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27 4.5 Formal Charges and Resonance

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 ICl 4 − .

A Lewis structure is shown. An iodine atom with two lone pairs of electrons is single bonded to four chlorine atoms, each of which has three lone pairs of electrons. Brackets surround the structure and there is a superscripted negative sign.

  • 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.
  • Subtract this number from the number of valence electrons for the neutral atom: I: 7 – 8 = –1Cl: 7 – 7 = 0The sum of the formal charges of all the atoms equals –1, which is identical to the charge of the ion (–1).

Check Your Learning Calculate the formal charge for each atom in the carbon monoxide molecule:

A Lewis structure is shown. A carbon atom with one lone pair of electrons is triple bonded to an oxygen with one lone pair of electrons.

Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen molecule BrCl 3 .

A Lewis structure is shown. A bromine atom with two lone pairs of electrons is single bonded to three chlorine atoms, each of which has three lone pairs of electrons.

  • Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
  • Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge:Br: 7 – 7 = 0Cl: 7 – 7 = 0All 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.

Check Your Learning Determine the formal charge for each atom in NCl 3 .

N: 0; all three Cl atoms: 0

A Lewis structure is shown. A nitrogen atom with one lone pair of electrons is single bonded to three chlorine atoms, each of which has three lone pairs of electrons.

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:

Three Lewis structures are shown. The left and right structures show a carbon atom double bonded to two oxygen atoms, each of which has two lone pairs of electrons. The center structure shows a carbon atom that is triple bonded to an oxygen atom with one lone pair of electrons and single bonded to an oxygen atom with three lone pairs of electrons. The third structure shows an oxygen atom double bonded to another oxygen atom with to lone pairs of electrons. The first oxygen atom is also double bonded to a carbon atom with two lone pairs of electrons.

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:

Two rows of structures and numbers are shown. The top row is labeled, “Structure” and depicts three Lewis structures and the bottom row is labeled, “Formal charge.” The left structure shows a carbon atom double bonded to a nitrogen atom with two lone electron pairs on one side and double bonded to a sulfur atom with two lone electron pairs on the other. The structure is surrounded by brackets and has a superscripted negative sign. Below this structure are the numbers negative one, zero, and zero. The middle structure shows a carbon atom with two lone pairs of electrons double bonded to a nitrogen atom that is double bonded to a sulfur atom with two lone electron pairs. The structure is surrounded by brackets and has a superscripted negative sign. Below this structure are the numbers negative two, positive one, and zero. The right structure shows a carbon atom with two lone electron pairs double bonded to a sulfur atom that is double bonded to a nitrogen atom with two lone electron pairs. The structure is surrounded by brackets and has a superscripted negative sign. Below this structure are the numbers negative two, positive two, and one.

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?

Two Lewis structures are shown with the word “or” in between them. The left structure depicts a nitrogen atom with two lone pairs of electrons double bonded to a nitrogen that is double bonded to an oxygen with two lone pairs of electrons. The right structure shows a nitrogen atom with two lone pairs of electrons double bonded to an oxygen atom that is double bonded to a nitrogen atom with two lone pairs of electrons.

Solution Determining formal charge yields the following:

Two Lewis structures are shown with the word “or” in between them. The left structure depicts a nitrogen atom with two lone pairs of electrons double bonded to a nitrogen atom that is double bonded to an oxygen atom with two lone pairs of electrons. The numbers negative one, positive one, and zero are written above this structure. The right structure shows a nitrogen atom with two lone pairs of electrons double bonded to an oxygen atom that is double bonded to a nitrogen atom with two lone pairs of electrons. The numbers negative one, positive two, and negative one are written above this structure.

The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:

A Lewis structure is shown. A nitrogen atom with two lone pairs of electrons is double bonded to a nitrogen atom that is double bonded to an oxygen atom with two lone pairs of electrons.

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 (NO 2 − ) ion?

Two Lewis structures are shown with the word “or” written between them. The left structure shows a nitrogen atom with two lone pairs of electrons double bonded to an oxygen atom with one lone pair of electrons that is single bonded to an oxygen atom with three lone pairs of electrons. Brackets surround this structure and there is a superscripted negative sign. The right structure shows an oxygen atom with two lone pairs of electrons double bonded to a nitrogen atom with one lone pair of electrons that is single bonded to an oxygen with three lone pairs of electrons. Brackets surround this structure and there is a superscripted negative sign.

You may have noticed that the nitrite anion in Example 3 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:

Two Lewis structures are shown. The left structure shows an oxygen atom with three lone pairs of electrons single bonded to a nitrogen atom with one lone pair of electrons that is double bonded to an oxygen with two lone pairs of electrons. Brackets surround this structure, and there is a superscripted negative sign. The right structure shows an oxygen atom with two lone pairs of electrons double bonded to a nitrogen atom with one lone pair of electrons that is single bonded to an oxygen atom with three lone pairs of electrons. Brackets surround this structure, and there is a superscripted negative sign.

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 − 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 − 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 − 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 NO 2 − ion is shown as:

Two Lewis structures are shown with a double headed arrow drawn between them. The left structure shows an oxygen atom with two lone pairs of electrons double bonded to a nitrogen atom with one lone pair of electrons that is single bonded to an oxygen atom with three lone pairs of electrons. Brackets surround this structure, and there is a superscripted negative sign. The right structure shows an oxygen atom with three lone pairs of electrons single bonded to a nitrogen atom with one lone pair of electrons that is double bonded to an oxygen atom with two lone pairs of electrons. Brackets surround this structure, and there is a superscripted negative sign.

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− , provides a second example of resonance:

Three Lewis structures are shown with double headed arrows in between. Each structure is surrounded by brackets, and each has a superscripted two negative sign. The left structure depicts a carbon atom bonded to three oxygen atoms. It is single bonded to two of these oxygen atoms, each of which has three lone pairs of electrons, and double bonded to the third, which has two lone pairs of electrons. The double bond is located between the lower left oxygen atom and the carbon atom. The central and right structures are the same as the first, but the position of the double bonded oxygen has moved to the lower right oxygen in the central structure and to the top oxygen in the right structure.

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

  • [latex]\text{formal charge} = \# \;\text{valence shell electrons (free atom)} \; - \;\# \;\text{lone pair electrons}\; - \frac{1}{2} \# \;\text{bonding electrons}[/latex]

Chemistry End of Chapter Exercises

(a) selenium dioxide, OSeO

(b) nitrate ion, NO 3 −

(c) nitric acid, HNO 3 (N is bonded to an OH group and two O atoms)

(d) benzene, C 6 H 6 :

A Lewis structure shows a hexagonal ring composed of six carbon atoms. They form single bonds to each another and single bonds to one hydrogen atom each.

(e) the formate ion:

A Lewis structure shows a carbon atom single bonded to two oxygen atoms and a hydrogen atom. The structure is surrounded by brackets and there is a superscripted negative sign.

(a) sulfur dioxide, SO 2

(b) carbonate ion, CO 3 2−

(c) hydrogen carbonate ion, HCO 3 − (C is bonded to an OH group and two O atoms)

(d) pyridine:

A Lewis structure depicts a hexagonal ring composed of five carbon atoms and one nitrogen atom. Each carbon atom is single bonded to a hydrogen atom.

(e) the allyl ion:

A Lewis structure shows a carbon atom single bonded to two hydrogen atoms and a second carbon atom. The second carbon atom is single bonded to a hydrogen atom and a third carbon atom. The third carbon atom is single bonded to two hydrogen atoms. The whole structure is surrounded by brackets, and there is a superscripted negative sign.

  • 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, NO 2 – .

Two Lewis structures are shown with a double headed arrow in between. The left structure shows a carbon atom single bonded to three hydrogen atoms and a second carbon atom. The second carbon is single bonded to two oxygen atoms. One of the oxygen atoms is single bonded to a hydrogen atom. The right structure, surrounded by brackets and with a superscripted negative sign, depicts a carbon atom single bonded to three hydrogen atoms and a second carbon atom. The second carbon atom is single bonded to two oxygen atoms.

  • 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.

(a) H 3 O +

(b) SO 4 2−

(e) H 2 O 2

  • Calculate the formal charge of chlorine in the molecules Cl 2 , BeCl 2 , and ClF 5 .

(d) SnCl 3 −

(e) H 2 CCH 2

(h) PO 4 3−

  • 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?
  • 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.

Two Lewis structures are shown, with the word “or” in between. The left structure shows a nitrogen atom single bonded to an oxygen atom with three lone pairs of electrons. It is also single bonded to a hydrogen atom and double bonded to an oxygen atom with two lone pairs of electrons. The right structure shows a hydrogen atom single bonded to an oxygen atom with two lone pairs of electrons. The oxygen atom is single bonded to a nitrogen atom which is double bonded to an oxygen atom with two lone pairs of electrons.

  • 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.

Answers to Chemistry End of Chapter Exercises

Two Lewis structures are shown with a double-headed arrow in between. The left structure shows a sulfur atom with a lone pair of electrons single bonded to the left to an oxygen atom with three lone pairs of electrons. The sulfur atom is also double bonded on the right to an oxygen atom with two lone pairs of electrons. The right structure depicts the same atoms, but this time the double bond is between the left oxygen and the sulfur atom. The lone pairs of electrons have also shifted to account for the change of bond types. The sulfur atom in the right structures, also has a third electron dot below it.

