Print

Chapter 5 – Covalent Bond

Introduction

We saw in Chapter 4 that ionic bonds are not directional and that ionic compounds exist as extended networks rather than individual molecules. In this chapter, we begin our study of molecular substances, substances that exist as discrete molecules. Covalent bonds are directional and covalently bound atoms form molecular substances.

5.1 The Covalent Bond

Introduction

Covalent bonds result from the overlap of orbitals and involve a sharing of pairs of electrons. In this lesson, we examine the nature of the covalent bond.

Prerequisites

Objectives

5.1-1. H–H Bond Video

The energy of interaction between two hydrogen atoms as a function of the distance between them.

  • Viewing the Video
  • View the video in this window by selecting the play button.
  • Use the video controls to view the video in full screen.
  • View the video in text format by scrolling down.
  • Jump to the exercises for this topic.

5.1-2. H–H Bond

A covalent bond results when atoms share one or more pairs of electrons. The electrons that are shared in a covalent bond are called bonding electrons or bonding pairs.
In Chapters 2 and 3, we discussed the energy of interaction between a nucleus and its electrons, which reduces their potential energy and is responsible for the existence of atoms. However, valence electrons can reduce their energy ever farther by interacting with more than one nucleus. Consider Figure 5.1, which shows the energy of interaction between two hydrogen atoms as a function of their separation, r.
This is a graph with energy on the y-axis and the distance between two atoms on the x-axis.  The energy decreases from near 0 on the y-axis and then increases reaching a limit of 0 energy as the distance approaches infinity.  r1 is at 0 energy and an infinite distance.  r2 is at a lower energy and smaller distance than r1.  r3 is at a lower energy and smaller distance than r2.  r4 is the minimum of the graph.  r5 is at a higher energy and smaller distance than r4.
Figure 5.1: H–H Bond Formation and Length
Energy of interaction between two H atoms as a function of r, the distance between the nuclei, which are represented by the small dots.

5.1-3. Bond Length and Bond Energy

The bond energy (D) is the energy required to break a mole of bonds in the gas phase.
The bond length is the separation between two bound atoms at the position of minimum energy.
The equilibrium distance between two bound nuclei in a molecule defines the bond length of their bond. The H–H bond length, which is 0.74 Å or 74 pm (p = 1 × 10–12), is a very short bond because the two atoms are so small that they must get very close to interact. The I–I bond length, which is 2.7 Å (270 pm), is very long because the two atoms are so large. Most bond lengths lie between 1 and 3 Å (100 and 300 pm).
This is a graph of the energy of interaction between two hydrogen atoms and reaches a minimum at -436 kJ/mol and 74 pm.  There is an additional line extending vertically from the minimum to 0 kJ/mol with the label D = 436 kJ/mol.
Figure 5.2
The energy of two bound atoms is lower than that of the separated atoms because the bonding electrons on each atom interact with both nuclei. Thus, energy is released when atoms bond. In the case of the H–H bond, 436 kJ is released when two moles of hydrogen atoms combine to form one mole H–H bonds. This means that the minimum energy of two interacting hydrogen atoms is –436 kJ/mol. Separating the two bound atoms requires breaking the bond, which requires an input of the same amount of energy that was released when the bond formed. The energy required to break or dissociate a bond in the gas phase is called the bond or dissociation energy and given the symbol D. The H–H bond energy in H2 is D = 436 kJ/mol and is an average bond energy, as most bond energies lie between 100 and 1000 kJ/mol. The I–I bond is very weak with a bond energy of only 151 kJ/mol.

5.2 Bond Polarity

Introduction

Although the bonding electrons are shared in a covalent bond, electronegativity differences between the bound atoms can result in unequal sharing. Bonds in which the electrons are not shared equally are said to be polar bonds, the topic of this section.

Prerequisites

Objectives

5.2-1. Polar vs. Nonpolar Bonds

If the electronegativities of two atoms in a bond are different, then the bond is polar.

  • Viewing the Video
  • View the video in this window by selecting the play button.
  • Use the video controls to view the video in full screen.
  • View the video in text format by scrolling down.
  • Jump to the exercises for this topic.

5.2-2. Polar Bonds and Bond Dipoles

Polar bonds have bond dipoles, which are represented with arrows pointing from the positive pole (less electronegative atom) toward the negative pole (more electronegative atom) with a line through the positive end.
In H2 the bound atoms have identical electronegativities, so the bonding electrons are shared equally. Such bonds are purely covalent bonds. However, if the electronegativities of the bound atoms differ, as in the case of HF, the bonding electrons are not shared equally as the more electronegative fluorine becomes electron rich and the less electronegative hydrogen becomes electron poor. The result is that negative charge is produced on the more electronegative atom and positive charge on the less electronegative atom. The charge that is produced in a covalent bond is only a part of what it would be if the bond were ionic, so it is said to be a partial charge, which is represented with a δ (delta). Thus, a covalent bond involving atoms of different electronegativities has two electrical poles (δ– and
δ+) and is said to have a bond dipole and is called a polar covalent bond.
Bond dipoles are frequently represented with arrows pointing from the positive pole toward the negative pole. A line is then drawn through the arrow so as to make a plus sign at the positive end. Although the length of the arrow is normally related to the strength of the dipole, we will use the arrow to show direction only, so relative lengths are not meaningful. No bond dipole exists for nonpolar molecules, so no arrow is used with them.
See  discussion above.
Figure 5.3
The direction of the bond dipole in HF is shown with the red arrow. Note that a line through the arrow makes a plus sign at the positive end of the dipole.

5.2-3. Charge Distribution

Although molecules are neutral, they do contain regions of charge due to the asymmetric charge distribution in polar bonds. We will use the convention of indicating regions of negative charge as red and regions of positive charge as blue. Regions of no charge will be shown as white. Consider the examples below.
the I-I bond
Figure 5.4a: Charge Distribution of an I–I Bond
Two iodine atoms have the same electronegativities, so the I–I bond is not polar and there is no asymmetric charge distribution. Consequently, the electron cloud is shown as white.
the HI bond
Figure 5.4b: Charge Distribution of an I–H Bond
Iodine is more electronegative than hydrogen, so the bonding electrons reside nearer the iodine. The increase in electron density is shown in red. The electrons move away from the hydrogen, giving it a slight positive charge shown in blue. The electronegativity difference is only 0.5, so the intensities of the colors are not very great. The bond dipole points from the positive (blue) region toward the negative (red) region.
the IF bond
Figure 5.4c: Charge Distribution of an I–F Bond
Iodine is less electronegative than fluorine, so the bonding electrons reside nearer the fluorine. Thus, F carries a partial negative charge, which is represented by the red color. The I atoms carries a partial positive charge as shown by the blue color.
Δχ = 1.3,
so the colors are slightly more intense than in HI.
the HF bond
Figure 5.4d: Charge Distribution of an H–F Bond
Fluorine is much more electronegative than hydrogen (Δχ = 1.8) so the regions are intensely colored.

