5 Resonance Structures and Formal Charge

Introduction

This section explores resonance structures for molecules and polyatomic ions and how the contribution of each resonance form can be assessed using formal charge. In many cases, more than one valid Lewis structure can be drawn for a given molecule. In these cases, the actual electronic structure is not well-represented by any single Lewis structure, but is well-represented by a weighted average of all contributing resonance structures. The section below provides a more detailed description of these topics, worked examples, practice problems and a glossary of important terms

Learning Objectives for Resonance Structures and Formal Charge

| Key Concepts and Summary | Key Equations | Glossary | End of Section Exercises |

Bond Order, Bond Length, and Bond Strength

The way that two nuclei share electrons in a bond affects quantities such as bond order, bond polarity, bond length, and bond strength. Bond order is the number of electron-pair bonds connecting two nuclei. The Lewis structure of formaldehyde, CH2O, that contributes the most to the bonding in the molecule is as follows:

Lewis structure for the molecule formaldehyde. A central carbon is double bonded to one oxygen atom and singly bonded to two hydrogen atoms.

Looking at the structure of formaldehyde we can see that there is a double bond between the central carbon atom and the oxygen atom giving a CO bond order of two. The carbon is singly bonded to each hydrogen atom, which would give each CH bond orders of one.

Bond order is an index of bond strength: the higher the bond order, the stronger the bond. Bond strength is a measured quantity: the energy (in kJ/mol) required to break a chemical bond, tabulated in Appendix G. The stronger the bond, the more energy that is required to break the bond. This means that a C=O double bond is stronger than a C-O single bond, and the C=O double bond requires more energy to break than a C-O single bond. Bond length is the equilibrium distance between two nuclei. Higher bond orders generally correlate with shorter bond distances. The C=O double bond is shorter than the C-O single bond.

We can apply these same concepts to carbon-carbon single, double, and triple bonds to compare how bond order, bond length and bond strength are related (Figure 1).

A table is shown for bond enthalpy and length. The first row is for a carbon carbon single bond. The bond order (number bond pairs) is 1. The bond enthalpy is 356 kilojoules per mole and the length is 154 picometers. The second row is for a carbon carbon double bond. The bond order is 2. The bond enthalpy is 598 kilojoules per mole. The bond length is 134 picometers. The third row is for a carbon carbon triple bond. The bond order is 3. The bond enthalpy is 813 kilojoule per mole and the length is 121 picometers.
Figure 1. comparison of carbon-carbon single, double, and triple bonds.

Resonance Structures

Following the five steps for drawing a Lewis structure we can determine a valid Lewis structure for NO2– :

Lewis structure for NO2-. Central nitrogen atom is singly bonded to one oxygen atom and double bonded to a second oxygen atom. Representation is placed in brackets with an overall minus charge.

If this representation is the only correct resonance structure, we would expect the molecule to be asymmetric, meaning the bond lengths between the central nitrogen and the oxygen atoms would be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. As we will see from the discussion of formal charge below, we would also expect that the electrons would be distributed such that the negative charge would be present on one oxygen atom. Experimental evidence, however, establishes that nitrite is symmetric and that both N–O bonds in NO2− have the same strength and length. The two N-O bonds and the two oxygen atoms in nitrite are equivalent in all chemical and physical properties.

It is not possible to write a single Lewis structure for NO2− which accurately represents the electronic structure. 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 a weighted average of the valid Lewis structures. Therefore, two valid Lewis structures must be drawn to represent the bonding in the nitrite ion, NO2–.

Two resonance structures for NO2- separated by a resonance arrow. In each structure the central nitrogen is singly bonded to one oxygen and double bonded to the second oxygen

The two headed arrow that connects two (or more) valid Lewis structures is important. It communicates that we’re talking about resonance structures and not a chemical reaction (which is signified by a single-headed arrow, →) or equilibrium (which is signified by two half-arrows, ⇌).