8. (a) H: 0, Cl: 0; (b) C: 0, F: 0; (c) P: 0, Cl 0; (d) P: 0, F: 0

10. Cl in Cl 2 : 0; Cl in BeCl 2 : 0; Cl in ClF 5 : 0

Two Lewis structures are shown with a double-headed arrow in between. The left structure shows an oxygen atom with one lone pair of electrons single bonded to an oxygen atom with three lone pairs of electrons. It is also double bonded to an oxygen atom with two lone pairs of electrons. The symbols and numbers below this structure read, “( 0 ), ( positive 1 ), ( negative 1 ).” The phrase, “Formal charge,” and a right-facing arrow lie to the left of this structure. The right structure appears as a mirror image of the left and the symbols and numbers below this structure read, “( negative 1 ), ( positive 1 ), ( 0 ).”

16. The structure that gives zero formal charges is consistent with the actual structure:

A Lewis structure shows a nitrogen atom with one lone pair of electrons single bonded to two hydrogen atoms and an oxygen atom which has two lone pairs of electrons. The oxygen atom is single bonded to a hydrogen atom.

General Chemistry 1 & 2 Copyright © 2016 by Rice University is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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  • 2.3 Formal Charges
  • Dedication and Preface
  • Why This Chapter?
  • 1.1 Atomic Structure: The Nucleus
  • 1.2 Atomic Structure: Orbitals
  • 1.3 Atomic Structure: Electron Configurations
  • 1.4 Development of Chemical Bonding Theory
  • 1.5 Describing Chemical Bonds: Valence Bond Theory
  • 1.6 sp 3 Hybrid Orbitals and the Structure of Methane
  • 1.7 sp 3 Hybrid Orbitals and the Structure of Ethane
  • 1.8 sp 2 Hybrid Orbitals and the Structure of Ethylene
  • 1.9 sp Hybrid Orbitals and the Structure of Acetylene
  • 1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur
  • 1.11 Describing Chemical Bonds: Molecular Orbital Theory
  • 1.12 Drawing Chemical Structures
  • Chemistry Matters—Organic Foods: Risk versus Benefit
  • Additional Problems
  • 2.1 Polar Covalent Bonds and Electronegativity
  • 2.2 Polar Covalent Bonds and Dipole Moments
  • 2.4 Resonance
  • 2.5 Rules for Resonance Forms
  • 2.6 Drawing Resonance Forms
  • 2.7 Acids and Bases: The Brønsted–Lowry Definition
  • 2.8 Acid and Base Strength
  • 2.9 Predicting Acid–Base Reactions from p K a Values
  • 2.10 Organic Acids and Organic Bases
  • 2.11 Acids and Bases: The Lewis Definition
  • 2.12 Noncovalent Interactions between Molecules
  • Chemistry Matters—Alkaloids: From Cocaine to Dental Anesthetics
  • 3.1 Functional Groups
  • 3.2 Alkanes and Alkane Isomers
  • 3.3 Alkyl Groups
  • 3.4 Naming Alkanes
  • 3.5 Properties of Alkanes
  • 3.6 Conformations of Ethane
  • 3.7 Conformations of Other Alkanes
  • Chemistry Matters—Gasoline
  • 4.1 Naming Cycloalkanes
  • 4.2 Cis–Trans Isomerism in Cycloalkanes
  • 4.3 Stability of Cycloalkanes: Ring Strain
  • 4.4 Conformations of Cycloalkanes
  • 4.5 Conformations of Cyclohexane
  • 4.6 Axial and Equatorial Bonds in Cyclohexane
  • 4.7 Conformations of Monosubstituted Cyclohexanes
  • 4.8 Conformations of Disubstituted Cyclohexanes
  • 4.9 Conformations of Polycyclic Molecules
  • Chemistry Matters—Molecular Mechanics
  • 5.1 Enantiomers and the Tetrahedral Carbon
  • 5.2 The Reason for Handedness in Molecules: Chirality
  • 5.3 Optical Activity
  • 5.4 Pasteur’s Discovery of Enantiomers
  • 5.5 Sequence Rules for Specifying Configuration
  • 5.6 Diastereomers
  • 5.7 Meso Compounds
  • 5.8 Racemic Mixtures and the Resolution of Enantiomers
  • 5.9 A Review of Isomerism
  • 5.10 Chirality at Nitrogen, Phosphorus, and Sulfur
  • 5.11 Prochirality
  • 5.12 Chirality in Nature and Chiral Environments
  • Chemistry Matters—Chiral Drugs
  • 6.1 Kinds of Organic Reactions
  • 6.2 How Organic Reactions Occur: Mechanisms
  • 6.3 Polar Reactions
  • 6.4 An Example of a Polar Reaction: Addition of HBr to Ethylene
  • 6.5 Using Curved Arrows in Polar Reaction Mechanisms
  • 6.6 Radical Reactions
  • 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes
  • 6.8 Describing a Reaction: Bond Dissociation Energies
  • 6.9 Describing a Reaction: Energy Diagrams and Transition States
  • 6.10 Describing a Reaction: Intermediates
  • 6.11 A Comparison Between Biological Reactions and Laboratory Reactions
  • Chemistry Matters—Where Do Drugs Come From?
  • 7.1 Industrial Preparation and Use of Alkenes
  • 7.2 Calculating the Degree of Unsaturation
  • 7.3 Naming Alkenes
  • 7.4 Cis–Trans Isomerism in Alkenes
  • 7.5 Alkene Stereochemistry and the E , Z Designation
  • 7.6 Stability of Alkenes
  • 7.7 Electrophilic Addition Reactions of Alkenes
  • 7.8 Orientation of Electrophilic Additions: Markovnikov’s Rule
  • 7.9 Carbocation Structure and Stability
  • 7.10 The Hammond Postulate
  • 7.11 Evidence for the Mechanism of Electrophilic Additions: Carbocation Rearrangements
  • Chemistry Matters—Bioprospecting: Hunting for Natural Products
  • 8.1 Preparing Alkenes: A Preview of Elimination Reactions
  • 8.2 Halogenation of Alkenes: Addition of X 2
  • 8.3 Halohydrins from Alkenes: Addition of HO-X
  • 8.4 Hydration of Alkenes: Addition of H 2 O by Oxymercuration
  • 8.5 Hydration of Alkenes: Addition of H 2 O by Hydroboration
  • 8.6 Reduction of Alkenes: Hydrogenation
  • 8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation
  • 8.8 Oxidation of Alkenes: Cleavage to Carbonyl Compounds
  • 8.9 Addition of Carbenes to Alkenes: Cyclopropane Synthesis
  • 8.10 Radical Additions to Alkenes: Chain-Growth Polymers
  • 8.11 Biological Additions of Radicals to Alkenes
  • 8.12 Reaction Stereochemistry: Addition of H 2 O to an Achiral Alkene
  • 8.13 Reaction Stereochemistry: Addition of H 2 O to a Chiral Alkene
  • Chemistry Matters—Terpenes: Naturally Occurring Alkenes
  • Summary of Reactions
  • 9.1 Naming Alkynes
  • 9.2 Preparation of Alkynes: Elimination Reactions of Dihalides
  • 9.3 Reactions of Alkynes: Addition of HX and X 2
  • 9.4 Hydration of Alkynes
  • 9.5 Reduction of Alkynes
  • 9.6 Oxidative Cleavage of Alkynes
  • 9.7 Alkyne Acidity: Formation of Acetylide Anions
  • 9.8 Alkylation of Acetylide Anions
  • 9.9 An Introduction to Organic Synthesis
  • Chemistry Matters—The Art of Organic Synthesis
  • 10.1 Names and Structures of Alkyl Halides
  • 10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation
  • 10.3 Preparing Alkyl Halides from Alkenes: Allylic Bromination
  • 10.4 Stability of the Allyl Radical: Resonance Revisited
  • 10.5 Preparing Alkyl Halides from Alcohols
  • 10.6 Reactions of Alkyl Halides: Grignard Reagents
  • 10.7 Organometallic Coupling Reactions
  • 10.8 Oxidation and Reduction in Organic Chemistry
  • Chemistry Matters—Naturally Occurring Organohalides
  • 11.1 The Discovery of Nucleophilic Substitution Reactions
  • 11.2 The S N 2 Reaction
  • 11.3 Characteristics of the S N 2 Reaction
  • 11.4 The S N 1 Reaction
  • 11.5 Characteristics of the S N 1 Reaction
  • 11.6 Biological Substitution Reactions
  • 11.7 Elimination Reactions: Zaitsev’s Rule
  • 11.8 The E2 Reaction and the Deuterium Isotope Effect
  • 11.9 The E2 Reaction and Cyclohexane Conformation
  • 11.10 The E1 and E1cB Reactions
  • 11.11 Biological Elimination Reactions
  • 11.12 A Summary of Reactivity: S N 1, S N 2, E1, E1cB, and E2
  • Chemistry Matters—Green Chemistry
  • 12.1 Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments
  • 12.2 Interpreting Mass Spectra
  • 12.3 Mass Spectrometry of Some Common Functional Groups
  • 12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments
  • 12.5 Spectroscopy and the Electromagnetic Spectrum
  • 12.6 Infrared Spectroscopy
  • 12.7 Interpreting Infrared Spectra
  • 12.8 Infrared Spectra of Some Common Functional Groups
  • Chemistry Matters—X-Ray Crystallography
  • 13.1 Nuclear Magnetic Resonance Spectroscopy
  • 13.2 The Nature of NMR Absorptions
  • 13.3 Chemical Shifts
  • 13.4 Chemical Shifts in 1 H NMR Spectroscopy
  • 13.5 Integration of 1 H NMR Absorptions: Proton Counting
  • 13.6 Spin–Spin Splitting in 1 H NMR Spectra
  • 13.7 1 H NMR Spectroscopy and Proton Equivalence
  • 13.8 More Complex Spin–Spin Splitting Patterns
  • 13.9 Uses of 1 H NMR Spectroscopy
  • 13.10 13 C NMR Spectroscopy: Signal Averaging and FT–NMR
  • 13.11 Characteristics of 13 C NMR Spectroscopy
  • 13.12 DEPT 13 C NMR Spectroscopy
  • 13.13 Uses of 13 C NMR Spectroscopy
  • Chemistry Matters—Magnetic Resonance Imaging (MRI)
  • 14.1 Stability of Conjugated Dienes: Molecular Orbital Theory
