8.3: Formal Charge, Resonance, and Bond Energy

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 Learning Objectives

• 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
• Describe the energetics of covalent and ionic bond formation and breakage
• Use the Born-Haber cycle to compute lattice energies for ionic compounds
• Use average covalent bond energies to estimate enthalpies of reaction

Formal Charge

Previously, we discussed how to write Lewis structures for molecules and polyatomic ions. In some cases, however, 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 of an atom as follows:

\[\textrm{formal charge = # valence shell electrons (free atom) − # lone pair electrons − # bonds} \nonumber \]

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 equal to the charge of the molecule or ion.

We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.

Example \(\PageIndex{1}\): Calculating Formal Charge from Lewis Structures

Assign formal charges to each atom in the ion \(\ce{ICl4-}\).

Solution

We divide the bonding electron pairs equally for all \(\ce{I–Cl}\) 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.

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 \(\PageIndex{1}\)

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

Answer

C −1, O +1

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. 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 = 0
• Cl: 7 – 7 = 0

All atoms in \(\ce{BrCl3}\) have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.

Exercise \(\PageIndex{2}\)

Determine the formal charge for each atom in \(\ce{NCl3}\).

Answer

N: 0; all three Cl atoms: 0

 

 

Using Formal Charge to Predict Molecular Structure

The arrangement of atoms in a molecule or ion is called its molecular structure. In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure—different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion.

Predicting Molecular Structure Guidelines

1. A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
2. If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
3. Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
4. 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, \(\ce{CO2}\). We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:

 

 

Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).  The other two structures both have formal charges on two of the atoms.

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: \(\ce{CNS^{–}}\), \(\ce{NCS^{–}}\), or \(\ce{CSN^{–}}\). The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:

 

 

Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).

Example \(\PageIndex{3}\): Using Formal Charge to Determine Molecular Structure

Nitrous oxide, N₂O, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the likely structure for nitrous oxide?

Solution Determining formal charge yields the following:

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

The number of atoms with formal charges are minimized (Guideline 2), and there is no formal charge larger than one (Guideline 2). This is again consistent with the preference for having the less electronegative atom in the central position.

Exercise \(\PageIndex{3}\)

Which is the most likely molecular structure for the nitrite (\(\ce{NO2-}\)) ion?

Answer

\(\ce{ONO^{–}}\)

Resonance

You may have noticed that the nitrite anion in Example \(\PageIndex{3}\) 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:

 

 

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 \(\ce{NO2-}\) have the same strength and length, and are identical in all other properties.

It is not possible to write a single Lewis structure for \(\ce{NO2-}\) 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 \(\ce{NO2-}\) 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 \(\ce{NO2-}\) ion is shown as:

 

 

We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).

The carbonate anion, \(\ce{CO3^2-}\), provides a second example of resonance:

 

 

One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three C–O bonds are exactly the same.

Bond Strength: Covalent Bonds

A bond’s strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms. In this section, you will learn about the bond strength of covalent bonds, and then compare that to the strength of ionic bonds, which is related to the lattice energy of a compound.

Stable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy; the stronger a bond, the greater the energy required to break it. The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, \(D_{X–Y}\), is defined as the standard enthalpy change for the endothermic reaction:

\[XY_{(g)}⟶X_{(g)}+Y_{(g)}\;\;\; D_{X−Y}=ΔH° \label{7.6.1}\]

For example, the bond energy of the pure covalent H–H bond, \(\Delta_{H–H}\), is 436 kJ per mole of H–H bonds broken:

\[H_{2(g)}⟶2H_{(g)} \;\;\; D_{H−H}=ΔH°=436kJ \label{EQ2}\]

Breaking a bond always require energy to be added to the molecule. Correspondingly, making a bond always releases energy.

Molecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four C–H bond energies in CH₄, 1660 kJ, is equal to the standard enthalpy change of the reaction:

The average C–H bond energy, \(D_{C–H}\), is 1660/4 = 415 kJ/mol because there are four moles of C–H bonds broken per mole of the reaction. The 415 kJ/mol average value for a C-H bond is listed in Table \(\PageIndex{1}\). 

The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Table \(\PageIndex{1}\) shows bond energies for common double and triple bonds.  Also notice that bond strength typically decreases going down a group in the periodic table.  For example; C–F is 439 kJ/mol, C–Cl is 330 kJ/mol, and C–Br is 275 kJ/mol.

 

Table \(\PageIndex{1}\): Bond Energies (kJ/mol)

	H	C	N	O	F	Cl	Br
H	436	415	390	464	569	432	370
C		345 (C-C) 611 (C=C) 837 (C≡C)	290 (C-N) 615 (C=N) 891 (C≡N)	350 (C-O) 741 (C=O) 1080 (C≡O)	439	330	275
N			160 (N-N) 418 (N=N) 946 (N≡)	200	270	200	245
O				140 (O-O) 498 (O=O)	160	205
F					160	255	235
Cl						243	220
Br							190

 

We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic.

• An exothermic reaction (ΔH negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants.
• An endothermic reaction (ΔH positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.

The enthalpy change, ΔH, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy “in”, positive sign) plus the energy released when all bonds are formed in the products (energy “out,” negative sign). This can be expressed mathematically in the following way:

\[\Delta H=\sum Bond Energy_{\text{bonds broken}}− \sum Bond Energy_{\text{bonds formed}} \]

In this expression, the symbol \(\Sigma\) means “the sum of” and the bond energy from Table \(\PageIndex{1}\) is in kilojoules per mole, which is always a positive number. The bond energy is for a type of bond, single, double, or triple, between to specific atoms. Enthalpies calculated this way must include all bonds broke and created in a chemical reaction.  Because bond energy values are averages for one type of bond in many different molecules, these calculations provide an estimate for the enthalpy of reaction.

