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Unformatted text preview: Chapter Nine CHEMICAL BONDING I: BASIC CONCEPTS
• Lewis Symbols
Ionic Bonding and the Lattice Energy
Covalent Bonding and Lewis Structures
The Concept of Resonance
Exceptions to the Octet Rule
Bond Dissociation Energy LEWIS SYMBOLS
1. Write the Lewis symbols for atoms and ions. Lewis Symbols. Ionic and covalent bonds are important types of chemical bonds. In this chapter you will
learn to use electron configurations and the periodic table to predict the type and the number of bonds an atom of
a particular element can form. Electron configurations are used to write the Lewis dot symbols of the
representative elements. A Lewis dot symbol of an element consists of the chemical symbol with one or more
dots placed around it. Each dot represents a valence electron. The orbital diagram and the Lewis symbol for the
fluorine atom are shown in Figure 9.1. Fluorine has 7 electrons in its outermost principal energy level (n = 2),
and therefore has 7 dots in its Lewis symbol. 1s 2s 2p Orbital diagram Lewis symbol ..
.. Figure 9.1 Orbital diagram and Lewis symbol for the fluorine atom. 1 64
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The Lewis dot symbols and the electron configurations of the outermost electrons of elements in the third
period are as follows:
Na· · Mg· .
· Al· .
· Si ·
˙ 3s1 3s2 3 s 2 3p1 ..
˙ 3 s 2 3p2 3 s 2 3p3 ..
˙˙ 3 s 2 3p4 3 s 2 3p5 3 s 2 3p6 For representative elements, the number of valence electrons is the same as the group number.
The representative metals form ions by losing all of their valence electrons. Lewis symbols can be used to
represent the formation of positive magnesium and aluminum ions as follows: · Mg· → Mg2+ + 2e–
· Al· → Al3+ + 3e–
The nonmetal elements form negative ions by acquiring electrons until they are isoelectronic with a noble gas
atom. Ions of nonmetal have 8 electrons in their valence shell. .
:S : + 2e– → :S : 2–
:Cl · + e– → :Cl : –
Sulfide and chloride ions, for example, have net charges of –2 and –1 because the atoms needed 2e– and 1e– ,
respectively, to become isoelectronic with the noble gas argon.
EXAMPLE 9.1 Lewis Dot Symbols
Write Lewis symbols for the following elements: a. Ca b. O, S and Se •Method of Solution
The Lewis symbol of an element consists of the element symbol surrounded by between 1 and 8 dots,
representing valence electrons.
a. Ca is in Group 2A. It has two electrons in the outermost energy level. These are its valence electrons.
The Lewis symbol is · Ca·
b. O, S, and Se are all in the same group, 6A. Answer: Each has 6 valence electrons. ..
EXAMPLE 9.2 Lewis Dot Symbols for Ions
Write Lewis symbols for the following ions: a. Ca2+ b . Se 2– •Method of Solution
a. Back Removing two electrons (dots) from · Ca· gives simply Ca2+ as the Lewis symbol of a calcium ion. No
electrons are shown around the Ca2+ symbol because this ion has lost its valence shell electrons. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 66 / Chemical Bonding I: Basic Concepts
b. .. Adding two electrons to · Se· gives :Se: 2– as the Lewis symbol for a selenide ion. ˙˙ ˙˙ _______________________________________________________________________________ EXERCISES
1. Write Lewis dot symbols for atoms of the following elements:
a. Mg b. Se c. Al d. Br e. Xe 2. Write Lewis dot symbols for the following ions:
a. K + b . S 2– c. N3– d. I– e. Sr2+ IONIC BONDING AND LATTICE ENERGY
3. Predict the relative strength of ionic bonds using Coulomb's law.
List three energy terms that influence the tendency for two elements to form an ionic compound.