5.2-4. Bond Type and Electronegativity Differences

Bond types vary continuously from purely covalent to polar covalent to ionic. There is no clear distinction between a polar covalent bond and an ionic one.
The value of δ in δ+X–Yδ ranges from 0 to 1 for a compound like H–Cl, depending upon the electronegativity difference between X and Y. As δ increases, the charge separation in the bond increases, and it gains ionic character. When δ is 0, the bond is purely covalent, and when δ is 1, the bond is purely ionic. Thus, we refer to the percent ionic character of a bond, which increases as the electronegativity difference between the bound atoms increases. There are no clear lines separating ionic and covalent bonds; they gradually merge into one another as shown in Figure 5.5. However, we will use the following for discussion.
line graph ionic character versus electronegativity difference
Figure 5.5
Percent Ionic Character of a Bond as a Function of the Electronegativity Difference Between the Bound Atoms
Bond types change continuously from covalent to polar covalent to ionic with no clear boundaries between the bond types. However, ionic bonds are favored by large differences in electronegativity, while covalent bonds are favored by small differences. Consequently, we will use the following generalizations in this course:

5.2-5. Exceptions

Assume that metal-nonmetal bonds are ionic and nonmetal-nonmetal bonds are covalent, but it is not always true.
The generalization that metal-nonmetal bonds are ionic and nonmetal-nonmetal bonds are covalent will aid our discussions, but it is not a hard and fast rule. As a result, metal-nonmetal bonds are often more covalent than ionic. For example, Δχ = 1.1 for the Ag–Cl bond, so it is only about 30% ionic, and Δχ = 0.7 for the Pb–Cl bond, so it is less than 15% ionic. Thus, not all metal-nonmetal bonds are ionic. In addition, the H–F bond is between nonmetals, so it will be referred to as covalent, but Δχ = 1.9 for the bond, which makes it about 60% ionic. Thus, not all nonmetal-nonmetal bonds are very covalent.

Metal χ
Ag 1.9
Sn 2.0
Hg 2.0
Tl 2.0
Pb 2.3
Table 5.1

5.2-6. Bond Type Exercises

Exercise 5.1:

Use only a Periodic Table to answer the following questions.
    Which is the most polar bond?
  • H–N The H atom is the less electronegative atom in each bond, so it will be the positive end of the bond dipole. Thus, the most polar bond is the one that has the most electronegative atom at the other end. N is not the most electronegative, so the N–H bond is not the most polar.
  • H–C The H atom is the less electronegative atom in each bond, so it will be the positive end of the bond dipole. Thus, the most polar bond is the one that has the most electronegative atom at the other end. C is the least electronegative, so the C–H bond is the least polar.
  • H–O The H atom is the less electronegative atom in each bond, so it will be the positive end of the bond dipole. Thus, the most polar bond is the one that has the most electronegative atom at the other end. O is the most electronegative, so the O–H bond is the most polar.
    Which is the most polar bond?
  • S–F The F atom is the most electronegative atom in each bond, so it will be the negative end of the bond dipole. Thus, the most polar bond is the one that has the least electronegative atom at the other end. S is the least electronegative, so the S–F bond is the most polar.
  • O–F The F atom is the most electronegative atom in each bond, so it will be the negative end of the bond dipole. Thus, the most polar bond is the one that has the least electronegative atom at the other end. O is the not the least electronegative, so the O–F bond is not the most polar.
  • F–F The two bound atoms are identical so this is a purely covalent bond. As such, it is the least polar bond in the list.
    The C–Cl bond is which of the following?
  • purely covalent Carbon and chlorine are both nonmetals with different electronegativities, so the C–Cl bond is not purely covalent.
  • polar covalent Carbon and chlorine are both nonmetals with different electronegativities, so the C–Cl bond is a polar covalent bond.
  • ionic Carbon and chlorine are both nonmetals, so the C–Cl bond is not an ionic bond.
    The Na–Cl bond is which of the following?
  • purely covalent Sodium is a metal and chlorine a nonmetal, so this bond cannot be purely covalent.
  • polar covalent Sodium is a metal and chlorine a nonmetal, so this bond is not polar covalent.
  • ionic Sodium is a metal and chlorine a nonmetal, so this bond is not polar covalent. Sodium is a 1A metal and chlorine a 7A nonmetal, so this bond is ionic.
    The Cl–Cl bond is which of the following?
  • purely covalent Bonds between atoms with identical electronegativities are purely covalent.
  • polar covalent Bonds between atoms with identical electronegativities are purely covalent.
  • ionic Bonds between atoms with identical electronegativities are purely covalent.
    The Sn–Cl bond is which of the following?
  • purely covalent Although tin is a metal, it lies well on the right side of the periodic table and is one of the metals that is an exception to the rule that metal-nonmetal bonds are ionic. The Sn–Cl bond is a polar covalent bond with only about 25% ionic character.
  • polar covalent Although tin is a metal, it lies well on the right side of the periodic table and is one of the metals that is an exception to the rule that metal-nonmetal bonds are ionic. The Sn–Cl bond is a polar covalent bond with only about 25% ionic character.
  • ionic Although tin is a metal, it lies well on the right side of the periodic table and is one of the metals that is an exception to the rule that metal-nonmetal bonds are ionic. The Sn–Cl bond is a polar covalent bond with only about 25% ionic character.

5.3 Naming Binary Covalent Compounds

Introduction

In this lesson, you will learn how to name binary covalent molecules.

Objectives

5.3-1. Naming Binary Compounds

Prefixes are used in the names of covalent compounds to indicate the number of each atom present.
Binary compounds are compounds that contain only two different elements. The name of a binary covalent compound consists of the name of the less electronegative element followed by the name of the more electronegative element with its ending changed to -ide. The number of atoms of each element present in a molecule is given by a Greek prefix except that the prefix "mono" is not used for the first atom in the formula. Thus, CO is carbon monoxide (not monocarbon monoxide), but N2F4 is dinitrogen tetrafluoride. Many compounds have common names. For example, H2O is water, NH3 is ammonia, and NO is nitric oxide. By convention, the less electronegative element is written first in the formula except if one of the elements is hydrogen. Many hydrogen-containing compounds are acids, and, by convention, hydrogens at the beginning of the formula are considered to be acidic hydrogens. Ammonia is written NH3 even though hydrogen is the less electronegative element, but dihydrogen sulfide is written as H2S because it is an acid. The following Greek prefixes indicate the number of each type of atom that appears in the formula of covalent compounds.

Number Prefix Example
1 mono CO - Carbon monoxide
2 di CO2 - Carbon dioxide
3 tri SO3 - Sulfur trioxide
4 tetra CCl4 - Carbon tetrachloride
5 penta PF5 - Phosphorus pentafluoride
6 hexa SF6 - Sulfur hexafluoride
7 hepta Cl2O7 - Dichlorine heptaoxide*
8 octa
9 nona
10 deca
*An 'a' at the end of a prefix is frequently omitted when it is followed by a vowel, so the name may also appear as 'Dichlorine heptoxide.'
Table 5.2: Prefixes Used in Naming Covalent Compounds

5.3-2. Naming Binary Covalent Compounds Exercise

Exercise 5.2:

Name each of the following compounds.
PF3 o_phosphorus trifluoride_s phosphorus trifluoride
Note: 'mono' is not normally used for the first element in the formula.
IF5 o_iodine pentafluoride_s iodine pentafluoride
Note: 'mono' is not normally used for the first element in the formula.
N2O3 o_dinitrogen trioxide_s dinitrogen trioxide
CuCl2 o_copper(II) chloride_s copper(II) chloride
Copper is a metal and chloride is a nonmetal, so CuCl2 is assumed to be ionic. Thus, it is named by the rules given in Chapter 4 for ionic substances.
SiCl4 o_silicon tetrachloride_s silicon tetrachloride
P2O5 o_diphosphorus pentoxide_s diphosphorus pentoxide
Note: when the name of the element begins with a vowel, the 'a' at the end of a prefix is omitted; i.e., it is not pentaoxide.
NO o_nitrogen monoxide_s nitrogen monoxide
This molecule is so common that it has a common name, nitric oxide.

5.3-3. More Naming Compounds Exercises

Exercise 5.3:

Name the following compounds.
N2F4 o_dinitrogen tetrafluoride_s N2F4 is dinitrogen tetrafluoride.
SO3 o_sulfur trioxide_s The mono prefix is not used for the first element in the formula, so SO3 is sulfur trioxide.
PbO2 o_lead(IV) oxide_s Lead is a metal, so the compound is treated as ionic not covalent. Thus, the number of oxygen atoms is indicated with the oxidation state of the lead atom. PbO2 is lead(IV) oxide. The oxidation state should be given with capital letters. All other letters should be lower case.
CS2 o_carbon disulfide_s The mono prefix is not used for the first atom, so CS2 is carbon disulfide.
SF6 o_sulfur hexafluoride_s The mono prefix is not used for the first atom, so SF6 is sulfur hexafluoride.
N2O4 o_dinitrogen tetroxide_s The 'a' at the end of a prefix is usually not used when the name of the element begins with a vowel, so N2O4 is dinitrogen tetroxide.