  • 14.2 Electrophilic Additions to Conjugated Dienes: Allylic Carbocations
  • 14.3 Kinetic versus Thermodynamic Control of Reactions
  • 14.4 The Diels–Alder Cycloaddition Reaction
  • 14.5 Characteristics of the Diels–Alder Reaction
  • 14.6 Diene Polymers: Natural and Synthetic Rubbers
  • 14.7 Ultraviolet Spectroscopy
  • 14.8 Interpreting Ultraviolet Spectra: The Effect of Conjugation
  • 14.9 Conjugation, Color, and the Chemistry of Vision
  • Chemistry Matters—Photolithography
  • 15.1 Naming Aromatic Compounds
  • 15.2 Structure and Stability of Benzene
  • 15.3 Aromaticity and the Hückel 4 n + 2 Rule
  • 15.4 Aromatic Ions
  • 15.5 Aromatic Heterocycles: Pyridine and Pyrrole
  • 15.6 Polycyclic Aromatic Compounds
  • 15.7 Spectroscopy of Aromatic Compounds
  • Chemistry Matters—Aspirin, NSAIDs, and COX-2 Inhibitors
  • 16.1 Electrophilic Aromatic Substitution Reactions: Bromination
  • 16.2 Other Aromatic Substitutions
  • 16.3 Alkylation and Acylation of Aromatic Rings: The Friedel–Crafts Reaction
  • 16.4 Substituent Effects in Electrophilic Substitutions
  • 16.5 Trisubstituted Benzenes: Additivity of Effects
  • 16.6 Nucleophilic Aromatic Substitution
  • 16.7 Benzyne
  • 16.8 Oxidation of Aromatic Compounds
  • 16.9 Reduction of Aromatic Compounds
  • 16.10 Synthesis of Polysubstituted Benzenes
  • Chemistry Matters—Combinatorial Chemistry
  • 17.1 Naming Alcohols and Phenols
  • 17.2 Properties of Alcohols and Phenols
  • 17.3 Preparation of Alcohols: A Review
  • 17.4 Alcohols from Carbonyl Compounds: Reduction
  • 17.5 Alcohols from Carbonyl Compounds: Grignard Reaction
  • 17.6 Reactions of Alcohols
  • 17.7 Oxidation of Alcohols
  • 17.8 Protection of Alcohols
  • 17.9 Phenols and Their Uses
  • 17.10 Reactions of Phenols
  • 17.11 Spectroscopy of Alcohols and Phenols
  • Chemistry Matters—Ethanol: Chemical, Drug, and Poison
  • 18.1 Names and Properties of Ethers
  • 18.2 Preparing Ethers
  • 18.3 Reactions of Ethers: Acidic Cleavage
  • 18.4 Cyclic Ethers: Epoxides
  • 18.5 Reactions of Epoxides: Ring-Opening
  • 18.6 Crown Ethers
  • 18.7 Thiols and Sulfides
  • 18.8 Spectroscopy of Ethers
  • Chemistry Matters—Epoxy Resins and Adhesives
  • Preview of Carbonyl Chemistry
  • 19.1 Naming Aldehydes and Ketones
  • 19.2 Preparing Aldehydes and Ketones
  • 19.3 Oxidation of Aldehydes and Ketones
  • 19.4 Nucleophilic Addition Reactions of Aldehydes and Ketones
  • 19.5 Nucleophilic Addition of H 2 O: Hydration
  • 19.6 Nucleophilic Addition of HCN: Cyanohydrin Formation
  • 19.7 Nucleophilic Addition of Hydride and Grignard Reagents: Alcohol Formation
  • 19.8 Nucleophilic Addition of Amines: Imine and Enamine Formation
  • 19.9 Nucleophilic Addition of Hydrazine: The Wolff–Kishner Reaction
  • 19.10 Nucleophilic Addition of Alcohols: Acetal Formation
  • 19.11 Nucleophilic Addition of Phosphorus Ylides: The Wittig Reaction
  • 19.12 Biological Reductions
  • 19.13 Conjugate Nucleophilic Addition to α , β ‑Unsaturated Aldehydes and Ketones
  • 19.14 Spectroscopy of Aldehydes and Ketones
  • Chemistry Matters—Enantioselective Synthesis
  • 20.1 Naming Carboxylic Acids and Nitriles
  • 20.2 Structure and Properties of Carboxylic Acids
  • 20.3 Biological Acids and the Henderson–Hasselbalch Equation
  • 20.4 Substituent Effects on Acidity
  • 20.5 Preparing Carboxylic Acids
  • 20.6 Reactions of Carboxylic Acids: An Overview
  • 20.7 Chemistry of Nitriles
  • 20.8 Spectroscopy of Carboxylic Acids and Nitriles
  • Chemistry Matters—Vitamin C
  • 21.1 Naming Carboxylic Acid Derivatives
  • 21.2 Nucleophilic Acyl Substitution Reactions
  • 21.3 Reactions of Carboxylic Acids
  • 21.4 Chemistry of Acid Halides
  • 21.5 Chemistry of Acid Anhydrides
  • 21.6 Chemistry of Esters
  • 21.7 Chemistry of Amides
  • 21.8 Chemistry of Thioesters and Acyl Phosphates: Biological Carboxylic Acid Derivatives
  • 21.9 Polyamides and Polyesters: Step-Growth Polymers
  • 21.10 Spectroscopy of Carboxylic Acid Derivatives
  • Chemistry Matters—β-Lactam Antibiotics
  • 22.1 Keto–Enol Tautomerism
  • 22.2 Reactivity of Enols: α -Substitution Reactions
  • 22.3 Alpha Halogenation of Aldehydes and Ketones
  • 22.4 Alpha Bromination of Carboxylic Acids
  • 22.5 Acidity of Alpha Hydrogen Atoms: Enolate Ion Formation
  • 22.6 Reactivity of Enolate Ions
  • 22.7 Alkylation of Enolate Ions
  • Chemistry Matters—Barbiturates
  • 23.1 Carbonyl Condensations: The Aldol Reaction
  • 23.2 Carbonyl Condensations versus Alpha Substitutions
  • 23.3 Dehydration of Aldol Products: Synthesis of Enones
  • 23.4 Using Aldol Reactions in Synthesis
  • 23.5 Mixed Aldol Reactions
  • 23.6 Intramolecular Aldol Reactions
  • 23.7 The Claisen Condensation Reaction
  • 23.8 Mixed Claisen Condensations
  • 23.9 Intramolecular Claisen Condensations: The Dieckmann Cyclization
  • 23.10 Conjugate Carbonyl Additions: The Michael Reaction
  • 23.11 Carbonyl Condensations with Enamines: The Stork Enamine Reaction
  • 23.12 The Robinson Annulation Reaction
  • 23.13 Some Biological Carbonyl Condensation Reactions
  • Chemistry Matters—A Prologue to Metabolism
  • 24.1 Naming Amines
  • 24.2 Structure and Properties of Amines
  • 24.3 Basicity of Amines
  • 24.4 Basicity of Arylamines
  • 24.5 Biological Amines and the Henderson–Hasselbalch Equation
  • 24.6 Synthesis of Amines
  • 24.7 Reactions of Amines
  • 24.8 Reactions of Arylamines
  • 24.9 Heterocyclic Amines
  • 24.10 Spectroscopy of Amines
  • Chemistry Matters—Green Chemistry II: Ionic Liquids
  • 25.1 Classification of Carbohydrates
  • 25.2 Representing Carbohydrate Stereochemistry: Fischer Projections
  • 25.3 D , L Sugars
  • 25.4 Configurations of the Aldoses
  • 25.5 Cyclic Structures of Monosaccharides: Anomers
  • 25.6 Reactions of Monosaccharides
  • 25.7 The Eight Essential Monosaccharides
  • 25.8 Disaccharides
  • 25.9 Polysaccharides and Their Synthesis
  • 25.10 Some Other Important Carbohydrates
  • Chemistry Matters—Sweetness
  • 26.1 Structures of Amino Acids
  • 26.2 Amino Acids and the Henderson–Hasselbalch Equation: Isoelectric Points
  • 26.3 Synthesis of Amino Acids
  • 26.4 Peptides and Proteins
  • 26.5 Amino Acid Analysis of Peptides
  • 26.6 Peptide Sequencing: The Edman Degradation
  • 26.7 Peptide Synthesis
  • 26.8 Automated Peptide Synthesis: The Merrifield Solid-Phase Method
  • 26.9 Protein Structure
  • 26.10 Enzymes and Coenzymes
  • 26.11 How Do Enzymes Work? Citrate Synthase
  • Chemistry Matters—The Protein Data Bank
  • 27.1 Waxes, Fats, and Oils
  • 27.3 Phospholipids
  • 27.4 Prostaglandins and Other Eicosanoids
  • 27.5 Terpenoids
  • 27.6 Steroids
  • 27.7 Biosynthesis of Steroids
  • Chemistry Matters—Saturated Fats, Cholesterol, and Heart Disease
  • 28.1 Nucleotides and Nucleic Acids
  • 28.2 Base Pairing in DNA
  • 28.3 Replication of DNA
  • 28.4 Transcription of DNA
  • 28.5 Translation of RNA: Protein Biosynthesis
  • 28.6 DNA Sequencing
  • 28.7 DNA Synthesis
  • 28.8 The Polymerase Chain Reaction
  • Chemistry Matters—DNA Fingerprinting
  • 29.1 An Overview of Metabolism and Biochemical Energy
  • 29.2 Catabolism of Triacylglycerols: The Fate of Glycerol
  • 29.3 Catabolism of Triacylglycerols: β -Oxidation
  • 29.4 Biosynthesis of Fatty Acids
  • 29.5 Catabolism of Carbohydrates: Glycolysis
  • 29.6 Conversion of Pyruvate to Acetyl CoA
  • 29.7 The Citric Acid Cycle
  • 29.8 Carbohydrate Biosynthesis: Gluconeogenesis
  • 29.9 Catabolism of Proteins: Deamination
  • 29.10 Some Conclusions about Biological Chemistry
  • Chemistry Matters—Statin Drugs
  • 30.1 Molecular Orbitals of Conjugated Pi Systems
  • 30.2 Electrocyclic Reactions
  • 30.3 Stereochemistry of Thermal Electrocyclic Reactions
  • 30.4 Photochemical Electrocyclic Reactions
  • 30.5 Cycloaddition Reactions
  • 30.6 Stereochemistry of Cycloadditions
  • 30.7 Sigmatropic Rearrangements
  • 30.8 Some Examples of Sigmatropic Rearrangements
  • 30.9 A Summary of Rules for Pericyclic Reactions
  • Chemistry Matters—Vitamin D, the Sunshine Vitamin
  • 31.1 Chain-Growth Polymers
  • 31.2 Stereochemistry of Polymerization: Ziegler–Natta Catalysts
  • 31.3 Copolymers
  • 31.4 Step-Growth Polymers
  • 31.5 Olefin Metathesis Polymerization
  • 31.6 Intramolecular Olefin Metathesis
  • 31.7 Polymer Structure and Physical Properties
  • Chemistry Matters—Degradable Polymers
  • A | Nomenclature of Polyfunctional Organic Compounds
  • B | Acidity Constants for Some Organic Compounds
  • C | Glossary
  • D | Periodic Table