Use the Born-Haber cycle to calculate the magnitude of the lattice energies of ionic solids. Coulomb's Law. The force that gives rise to the ionic bond is the electrical attraction existing between a
positive ion and a negative ion. Chemical bonding results when the energy of two interacting atoms (or ions) is
lowered. Coulomb's law states that the potential energy (E) of interaction of two ions is directly proportional to
the product of their charges and inversely proportional to the distance between them.
where Q1 and Q2 are the charges of the two ions, r is their distance of separation, and k is a proportionality
constant (its value will not be needed). When one ion is positive and the other negative, E will be negative.
Thus bringing two oppositely charged particles closer together lowers their energy. The lower the value of the
potential energy, the more stable is the pair of ions.
The factors that govern the stability of ion pairs are the magnitude of their charges, and the distance between
the ion centers. The distance between ionic centers (r) is the sum of the ionic radii of the individual ions.
T he energy of attraction between two oppositely charged ions depends:
•Directly on the magnitude of the ion charges. The greater the ion charges, the greater the energy. The energy of
interaction of Mg 2+ and O2– ions is much greater than that of Na+ and Cl– ions.
•Inversely on the distance between ion centers. The distance between ions depends on the sizes of the ions
involved. The energy decreases as the ion size increases. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 67
Lattice Energy. In an ionic crystal the cations and anions are arranged in an orderly three-dimensional
array, as illustrated in Figure 2.10 of the textbook. In a sodium chloride or rock salt type of crystal, the anion
has six nearest neighbor cations. The ions are packed in such a way as to maximize attraction and minimize
The lattice energy provides a measure of the attraction between ions and the strength of the ionic bond. The
lattice energy is the energy required to separate the ions in 1 mole of a solid ionic compound into gaseous
ions. Gaseous ions are far enough apart from one another that they do not interact. For example, the lattice
energy of NaF(s) is equal to 908 kJ/mol. This means 908 kJ are required to vaporize one mole of NaF(s) and
form one mole of Na+ ions and one mole of F– ions in the gas phase.
NaF(s) → Na+(g) + F– (g) lattice energy = 908 kJ The lattice energy is related to the stability of the ionic solid. The larger the lattice energy, the more stable the
The value of the lattice energy depends on the charges of the ions and the ionic radii, in accordance with
Coulomb's law. This can be seen in Table 9.1 by comparing the sodium halides. As the sum of the two ionic
radii increases in going from NaF to NaI, the separation of ionic centers (r = r1 + r2 ) increases and the lattice
energy decreases. The effect of ionic charge on lattice energy can be seen by comparing the lattice energies of
NaF and CaO in Table 9.1. The larger value for CaO is due to the stronger attraction of doubly charged cations
(Ca2+ ) for doubly charged anions (O2– ).
The lattice energies are reflected in the melting points of ionic crystals. During melting the ions gain
enough kinetic energy to overcome the potential energy of attraction, and they move away from each other. The
higher the melting point, the more energy the ions need to separate from one another. Therefore as the lattice
energy increases so does the melting point.
T a b l e 9 . 1 Lattice Energies of Several Ionic Compounds
r1 + r2 *
* See Table 8.4 for individual ionic radii values. Factors Favoring Formation of Ionic Bonds. It is important to identify any properties of atoms
that affect their ability to form ionic compounds. In the formation of ions from atoms, the metal atom loses an
electron and the nonmetal atom gains an electron.
Li(g) → Li +(g) + e–
e– + F(g) → F – (g)
The overall change is
Li(g) + F(g) → Li +(g) + F– (g)
One factor favoring the formation of an ionic compound is the ease with which the metal atom loses an
electron. A second factor is the tendency of the nonmetal atom to gain an electron. Thus ionic compounds tend
to form between elements of low ionization energy and those of high electron affinity. The alkali metals and Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 68 / Chemical Bonding I: Basic Concepts
alkaline earth metals have low ionization energies. They tend to form ionic compounds with the halogens and
Group 6A elements both of which have high electron affinities.
A third factor is the lattice energy. Electrostatic attraction of the ions results in large amounts of energy
being released when two kinds of gaseous ions are brought together to form a crystal lattice.
