative charge, and central oxygen carries +1 formal charge as shown in the figure below. O3 lewis structure formal charges O3 lewis structure lone pairs
Formal charge on right Oxygen = Valence electrons - Nonbonding electrons - (Bonding electrons)/2 = 6 - 6 - (2/2) = 1-. So the formal charge on right oxygen atom is 1-. Now let's put all these charges on the lewis dot structure of O3. So there is overall 0 charge left on the entire molecule. This indicates that the O3 (Ozone) has 0 charge.
- The formal charge on oxygen atom 3 = 6 - 2 2 - 6 = -1 - The formal charge on oxygen atom 3 is '-1'. - Now the total formal charge of the ozone = 0 + 1 - 1 = 0 - Therefore the formal charge of ozone is '0'. Note: The formal charge of a molecule is equal to the sum of the formal charge of all the individual atoms present in the given molecule.
Formal charge in O 3 ( Ozone): In an O 3 molecule, the formal charge on the middle oxygen atom ( 2) is + 1. F. C = 6 - 2 - 1 2 ( 6) F. C = 6 - 5 F. C = 1 In an O 3 molecule, the formal charge on the left oxygen atom ( 3) is - 1. F. C = 6 - 6 - 1 2 ( 2) F. C = 6 - 7 F. C = - 1 In an O 3 molecule, the formal charge on the right oxygen atom ( 1) is 0.
In BH4, the formal charge of hydrogen is 1-(0+1), resulting in a formal charge of 0. Example 2: Calculate the formal charge on the following: O atoms of O3. Cl atom in HClO4- ion. S atom in HSO4- ion. Ans: We are showing how to find a formal charge of the species mentioned. Formal charge on O1: 6 - 6/2 - 2 = +1. Formal charge on O2: 6 - 4 ...
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 Z eff, the number of protons that an electron in the valence orbital "sees" due to screening by inner-shell ...
Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen molecule BrCl 3. Solution. Step 1. Assign one of the electrons in each Br-Cl bond to the Br atom and one to the Cl atom in that bond: Step 2. Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven ...
Answer: C −1, O +1. Calculating Formal Charge from Lewis Structures. Assign formal charges to each atom in the interhalogen molecule BrCl 3. Solution. Assign one of the electrons in each Br-Cl bond to the Br atom and one to the Cl atom in that bond: Assign the lone pairs to their atom.
Typically a Lewis structure of O= O¨+−O −, would be depicted. Going from left O to right O and including the 2 inner core electrons on each atom, there are 8,8 and 8 electrons around each oxygen centre, resulting in formal charges of 0,+1, and −1, respectively. Of course, you can draw the other resonance structure, but the Lewis ...