2.3 • Formal Charges

Closely related to the ideas of bond polarity and dipole moment is the assignment of formal charges to specific atoms within a molecule, particularly atoms that have an apparently “abnormal” number of bonds. Look at dimethyl sulfoxide (CH 3 SOCH 3 ), for instance, a solvent commonly used for preserving biological cell lines at low temperature. The sulfur atom in dimethyl sulfoxide has three bonds rather than the usual two and has a formal positive charge. The oxygen atom, by contrast, has one bond rather than the usual two and has a formal negative charge. Note that an electrostatic potential map of dimethyl sulfoxide shows the oxygen as negative (red) and the sulfur as relatively positive (blue), in accordance with the formal charges.

Formal charges, as the name suggests, are a formalism and don’t imply the presence of actual ionic charges in a molecule. Instead, they’re a device for electron “bookkeeping” and can be thought of in the following way: A typical covalent bond is formed when each atom donates one electron. Although the bonding electrons are shared by both atoms, each atom can still be considered to “own” one electron for bookkeeping purposes. In methane, for instance, the carbon atom owns one electron in each of the four C–H bonds. Because a neutral, isolated carbon atom has four valence electrons, and because the carbon atom in methane still owns four, the methane carbon atom is neutral and has no formal charge.

The same is true for the nitrogen atom in ammonia, which has three covalent N–H bonds and two nonbonding electrons (a lone pair). Atomic nitrogen has five valence electrons, and the ammonia nitrogen also has five—one in each of three shared N–H bonds plus two in the lone pair. Thus, the nitrogen atom in ammonia has no formal charge.

The situation is different in dimethyl sulfoxide. Atomic sulfur has six valence electrons, but the dimethyl sulfoxide sulfur owns only five—one in each of the two S–C single bonds, one in the S–O single bond, and two in a lone pair. Thus, the sulfur atom has formally lost an electron and therefore has a positive formal charge. A similar calculation for the oxygen atom shows that it has formally gained an electron and has a negative charge. Atomic oxygen has six valence electrons, but the oxygen in dimethyl sulfoxide has seven—one in the O–S bond and two in each of three lone pairs. Thus, the oxygen has formally gained an electron and has a negative formal charge.

To express the calculations in a general way, the formal charge on an atom is equal to the number of valence electrons in a neutral, isolated atom minus the number of electrons owned by that bonded atom in a molecule. The number of electrons in the bonded atom, in turn, is equal to half the number of bonding electrons plus the nonbonding, lone-pair electrons.

A summary of commonly encountered formal charges and the bonding situations in which they occur is given in Table 2.2 . Although only a bookkeeping device, formal charges often give clues about chemical reactivity, so it’s helpful to be able to identify and calculate them correctly.

Organic phosphate groups occur commonly in biological molecules. Calculate formal charges on the four O atoms in the methyl phosphate dianion.

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Core Concepts in Chemistry

Concepts in Chemistry

Formal charge: what is it and how to determine it

Formal charge: what is it and how to determine it

posted on October 1, 2019

Formal charge is the charge we assign to a bonded atom if the bonding electrons were shared equally between the bonded atoms.

Why do we have to assign formal charges?

Sometimes we can write more than one Lewis structure for a particular ion or molecule. When that happens, we usually assign formal charges to the bonded atoms to help determine the correct Lewis structure.

To determine the formal charge for an atom, we usually follow these rules:

  • Assign all lone pairs of electrons to the atom on which we find them
  • Assign half of the bonding electrons to each atom in the bond

After applying the rules outlined above to each atom in the Lewis structure, we will then use the following formula to calculate the formal charge of each atom:

how to assign formal charges

Once we add all the formal charges for the atoms in the Lewis structure, we should get a value equal to the actual charge of the molecule or ion. If it is a neutral molecule, then the sum of all the formal charges must equal zero. If it is a molecular ion, then the sum of all the formal charges must equal the ionic charge.

How to decide the correct Lewis structure after assigning formal charges

After assigning formal charges, we again apply the following rules to identify the correct Lewis structure:

  • Negative formal charge should be on the most electronegative atom
  • Like charges should not be on adjacent atoms

Now, let’s apply the above rules to predict the best Lewis structure for the molecule,

Hydrogen Cyanide (HCN).

To draw the Lewis structure for HCN , we will first calculate the total number of valence electrons. If you look on the periodic table, you will notice that H has one valence electron, C has 4 , and N has 5 . If you sum all these valence electrons, you will get 10 . Next, we identify the central atom. But we know that hydrogen can form only a single bond, this means H cannot be the central atom. Since H cannot be the central atom, it follows that only C or N can be the central atom. If carbon is the central atom, then the bond skeleton of its Lewis structure will appear as: 

Bond skeleton with C in the center

If N is the central atom, then the bond skeleton of its Lewis structure will appear as:

Bond skeleton with N in center

Since the bond skeleton consists of 4 electrons, it follows that we have 6 electrons ( 10 – 4 ) leftover to distribute. To distribute the 6 electrons, we start from the outer atoms, but since H can have only 2 valence electrons and it does have its 2 valence electrons, we move on to N in the bond skeleton (a). As you can see, N needs only 6 more electrons to have an octet, so we give all the 6 electrons to it. Therefore, the Lewis structure of (a) will appear as:

N has an octet but C does not. Move lone pairs from N to form multiple bonds with C

However, notice C arbon has only 4 electrons, but needs 4 more to satisfy the octet rule. This means, we must move two lone pairs of electrons from N to form multiple bonds between C and N . If we do, the completed Lewis structure with carbon in center will appear as:

Complete Lewis structure with C in center

And the completed Lewis structure with N as the central atom (b) will appear as:

Complete Lewis structure N in center

So, the next question is which of the Lewis structures is the correct one?

To determine that, we must assign formal charges. So, let’s start with Lewis structure (e). To determine the formal charge of H , we must first figure out how many electrons it owns in the Lewis structure. From the rules outlined above, we do know it owns half of the shared electrons between it and C . And since there is a single bond between it and C , it follows that H owns only 1 electron ( 2 electrons divided by 2 ). Now, to determine the formal charge of H, we will simply subtract 1 from the valence electron of H predicted by the periodic table. If we do, we will get: 1-1 = 0 . Therefore, the formal charge of H is zero.

Similarly, formal charge of C will be: 4 – 4 = 0 .  And formal charge of N will be: 5-5 = 0 (recall to count the lone pairs on N)

If we include the formal charges to Lewis structure (e), it will appear as:

Formal charges are all equal to zero.  As a result, that's the correct Lewis structure

If we calculate the formal charge for Lewis structure (f), we will get:

Carbon, which is less electronegative than N has negative 1 formal charge. That makes the Lewis structure unstable

Now, if we look at Lewis structures (e) and (f) with formal charges, we can predict with reason that structure (e) should be stable. We think so because all the atoms in (f) have a formal charge of zero. However, in structure (f) notice that N has a formal charge of 1+ , while C has a formal charge of 1- , but N is more electronegative than carbon. Since the negative charge should reside on the most electronegative atom, if follows that Lewis structure (f) is incorrect (unstable).