Li+(g) + F– (g) → LiF(s)
In general, the smaller the ionic radius and the greater the ionic charge, the greater the lattice energy. Calculation of the Lattice Energy Using the Born-Haber Cycle. Lattice energies cannot be
measured directly and must be calculated using Hess's law. In Chapter 6 you learned that if a reaction can be
broken down into a series of steps, the overall enthalpy of reaction is equal to the sum of the enthalpy changes
for the individual steps. The series of steps used to calculate the lattice energies of ionic solids is called the
Born-Haber cycle. Each step is one you've seen previously in relation to the properties of atoms. Here we will
illustrate the calculation of the lattice energy for potassium chloride.
KCl(s) → K+(g) +Cl – (g) lattice energy = ? The Born-Haber cycle starts by taking the overall equation to be the reaction in which the ionic compound
is formed from the elements in their standard states. In this example the enthalpy change is the same as the
enthalpy of formation of potassium chloride from potassium and chlorine.
o 1 K(s) + 2 C l 2 (g) → KCl(s) o ∆ Hoverall = ∆ Hf (KCl) This reaction is then envisioned to occur by a number of steps.
1. 2. First potassium sublimes:
K(s) → K(g) o ∆ H1 = ∆ Hsub Next, diatomic chlorine is dissociated into Cl atoms. The required energy is the bond dissociation energy for
1/2 mole of Cl—Cl bonds.
C l 2 (g) → Cl(g)
2 1 ∆ H2 = 2 BE(Cl—Cl) Up to now the focus has been on making the atoms of the elements potassium and chlorine. Next these will be
made into the appropriate ions.
3. Now ionize 1 mole of potassium atoms:
o K(g) → K+(g) + e–
4. ∆ H3 = I To form a Cl– ion, an electron is added to the chlorine atom. This will release an energy equal to the
electron affinity of chlorine, EA. Because energy is given off, ∆ H has the opposite sign of the electron
o Cl(g) + e – → C l – (g)
5. ∆ H4 = –EA Finally, 1 mole of gaseous K+ ions, and 1 mole of gaseous Cl– ions are combined to make 1 mole of
KCl(s). An amount of energy equal to the lattice energy will be released. The lattice energy must have the
same magnitude as ∆ H5 , but an opposite sign.
o K+(g) +Cl – (g) → KCl(s) ∆ H5 = – lattice energy The summation of these changes gives the overall reaction above.
o 1 K(s) + 2 C l 2 (g) → KCl(s) Back Forward Main Menu o ∆ Hoverall = ∆ Hf (KCl) TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 69
Therefore, in general:
o 1 ∆ Hf (KCl) = ∆ Hsub + 2 BE + I(metal element) – EA(nonmetal) – lattice energy or o o o o o o ∆ Hoverall = ∆ H1 + ∆ H2 + ∆ H3 + ∆ H4 + ∆ H5
o where ∆ H5 = –lattice energy
o o o o o o –lattice energy = ∆ H5 = ∆ Hoverall – [∆ H1 + ∆ H2 + ∆ H3 + ∆ H4 _______________________________________________________________________________
EXAMPLE 9.3 Predicting Melting Points
Which member of the pair will have the higher melting point?
a. NaCl or CaO
b. NaCl or NaI
•Method of Solution
b. According to Coulomb's law, the doubly charged ions in CaO will attract each other more strongly than do
the singly charged ions in NaCl. Answer: CaO will have a higher melting temperature.