The nitrite ion is an example of equivalent resonance, which means that the two bonds are identical. Both NO bonds have the same atoms involved, the same length, and require the same energy added to break them. Thus, the two NO bonds are identical. The actual distribution of electrons in each of the NO bonds in NO2− is the weighted average of a double bond and a single bond. Since the bonds are equivalent, they are equally weighted, so each connection between nitrogen and oxygen has an N-O bond order of 1.5.

The actual electronic structure of the molecule (the weighted average of the resonance forms) is represented by a resonance hybrid of the individual resonance forms. Thus, the electronic structure of the NO2− ion is shown as:

Resonance hybrid for NO2-. Nitrogen is the central atom and is singly bonded to two oxygen atoms. Each oxygen also has a second dashed bond between them and the central nitrogen.

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 weighted average of that shown by all resonance forms.

Notice that the atoms did not change position in the nitrite example of resonance. If atom positions change with respect to one another, then this is not an example of resonance. Instead, the molecules are isomers involved in a chemical change, and that will be explored in future courses.

The carbonate anion, CO32−, provides a second example of a polyatomic ion with equivalent resonance or equally weighted resonance structures:

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 Lewis structure could be drawn with the double bond between carbon and any one of the three oxygen atoms. This gives rise to three equivalent resonance forms of the carbonate ion. Because we can write three equivalent resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the equally weighted average of the three structures. These three structures highlight the symmetric bonding and distribution of electrons present in the carbonate ion. Again, experimental evidence establishes the symmetry of carbonate and shows that all three CO bonds are equivalent (bond length and bond strength) and that each oxygen atom is chemically equivalent.

Bond Order Revisited

As we described at the beginning of this section, bond order is the number of electron-pair bonds connecting two atoms. Using the carbonate ion, CO32- as an example, we already know the possible resonance structures for this ion. If we would like to determine the bond order between the central carbon and the top oxygen atom (labeled with a number one in the image below), we can calculate that CO bond order in each resonance structure and then divide by the total number of resonance structures. 
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. The top oxygen in each resonance structure is labeled with a 1, the bottom left oxygen is labeled with a 2, and the bottom right resonance structure is labeled with a three.

So for determining the bond order between carbon and oxygen number one the calculation would be as follows:

bond order for oxygen one = [latex]\frac{1+1+2}{3} = \frac{4}{3}[/latex]

and for oxygen number two:

bond order for oxygen two = [latex]\frac{2+1+1}{3} = \frac{4}{3}[/latex]

and for oxygen number three:

bond order for oxygen three = [latex]\frac{1+2+1}{3} = \frac{4}{3}[/latex]

This gives the bond order between the central carbon and each oxygen atom as being 4/3. This means that in the resonance hybrid each bond between carbon and oxygen has bond character that is between that of a single bond and a double bond, which we have proven experimentally. The resonance hybrid for CO32- is provided below:

Resonance Hybrid for CO3 2-. The central carbon is singly bonded to each oxygen. Each oxygen also has a dashed line to represent that the bond is between that of a single bond and a double bond.

As indicated by the resonance hybrid and the bond order calculation, each bond between carbon and oxygen has a total bond order between that of a single bond (bond order = 1) and a double bond (bond order = 2).  This is consistent with all of the experimental observations of the bond lengths and the reactivity of each atom, as well as theoretical predictions of the electronic structure. The nature of each of the bonds in a double bond will be explored in a later module.

All of the resonance structures we’ve considered up to this point have been equivalent to one another. That is, the location of the double bond had an equal likelihood of being associated with any of the oxygen atoms in NO2– or CO32- so each resonance structure has equal stability. There are molecules where one resonance structure may not be as stable as another while still satisfying the octet rule. We are able to compare these resonance structures to one another using a concept known as formal charge.

 

The online Lewis Structure Maker includes many examples to practice drawing resonance structures.

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 half the number of bonding electrons (distribute them evenly to each of the two atoms that are bonded).

Thus, we calculate formal charge as follows:

An equation is shown. The formal charge is equal to the number of valence electrons on the neutral atom minus the open parenthesis number of unshared electrons plus one half the number of bonding electrons close parenthesis.