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Chemical Bonding and Molecular Geometry

Formal Charges and Resonance

OpenStaxCollege

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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 \({\text{ICl}}_{4}{}^{\text{−}}.\)

A Lewis structure is shown. An iodine atom with two lone pairs of electrons is single bonded to four chlorine atoms, each of which has three lone pairs of electrons. Brackets surround the structure and there is a superscripted negative sign.

  • 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:

A Lewis structure is shown. A carbon atom with one lone pair of electrons is triple bonded to an oxygen with one lone pair of electrons.

Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen molecule BrCl 3 .

A Lewis structure is shown. A bromine atom with two lone pairs of electrons is single bonded to three chlorine atoms, each of which has three lone pairs of electrons.

  • 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

A Lewis structure is shown. A nitrogen atom with one lone pair of electrons is single bonded to three chlorine atoms, each of which has three lone pairs of electrons.

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:

Three Lewis structures are shown. The left and right structures show a carbon atom double bonded to two oxygen atoms, each of which has two lone pairs of electrons. The center structure shows a carbon atom that is triple bonded to an oxygen atom with one lone pair of electrons and single bonded to an oxygen atom with three lone pairs of electrons. The third structure shows an oxygen atom double bonded to another oxygen atom with to lone pairs of electrons. The first oxygen atom is also double bonded to a carbon atom with two lone pairs of electrons.

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:

Two rows of structures and numbers are shown. The top row is labeled, “Structure” and depicts three Lewis structures and the bottom row is labeled, “Formal charge.” The left structure shows a carbon atom double bonded to a nitrogen atom with two lone electron pairs on one side and double bonded to a sulfur atom with two lone electron pairs on the other. The structure is surrounded by brackets and has a superscripted negative sign. Below this structure are the numbers negative one, zero, and zero. The middle structure shows a carbon atom with two lone pairs of electrons double bonded to a nitrogen atom that is double bonded to a sulfur atom with two lone electron pairs. The structure is surrounded by brackets and has a superscripted negative sign. Below this structure are the numbers negative two, positive one, and zero. The right structure shows a carbon atom with two lone electron pairs double bonded to a sulfur atom that is double bonded to a nitrogen atom with two lone electron pairs. The structure is surrounded by brackets and has a superscripted negative sign. Below this structure are the numbers negative two, positive two, and one.

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?

Two Lewis structures are shown with the word “or” in between them. The left structure depicts a nitrogen atom with two lone pairs of electrons double bonded to a nitrogen that is double bonded to an oxygen with two lone pairs of electrons. The right structure shows a nitrogen atom with two lone pairs of electrons double bonded to an oxygen atom that is double bonded to a nitrogen atom with two lone pairs of electrons.

Solution Determining formal charge yields the following:

Two Lewis structures are shown with the word “or” in between them. The left structure depicts a nitrogen atom with two lone pairs of electrons double bonded to a nitrogen atom that is double bonded to an oxygen atom with two lone pairs of electrons. The numbers negative one, positive one, and zero are written above this structure. The right structure shows a nitrogen atom with two lone pairs of electrons double bonded to an oxygen atom that is double bonded to a nitrogen atom with two lone pairs of electrons. The numbers negative one, positive two, and negative one are written above this structure.

The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:

A Lewis structure is shown. A nitrogen atom with two lone pairs of electrons is double bonded to a nitrogen atom that is double bonded to an oxygen atom with two lone pairs of electrons.

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?

Two Lewis structures are shown with the word “or” written between them. The left structure shows a nitrogen atom with two lone pairs of electrons double bonded to an oxygen atom with one lone pair of electrons that is single bonded to an oxygen atom with three lone pairs of electrons. Brackets surround this structure and there is a superscripted negative sign. The right structure shows an oxygen atom with two lone pairs of electrons double bonded to a nitrogen atom with one lone pair of electrons that is single bonded to an oxygen with three lone pairs of electrons. Brackets surround this structure and there is a superscripted negative sign.

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:

Two Lewis structures are shown. The left structure shows an oxygen atom with three lone pairs of electrons single bonded to a nitrogen atom with one lone pair of electrons that is double bonded to an oxygen with two lone pairs of electrons. Brackets surround this structure, and there is a superscripted negative sign. The right structure shows an oxygen atom with two lone pairs of electrons double bonded to a nitrogen atom with one lone pair of electrons that is single bonded to an oxygen atom with three lone pairs of electrons. Brackets surround this structure, and there is a superscripted negative sign.

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:

Two Lewis structures are shown with a double headed arrow drawn between them. The left structure shows an oxygen atom with two lone pairs of electrons double bonded to a nitrogen atom with one lone pair of electrons that is single bonded to an oxygen atom with three lone pairs of electrons. Brackets surround this structure, and there is a superscripted negative sign. The right structure shows an oxygen atom with three lone pairs of electrons single bonded to a nitrogen atom with one lone pair of electrons that is double bonded to an oxygen atom with two lone pairs of electrons. Brackets surround this structure, and there is a superscripted negative sign.

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:

Three Lewis structures are shown with double headed arrows in between. Each structure is surrounded by brackets, and each has a superscripted two negative sign. The left structure depicts a carbon atom bonded to three oxygen atoms. It is single bonded to two of these oxygen atoms, each of which has three lone pairs of electrons, and double bonded to the third, which has two lone pairs of electrons. The double bond is located between the lower left oxygen atom and the carbon atom. The central and right structures are the same as the first, but the position of the double bonded oxygen has moved to the lower right oxygen in the central structure and to the top oxygen in the right structure.

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 :

A Lewis structure shows a hexagonal ring composed of six carbon atoms. They form single bonds to each another and single bonds to one hydrogen atom each.

(e) the formate ion:

A Lewis structure shows a carbon atom single bonded to two oxygen atoms and a hydrogen atom. The structure is surrounded by brackets and there is a superscripted negative sign.

(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:

A Lewis structure depicts a hexagonal ring composed of five carbon atoms and one nitrogen atom. Each carbon atom is single bonded to a hydrogen atom.

(e) the allyl ion:

A Lewis structure shows a carbon atom single bonded to two hydrogen atoms and a second carbon atom. The second carbon atom is single bonded to a hydrogen atom and a third carbon atom. The third carbon atom is single bonded to two hydrogen atoms. The whole structure is surrounded by brackets, and there is a superscripted negative sign.

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, \({\text{NO}}_{\text{2}}{}^{\text{–}}\text{.}\)

Two pairs of Lewis structures are shown with a double-headed arrow in between each pair. The left structure of the first pair shows a nitrogen atom with one lone pair of electrons single bonded to an oxygen atom with three lone pairs of electrons. It is also double bonded to an oxygen with two lone pairs of electrons. The right image of this pair depicts the mirror image of the left. Both images are surrounded by brackets and a superscripted negative sign. They are labeled, “For N O subscript two superscript negative sign.” The left structure of the second pair shows an oxygen atom with one lone pair of electrons single bonded to an oxygen atom with three lone pairs of electrons. It is also double bonded to an oxygen atom with two lone pairs of electrons. The right structure appears as a mirror image of the left. These structures are labeled, “For O subscript three.”

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:

Two Lewis structures are shown with a double headed arrow in between. The left structure shows a carbon atom single bonded to three hydrogen atoms and a second carbon atom. The second carbon is single bonded to two oxygen atoms. One of the oxygen atoms is single bonded to a hydrogen atom. The right structure, surrounded by brackets and with a superscripted negative sign, depicts a carbon atom single bonded to three hydrogen atoms and a second carbon atom. The second carbon atom is single bonded to two oxygen atoms.

Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.

This structure shows a carbon atom double bonded to two oxygen atoms, each of which has two lone pairs of electrons.

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) \({\text{NO}}_{3}{}^{\text{−}}\)

Two Lewis structures are shown with a double-headed arrow in between. The left structure shows an oxygen atom with one lone pair of electrons single bonded to an oxygen atom with three lone pairs of electrons. It is also double bonded to an oxygen atom with two lone pairs of electrons. The symbols and numbers below this structure read, “( 0 ), ( positive 1 ), ( negative 1 ).” The phrase, “Formal charge,” and a right-facing arrow lie to the left of this structure. The right structure appears as a mirror image of the left and the symbols and numbers below this structure read, “( negative 1 ), ( positive 1 ), ( 0 ).”

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:

A Lewis structure shows a nitrogen atom with one lone pair of electrons single bonded to two hydrogen atoms and an oxygen atom which has two lone pairs of electrons. The oxygen atom is single bonded to a hydrogen atom.

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.

A Lewis structure shows a nitrogen atom with one lone pair of electrons single bonded to three fluorine atoms, each with three lone pairs of electrons.

Which of the following structures would we expect for nitrous acid? Determine the formal charges:

Two Lewis structures are shown, with the word “or” in between. The left structure shows a nitrogen atom single bonded to an oxygen atom with three lone pairs of electrons. It is also single bonded to a hydrogen atom and double bonded to an oxygen atom with two lone pairs of electrons. The right structure shows a hydrogen atom single bonded to an oxygen atom with two lone pairs of electrons. The oxygen atom is single bonded to a nitrogen atom which is double bonded to an oxygen atom with two lone pairs of electrons.

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.

A Lewis structure shows a hydrogen atom single bonded to an oxygen atom with two lone pairs of electrons. The oxygen atom is single bonded to a sulfur atom. The sulfur atom is double bonded to two oxygen atoms, each of which have three lone pairs of electrons, and single bonded to an oxygen atom with two lone pairs of electrons. This oxygen atom is single bonded to a hydrogen atom.