According to Coulomb's law, the closer the centers of two ions can approach each other the stronger the
attraction between them. The sum of the ionic radii in NaCl is smaller than in NaI. Answer: NaCl will
have the higher melting point. _______________________________________________________________________________
EXAMPLE 9.4 The Born-Haber Cycle
Given the following data calculate the lattice energy of potassium chloride:
Enthalpy of sublimation of potassium
Dissociation energy (BE) of Cl—Cl
Ionization energy (I) of K
Electron affinity (EA) of Cl
∆ Hf KCl(s) = 90.0 kJ
= 242.7 kJ
= 419 kJ
= +349 kJ
= –435.9 kJ •Method of Solution
o From the discussion in the above section, we can calculate the lattice energy (or –∆ H5 ) from the equation:
o o o o o o ∆ H5 = ∆ Hoverall – [∆ H1 + ∆ H2 + ∆ H3 + ∆ H4 Next we identify the appropriate energy changes:
o K(s) → K(g) ∆ H1 = ∆ Hsub = 90.0 kJ
C l 2 (g) → Cl(g)
2 1 ∆ H2 = 2 BE = 121.4 kJ
o K(g) → K+(g) + e– ∆ H3 = I = 419 kJ Cl(g) + e – → C l – (g) ∆ H4 = –EA = –349 kJ o 1 K(s) + 2 C l 2 (g) → KCl(s) Back Forward Main Menu TOC o ∆ Hoverall = – 435.9 kJ Study Guide TOC Textbook Website MHHE Website 1 70 / Chemical Bonding I: Basic Concepts
Next substitute into the equation:
o o o o o o ∆ H5 = ∆ Hoverall – [∆ H1 + ∆ H2 + ∆ H3 + ∆ H4 o ∆ H5 = – 435.9 kJ – [90.0 kJ + 121.4 kJ + 419 kJ + (–349 kJ)]
o ∆ H5 = – 435.9 kJ – 281.4 kJ = –717 kJ
lattice energy = 717 kJ
3. Write a chemical equation for the process that corresponds to the lattice energy of MgO. 4. Which member of the pair will have the higher melting point?
a. KCl or CaCl2
b. RbI or NaI 5. Which member of the pair will have the higher lattice energy?
a. NaCl or CaO
b. KI or KCl 6. Calculate the lattice energy of sodium bromide from the following information.
∆ Hsub (Na) = 109 kJ
Ionization Energy (Na) = 496 kJ
Dissociation Energy (Br—Br) = 192 kJ
Electron Affinity (Br) = 324 kJ
∆ Hf (NaBr) = –359 kJ kJ COVALENT BONDING AND LEWIS STRUCTURES STUDY OBJECTIVES
2. State the octet rule and describe its basis.
Describe covalent bonding in molecules by drawing their Lewis structures. Octet Rule. The Lewis symbols of the noble gas elements show eight electrons corresponding to filled s
and p subshells. In our study of the periodic table we saw that the outer electron configuration is related to the
chemical and physical properties of an element. G. N. Lewis reasoned that when atoms enter into chemical
combination they become more stable. He proposed that atoms gain or lose electrons until they have the same
number of valence electrons as noble gas atoms, that is, eight. The o ctet rule states that when forming bonds
atoms of the representative elements tend to gain, lose, or, share electrons until they have eight electrons in the
Molecules are held together by bonds resulting from the sharing of electrons between two atoms in a
manner that is consistent with the octet rule. A simple c ovalent bond is formed when two atoms in a
molecule share a pair of electrons. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 71
The formation of a covalent bond in hydrogen chloride can be represented with Lewis structures: H· ..
+ · Cl : → H : Cl : or H—Cl :
˙˙ where the dash represents a covalent bond, or a pair of electrons shared by both the H atom and the Cl atom. By
sharing the electron pair, the Cl atom has eight valence shell electrons. The stability of this bond results from
both atoms acquiring a noble gas configuration. Notice that hydrogen is an exception to the octet rule. Rather
than achieving an octet, it needs only two electrons to achieve a filled outer energy level. The H atom becomes
isoelectronic with helium. The electron pairs on the Cl atom that are not involved in bonding are called l one
pairs , u nshared pairs , and n onbonding electrons . The sharing of valence electrons in methane and
carbon tetrachloride is shown in Figure 9.2. The circles represent the valence shells of the atoms. They help to
point out that each atom achieves an octet of valence electrons by sharing one or more pairs of electrons.