5.4 Lewis Symbols of the Elements

Introduction

A Lewis structure shows the placement of the valence electrons around each of the atoms in a molecule. They are important because they allow us to predict the shapes of molecules, the types and relative strengths of the bonds, and regions of high and low electron density within the molecule. In this section, we introduce Lewis symbols for the elements.

Prerequisites

Objectives

5.4-1. Lewis Symbols

The structures of covalently bound molecules and ions depend upon the distribution of the valence electrons. The distribution of valence electrons can be represented by Lewis structures. The Lewis symbol of an atom shows the valence electrons spread in four different regions and remaining unpaired until each region has at least one electron. Lewis symbols represent the bonding atom, not the isolated atom. The number of valence electrons for a main group element is simply the element's group number, so the electron distribution given in the Lewis symbols is the same for all atoms in a group.
main group lewis symbols
Figure 5.6: Lewis Symbols
Lewis Symbol representation of the main group elements.

5.5 Lewis Structures of Diatomic Molecules

Introduction

We use diatomic molecules to introduce Lewis structures of molecules and to introduce some new terms.

Objectives

5.5-1. Octet Rule

Lewis structures almost always show eight valence electrons around each atom. An exception is hydrogen, which has only two.
Filled or closed shells are very stable and atoms strive toward these configurations when they bond. In Chapter 4, we saw that atoms can attain these configurations by becoming ions: metals lose their valence electrons to obtain the closed shell configuration of the preceding noble gas, while nonmetals gain electrons to obtain the closed shell configuration of the following noble gas. Nonmetals can also achieve closed shells by sharing electrons. A closed shell for a nonmetal consists of eight electrons (2 s and 6 p electrons). Thus, nonmetals strive to obtain eight electrons, which is called an octet, in their valence shell. This tendency is summarized by the octet rule. The octet rule provides a guide to drawing the Lewis structures of compounds. As indicated by "almost always," there are some exceptions. The most common exception is the hydrogen atom. A closed shell for hydrogen consists of two, not eight, electrons, so hydrogen atoms attain only a duet of electrons. Thus, one and only one bond is drawn to a hydrogen atom. Other exceptions occur, but they are rare and usually result when the molecule has an odd number of electrons or contains an atom of low electronegativity such as boron.

5.5-2. Chlorine

See discussion above.
Figure 5.7: Lewis Structure of Cl2
The yellow and red circles are not part of the Lewis structure and are used here only to distinguish between the two sets of valence electrons and to highlight the bonding pair shown in blue.

5.5-3. Shared Pairs

The number of shared pairs in a molecule can be determined from the number of valence electrons available from the atoms.
The number of shared pairs (bonding pairs) in a molecule is an important property of a molecule because it gives us information about how the atoms bond to one another. Thus, we will have occasion to determine it frequently. The number of electrons that must be shared equals the number of electrons required to satisfy the octet rule for every atom without sharing minus the number of valence electrons that are available. The number of shared pairs is half of the number of shared electrons. Mathematically, the relationship is expressed as
SP =
1
2
(ER − VE)
Exercise 5.4:

Determine the number of shared pairs in each of the following.
F2
ER = 16_0__ ER = (8 e/atom)(2 atoms)= 16 e
VE = 14_0__ F is a 7A nonmetal, so it has 7 valence electrons. VE = (7 e/atom)(2 atoms)= 14 e
SP = 1_0__ SP =
1
2
(16 e − 14 e) = 1 shared pair
O2
ER = 16_0__ ER = (8 e/atom)(2 atoms)= 16 e
VE = 12___ O is a 6A nonmetal, so it has 6 valence electrons. VE = (6 e/atom)(2 atoms)= 12 e
SP = 2_0__
SP =
1
2
(16 e − 12 e) = 2 shared pairs
N2
ER = 16_0__ ER = (8 e/atom)(2 atoms)= 16 e
VE = 10___ N is a 5A nonmetal, so it has 5 valence electrons. VE = (5 e/atom)(2 atoms)= 10 e
SP = 3_0__
SP =
1
2
(16 e − 10 e) = 3 shared pairs

5.5-4. Lewis Structures of Three Molecules

Bonding pairs are shown as lines in a Lewis structure, but lone pairs are shown as two dots.
In the previous exercise, we determined the number of shared pairs (SP) in three diatomic molecules (F2, O2, and N2). Each atom obeys the octet rule, so the Lewis structures are drawn by first placing the correct number of shared pairs, then placing enough lone pairs around each atom to assure that each has an octet. In addition, bonding (shared) pairs of electrons are normally shown as lines rather than dots. Thus, the Lewis structures of these three diatomic molecules would be represented as follows:

Molecule F2 O2 N2
SP 1 2 3
Lewis Structure Lewis structure of F2 Lewis structure of O2 Lewis structure of N2
Table 5.3: Lewis Structures of F2, O2, and N2
Thus, each F atom has three lone pairs, each O atom has two, and each N atom has only one lone pair.

5.5-5. Bond Order

The number of bonding pairs in a bond is called its bond order.
The bond order of a bond is the number of shared pairs of electrons in the bond (the number of lines). The bond order is frequently given by indicating a single, double, or triple bond.

Bond Bond Order Name
F–F 1 single bond
O=O 2 double bond
N≡N 3 triple bond
Table 5.4

5.5-6. Some Bond Lengths and Strengths

Bonds get shorter and stronger as their bond order increases.
The number of electron pairs in a bond dictates the strength and length of the bond. Thus, increasing the X–Y bond order strengthens and shortens the X–Y bond. This is evident in the table of average bond energies and bond lengths of bonds involving C, N, and O that is shown below. The bond length also depends on the bound atoms, so this generalization can be applied only to bonds between the same two atoms. That is; X–Y bonds can be compared to one another as can X–Z bonds, but X–Y bonds cannot be compared with X–Z bonds.

Bond Bond Energy
(kJ/mol)
Bond Length
(Å)
     Bond Bond Energy
(kJ/mol)
Bond Length
(Å)
     Bond Bond Energy
(kJ/mol)
Bond Length
(Å)
C–O 358 1.43      C–C 347 1.54      N–N 163 1.47
C=O 799 1.23      C=C 612 1.33      N=N 418 1.24
C≡O 1072 1.13      C≡C 820 1.20      N≡N 941 1.10
Table 5.5: Bond Strength and Length as a Function of Bond Order

5.6 Determining Lewis Structures

Introduction

We now extend the previous lesson to include molecules other than diatomic molecules.

Objectives

5.6-1. Determining Lewis Formulas

To determine the Lewis structure of a molecule:
1
Determine the number of shared pairs as before:
SP = 1/2(ER − VE)
where
  • ER = electrons required = 8 times the number of nonhydrogen atoms + 2 times the number of hydrogens.
  • VE = valence electrons = sum of the valence electrons on all of the atoms in the molecule and any charge on an ion. The number of valence electrons is increased by the negative charge of an anion and decreased by positive charge on a cation. VE is the number of electrons that must be shown in the final Lewis structure.
2
Draw the skeleton of the molecule (connect all of the atoms with single lines), then determine how many more shared pairs must be added and add them.
3
Add lone pairs to atoms to assure that each nonhydrogen atom has an octet.
4
Check your structure to be certain that each atom has an octet and that the number of electrons shown in the structure is the same as the number of valence electrons determined above.
Lewis structures should never show double or triple bonds to hydrogen or to a halogen. More than one single bond must be drawn to a halogen when it is the central atom as in ClO41–, but a double or triple bond should not be used. We explain why in the section on formal charge.