Formal Charges and Resonance Copyright © 2014 by OpenStaxCollege is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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: May 22nd, 2023 |

How To Calculate Formal Charge

To calculate the formal charge of an atom, we start by:

  • evaluating the number of valence electrons ( VE ) the neutral atom has (e.g. 3 for boron, 4 for carbon, 5 for nitrogen, and so on).  (note: this is also equivalent to the effective nuclear charge 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:

FC = VE – NBE – B

  • VE corresponds to the number of electrons around the neutral atom (3 for boron, 4 for carbon, 5 for nitrogen, 6 for oxygen, 7 for fluorine)
  • NBE corresponds to the number of non-bonded electrons around the atom (2 for a lone pair, 1 for a singly-occupied orbital, 0 for an empty orbital)
  • B is the number of  bonds around the atom (equivalent to half the number of bonding electrons)

It’s called “ formal ” charge because it assumes that all bonding electrons are shared equally . It doesn’t account for electronegativity differences (i.e. dipoles).

For that reason formal charge isn’t always a good guide to where the electrons actually are in a molecule and can be an unreliable guide to reactivity. We’ll have more to say on that below .

2. Simple Examples For First-Row Elements

When all the lone pairs are drawn out for you, calculating formal charge is fairly straightforward.

Let’s work through the first example in the quiz below.

  • In the hydronium ion (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.

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"
  • 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
  • Learning New Reactions: How Do The Electrons Move?
  • 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
  • Equilibrium and Energy Relationships
  • What's a Transition State?
  • Hammond's Postulate
  • Learning Organic Chemistry Reactions: A Checklist (PDF)
  • 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
  • How To Draw The Enantiomer Of A Chiral Molecule
  • How To Draw A Bond Rotation
  • 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
  • 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
  • SN1 vs E1 and SN2 vs E2 : The Temperature
  • Deciding SN1/SN2/E1/E2 - The Solvent
  • 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
  • Stereoselective and Stereospecific Reactions
  • Regioselectivity In Alkene Addition Reactions
  • Stereoselectivity In Alkene Addition Reactions: Syn vs Anti Addition
  • Hydrohalogenation of Alkenes and Markovnikov's Rule
  • Hydration of Alkenes With Aqueous Acid
  • Rearrangements in Alkene Addition Reactions
  • Halogenation of Alkenes and Halohydrin Formation
  • Oxymercuration Demercuration of Alkenes
  • Hydroboration Oxidation of Alkenes
  • m-CPBA (meta-chloroperoxybenzoic acid)
  • OsO4 (Osmium Tetroxide) for Dihydroxylation of Alkenes
  • Palladium on Carbon (Pd/C) for Catalytic Hydrogenation of Alkenes
  • Cyclopropanation of Alkenes
  • A Fourth Alkene Addition Pattern - Free Radical Addition
  • Alkene Reactions: Ozonolysis
  • Summary: Three Key Families Of Alkene Reaction Mechanisms
  • 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
  • Partial Reduction of Alkynes With Na/NH3 To Obtain Trans Alkenes
  • Alkyne Hydroboration With "R2BH"
  • Hydration and Oxymercuration of Alkynes
  • Hydrohalogenation of Alkynes
  • 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
  • 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
  • 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
  • Synthesis Problems Involving Grignard Reagents
  • 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 Michael Addition Reaction and Conjugate Addition
  • 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

  • A Gallery of Some Interesting Molecules From Nature
  • 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
  • On Cats, Part 4: Enantiocats
  • On Cats, Part 6: Stereocenters
  • Organic Chemistry Is Shit
  • The Organic Chemistry Behind "The Pill"
  • Maybe they should call them, "Formal Wins" ?
  • Why Do Organic Chemists Use Kilocalories?
  • The Principle of Least Effort
  • Organic Chemistry GIFS - Resonance Forms
  • Reproducibility In Organic Chemistry
  • What Holds The Nucleus Together?
  • How Reactions Are Like Music
  • Organic Chemistry and the New MCAT

26 Organic Chemistry Tips and Tricks

  • Draw The Ugly Version First
  • Organic Chemistry Study Tips: Learn the Trends
  • The 8 Types of Arrows In Organic Chemistry, Explained
  • Top 10 Skills To Master Before An Organic Chemistry 2 Final
  • Common Mistakes with Carbonyls: Carboxylic Acids... Are Acids!
  • Planning Organic Synthesis With "Reaction Maps"
  • Alkene Addition Pattern #1: The "Carbocation Pathway"
  • Alkene Addition Pattern #2: The "Three-Membered Ring" Pathway
  • Alkene Addition Pattern #3: The "Concerted" Pathway
  • Number Your Carbons!
  • The 4 Major Classes of Reactions in Org 1
  • How (and why) electrons flow
  • Grossman's Rule
  • Three Exam Tips
  • A 3-Step Method For Thinking Through Synthesis Problems
  • Putting It Together
  • Putting Diels-Alder Products in Perspective
  • The Ups and Downs of Cyclohexanes
  • The Most Annoying Exceptions in Org 1 (Part 1)
  • The Most Annoying Exceptions in Org 1 (Part 2)
  • The Marriage May Be Bad, But the Divorce Still Costs Money
  • 9 Nomenclature Conventions To Know
  • Nucleophile attacks Electrophile

27 Case Studies of Successful O-Chem Students

  • Success Stories: How Corina Got The The "Hard" Professor - And Got An A+ Anyway
  • How Helena Aced Organic Chemistry
  • From a "Drop" To B+ in Org 2 – How A Hard Working Student Turned It Around
  • How Serge Aced Organic Chemistry
  • Success Stories: How Zach Aced Organic Chemistry 1
  • Success Stories: How Kari Went From C– to B+
  • How Esther Bounced Back From a "C" To Get A's In Organic Chemistry 1 And 2
  • How Tyrell Got The Highest Grade In Her Organic Chemistry Course
  • This Is Why Students Use Flashcards
  • Success Stories: How Stu Aced Organic Chemistry
  • How John Pulled Up His Organic Chemistry Exam Grades
  • Success Stories: How Nathan Aced Organic Chemistry (Without It Taking Over His Life)
  • How Chris Aced Org 1 and Org 2
  • Interview: How Jay Got an A+ In Organic Chemistry
  • How to Do Well in Organic Chemistry: One Student's Advice
  • "America's Top TA" Shares His Secrets For Teaching O-Chem
  • "Organic Chemistry Is Like..." - A Few Metaphors
  • How To Do Well In Organic Chemistry: Advice From A Tutor
  • Guest post: "I went from being afraid of tests to actually looking forward to them".

Comment section

60 thoughts on “ a key skill: how to calculate formal charge ”.

Hello, thanks for your wonderful posts on organic chemistry. It reallys helps me to recap org chem and I really like how you explain all these topics with a bit of humor.

That said, I think in this posts may be some typos: I think there are two typos in the solution of the last quiz of chapter 3 (ID 2310): (a) In the third task [C3H7N] of the quiz, there is just one electron on the negative charged carbon. Shouldn’t there be two electrons? (b) And in the fourth task [O-CH2] the sign of the formal charge of the carbon atom should be +1 (in the calculation). (c) Note 4 says “(…) of various compounds of first-row elements.” Aren’t the shown elements in the picture from the second row of the periodic table?

Thank you very much!

  • Pingback: A Key Skill: How to Calculate Formal Charge | Straight A Mindset

Your explanations and examples were clear and easy to understand. I appreciate the detailed step-by-step instructions, which made it easy to follow along and understand the concept. Thank you for taking the time to create this helpful resource

I think for Quiz ID: 2310, the formal charge for the carbon in the fourth molecule should be +1 instead of -1.

Fixed. Thanks for the spot!

Thank you so much sir. Finally i understood how to calculate the formal charge

Nice simple explanation

Great teaching , can I know where did u studied ??

Hi I am extremely confused. The two formulas for calculating FC that you provided are not the same and don’t produce the same results when I tried them out.

Formal charge = [# of valence electrons] – [electrons in lone pairs + 1/2 the number of bonding electrons]

Formal Charge = [# of valence electrons on atom] – [non-bonded electrons + number of bonds].

They do not produce the same result… If I have the formula BH4, and use the first formula provided to find FC of B, I would get:

(3) – (0 + 2) = +1

Using the second formula provided:

(3) – (0+4) = -1

Aren’t these formulas supposed to produce the same results? I am quite confused and I don’t know if I missed something.

Ah. I should have been more clear. The number of bonding electrons in BH4 equals 8, since each bond has two electrons and there are 4 B-H bonds. Half of this number equals 4. This should give you the same answer. I have updated the post to make this more explicit.

  • Pingback: Como posso calcular a cobrança formal? – CorujaSabia

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.

  • Pingback: ¿Cómo puedo calcular el cargo formal? – ElbuhoSabio

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.

  • Pingback: Come posso calcolare l’addebito formale? – GufoSaggio

YOU ARE THE BEST. I GOT THE HIGHEST MARK IN MY FIRST QUIZ, AND I KNOW THAT THROUGH THIS I WILL GET THE BEST IN MIDTERM AND FINAL. I want you guys to go on youtube and follow the steps. THANK YOU VERY MUCH.

I remember learning that in the cyanide ion, the carbon is nucleophilic because the formal negative charge is on carbon, not nitrogen, despite nitrogen being more electronegative. So I think a different explanation could me more accurate, but I’m not sure how to properly address it. I better keep reading.

In cyanide ion, there are two lone pairs – one on carbon, one on nitrogen. The lone pair on carbon is more nucleophilic because it is less tightly held (the atom is less electronegative than nitrogen). On all the examples I show that are negatively charged (eg BH4(-) ) there isn’t a lone pair to complicate questions of nucleophilicity.