: Cl :
: C : Cl :
H: C : H
H : Cl :
.. Figure 9.2 Sharing of electron pairs in CH4 and CCl 4 .
In some cases two or three pairs of electrons are shared by two atoms in order to reach an octet. In these
molecules, multiple bonds exist. A double bond is a covalent bond in which two pairs of electrons are shared
between two atoms, as between C and O in formaldehyde.
H .. . .
H : C :: O
˙˙ H2 CO (formaldehyde) or H
O ˙˙ In general, atoms joined by a double bond lie closer together than atoms joined by a single bond. The C
bond length is shorter than the C—O bond length.
Nitrogen molecules (N2 ) contain a triple bond. :N ::: N: or :N O N: Lewis Structures. Lewis structures represent the covalent bonding and location of unshared electron pairs
within molecules and polyatomic ions. The steps for writing Lewis structures are as follows.
3. 4. Back Arrange the atoms in a reasonable skeletal form, placing the unique atom in the center. Determine what
atoms are bonded to each other.
Count the valence electrons. For polyatomic anions remember to add one electron for each unit of charge.
Connect the central atom to the surrounding atoms with single bonds. First add unshared pairs to the atoms
bonded to the central atom to complete their octets (except for hydrogen, of course). Then, keeping in mind
that the maximum number of electrons is the number counted in step 2, add the remaining unshared pairs to
the central atom.
If the octet rule is satisfied for each atom, and the total number of electrons is correct, stop here since the
structure can be considered correct. If the octet rule is not met for the central atom, go on to step 5. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 72 / Chemical Bonding I: Basic Concepts
5. 6. In some cases there is a shortage of valence electrons. To complete the octet of the central atom we must
write double or triple bonds between the central atom and the surrounding atoms. To make a double bond,
move one of the unshared pairs from the surrounding atom to make the additional bond.
Repeat steps 4 and 5.
For application of this procedure see Example 9.5 and 9.6. _______________________________________________________________________________
EXAMPLE 9.5 Drawing a Lewis Structure
Draw the Lewis structure for hydrazine, N2 H4 . How many unshared electron pairs (lone pairs) are there on each
•Method of Solution
1. Arrange the atoms in a reasonable skeletal form. H atoms form only one bond and so must be located on
the outside of the atom.
HH 2. Count the valence electrons. Each N atom has 5 valence electrons and each H atom has 1. There are 2(5) +
4(1) = 14 valence electrons. 3. Connect the atoms with single bonds:
Normally we would add unshared pairs to complete all octets of surrounding atoms, but in this case the H
atoms only need the two electrons shared in the bond to the N atom. Counting the number of electrons
used; 5 pairs = 10 valence electrons. Now add unshared pairs to complete the octets of the N atoms. .. .. H—N—N—H |
4. | H H Count the electrons: 7 pairs = 14 valence electrons. This is the same number as given in step 2. _______________________________________________________________________________ EXERCISES
7. 8. Back How many lone pairs are on the underlined atoms in the following compounds?
a. P H3
b . S Cl 2 c . H 2 C O
Write the Lewis structures for the following species.
b . NCl 3 c . CF 2 Cl 2 Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 73 ELECTRONEGATIVITY
3. Describe the trends in electronegativity within the periodic table.
Explain what is meant by a polar bond and predict the relative polarity of bonds.
Classify bonds in given substances as ionic, polar covalent, or covalent. Electronegativity. Chemical bonds are rarely purely covalent or completely ionic. Rather, most bonds
exhibit some characteristics of both. In the previous chapter we saw that atoms of the elements exhibit varying
tendencies in their ability to attract and hold free electrons in the gas phase. In other words, electron affinity
values show periodic variations. The term e lectronegativity is used to describe the ability of an atom within
a molecule to attract a shared electron pair toward itself.
Linus Pauling developed a method for determining the relative electronegativities of the elements. These
values are given in Figure 9.8 (text). Pauling assigned the value 1.0 to Li and 4.0 to F. The values for second
and third row elements are given in Table 9.2. Electronegativity values exhibit periodic behavior. In general,
electronegativities increase from left to right across a period, and decrease within a group from top to bottom as
shown in Figure 9.3.