5.6-2. Determining Shared Pairs and Lone Pairs Exercise

Exercise 5.5:

Determine the number of shared pairs (SP) and the number of lone pairs (LP) around the central atom (the atom that appears first in the formula) in each of the following.
SO2
ER = 24_0__ ER = (8 e/atom)(3 atoms)= 24 e
VE = 18_0__ S and O are each 6A nonmetals, so they each have 6 valence electrons. VE = (6 e/atom)(3 atoms)= 18 e
SP = 3_0__ SP = 1/2(24 e – 18 e) = 3 shared pairs
LP = 1_0__ The sulfur atom has 3 shared pairs around it, which is 6 electrons. Thus, it needs two more electrons (1 lone pair) to obtain an octet.
NH3
ER = 14_0__ ER = (8 e/atom)(1 atoms) + (2 e/H atom)(3 H atoms) = 14 e
VE = 8_0__ N has 5 valence electrons and each H has 1 valence electron. VE = (5 e/N)(1 N) + (1 e/H)(3 H) = 8 e
SP = 3_0__ SP = 1/2(14 e – 8 e) = 3 shared pairs
LP = 1_0__ The nitrogen atom has 3 shared pairs around it, which is 6 electrons. Thus, it needs two more electrons (1 lone pair) to obtain an octet.
OF2
ER = 24_0__ ER = (8 e/atom)(3 atoms)= 24 e
VE = 20_0__ O is 6A and F is a 7A. VE = (6e/O)(1O) + (7e/F)(2F) = 20 e
SP = 2_0__ SP = 1/2(24 e – 20 e) = 2 shared pairs
LP = 2_0__ The oxygen atom has 2 shared pairs around it, which is 4 electrons. Thus, it needs four more electrons (2 lone pairs) to obtain an octet.
CO32–
ER = 32_0__ ER = (8 e/atom)(4 atoms)= 32 e
VE = 24_0__ C is a 4A and O a 6A nonmetal. In addition, the –2 charge on the ion means that the ion has an additional two electrons. VE = (1C)(4e/C) + (3O)(6e/O) + 2e = 24 e
SP = 4_0__ SP = 1/2(32 e – 24 e) = 4 shared pairs
LP = 0_0__ The carbon atom has 4 shared pairs around it, which is 8 electrons. Thus, it needs no more electrons (0 lone pairs) to obtain an octet.
SO32–
ER = 32_0__ ER = (8 e/atom)(4 atoms)= 32 e
VE = 26_0__ S and O a 6A nonmetals. In addition, the –2 charge on the ion means that the ion has an additional two electrons. VE = (4 atoms)(6e/atom) + 2e = 26 e
SP = 3_0__ SP = 1/2(32 e – 26 e) = 3 shared pairs
LP = 1_0__ The sulfur atom has 3 shared pairs around it, which is 6 electrons. Thus, it needs two more electrons (1 lone pair) to obtain an octet.

5.6-3. Lewis Structure of CO Exercise

Exercise 5.6:

Draw the Lewis structure for carbon monoxide, CO.
1. Determine the number of electrons required with no sharing (ER).
16_0__ ER = 2 atoms × 8 electrons each = 16 electrons needed
2. Determine the number of valence electrons (VE).
10_0__ C is in Group 4a and O is in Group 6A. VE = 4 from C + 6 from O = 10 valence electrons
3. Determine the number of shared pairs (SP).
3_0__
SP =
1
2
(16 − 10) = 3 shared pairs
4. What is the C–O bond order in CO?
3_0__ The C–O bond contains three shared pairs, so the bond order is 3.

5.6-4. Lewis structure of CO2 Exercise

Exercise 5.7:

Draw the Lewis structure for carbon monoxide, CO2.
1. Determine the number of electrons required with no sharing (ER).
24_0__ ER = 3 atoms × 8 electrons each = 24 electrons needed
2. Determine the number of valence electrons (VE).
16_0__ C is in Group 4A and O is in Group 6A. VE = group number of C + 2(group number of O) = (1 × 4) + (2 × 6) = 16 valence electrons
3. Determine the number of shared pairs (SP).
4_0__
SP =
1
2
(24 − 16) = 4 shared pairs
4. What is the C–O bond order in CO2?
2_0__ The four shared pairs are divided between two bonds, so the bond order is 2.

5.7 Resonance

Introduction

In some instances, more than one Lewis structure can be drawn for a molecule. The different structures are called resonance structures.

Objectives

5.7-1. Resonance Structures

Resonance structures are Lewis structures that differ only in the placement of the electron pairs.
Resonance structures are Lewis structures that differ only in the placement of the electron pairs. Curved arrows are used to show how the electrons can be moved to convert from one resonance structure into another. We will also use curved arrows in Chapters 12 and 13 to show how acid-base reactions take place, so take a moment to familiarize yourself with their meaning. As shown in the table, they are used in only two ways.

Representation Direction Action
arrow from lone pair to a bond lone pair bond the lone pair becomes a bonding pair
arrow from a bond to an atom bond atom a bonding pair becomes a lone pair
Table 5.6

5.7-2. Resonance forms of CO2

curved arrows in CO2 resonance
Figure 5.8
View the animation to see how curved arrows are used to explain the electron movement required to attain the three resonance structures of CO2.

  • Viewing the Video
  • View the video in this window by selecting the play button.
  • Use the video controls to view the video in full screen.

5.7-3. The Number of Resonance Structures

The most common form of resonance is one in which a multiple bond can be drawn in any of several positions. Follow these steps to determine the number of such resonance forms that are possible. Determine the number of shared pairs.
1
Determine the number of electrons required with no sharing (ER).
2
Determine the number of valence electrons (VE).
3
Determine the number of shared pairs (SP).
Resonance structures form when the number of shared pairs exceeds the number of identical bonding regions in which they can be placed. Resonance structures for molecules or ions having a single central atom occur when the number of shared pairs exceeds the number of regions by one. In this case, the number of structures equals the number of bonding regions in which the extra shared pair can be placed. The example of SO2 is examined in the following exercise.

5.7-4. Resonance in SO2 Exercise

Exercise 5.8:

ER = 24_0__ ER = 3 atoms × 8 electrons each = 24 electrons needed
VE = 18_0__ VE = 6 from S + 2(6) from O = 18 valence electrons
SP = 3_0__
SP =
1
2
(24 − 18) = 3 shared pairs
Based on the bonding regions, how many resonance structures can be drawn for SO2?
2_0__ There are 3 shared pairs (bonds) but there are only 2 bonding regions. One bonding region must contain two of the bonds while the other contains a single bond. Since the double bond can be placed in either bonding region, there are two resonance forms of SO2.