Consider the following reaction:

\[\ce{H_{2(g)} + Cl_{2(g)}⟶2HCl_{(g)}} \label{EQ4}\]

or

\[\ce{H–H_{(g)} + Cl–Cl_{(g)}⟶2H–Cl_{(g)}} \label{\EQ5}\]

To form two moles of HCl, one mole of H–H bonds and one mole of Cl–Cl bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the H–H bond (436 kJ/mol) and the Cl–Cl bond (243 kJ/mol). During the reaction, two moles of H–Cl bonds form (bond energy = 432 kJ/mol) to release 2 × 432 kJ; or 864 kJ of energy.

\[\begin {align*}
ΔH&= \sum \mathrm{D_{bonds\: broken}}− \sum \mathrm{D_{bonds\: formed}}\\[4pt]
&=\mathrm{[D_{H−H}+D_{Cl−Cl}]−2D_{H−Cl}}\\[4pt]
&=\mathrm{[436+243]−2(432)=−185\:kJ}
\end {align*}\]

This excess energy is released as heat, so the reaction is exothermic.  The value calculated -864 kJ is for the formation of 2 moles of HCl, so this compares with two times the standard molar enthalpy of formation of HCl(g), \(ΔH^\circ_\ce f\).  –92.3 kJ/mol x 2 mol = –184.6 kJ, which agrees with found using bond energies.

Example \(\PageIndex{4}\): Using Bond Energies to Approximate Enthalpy Changes

Methanol, CH₃OH, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO, and hydrogen, H₂, from which methanol can be produced. Using the bond energies in Table \(\PageIndex{1}\), calculate the approximate enthalpy change, ΔH, for the reaction here:

\[CO_{(g)}+2H2_{(g)}⟶CH_3OH_{(g)}\]

Solution

First, we need to write the Lewis structures of the reactants and the products:

From this, we see that ΔH for this reaction involves the energy required to break a C–O triple bond and two H–H single bonds, as well as the energy produced by the formation of three C–H single bonds, a C–O single bond, and an O–H single bond. We can express this as follows (via Equation \ref{EQ3}):

\[\begin {align*}
ΔH&= \sum D_{bonds\: broken}− \sum D_{bonds\: formed}\\
ΔH&=\mathrm{[D_{C≡O}+2(D_{H−H})]−[3(D_{C−H})+D_{C−O}+D_{O−H}]}
\end {align*}\]

Using the bond energy values in Table \(\PageIndex{1}\), we obtain:

\[\begin {align*}
ΔH&=[1080+2(436)]−[3(415)+350+464]\\
&=\ce{−107\:kJ}
\end {align*}\]

We can compare this value to the value calculated based on \(ΔH^\circ_\ce f\) data from Appendix G:

\[\begin {align*}
ΔH&=[ΔH^\circ_{\ce f}\ce{CH3OH}(g)]−[ΔH^\circ_{\ce f}\ce{CO}(g)+2×ΔH^\circ_{\ce f}\ce{H2}]\\
&=[−201.0]−[−110.52+2×0]\\
&=\mathrm{−90.5\:kJ}
\end {align*}\]

Note that there is a fairly significant gap between the values calculated using the two different methods. This occurs because D values are the average of different bond strengths; therefore, they often give only rough agreement with other data.

Exercise \(\PageIndex{4}\)

Ethyl alcohol, CH₃CH₂OH, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction:

Using the bond energies in Table \(\PageIndex{1}\), calculate an approximate enthalpy change, ΔH, for this reaction.

Answer

–35 kJ

Bond Length

Table \(\PageIndex{2}\) shows average bond lengths for some common covalent bonds with carbon, nitrogen, and oxygen.

 

Table \(\PageIndex{2}\): Bond Length (pm)

	C	N	O
C	154 (C-C) 134 (C=C) 120 (C≡C)	143 (C-N) 138 (C=N) 116 (C≡N)	143 (C-O) 123(C=O) 113(C≡O)

 

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).

The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound.

Key Equations

• Bond energy for a diatomic molecule: \(\ce{XY}(g)⟶\ce{X}(g)+\ce{Y}(g)\hspace{20px}\ce{D_{X–Y}}=ΔH°\)
• Enthalpy change: ΔH = ƩD_{bonds broken} – ƩD_{bonds formed}
• Lattice energy for a solid MX: \(\ce{MX}(s)⟶\ce M^{n+}(g)+\ce X^{n−}(g)\hspace{20px}ΔH_\ce{lattice}\)
• Lattice energy for an ionic crystal: \(ΔH_\ce{lattice}=\mathrm{\dfrac{C(Z^+)(Z^-)}{R_o}}\)
• \(\textrm{formal charge = # valence shell electrons (free atom) − # one pair electrons − }\dfrac{1}{2}\textrm{ # bonding electrons}\)

Glossary

bond energy

(also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance.

formal charge

charge that would result on an atom by taking the number of valence electrons on the neutral atom and subtracting the nonbonding electrons and the number of bonds (one-half of the bonding electrons)

lattice energy (ΔH_lattice)

energy required to separate one mole of an ionic solid into its component gaseous ions

molecular structure

arrangement of atoms in a molecule or ion

resonance

situation in which one Lewis structure is insufficient to describe the bonding in a molecule and the average of multiple structures is observed

resonance forms

two or more Lewis structures that have the same arrangement of atoms but different arrangements of electrons

resonance hybrid

average of the resonance forms shown by the individual Lewis structures

Contributors and Attributions

• Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).