Table 9.2 Electronegativities of Second- and Third-Row Representative Elements.
Second Row Li
4.0 Third Row Na
3.0 Decreasing Electronegativity Increasing Electronegativity Figure 9.3 Periodic trends in electronegativity. As a consequence of the differing abilities of atoms in a bond to attract the shared electron pair, most
electron pairs are not shared equally. This imbalance causes an electron pair to shift slightly toward the more
electronegative atom, giving rise to a polar covalent bond. In the HCl molecule, for instance, the
electronegativity χ of Cl is 3.0, and for H it is 2.1. The electronegativity difference ∆χ is:
∆χ = χCl – χH
= 3.0 – 2.1 = 0.9 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 74 / Chemical Bonding I: Basic Concepts
The chlorine atom with its higher electronegativity attracts the electron pair more strongly. This makes the Cl
atom slightly negative and H slightly positive.
Here δ denotes a partial charge, that is, a charge less than 1.0, as it would be in an ion.
Pure covalent bonding, which is the equal sharing of electron pairs, occurs only in homonuclear diatomic
molecules. Examples are H2 , N 2 , and Cl 2 . In a diatomic molecule with both atoms the same, ∆χ must be zero,
and the bonding electron pair is shared equally. Bonds of this type are described as n onpolar covalent
bonds , or p ure covalent bonds.
In bonds involving different atoms, the electronegativity difference will depend on the individual
electronegativities. Bonds between atoms such that ∆χ < 2.0 are classified as p olar covalent bonds , or
simply polar bonds. When ∆χ ≥ 2.0 a bond is mostly i onic , for in this case one atom so outdoes the other at
attracting electron pairs that electrons can be considered to be completely transferred to the more electronegative
Here we see that nonpolar covalent bonds and completely ionic bonds represent extreme situations in
bonding. To refer to a bond as being "ionic" or "covalent" is an oversimplification. Sometimes the term percent
ionic character is used to describe the polar nature of a bond. A pure covalent bond has zero percent ionic
character, while a purely ionic bond would be described as having 100% ionic character. Bonds between Group
1A or 2A metals and the halogens are classified as ionic because of the large electronegativity differences. See
Table 9.3 for examples.
Table 9.3 Bond Character of Some Common Bonds
Ionic Compounds ∆χ
Polar Covalent Bonds
EXAMPLE 9.6 Bond Polarity
Arranging the following bonds in order of increasing ionic character: C—O, C—H, and O—H.
•Method of Solution
As the electronegativity difference increases the bond becomes more polar and its ionic character increases. Using
Figure 9.8 (textbook), we can determine the differences.
for C—O ∆χ = 3.5 – 2.5 = 1.0 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 75
for C—H ∆χ = 2.5 – 2.1 = 0.4
for O—H ∆χ = 3.5 – 2.1 = 1.4
The ionic character increases in the order
C—H < C—O < O—H
The electronegativity of hydrogen is unlike that of the other elements of Group 1A. In terms of
electronegativity, hydrogen is similar to the nonmetal elements boron and carbon. Bonds of H to nonmetal
atoms are polar covalent bonds, rather than ionic bonds, such as the bonds in LiCl and NaCl.
EXAMPLE 9.7 Electronegativity Trends
Using the trends within the periodic table, determine which of the following is the most electronegative
element: As, Se, or S.
•Method of Solution
Se and As are in the same period, and so the one further to the right has the higher electronegativity. That one is
Se. Now compare S and Se. They are in the same group. The one nearer the top of the group has the greater
electronegativity which is sulfur.
EXAMPLE 9.8 Types of Bonds
For the following pairs of elements label all bonds between them as ionic, polar covalent, or pure covalent.
a. Rb and Br
b. S and S
c. C and N
•Method of Solution
c. Bond type depends on electronegativity differences. The value of ∆χ for a Rb—Br bond is 2.0. Answer:
RbBr is ionic.