5.7-5. Bond Orders in Bonds Involved in Resonance

When more than one identical resonance structures can be drawn for a molecule, they all contribute equally to the bonding picture. Fractional bond orders result in these cases as one bonding pair is shared in more than one region.
resonance structures of SO2
Figure 5.9: Resonance in SO2
Note that a line with arrows on both ends is used to indicate resonance structures.
When the different resonance forms are equivalent, as is the case in SO2, each form contributes equally to the bonding. When more than one resonance form is important in the bonding, then the bonding in the molecule is the average of the resonance forms. The most common cases of resonance arise when a multiple bond can be placed in more than one equivalent position. When resonance is important, one bonding pair is shared by (spread over) more than one bond, and the bond order is determined as the following:
( 5.1 )
bond order =
number of shared pairs
number of bonding regions
where only the bonding regions in which the double bond can be placed are considered. The two resonance structures of SO2 are equivalent, so the bonding is the average of both forms and the double bond is shared equally by the two S–O bonds. The result is that neither bond is a single bond and neither is a double bond. The two sulfur-oxygen bonds in sulfur dioxide are of equal length, both being shorter than a S–O single bond but longer than a S=O double bond.
Example:
The bond order of each of the bonds in SO2 involved in the resonance is
BO =
(3 shared pairs)
(2 bonds)
= 1.5
Bonding electrons that are shared among more than two atoms are sometimes represented by dotted lines in the Lewis structure. The following shows that the two bonds are identical and have bond orders of 3/2.
bond order in SO2
Figure 5.10

5.7-6. Bond Order is a Bond Property, Not a Molecular Property

Bond orders apply to bonds, not molecules.
A common misconception among students is that molecules have bond orders. They do not; bonds have bond orders. This misconception arises because all of the bonds are identical in so many of our examples. Thus, it is incorrect to state that the bond order of SO2 is 1.5; rather, it should be stated that the bond order of each bond in SO2 is 1.5. Resonance is important when a multiple bond can be placed in more than one position. In these cases, the bond order of each bond that is affected by the resonance is given by Equation 5.1
bond order =
number of shared pairs
number of bonding regions
. However, Equation 5.1 applies only to the bonding regions involved in the resonance. This distinction is the focus of the following exercise.

5.7-7. Bond Order Exercise

Exercise 5.9:

The Lewis structure of each of the following skeletal structures requires one more shared pair. Decide how to place the shared pair and then determine the bond order of the highlighted bond.
Lewis structure for exercise
2_0__ Hydrogen atoms can only have two electrons (one shared pair), so the fourth shared pair cannot be placed in the C–H bonds. Thus, it must go into the C–O bond, which makes the C–O bond a double bond; i.e., its bond order is 2.
Lewis structure for exercise
1.5_0__ Hydrogen atoms can only have two electrons (one shared pair), so the fourth shared pair cannot be placed in the C–H bonds. The carbon bound to the H atoms already has an octet. The two C–O bonds are identical, so the double bond can be placed in either, which results in two resonance structures. Thus, the C–O bond order is 1.5.

5.8 Formal Charge and Oxidation State

Introduction

In Chapter 4, we showed how to determine oxidation states of the atoms in a substance without knowledge of the Lewis structure of the substance. However, electron counting can be facilitated with a Lewis structure. By comparing the number of electrons around an atom in a Lewis structure with the number of valence electrons in the unbound atom, we can assign a charge to the atom in a molecule or ion. Oxidation states assign charge by assuming that the bonds are ionic. In this section, we introduce formal charge, which assigns charge by assuming that the bonds are covalent.

Prerequisites

Objectives

5.8-1. Regions of Charge in Molecules

The charge on an atom in a molecule lies between its formal charge and its oxidation state. It lies closer to the oxidation state when the bonds are very polar, but closer to the formal charge if they are not.
The charge distribution in bonds is usually asymmetric because the bonds are polar. This asymmetry introduces regions of charge in the molecule. These regions greatly influence both the physical and chemical properties of the molecule. Determining where the charge resides is an important challenge to the chemist. In this section, we show how to determine those regions in the two extremes of purely ionic and purely covalent bonding. The actual situation depends upon the polarity of the bonds and will lie somewhere between the two limits. The charge that is assigned to an atom in a molecule equals the number of valence electrons in the free atom minus the number of valence electrons assigned to the atom in the molecule. The number of valence electrons in the free atom (VE) is given by the atom's group number. The number of electrons assigned to the atom in the molecule is the number of nonbonding electrons (NB) plus those bonding electrons (BE) that are assigned to it. Thus, the charge on atom A is given as
qA = VE − (NB + aBE)
where (NB + aBE) is the number of electrons (nonbonding and bonding) that are assigned to the atom. The manner in which a is determined and the name of the resulting charge depend upon the type of bonding that is assumed.

5.8-2. Oxidation State vs. Formal Charge Videos

Determining Formal Charges

  • Viewing the Video
  • View the video in this window by selecting the play button.
  • Use the video controls to view the video in full screen.
  • View the video in text format by scrolling down.
  • Jump to the exercises for this topic.

Determining Oxidation States

  • Viewing the Video
  • View the video in this window by selecting the play button.
  • Use the video controls to view the video in full screen.
  • View the video in text format by scrolling down.
  • Jump to the exercises for this topic.

5.8-3. Formal Charge

Formal charge is the charge an atom would have if the bonds were purely covalent. It can be determined by electron counting.
The formal charge is the charge an atom would have if the bonds were purely covalent. This means that one electron from each bonding pair is assigned to each atom in the bond. Consequently, only half of the bonding electrons around an atom are assigned to it. That is,
a =
1
2
,
and the formal charge on atom A is
FCA = VE −
NB +
1
2
BE
Each bond contains two electrons, so one-half of the bonding electrons equals the number of bonds (B). Thus, the formal charge on atom A can also be expressed by
FCA = VE − (NB + B) = VE − NB − B
The formal charge on an atom is placed in a circle near the atom. Zero formal charges are not shown. Thus, formal charges of +1, –1, +2, and –2 would be represented by the following.
formal charge representation
Figure 5.11
Exercise 5.10:

Assign electrons to each of the following and determine their formal charge.
Valence Electrons 6_0__ Oxygen is a 6A nonmetal, so it has 6 valence electrons. This O has 2 lone pairs and 2 bonding pairs, so it is assigned 4 nonbonding electrons and 2 of the bonding electrons for a total of 6 electrons. Its formal charge is then
6 − 6 = 0.
6_0__ Sulfur is a 6A nonmetal, so it has 6 valence electrons. This S has 1 lone pair and 3 bonding pairs, so it is assigned 2 nonbonding electrons and 3 of the bonding electrons for a total of 5 electrons. Its formal charge is then
6 − 5 = +1.
6_0__ Oxygen is a 6A nonmetal, so it has 6 valence electrons. This O has 3 lone pairs and 1 bond, so it is assigned 6 nonbonding and 1 bonding electron for a total of 7 electrons. Its formal charge is then
6 − 7 = −1.
Non-Bonding         4_0__ Oxygen is a 6A nonmetal, so it has 6 valence electrons. This O has 2 lone pairs and 2 bonding pairs, so it is assigned 4 nonbonding electrons and 2 of the bonding electrons for a total of 6 electrons. Its formal charge is then
6 − 6 = 0.
2_0__ Sulfur is a 6A nonmetal, so it has 6 valence electrons. This S has 1 lone pair and 3 bonding pairs, so it is assigned 2 nonbonding electrons and 3 of the bonding electrons for a total of 5 electrons. Its formal charge is then
6 − 5 = +1.
6_0__ Oxygen is a 6A nonmetal, so it has 6 valence electrons. This O has 3 lone pairs and 1 bond, so it is assigned 6 nonbonding and 1 bonding electron for a total of 7 electrons. Its formal charge is then
6 − 7 = −1.
Bonding                 2_0__ Oxygen is a 6A nonmetal, so it has 6 valence electrons. This O has 2 lone pairs and 2 bonding pairs, so it is assigned 4 nonbonding electrons and 2 of the bonding electrons for a total of 6 electrons. Its formal charge is then
6 − 6 = 0.
3_0__ Sulfur is a 6A nonmetal, so it has 6 valence electrons. This S has 1 lone pair and 3 bonding pairs, so it is assigned 2 nonbonding electrons and 3 of the bonding electrons for a total of 5 electrons. Its formal charge is then
6 − 5 = +1.
1_0__ Oxygen is a 6A nonmetal, so it has 6 valence electrons. This O has 3 lone pairs and 1 bond, so it is assigned 6 nonbonding and 1 bonding electron for a total of 7 electrons. Its formal charge is then
6 − 7 = −1.
Formal Charge       0_0__ Oxygen is a 6A nonmetal, so it has 6 valence electrons. This O has 2 lone pairs and 2 bonding pairs, so it is assigned 4 nonbonding electrons and 2 of the bonding electrons for a total of 6 electrons. Its formal charge is then
6 − 6 = 0.
1_0__ Sulfur is a 6A nonmetal, so it has 6 valence electrons. This S has 1 lone pair and 3 bonding pairs, so it is assigned 2 nonbonding electrons and 3 of the bonding electrons for a total of 5 electrons. Its formal charge is then
6 − 5 = +1.
-1_0__ Oxygen is a 6A nonmetal, so it has 6 valence electrons. This O has 3 lone pairs and 1 bond, so it is assigned 6 nonbonding and 1 bonding electron for a total of 7 electrons. Its formal charge is then
6 − 7 = −1.
The Lewis structure of SO2 that includes the formal charge is shown in the following section.