By subtracting one-half the number of bonding electrons, we essentially assign half the bonding electrons to each atom. This implicitly assumes that these bonding electrons are shared relatively equally, as in covalent bonding. Since each bond represents two electrons, we can also count the number of bonds connected to that atom in the Lewis structure and the equation becomes:

An equation is shown. The formal charge is equal to the number of valence electrons on the neutral atom minus the open parenthesis number of unshared electrons plus number of bonds close parenthesis.

The previous two equations are both common ways of calculating formal charge and will yield the same answer. The sum of the formal charges on all atoms in a neutral molecule is zero; the sum of the formal charges on a polyatomic ion is the charge on 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. We can compare three different ways of thinking about the H-Cl bond using different concepts covered so far (as summarized in Figure 2). The calculated charge distribution illustrates the polar bond observed experimentally.

Three depictions of hydrochloric acid are shown. In the one on the left, titled “oxidation number”, there is a white hydrogen atom sphere on the left bonded to a green chlorine atom sphere. Above the H is a +1 and above the Cl is a -1. A label below says “complete electron transfer: ionic bonding”. The central depiction is titled “formal charge”. The same visual of the white hydrogen atom sphere bonded to the green chlorine atom sphere is shown. Above the H, the formal charge is indicated to be zero. Above the Cl, the formal charge is indicated to be zero. A label below says “no electron transfer: non polar covalent bond”. The right depiction is titled “calculated charge distribution”. In this one, we have a oval shape with the left half smaller than the right half. The left half is shaded blue, then there is a gradient to red on the right half. Above the left half is a +0.2 partial positive label and above the right side is a -0.2 partial negative label. The label below says “actual partial electron transfer”.
Figure 2. Comparison of oxidation number, formal charge and actual partial electron transfer.

Example 1

Calculating Formal Charge from Lewis Structures
Assign formal charges to each atom in the interhalogen ion ICl4−.

Solution

  1. We divide the bonding electron pairs equally for all I–Cl bonds:

Resonance Structure for ICl4. The central iodine atom is singly bonded to four chlorine atoms. The central iodine atom also has two lone pairs.

  1. We assign lone pairs of electrons to their atoms. Each Cl atom now has 6 unshared electrons and 2 bonding electrons, and the I atom has 4 unshared electrons and 8 bonding electrons.
  2. Subtract the number of unshared electrons + ½ number of bonding electrons from the number of valence electrons for the neutral atom:
    I: 7 – (4 + ½ (8)) = –1
    Cl: 7 – (6 + ½ (2)) = 0
    The sum of the formal charges of all the atoms equals –1, which is identical to the charge of the ion (–1).
    The formal charges for each atom are drawn next to them in red for the final Lewis structure provided below.

Lewis structure for ICl4 with formal charges written next to each atom in red. The only non-zero formal charge is minus 1 written next to the central iodine.

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

The right structure of this pair shows a carbon atom with one lone pair of electrons triple bonded to an oxygen with one lone pair of electrons.

Answer:

C −1, O +1
The formal charges for each atom are drawn next to them in red for the final Lewis structure provided below.
Lewis structure for CO. Carbon is triple bonded to the oxygen atom. Each atom has a lone pair. This gives the carbon atom a formal charge of minus 1, and the oxygen a formal charge of plus 1.

Example 2

Calculating Formal Charge from Lewis Structures
Assign formal charges to each atom in the interhalogen molecule BrCl3.

Solution

  1. Assign one of the electrons in each Br–Cl bond to the Br atom and one to the Cl atom in that bond:

Lewis structure for BrCl3 with the central bromine atom single bonded to three chlorine atoms. The central bromine has two lone pairs, and the chlorine atoms have three.

  1. Assign the lone pairs to their atom. Now each Cl atom has 6 unshared electrons and 2 bonding electrons, and the Br atom has 4 unshared electrons and 6 bonding electrons.
  2. Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge:
    Br: 7 – (4 + ½ (6)) = 0
    Cl: 7 – (6 + ½ (2)) = 0
    All atoms in BrCl3 have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.
    The formal charges for each atom are drawn next to them in red for the final Lewis structure provided below.