This really helped for neutral covalent molecules. However, I’m having trouble applying this technique for molecules with an overall charge other than 0. For instance, in (ClO2)- , the formal charge of Cl should be 1. However, with your equation the charge should be 0. With the conventional equation, the charge is indeed 1.

I’d appreciate it if you replied sooner rather than later, as I do have a chemistry midterm on Friday. I’m quite confused with formal charges :)

Thanks for the study guide.

This method is wrong For CH3 , the valence eloctron is 4 , no : of bonds is 3 and no of non bonded electrons is 1 Then by this equation

F.C= 4-(1+3) = 0 but here it is given as +1

That analysis would be accurate for the methyl radical. However it fails for the methyl carbocation.

That example referred to the carbocation. For the methyl radical, the formal charge is indeed zero.

This was so helpful n the best explanation about the topic…

Thanks for the easy approach.

But when I used this formula it works. Thus #valence electrons_#lone pair__#1/2.bond pairs

Thanks for the easy approach. I have a problem in finding the FC on each O atom in ozone. Can you help me with that ASAP?

The FC on central atom would be +1 because [6-(2+3)] FC on O atom with coordinate bond would be: -1 because [6-(6+1)]. FC on O atom with double bond is: 0 because [6-(4+2)].

Hope I solved your question!

Thank u very much my exam is today and i wouldn’t pass without this information

AM REALLY LOST NOW ON THAT EXAMPLE OF CH3 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 .

I am beryllium and i got offended!!!!!!……..LOL Just kidding…….BTW, I found this article very useful.Thanks!!!!!!!!!!

what does it means if we determine a molecule with zero charge ?

It’s neutral!

you said that non bonded electrons in carbon is 2, but how ? because i see it as only 1 because out of the 4 valence electrons in carbon, three are paired with hydrogen so it’s only 1 left

If the charge is -1, there must be an “extra” electron on carbon – this is why there’s a lone pair. If there was only one electron, it would be neutral.

This works! I would take your class with organic chemistry if you are a professor. I am taking chemistry 2 now. Organic is next. Thank you so much!

Thank you very very more for the simple explanation! Unbelievably easy and saves so much time!!!!!!

Thank you!!! this was awesome, I’m a junior in chemistry and this finally answered all my questions about formal charge :)

Glad it was helpful Haley!

If formal charges bear no resemblance to reality, what are their significance?

I hope the post doesn’t get interpreted as “formal charges have no significance”. If it does I will have to change some of the wording.

What I mean to get across is that formal charges assigned to atoms do not *always* accurately depict electron density on that atom, and one has to be careful.

In other words, formal charge and electron density are two different things and they do not always overlap.

Formal charge is a book-keeping device, where we count electrons and assign a full charge to one or more of the atoms on a molecule or ion. Electron density, on the other hand, is a measurement of where the electrons actually are (or aren’t) on a species, and those charges can be fractional or partial charges.

First of all, the charge itself is very real. The ions NH4+ , HO-, H3O+ and so on actually do bear a single charge. The thing to remember is that from a charge density perspective, that charge might be distributed over multiple atoms. Take an ion like H3O+, for example. H3O *does* bear a charge of +1,

However, if one thinks about where the electrons are in H3O+, one realizes that oxygen is more electronegative than hydrogen, and is actually “taking’ electrons from each hydrogen. If you look at an electron density map of H3O+ , one will see that the positive charge is distributed on the three hydrogens, and the oxygen actually bears a slight negative charge. There’s a nice map here.

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?

There are meny compounds which bears various structure among these which one is more stable or less energetic is it possible to predicu from the formal charge calculation?

Hey great explanation. I have a question though. Why is the FC commonly +/- 1? Could you give me an example when the FC is not +/- 1? Thanks.

Sure, try oxygen with no bonds and a full octet of electrons.

Great!i can use this for my exam!thanks!

Shouldn’t the formal charge of CH3 be -1? I was just wondering because in your example its +1 and in the chart its -1.

In the question.. its mentioned that CH3 without any lone pairs.. which means the valence would be 4 but there will not be any (2electrons) lone pairs left.. Hence it will be (4-)-(0+3)= 1

In CH3 i think FC on C should be -1 as carbon valency is 4 it has already bonded with 3 hydrogen atom one electron is left free on carbon to get bond with or share with one electron H hence, number of non bonded electrons lone pair of electrons is considered as 2. 4-(2+3) = -1. In your case if we take 0 than valency of c is not satisfied.

thank you for collaboration of formal charge

The answer to the question in the post above is “carbenes” – they have two substitutents, one pair of electrons, and an empty p orbital – so a total of four electrons “to itself”, making it neutral.

thank you for excellent explanation

Glad you found it useful Peter!

Very good explanation.I finally understood how to calculate the formal charge,was having some trouble with it.Thanks:)

Glad you found it helpful.

nice, concise explanation

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

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Read Chemistry

Formal charge: definition, formula, calculation, examples.

Organic Chemistry

– In this subject, we will discuss the Formal Charge: Definition, Formula, Calculation, Examples

Formal Charge: Definition, Formula, Calculation, Examples

Formal Charge and How To Calculate

– Many Lewis structures are incomplete until we decide whether any of their atoms have a formal charge.

– Calculating the formal charge on an atom in a Lewis structure is simply a bookkeeping method for its valence electrons.

– First, we examine each atom and, using the periodic table , we determine how many valence electrons it would have if it were an atom not bonded to any other atoms.

– This is equal to the group number of the atom in the periodic table.

– For hydrogen, this number equals 1, for carbon it equals 4, for nitrogen it equals 5, and for oxygen, it equals 6.

– Next, we examine the atom in the Lewis structure and we assign the valence electrons in the following way:

– We assign to each atom half of the electrons it is sharing with another atom and all of its unshared (lone) electron pairs.

– Then we do the following calculation for the atom:

– Formal charge = number of valence electrons – 1/2 number of shared electrons – number of unshared electrons.

how to assign formal charges

– It is important to note, too, that the arithmetic sum of all the formal charges in a molecule or ion will equal the overall charge on the molecule or ion.

Examples of Calculating Formal Charge

– Let us consider several examples showing how this is done.

The Ammonium Ion (NH 4 + )

– As we see below, the ammonium ion has no unshared electron pairs.

– We divide all of the electrons in bonds equally between the atoms that share them. Thus, each hydrogen is assigned one electron.

– We subtract this from one (the number of valence electrons in a hydrogen atom) to give each hydrogen atom a formal charge of zero.

– The nitrogen atom is assigned four electrons (one from each bond).

– We subtract four from five (the number of valence electrons in a nitrogen atom) to give the nitrogen a formal charge of +1.

Formal Charge: Definition, Formula, Calculation, Examples

The Nitrate Ion (NO 3 – )

– Let us next consider the nitrate ion (NO 3 – ), an ion that has oxygen atoms with unshared electron pairs.

– Here we find that the nitrogen atom has a formal charge of +1, that two oxygen atoms have formal charges of -1, and that one oxygen has a formal charge equal to 0.

Formal Charge: Definition, Formula, Calculation, Examples

Water and Ammonia

– The sum of the formal charges on each atom making up a molecule must be zero.

– Consider the following examples:

Formal Charge: Definition, Formula, Calculation, Examples

Practice Problem(1): Write a Lewis structure for each of the following negative ions, and assign the formal negative charge to the correct atom:

how to assign formal charges

A Summary of Formal Charge

– With this background, it should now be clear that each time an oxygen atom of the type:

how to assign formal charges

  appears in a molecule or ion, it will have a formal charge of -1, and each time an oxygen atom of the type :

how to assign formal charges

  appears, it will have a formal charge of 0.

– Similarly,

how to assign formal charges

– These and other common structures are summarized in this Table:

Formal Charge: Definition, Formula, Calculation, Examples

Practice Problem(2): Assign the proper formal charge to the colored atom in each of the following structures:

Formal Charge: Definition, Formula, Calculation, Examples

Reference:   Organic chemistry / T.W. Graham Solomons, Craig B.Fryhle, Scott A. Snyder, / ( eleventh edition) / 2014.

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1.4: Formal Charge

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Chemical reactions occur via attraction and donation of electrons. Looking at the structure of a molecule can help us to understand or to predict the behavior of that compound. One of the tools that we will eventually use to understand reactivity is formal charge. That is because reactivity has to do with the reorganization of electrons between atoms. New chemical bonds are formed by sharing electrons. Old chemical bonds are broken when one atom takes the bonding electrons away from another atom.

Formal charge can help us to understand the behavior of carbon monoxide, \(CO\). When exposed to transition metal cations such as the iron in hemoglobin ( Fe 2 + ), the carbon is attracted to and binds to the metal. In the case of hemoglobin, because the carbon monoxide binds very strongly to the iron, the CO blocks the position where oxygen would normally be bound and carbon monoxide poisoning results.

  • formal charge can help us predict how a molecule behaves
  • atoms with positive formal charges often attract electrons
  • atoms with negative formal charges often donate electrons

Why does the molecule behave in this way? There are actually a number of reasons. However, the fact that the carbon is attracted to a metal cation begs the question: Is the carbon an anion? Yes, in a sense. In a Lewis structure of the compound, the carbon has a formal negative charge. You will see why below.

Formal charges are an important book-keeping device that we use in Lewis structures. They tell us if one atom is donating extra electrons to another to give it an octet. If an atom needs to donate more electrons than normal in order for everyone to get an octet, it will have a positive formal charge. If an atom donates fewer electrons than normal and everyone still has an octet, it must be getting extra electrons from somewhere else. It will have a negative formal charge.