5.8-4. Lewis Structure of SO2 with Formal Charge

In the previous exercise, we showed that the sulfur atom in SO2 has positive formal charge, the oxygen with the single bond had negative formal charge, and the oxygen with the double bond had no formal charge. Thus, the Lewis structures of the two resonance forms that show formal charge are shown in Figure 5.12.
formal charge in SO2
Figure 5.12: Formal Charge in SO2

5.8-5. Formal Charge and the Number of Bonds

Zero formal charge results when the number of bonds to an atom equals the number of unpaired electrons in its Lewis symbol.
The formal charge of an atom depends upon the number of bonds to the atom—the greater the number of bonds, the more positive the formal charge. Note that when counting bonds, double and triple bonds count as two and three bonds, respectively. The following figures give the formal charge of some common elements as a function of the number of bonds in which it is involved. The number of bonds resulting in zero formal charge is bolded.
Lewis symbol of carbon
Figure 5.13
Sharing the unpaired electrons in the Lewis symbol results in zero formal charge for an atom. The Lewis symbol of carbon shows four unpaired electrons, so carbon has zero formal charge when it is involved in four bonds. Nonzero formal charge is hardly ever assigned to carbon, so a very good rule to use in drawing Lewis structures is that carbon atoms always have four bonds.
Lewis symbol of nitrogen
Figure 5.14
The Lewis symbol of nitrogen shows three unpaired electrons, so nitrogen has zero formal charge when it is involved in three bonds. Thus, nitrogen usually has three bonds to it, but it is also frequently found with four bonds and positive formal charge, e.g., NH41+, the ammonium ion. Occasionally, it has only two bonds, in which case it carries a negative formal charge.
Lewis symbol of oxygen
Figure 5.15
The Lewis symbol of oxygen shows two unpaired electrons, so oxygen has zero formal charge when it is involved in two bonds. Thus, oxygen usually has two bonds to it, but it is also frequently found with one bond and negative formal charge, e.g., OH1–, the hydroxide ion. Occasionally, it has three bonds (H3O1+), in which case it carries a positive formal charge.
Lewis symbol of fluorine
Figure 5.16
The Lewis symbol of fluorine shows only one unpaired electron, so fluorine has zero formal charge when it is involved in only one bond. Adding a second bond to a fluorine atom would produce a positive formal charge, but F is the most electronegative atom, so positive formal charge is never placed on it. Thus, F has one and only one bond in all of its compounds.
oxoanions
Figure 5.17
The other halogens can have more than one bond in situations where they are bound to a number of more electronegative atoms (O or F). However, a good rule to follow is that double and triple bonds are not drawn to halogens because they place positive formal charge on these relatively electronegative elements.

5.8-6. Structure and Formal Charge

Lewis structures that minimize formal charge are preferred.
Formal charge can be used to determine the best Lewis structure in cases where more than one Lewis structure can be drawn because charge separation (creating centers of positive and negative charge) requires energy (Coulomb's law). Thus, Lewis structures with no formal charge are favored over those that do contain formal charge. The best Lewis structure is the one in which
Lewis structures for exercise
Figure 5.18: Using Formal Charge to Choose between Resonance Structures of CO2
Although two Lewis structures can be drawn for CO2, only structure a is important because it contains less formal charge.

5.8-7. Formal Charge Exercise

Exercise 5.11:

Determine the formal charges on each atom in the following Lewis structures of FCHO.
Structure a
O = 0_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 2 4 6 – 2 – 4 = 0
H 1 1 0 1 – 1 – 0 = 0
C 4 4 0 4 – 4 – 0 = 0
F 7 1 6 7 – 1 – 6 = 0
H = 0_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 2 4 6 – 2 – 4 = 0
H 1 1 0 1 – 1 – 0 = 0
C 4 4 0 4 – 4 – 0 = 0
F 7 1 6 7 – 1 – 6 = 0
C = 0_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 2 4 6 – 2 – 4 = 0
H 1 1 0 1 – 1 – 0 = 0
C 4 4 0 4 – 4 – 0 = 0
F 7 1 6 7 – 1 – 6 = 0
F = 0_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 2 4 6 – 2 – 4 = 0
H 1 1 0 1 – 1 – 0 = 0
C 4 4 0 4 – 4 – 0 = 0
F 7 1 6 7 – 1 – 6 = 0
Structure b
O = -1_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 1 6 6 – 1 – 6 = –1
H 1 1 0 1 – 1 – 0 = 0
C 4 4 0 4 – 4 – 0 = 0
F 7 2 4 7 – 2 – 4 = +1
H = 0_0__ lewis structure for exercise
Atom VE BE NB VE – BE – NB = FC
O 6 1 6 6 – 1 – 6 = –1
H 1 1 0 1 – 1 – 0 = 0
C 4 4 0 4 – 4 – 0 = 0
F 7 2 4 7 – 2 – 4 = +1
C = 0_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 1 6 6 – 1 – 6 = –1
H 1 1 0 1 – 1 – 0 = 0
C 4 4 0 4 – 4 – 0 = 0
F 7 2 4 7 – 2 – 4 = +1
F = +1_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 1 6 6 – 1 – 6 = –1
H 1 1 0 1 – 1 – 0 = 0
C 4 4 0 4 – 4 – 0 = 0
F 7 2 4 7 – 2 – 4 = +1
Structure c
O = 0_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 2 4 6 – 2 – 4 = 0
H 1 1 0 1 – 1 – 0 = 0
C 4 3 2 4 – 3 – 2 = –1
F 7 2 4 7 – 2 – 4 = +1
H = 0_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 2 4 6 – 2 – 4 = 0
H 1 1 0 1 – 1 – 0 = 0
C 4 3 2 4 – 3 – 2 = –1
F 7 2 4 7 – 2 – 4 = +1
C = -1_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 2 4 6 – 2 – 4 = 0
H 1 1 0 1 – 1 – 0 = 0
C 4 3 2 4 – 3 – 2 = –1
F 7 2 4 7 – 2 – 4 = +1
F = +1_0__ lewis structure for exercise
atom VE BE NB VE – BE – NB = FC
O 6 2 4 6 – 2 – 4 = 0
H 1 1 0 1 – 1 – 0 = 0
C 4 3 2 4 – 3 – 2 = –1
F 7 2 4 7 – 2 – 4 = +1
Note that the oxygen atom in the last structure has both an C–O and an O–F bond.
    Based on the formal charges, the preferred structure is which of the following?
  • Structure a
  • Structure b Structure a is preferred because it has no formal charge. Note that there are two bonds to F in structures b and c (double and two singles), which gives this very electronegative element a positive oxidation state. Double bonds are never drawn to terminal halogens because double bonds give these electronegative elements positive formal charge.
  • Structure c Structure a is preferred because it has no formal charge. Note that there are two bonds to F in structures b and c (double and two singles), which gives this very electronegative element a positive oxidation state. Double bonds are never drawn to terminal halogens because double bonds give these electronegative elements positive formal charge.