For S—S, ∆χ = 0.0, and so the bond is a pure covalent bond.
In the periodic table carbon and nitrogen are adjacent to each other in period 2. The one on the right is N; it
is more electronegative and so we expect a polar covalent bond. Electronegativity values give ∆χ = 0.5. _______________________________________________________________________________ EXERCISES
9. Which atom is the most electronegative?
Li Cs P As
10. List the following bonds in order of increasing ionic character:
N—O, Na—O, O—O, S—O
11. Classify the O—H bond in CH3 OH as ionic, polar covalent, or nonpolar covalent. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 76 / Chemical Bonding I: Basic Concepts
12. Which of the following is a nonpolar covalent bond (pure covalent)?
H—Cl Li—Br Se—Br Br—Br
13. Classify the following bonds as ionic, polar covalent, or nonpolar covalent.
Se—Cl Al—Cl K—F Cl—Cl FORMAL CHARGE
2. Define the formal charge of an atom in a molecule.
Determine the formal charges of atoms in a Lewis structure. Formal Charges on Lewis Structures. The concept of formal charge provides a rational basis for
choosing the more plausible Lewis structure from among several possibilities. The f ormal charge is the
charge that an atom seems to have in a Lewis structure. When determining the formal charge all nonbonding
electrons count as belonging entirely to the atom in which they are found. All bonding electrons are divided
equally between the bonded atoms. Thus the formal charge of an atom in a Lewis structure is the number of
valence electrons in an isolated atom minus the number of electrons assigned to that atom in a molecule. The
formula for the formal charge of an atom is
charge = n umber of – n umber of
– valence electrons nonbonding electrons 1
2 n umber of bonding electrons It will be good to keep in mind that the formal charge is really more a property of a structural formula than that
of the species the formula represents. Formal charges do not indicate actual charges on atoms in the real
Two possible Lewis structures for BF3 are ..
: F —B—F :
˙˙ | ˙˙
: F —B F
˙˙ | ˙˙
:F: and (1) ˙˙
(2) The formal charges in (1) are:
The boron atom:
The fluorine atom: formal charge = 3 – 0 – 1/2 (6) = 0
formal charge = 7 – 6 – 1/2 (2) = 0 The formal charges in (2) are:
The boron atom:
The fluorine atom (double bonded):
The other fluorine atoms: Back Forward Main Menu TOC formal charge = 3 – 0 – 1/2 (8) = –1
formal charge = 7 – 4 – 1/2 (4) = +1
formal charge = 7 – 6 – 1/2 (2) = 0 Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 77
The rule that is used to establish the more plausible structure is: A Lewis structure in which there are no formal
charges is preferred over one where formal charges are present. Therefore structure (1) is preferred over structure
When formal charges are assigned in a Lewis structure, the sum of the formal charges must be zero in a
neutral molecule. For a polyatomic ion the formal charges must add up to the charge of the ion. For the chlorite
ion (ClO 2 ), for instance: .. .. . .
[:O—Cl —O:] –
˙˙ ˙˙ ˙˙
the sum of the formal charges is –1: .– .. ..
[:O—Cl —O:] –
˙˙ ˙˙ ˙˙
The example brings out another feature of the formal charge concept. The most plausible Lewis structures
will be those with negative formal charges on the more electronegative atoms. Since oxygen is more
electronegative than chlorine, the formal negative charges on oxygen in this structure are more reasonable than
in some other structure that would place a positive charge on the oxygen atoms. _____________________________________________________________________
EXAMPLE 9.9 Assigning Formal Charges
Assign formal charges to the atoms in the following Lewis structures: a. :C O: .. .. . .