Lewis structure for BrCl3 with the formal charge of each atom written next to them in red. The formal charge of every atom is zero.

Check Your Learning
Determine the formal charge for each atom in NCl3.

Answer:

N: 0; all three Cl atoms: 0

The formal charges for each atom are drawn next to them in red for the final Lewis structure provided below.

Lewis structure for NCl3. Central nitrogen atom is single bonded to three chlorine atoms. Each chlorine atom has three lone pairs, and the nitrogen has one lone pair. The formal charge for each atom is zero.

Using Formal Charge to Predict Resonance Structure Contributions

As discussed previously, molecules often have more than one valid Lewis structure. These resonance structures contribute to the overall resonance hybrid for a given molecule. However, not all resonance structures contribute equally to the bonding of a molecule. The most stable individual resonance structure (and thus the largest contributor to the resonance hybrid) tends to:

  1. Minimize formal charges on atoms.
  2. Place negative formal charges on more electronegative atoms.

To see how these guidelines apply, let us consider some possible structures for carbon dioxide, CO2, with carbon as the central atom. We can draw three possibilities for the structure:

Three possible resonance structures for C O 2. On the left a central carbon is double bonded to an oxygen on the left (which has 2 lone pairs) and double bonded to an oxygen on the right (which has 2 lone pairs). In the central resonance structure, the central carbon is triple bonded to the oxygen on the left (which has 1 lone pair) and single bonded to the oxygen on the right (which has 3 lone pairs). In the resonance structure on the right, the central carbon is single bonded to the oxygen on the left (which has 3 lone pairs) and triple bonded to the oxygen on the right (which has one lone pair).

Assigning formal charge to each atom gives us the following values:

Three separate resonance structures for CO2 are displayed left to right. The left structure has the central carbon double bonding to each of the two periphery oxygen atoms. The formal charges for each of these atoms is zero. The central resonance structure has the central carbon triple bonding to the oxygen on the left, and single bonding to the oxygen on the right. The formal charges for these atoms are +1, 0, and -1 going left to right. The rightmost resonance structure has the central carbon triple bonding to the oxygen on the right and single bonding to the oxygen on the left. The formal charges for these atoms are -1,0, and +1 going left to right.

Computational chemistry predicts that the left structure (which minimizes formal charge since all atoms have a formal charge of zero) contributes the greatest to the bonding in CO2 (~50%). The other two structures contribute less to the bonding structure (~25% each). Thus, CO2 has non-equivalent resonance structures. Formal charges help us estimate the relative contributions by each resonance structure when non-equivalent resonance structures contribute to the resonance hybrid.

Commonly Found “Building Blocks” of Molecules with Zero Formal Charge

Many commonly found elements in molecules have zero formal charge. By counting the number of electrons and bonds around an element and whether that gives the element a formal charge of zero will help us quickly recognize how that impacts the stability and reactivity of the molecule as a whole. The chart below summarizes some common elements and their bonding patterns that give a zero formal charge.

Example 3

Using Formal Charge to Determine Major Resonance 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 structure is the major contributor to the resonance hybrid?

Three possible resonance structures for N2O. On the left a central nitrogen is triple bonded to the oxygen on the right, and single bonded to the nitrogen on the left. In the central resonance structure the central nitrogen is double bonded both to the nitrogen on the left and the oxygen on the right. In the right resonance structure the central nitrogen is triple bonded to the nitrogen on the left, and single bonded to the oxygen on the right.

Solution
Determining formal charge yields the following:

Three possible resonance structures for N2O. On the left a central nitrogen is triple bonded to the oxygen on the right, and single bonded to the nitrogen on the left. In the central resonance structure the central nitrogen is double bonded both to the nitrogen on the left and the oxygen on the right. In the right resonance structure the central nitrogen is triple bonded to the nitrogen on the left, and single bonded to the oxygen on the right. The leftmost resonance structure has formal charges of -2, +1, +1. The central resonance structure has formal charges of -1,+1. The rightmost resonance structure has formal charges of 0, -1, +1

The structure in which the oxygen carries the negative formal charge is the major resonance contributor since oxygen is more electronegative than nitrogen. Additionally the number of atoms with formal charges are minimized since one atom has a formal charge of 0.