5.8-8. Oxidation States

Oxidation state is the charge the atom would have if the bonds were ionic. It can be determined by electron counting.
The oxidation state is the charge the atom would have if the bonds were ionic. This means that all of the bonding electrons in each bond are assigned to the more electronegative atom. That is, a = 0 or 1, and the oxidation state of atom A is
OXA = VE −
NB +
aBE
where the sum is over all of the bonds to atom A and a is either 0 or 1 for each bond depending upon whether atom A is less or more electronegative than the other atom in the bond. In the case where the two atoms are identical, neither atom is more electronegative, so
a =
1
2
in this special case; i.e., the bond must be assumed to be covalent when the two electronegativities are identical. Thus, the value of a in each bond is determined as follows:
Oxidation states determined with the method outlined above may differ from those determined using the guidelines presented in Section 4.4 because the latter method yields the average oxidation state of all of the atoms if the element being considered, while the method above applies only to the one atom being considered.

5.8-9. Counting Electrons for Oxidation State Exercise

Exercise 5.12:

Determine the oxidation state of each atom in CO2.
Valence Electrons 6___ This O atom has two lone pairs and is involved in two bonds. O is more electronegative than C, so all of the bonding electrons are assigned to the O (i.e.,
a = 1).
Consequently, 4 nonbonding and 4 bonding electrons are assigned to O for a total of 8 valence electrons. The neutral atom has 6 valence electrons, so its oxidation state is
+6 − 8 = −2,
which is the same as is given in the priority rules given in Chapter 4. Oxygen is the second most electronegative element, so the bonding pairs are assigned to it except when it bonds to F or another O (peroxide).
4___ This C atom has no lone pairs and is involved in 4 bonds. C is less electronegative than O, so it is assigned none of the bonding electrons, giving it no valence electrons in the compound. A neutral carbon atom has four valence electrons, so its oxidation state is
4 − 0 = +4,
which is the same oxidation state that it would have using the rules given in Chapter 4.
6___ This O atom has two lone pairs and is involved in two bonds. O is more electronegative than C, so all of the bonding electrons are assigned to the O (i.e.,
a = 1).
Consequently, 4 nonbonding and 4 bonding electrons are assigned to O for a total of 8 valence electrons. The neutral atom has 6 valence electrons, so its oxidation state is
+6 − 8 = −2,
which is the same as is given in the priority rules given in Chapter 4. Oxygen is the second most electronegative element, so the bonding pairs are assigned to it except when it bonds to F or another O (peroxide).
Non-Bonding         4___ This O atom has two lone pairs and is involved in two bonds. O is more electronegative than C, so all of the bonding electrons are assigned to the O (i.e.,
a = 1).
Consequently, 4 nonbonding and 4 bonding electrons are assigned to O for a total of 8 valence electrons. The neutral atom has 6 valence electrons, so its oxidation state is
+6 − 8 = −2,
which is the same as is given in the priority rules given in Chapter 4. Oxygen is the second most electronegative element, so the bonding pairs are assigned to it except when it bonds to F or another O (peroxide).
0___ This C atom has no lone pairs and is involved in 4 bonds. C is less electronegative than O, so it is assigned none of the bonding electrons, giving it no valence electrons in the compound. A neutral carbon atom has four valence electrons, so its oxidation state is
4 − 0 = +4,
which is the same oxidation state that it would have using the rules given in Chapter 4.
4___ This O atom has two lone pairs and is involved in two bonds. O is more electronegative than C, so all of the bonding electrons are assigned to the O (i.e.,
a = 1).
Consequently, 4 nonbonding and 4 bonding electrons are assigned to O for a total of 8 valence electrons. The neutral atom has 6 valence electrons, so its oxidation state is
+6 − 8 = −2,
which is the same as is given in the priority rules given in Chapter 4. Oxygen is the second most electronegative element, so the bonding pairs are assigned to it except when it bonds to F or another O (peroxide).
Bonding                 4___ This O atom has two lone pairs and is involved in two bonds. O is more electronegative than C, so all of the bonding electrons are assigned to the O (i.e.,
a = 1).
Consequently, 4 nonbonding and 4 bonding electrons are assigned to O for a total of 8 valence electrons. The neutral atom has 6 valence electrons, so its oxidation state is
+6 − 8 = −2,
which is the same as is given in the priority rules given in Chapter 4. Oxygen is the second most electronegative element, so the bonding pairs are assigned to it except when it bonds to F or another O (peroxide).
0___ This C atom has no lone pairs and is involved in 4 bonds. C is less electronegative than O, so it is assigned none of the bonding electrons, giving it no valence electrons in the compound. A neutral carbon atom has four valence electrons, so its oxidation state is
4 − 0 = +4,
which is the same oxidation state that it would have using the rules given in Chapter 4.
4___ This O atom has two lone pairs and is involved in two bonds. O is more electronegative than C, so all of the bonding electrons are assigned to the O (i.e.,
a = 1).
Consequently, 4 nonbonding and 4 bonding electrons are assigned to O for a total of 8 valence electrons. The neutral atom has 6 valence electrons, so its oxidation state is
+6 − 8 = −2,
which is the same as is given in the priority rules given in Chapter 4. Oxygen is the second most electronegative element, so the bonding pairs are assigned to it except when it bonds to F or another O (peroxide).
Oxidation State      -2___ This O atom has two lone pairs and is involved in two bonds. O is more electronegative than C, so all of the bonding electrons are assigned to the O (i.e.,
a = 1).
Consequently, 4 nonbonding and 4 bonding electrons are assigned to O for a total of 8 valence electrons. The neutral atom has 6 valence electrons, so its oxidation state is
+6 − 8 = −2,
which is the same as is given in the priority rules given in Chapter 4. Oxygen is the second most electronegative element, so the bonding pairs are assigned to it except when it bonds to F or another O (peroxide).
+4___ This C atom has no lone pairs and is involved in 4 bonds. C is less electronegative than O, so it is assigned none of the bonding electrons, giving it no valence electrons in the compound. A neutral carbon atom has four valence electrons, so its oxidation state is
4 − 0 = +4,
which is the same oxidation state that it would have using the rules given in Chapter 4.
-2___ This O atom has two lone pairs and is involved in two bonds. O is more electronegative than C, so all of the bonding electrons are assigned to the O (i.e.,
a = 1).
Consequently, 4 nonbonding and 4 bonding electrons are assigned to O for a total of 8 valence electrons. The neutral atom has 6 valence electrons, so its oxidation state is
+6 − 8 = −2,
which is the same as is given in the priority rules given in Chapter 4. Oxygen is the second most electronegative element, so the bonding pairs are assigned to it except when it bonds to F or another O (peroxide).

5.8-10. Difference in Oxidation State Methods

The oxidation state of an atom, determined from the rules in Chapter 4, yields the average oxidation states of all the atoms of the element be considered, while the electron counting procedure outlined in this section produces the oxidation states of the individual atom being considered.
The oxidation states determined for the carbon and oxygen atoms in CO2 are the same whether the method used in Section 4.4 or electron counting is used because the oxidation states of the two oxygen atoms are identical. In cases where the oxidation states of two or more different atoms of the same element are different, the two methods may yield different results because the oxidation state determined by the method given in Section 4.4 is the average of the oxidation states, while that determined by electron counting applies only to the individual atom being considered. Two examples where the two methods differ are given in the following exercise.