  • formal charge is often present if the atom does not have its usual number of bonds
  • valence rules can act as flags to alert you that formal charges are present

To help us think about formal charges, let's look at a few small molecules that all contain carbon-oxygen multiple bonds but that are slightly different from each other.

Example 1: Formaldehyde

Formaldehyde (CH 2 O) is a chemical that is used to preserve tissues; you may be familiar with its odor from anatomy lab.

Look at the structure of formaldehyde. Oxygen has a normal valence of two, and it has two bonds in formaldehyde, so there is no formal charge on the oxygen. Carbon has a normal valence of four, and it has four bonds here. There is no formal charge on carbon. There are no formal charges on the hydrogens either.

Example 2: Carbon Monoxide

Carbon monoxide results from burning fossil fuels; it is also an important industrial chemical used in manufacturing detergents. Carbon monoxide has a structure that is very similar to formaldehyde. It does not have any hydrogens, though. With ten electrons total, the only way to get an octet on both atoms is to make three bonds between carbon and oxygen.

Oxygen has normal valence two, but here it is making three bonds. It is sharing an extra pair of its electrons with carbon to make that third bond. If it is sharing a pair of electrons, we can think of it keeping one for itself and giving the other to carbon. Since it gives one of its electrons to carbon, it has formal charge +1.

Carbon has normal valence four, but here it is only making three bonds, even though it has an octet. How did it get an octet with only three bonds? It got an extra electron from somewhere (the oxygen). It has formal charge -1.

Notice that overall the carbon monoxide molecule is neutral. Oxygen has a plus charge and carbon has a minus charge. These charges cancel to give an overall neutral molecule.

What we are really doing when we assign formal charge is comparing how many electrons the atom brought with it from the periodic table to how many it has now. If the atom brought four electrons of its own and it is now sharing eight, things are even. It brought four to share and got four from its neighbors in an even trade. If it only brought three of its own and is now sharing eight, it got more electrons than it gave, and it will have a negative charge.

To determine formal charge:

  • check the number of electrons on the atom in the periodic table
  • check the number of electrons entirely owned by the atom in the molecule; this is different than looking for an octet
  • "entirely owned" electrons include any electrons in lone pairs, since they belong completely to one atom
  • "entirely owned" electrons also include half of the electrons in the bonds to the atom, since it is sharing each of those pairs with other atoms.

Compare the number of entirely owned valence electrons in the periodic table to those entirely owned by the atom in the molecule.

  • if the number of entirely owned electrons on the atom in the molecule is higher than in the periodic table, the atom has a negative charge
  • if the number of entirely owned electrons on the atom in the molecule is lower than in the periodic table, the atom has a positive charge
  • the formal charge is additive: if the atom has two extra electrons in the molecule, it has a two minus charge. If it is two short, it has a two plus charge.

Remember, electron counting to determine an octet counts all of the bonding and nonbonding electrons equally. It is done simply to determine whether the atom has a noble gas configuration right now. Electron counting to determine formal charge is done to keep track of who has given electrons to whom when making the molecule. If, in getting to an octet, atoms have received more electrons than they have given, their electron/proton ratio has changed, and they become charged.

Example 3: Carbonate Ion

Calcium carbonate is found in limestone and chalk, for instance.

Problem IM5.1

Draw Lewis or Kekule structures for the following molecules, remembering to include formal charges, if any (and notice that some of these molecules are ions):

  • CH 3 CO 2 -

Problem IM5.2.

Given the structures below, assign any missing formal charges.

Problem IM5.3.

Given the structures below, draw in the missing electrons, if any.

IM5pt3q.png

Contributors

Chris P Schaller, Ph.D. , (College of Saint Benedict / Saint John's University)

Further Reading

MasterOrganicChemistry

How to Calculate Formal Charge

Common Mistakes: Formal Charge Can Mislead

Khan Academy

Formal Charge I

Formal Charge II

Carey 4th Ed Online

Lewis Structures and Formal Charge

Formal Charge Tutorial (with problems)

Formal Charge

Berkeley Video on Formal Charge

Formal Charge Video

How To Determine Formal Charges

IMAGES

  1. How To Find Formal Charge In Lewis Structure

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  2. Solved Hint n 16 of 32 > Assign formal charges to each atom

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  3. Ch 1 : Formal charges

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  4. Solved 1. Assign formal charges to the elements in each of

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  5. Solved: Assign formal charges to all atoms in the following Lewis

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  6. Solved Assign formal charges to these structures. Assume

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COMMENTS

  1. Formal Charge

    The formal charge of any atom in a molecule can be calculated by the following equation: FC = V − N − B 2 (1) (1) F C = V − N − B 2. where V is the number of valence electrons of the neutral atom in isolation (in its ground state); N is the number of non-bonding valence electrons on this atom in the molecule; and B is the total number ...

  2. 7.4 Formal Charges and Resonance

    Assign formal charges to each atom in the interhalogen ion ICl4−. ICl 4 −. Solution Step 1. We divide the bonding electron pairs equally for all I-Cl bonds: 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.

  3. 2.3: Formal Charges

    Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons. Given: chemical species. Asked for: Lewis electron structures, formal charges, and preferred arrangement. Strategy: A Use the step-by-step procedure to write two plausible Lewis electron structures for SCN −.

  4. Formal Charges and Lewis Structures

    2 Share 114 views 4 months ago General Chemistry Continuing our series on Lewis Dot structures, this video will explore how to assign formal charges to different atoms within a Lewis structure...

  5. 4.3: Formal Charge and Oxidation State

    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). Exercise 4.3.1 4.3. 1. Calculate the formal charge for each atom in the carbon monoxide molecule:

  6. Formal charge and dot structures (video)

    Learn how to calculate formal charge and draw dot structures for molecules using the octet rule and the VSEPR model. See examples of how to choose the more stable dot structure and how to use formal charges to evaluate resonance structures. Watch a video and answer questions from other viewers.

  7. Formal charge (video)

    We can calculate an atom's formal charge using the equation FC = VE - [LPE - ½ (BE)], where VE = the number of valence electrons on the free atom, LPE = the number of lone pair electrons on the atom in the molecule, and BE = the number of bonding (shared) electrons around the atom in the molecule. Created by Sal Khan. Questions Tips & Thanks

  8. 4.5 Formal Charges and Resonance

    Calculating Formal Charge from Lewis Structures Assign formal charges to each atom in the interhalogen ion ICl 4 −. Solution. We divide the bonding electron pairs equally for all I-Cl bonds: 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.

  9. Formal Charges

    To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules: Nonbonding electrons are assigned to the atom on which they are located. Bonding electrons are divided equally between the bonded atoms. For each atom, we then compute a formal charge:

  10. 2.3 Formal Charges

    2.3 • Formal Charges Closely related to the ideas of bond polarity and dipole moment is the assignment of formal charges to specific atoms within a molecule, particularly atoms that have an apparently "abnormal" number of bonds. Look at dimethyl sulfoxide (CH 3 SOCH 3), for instance, a solvent commonly used for preserving biological cell lines at low temperature.

  11. Formal Charges: Calculating Formal Charge

    A step-by-step description on how to calculate formal charges. Formal charges are important because they allow us to predict which Lewis structure is the mo...

  12. Formal charge: what is it and how to determine it

    Formal charge is the charge we assign to a bonded atom if the bonding electrons were shared equally between the bonded atoms. Why do we have to assign formal charges? Sometimes we can write more than one Lewis structure for a particular ion or molecule.

  13. Formal Charges and Resonance

    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 Calculate the formal charge for each atom in the carbon monoxide molecule: Answer: C −1, O +1 Calculating Formal Charge from Lewis Structures

  14. How To Calculate Formal Charge

    The simplest way to write the formula for formal charge ( 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) 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)

  15. Formal Charges and Resonance

    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:

  16. How To Calculate The Formal Charge of an Atom

    10K 830K views 6 years ago New AP & General Chemistry Video Playlist This chemistry video tutorial provides a basic introduction into how to calculate the formal charge of an atom or element in a...

  17. 6.5: Formal Charges and Resonance

    Each Cl atom now has seven electrons assigned to it, and the I atom has eight. 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). Exercise 6.5.1 6.5. 1.

  18. How to Assign Formal Charges to Each Atom in a Dot Structure

    Steps to Assign Formal Charges to Each Atom in a Dot Structure Step 1: Read through the provided information, and sketch the dot structure of each molecule. Step 2: For each atom, determine the...

  19. 10.7: Formal Charges

    Example \(\PageIndex{2}\): Calculating Formal Charge from Lewis Structures. Assign formal charges to each atom in the interhalogen molecule \(\ce{BrCl3}\). 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.

  20. Formal Charge

    This organic chemistry video tutorial explains how to calculate the formal of an atom in a molecule using a simple formula. Organic Chemistry - Basic Introd...

  21. Formal Charge: Definition, Formula, Calculation, Examples

    - Calculating the formal charge on an atom in a Lewis structure is simply a bookkeeping method for its valence electrons. - First, we examine each atom and, using the periodic table, we determine how many valence electrons it would have if it were an atom not bonded to any other atoms. - This is equal to the group number of the atom in the periodic table.

  22. 1.4: Formal Charge

    Formal charge can help us to understand the behavior of carbon monoxide, CO C O. When exposed to transition metal cations such as the iron in hemoglobin ( Fe 2+), the carbon is attracted to and binds to the metal. In the case of hemoglobin, because the carbon monoxide binds very strongly to the iron, the CO blocks the position where oxygen ...