The major resonance structure for N2O. The central nitrogen is triple bonded to the nitrogen on the left and single bonded to the oxygen on the right.

Check Your Learning
Using formal charge, rationalize which structure contributes the most for the thiocyanate (SCN−) ion?

Three resonance structure for SCN- written left to right. In the leftmost resonance structure carbon is in the center and singly bonded to the sulfur atom and triple bonded to the nitrogen atom. In the central resonance structure the carbon is double bonded to the sulfur atom and double bonded to the nitrogen atom. In the rightmost structure the central carbon is triple bonded to the sulfur and singly bonded to the nitrogen atom.

Answer:

First, assign formal charges:

Three separate resonance structures for S C N are displayed left to right. The left structure has the central carbon single bonded to a sulfur on the left (which has 3 lone pairs) and triple bonded to a nitrogen on the right (which has 1 lone pair). There are brackets around the structure and a superscript negative. The formal charges are -1 for S, 0 for C and 0 for N. The central resonance structure has the central carbon double bonded to sulfur on the left (which has 2 lone pairs) and double bonded to nitrogen on the right (which has 2 lone pairs). There are brackets around the structure and a superscript negative. The formal charges are 0 for S, 0 for C and -1 for N. The right resonance structure has the central carbon triple bonded to sulfur on the left (which has 1 lone pairs) and single bonded to nitrogen on the right (which has 3 lone pairs). There are brackets around the structure and a superscript negative. The formal charges are +1 for S, 0 for C and -2 for N.

Major contributor:

Major resonance structure for SCN-. Carbon is in the center and is singly bonded to the sulfur atom and triple bonded to the nitrogen atom. The formal charge for the sulfur is -1, and both carbon and nitrogen have a formal charge of 0.

In this resonance form of the thiocyanate ion, the formal charges are as follows:

S: 0

C: 0

N: -1

Since nitrogen is more electronegative than sulfur, placing the negative formal charge on nitrogen is favorable compared to the other two options above.

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) and does not possess an electronic structure depicted by an individual resonance form.

Key Equations

  • formal charge = number of valence shell electrons (free atom) − number of lone pair electrons − [latex]\frac{1}{2}[/latex] number of bonding electrons

Glossary

bond order
The number of electron-pair bonds connecting two atoms
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)
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

Chemistry End of Section Exercises

  1. Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
    1. sulfur dioxide, SO2
    2. carbonate ion, CO32−
    3. hydrogen carbonate ion, HCO3− (C is bonded to an OH group and two O atoms)
    4. 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.
    5. 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.
  2. Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, NO2–.
  3. In terms of the bonds present, explain why acetic acid, CH3CO2H, 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.
  4. Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.
    1. CO2
    2. CO
  5. 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.
  6. Determine the formal charge of each element in the following:
    1. HCl
    2. CF4
    3. PCl3
    4. PF5
  7. Calculate the formal charge of chlorine in the molecules Cl2, BeCl2, and ClF5.
  8. Draw all important or highly contributing resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures:
    1. O3
    2. SO2
    3. NO2−
    4. NO3−
  9. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?
  10. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH?
  11. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO?
  12. Draw a structure of hydroxylamine, H3NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges?
  13. 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.
  14. 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.
  15. 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, H2SO4, which has two oxygen atoms and two OH groups bonded to the sulfur.