5.8-11. Differences in Oxidation State Methods Exercise

Exercise 5.13:

What is the average oxidation state of the C atoms in C2H4O2 as determined by the rules outlined in Section 4.4?
OXC = 0_0__ Rule 4 states that H is +1 and Rule 5 states that O is –2. Acetic acid is a neutral molecule, so the oxidation state of carbon is determined to be
2x + 4(1) +2(−2) = 0,
which yields
x = 0.
The average oxidation state of the two carbon atoms in C2H4O2 is 0.
Use the method described in this chapter to determine the oxidation state of each carbon atom in acetic acid, which has the following Lewis structure:
Lewis structure for exercise
OXCA = -3_0__ C is more electronegative than H, so all six of the C–H bonding electrons are assigned to the carbon atom. Half of the bonding electrons in bonds between identical atoms are assigned to each atom, so one of the C–C bonding electrons is assigned to CA. Thus CA is assigned seven valence electrons. Carbon atoms have only four valence electrons so the oxidation state on CA
= 4 − 7 = −3.
OXCB = +3_0__ O is more electronegative than C, so all six of the C–O bonding electrons are assigned to the oxygen atom. Half of the bonding electrons in bonds between identical atoms are assigned to each atom, so one of the C–C bonding electrons is assigned to CB. Thus CB is assigned only one valence electron. Carbon atoms have only four valence electrons so the oxidation state on CB = 4 – 1 = +3. One carbon atom is +3 and the other is –3. The average of the two oxidation states is 0, in agreement with the result obtained using the method of Chapter 4.
What is the average oxidation state of the C atoms in C3H6O? Express your answer in decimal form.
OXC = -1.3333_.0333333__ Rule 4 states that H is +1 and Rule 5 states that O is –2. Acetone is a neutral molecule, so the oxidation state of carbon is determined to be
3x + 6(1) + 1(−2) = 0,
which yields
x = −4/3 = −1.3.
The average oxidation state of the two carbon atoms in C3H6O is –1.3. Note that, although the average is not an integer, the oxidation states of the individual atoms must be integers.
Use the method described in this chapter to determine the oxidation state of each carbon atom in C3H6O, which has the following Lewis structure:
Lewis structure for exercise
OXCA = -3_0__ C is more electronegative than H, so all six of the C–H bonding electrons are assigned to the carbon atom. Half of the bonding electrons in bonds between identical atoms are assigned to each atom, so one of the C–C bonding electrons is assigned to CA. Thus CA is assigned seven valence electrons. Carbon atoms have only four valence electrons so the oxidation state on CA = 4 – 7 = –3.
OXCB = +2_0__ O is more electronegative than C, so all four of the C–O bonding electrons are assigned to the oxygen atom. Half of the bonding electrons in bonds between identical atoms are assigned to each atom, so one of the bonding electrons in each C–C bond is assigned to CB. Thus, CB is assigned two valence electrons. Carbon atoms have only four valence electrons so the oxidation state on CB = 4 – 2 = +2. One carbon atom is +2 and the other two are –3 each. The average of the three oxidation states is –4/3, in agreement with the result obtained using the method of Chapter 4.

5.9 Practice with Lewis Structures

Introduction

Practice drawing Lewis structures in this section.

5.9-1. Lewis Structure of SO3 Exercise

Exercise 5.14:

Draw the Lewis structure of SO3, and indicate all nonzero formal charge.
ER = 32_0__ ER = 4 atoms × 8 electrons/atom = 32 electrons required
VE = 24_0__ VE = 6 from S + (3 × 6) from O = 24 valence electrons
SP = 4_0__
SP =
1
2
(32 − 24) = 4 shared pairs
How many resonance structures are there?
3_0__ The resonance structures of SO3 are: SO3 resonance structures
What is the formal charge on the sulfur atom?
+2_0__ There are 4 shared pairs and no lone pairs on the sulfur, and one electron per shared pair is assigned to the sulfur, so it is assigned a total of four electrons. Sulfur is in Group 6A, so the formal charge
6 − 4 = +2.

5.9-2. Lewis Structure of SO32– Exercise

Exercise 5.15:

Draw the Lewis structure of the following molecules and ions. Indicate all nonzero formal charge.
When dealing with ions, remember that negative charge increases the number of valence electrons, while positive charge decreases the number of valence electrons.
ER = 32_0__ ER = 4 atoms × 8 electrons/atom = 32 electrons required
VE = 26_0__ VE = 6 from S + (3 × 6) from O +2 for the charge = 26 valence electrons
SP = 3_0__
SP =
1
2
(32 − 26) = 3 shared pairs
What is the formal charge on the sulfur atom?
+1_0__ There are 3 shared pairs and one lone pair on the sulfur. One electron per shared pair and both electrons in the lone pair are assigned to the sulfur, so it is assigned a total of 5 electrons. Sulfur is in Group 6A, so the formal charge
6 − 5 = +1.

5.9-3. Lewis Structure of N31– Exercise

Exercise 5.16:

Draw the Lewis structure of N31–, indicating the formal charge on each atom.
ER = 24_0__ ER = electrons required = 8 times the number of non-hydrogen atoms + 2 times the number of hydrogens.
VE = 16_0__ VE = valence electrons = sum of the valence electrons on all of the atoms in the molecule. This is the number of electrons that must be shown in the final Lewis structure.
SP = 4_0__
SP = shared pairs =
1
2
(ER − VE)
What are the formal charges on the central and terminal N atoms?
central +1_0__ There are 4 shared pairs and no lone pairs on the central N atom and one electron per shared pair is assigned to it, so it is assigned a total of 4 electrons. Nitrogen is in Group 5A, so the formal charge
5 − 4 = +1.
terminal -1_0__ There are 2 shared pairs and two lone pairs on each terminal N atom. One electron per shared pair and all of the electrons in the lone pairs are assigned to each atom, so each is assigned a total of 6 electrons. Nitrogen is in Group 5A, so the formal charge
5 − 6 = −1.

5.9-4. Lewis Structure of H2O Exercise

Exercise 5.17:

Draw the Lewis structure of H2O, indicating the formal charge on each atom.
ER = 12_0__ ER = electrons required = 8 times the number of non-hydrogen atoms + 2 times the number of hydrogens.
VE = 8_0__ VE = valence electrons = sum of the valence electrons on all of the atoms in the molecule. This is the number of electrons that must be shown in the final Lewis structure.
SP = 2_0__
SP = shared pairs =
1
2
(ER − VE)
What is the formal charge on the O atom?
0___ There are 2 shared pairs and 2 lone pairs on the O atom. One electron per shared pair and all of the electrons in the lone pairs are assigned to it, so it is assigned a total of 6 electrons. Oxygen is in Group 6A, so the formal charge
6 − 6 = 0.

5.9-5. Lewis Structure of SO2 Exercise

Exercise 5.18:

Draw the Lewis structure of SO2, indicating the formal charge on each atom.
ER = 24_0__ ER = electrons required = 8 times the number of non-hydrogen atoms + 2 times the number of hydrogens.
VE = 18_0__ VE = valence electrons = sum of the valence electrons on all of the atoms in the molecule. This is the number of electrons that must be shown in the final Lewis structure.
SP = 3_0__
SP = shared pairs =
1
2
(ER − VE)
What is the formal charge on the S atom?
+1___ There are 3 shared pairs and 1 lone pairs on the S atom. One electron per shared pair and all of the electrons in the lone pairs are assigned to it, so it is assigned a total of 5 electrons. Sulfur is in Group 6A, so the formal charge
6 − 5 = +1.

5.9-6. Lewis Structure of NO31– Exercise

Exercise 5.19:

Draw the Lewis structure of NO31–, indicating the formal charge on each atom.
ER = 32_0__ ER = electrons required = 8 times the number of non-hydrogen atoms + 2 times the number of hydrogens.
VE = 24_0__ VE = valence electrons = sum of the valence electrons on all of the atoms in the molecule. This is the number of electrons that must be shown in the final Lewis structure.
SP = 4_0__
SP = shared pairs =
1
2
(ER − VE)
What is the formal charge on the N atom?
+1___ There are 4 shared pairs and no lone pairs on the N atom, and one electron per shared pair is assigned to it, so it is assigned a total of 4 electrons. Nitrogen is in Group 5A, so the formal charge
5 − 4 = +1.

5.10 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.