Answers to Chemistry End of Section Exercises

  1. (a)
    (b)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.
    (c)Two Lewis structures are shown, with a double-headed arrow in between, each surrounded by brackets and a superscripted negative sign. The left structure depicts a carbon atom bonded to three oxygen atoms. It is single bonded to one of these oxygen atoms, which has three lone pairs of electrons, and double bonded to the other two, which have two lone pairs of electrons. One of the double bonded oxygen atoms also has a single bond to a hydrogen atom. The right structure is the same as the first, but there is only one double bonded oxygen. The oxygen with the single bonded hydrogen now has a single bond to the carbon atom. The lone pairs of electrons have also changed to correspond with the bonds.
    (d)Two Lewis structures are shown with a double-headed arrow in between. The left structure depicts a hexagonal ring composed of five carbon atoms, each single bonded to a hydrogen atom, and one nitrogen atom that has a lone pair of electrons. The ring has alternating single and double bonds. The right structure is the same as the first, but each double bond has rotated to a new position.
    (e)
  3. Acetic acid has a C=O double bond and a C-O single bond. The acetate ion has only one type of carbon-oxygen bond, as illustrated by its resonance structures. The negative charge is delocalized on both the oxygen atoms and both carbon-oxygen bonds are of the same length.
  4. (a); (b)The right structure of this pair shows a carbon atom with one lone pair of electrons triple bonded to an oxygen with one lone pair of electrons.; CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO2 has double bonds.
  5. hydrogen carbonate ion: hydrogen peroxide:
  6. (a) H: 0, Cl: 0; (b) C: 0, F: 0; (c) P: 0, Cl 0; (d) P: 0, F: 0
  7. Cl in Cl2: 0; Cl in BeCl2: 0; Cl in ClF5: 0
  8. (a)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 ).”;
    (b)Two Lewis structures are shown, with a double-headed arrow in between. The left structure shows a sulfur atom with one lone pair of electrons single bonded to an oxygen atom with three lone pairs of electrons. The sulfur atom also double bonded to an oxygen atom with two lone pairs of electrons. The symbols and numbers below this structure read, “( negative 1 ), ( positive 1 ), ( 0 ).” The right structure appears as a mirror image of the left and the symbols and numbers below this structure read, “( 0 ), ( positive 1 ), ( negative 1 ).”;
    (c)[Two Lewis structures are shown, with brackets surrounding each with a superscripted negative sign and a double ended arrow in between. The left structure shows a nitrogen atom with one lone pair of electrons single bonded to an oxygen atom with three lone pairs of electrons and double bonded to an oxygen atom with two lone pairs of electrons. The symbols and numbers below this structure read “open parenthesis, 0, close parenthesis, open parenthesis, 0, close parenthesis, open parenthesis, negative 1, close parenthesis. The right structure appears as a mirror image of the left and the symbols and numbers below this structure read “open parenthesis, negative 1, close parenthesis, open parenthesis, 0, close parenthesis, open parenthesis, 0, close parenthesis.];
    (d)[Three Lewis structures are shown, with brackets surrounding each with a superscripted negative sign and a double ended arrow in between. The left structure shows a nitrogen atom single bonded to two oxygen atoms, each with three lone pairs of electrons and double bonded to an oxygen atom with two lone pairs of electrons. The single bonded oxygen atoms are labeled, from the top of the structure and going clockwise, “open parenthesis, negative 1, close parenthesis, open parenthesis, positive 1, close parenthesis”. The symbols and numbers below this structure read “open parenthesis, 0, close parenthesis, open parenthesis, negative 1, close parenthesis. The middle structure shows a nitrogen atom single bonded to two oxygen atoms, each with three lone pairs of electrons, one of which is labeled “open parenthesis, positive 1, close parenthesis” and double bonded to an oxygen atom with two lone pairs of electrons labeled “open parenthesis, 0, close parenthesis”. The symbols and numbers below this structure read “open parenthesis, negative 1, close parenthesis, open parenthesis, negative 1, close parenthesis. The right structure shows a nitrogen atom single bonded to two oxygen atoms, each with three lone pairs of electrons and double bonded to an oxygen atom with two lone pairs of electrons. One of the single bonded oxygen atoms is labeled, “open parenthesis, negative 1, close parenthesis while the double bonded oxygen is labeled, “open parenthesis, positive 1, close parenthesis”. The symbols and numbers below this structure read “open parenthesis, negative 1, close parenthesis” and “open parenthesis, 0, close parenthesis”.]
  9. ClNO
  10. HOCl
  11. OSO
  12. 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.
  13. NF3, 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., N: 0, F: 0
  14. , H: 0, O: 0, N: 0
  15. 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.

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