b. O S —O:
•Method of Solution
The formula used to calculate the formal charge of an atom is:
charge = n umber of – n umber of
– valence electrons nonbonding electrons 1
2 n umber of bonding electrons a. The carbon atom:
The oxygen atom: formal charge = 4 – 2 – 1/2 (6) = –1
formal charge = 6 – 2 – 1/2 (6) = +1 b. The sulfur atom:
formal charge = 6 – 2 – 1/2 (6) = +1
The oxygen atom on the right: formal charge = 6 – 6 – 1/2 (2) = –1
The oxygen atom on the left:
formal charge = 6 – 4 – 1/2 (4) = 0 •Comment
Some chemists don't approve of the CO structure given in part (a) because it places a positive formal charge on
the more electronegative oxygen atom. ____________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 78 / Chemical Bonding I: Basic Concepts
14. Assign formal charges to the atoms in the following Lewis structures: .. .. .. .. b. O a. H—N—H
H S O ˙˙ ˙˙ 15. Assign formal charges to the atoms in the following possible Lewis structures.
C—O a. : O : ˙˙ ..
˙˙ Be ..
˙˙ THE CONCEPT OF RESONANCE
2. Draw resonance structures for molecules or polyatomic ions.
Determine the formal charges on atoms in a resonance structure. Resonance Structures. It sometimes happens that a satisfactory electron dot structure for a molecule or
polyatomic ion cannot be drawn. When such a situation arises a special procedure is invoked to arrive at a Lewis
structure. For the nitrite ion, for instance, the following structure shows the correct number of valence electrons
and satisfies the octet rule. The brackets are used to indicate that the –1 charge belongs to the entire nitrite ion,
and not to just one atom in the structure. . . .. ..
O] – ˙˙ ˙˙ [:O—N However, the structure does not accurately represent what is known about the bond lengths of the N—O bonds
in NO2 . Both bond lengths are known to be the same, whereas according to the structure we expect the double
bond to be shorter than the single bond. An N—O single bond length should be about 136 pm, and an N O
double bond length about 122 pm. However, the two bond lengths in NO2 are equal and are intermediate
between these two values.
It turns out to be impossible to draw a single satisfactory Lewis structure for NO2 . Chemists have gotten
around this problem by the concept of resonance. First draw two structures for NO2 that reflect different choices
of electron arrangements. .. .. [:O—N
˙˙ .. –
.. .. . .
O] ←→ [O N—O:] –
˙˙ The nitrite ion is not adequately represented by either structure; but it may be described by a composite of these
structures. This composite structure cannot be drawn by the rules for writing Lewis structures. Each of the
structures that contributes to the composite structure is called a c ontributing or r esonance structure . The
symbol ‚ ←→ indicates that the structures shown are resonance structures. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 79
– In applying the concept of resonance we assume that NO2 is a composite or an average of the two
structures. Thus, the N—O bonds are intermediate between single and double bonds. The term resonance was
perhaps a poor choice because the word implies to some that the real molecule flips from one structure to the
other. This is not what is implied here. The point is that the properties of NO2 (and of other examples) cannot
be accounted for by a single Lewis structure. The term r esonance means the need for two or more Lewis
stuctures to represent a particular molecule.
These structures have the same placement of atoms but different positions of electron pairs. The electron
pairs do not actually move from one position to another within the molecule. Therefore resonance structures are
generated by moving electron pairs within the same skeletal structure.
EXAMPLE 9.10 Drawing a Lewis Structure with Resonance
Draw three resonance structures for N2 O. The skeletal structure is N—N—O.
•Method of Solution
1. Arrange the atoms
N—N—O 2. Count the valence electrons.
2(5) + 6 = 16 electrons 3. Add unshared pairs to the terminal atoms: ..
4. Count the electron pairs used. 8 electron pairs = 16 electrons. There are not enough electrons to add any to
the central atom. 5. Move one of the unshared pairs from the terminal N atom to make a double bond, and move an unshared
pair from the terminal O atom to make a double bond. ..
N O: This is a satisfactory structure because it completes the octets of all three atoms and uses the correct number
of electrons. Additional resonance structures can be generated by moving electron pairs in such a way that the
octet rule for each atom is always satisfied. The positions of the atoms in N2 O cannot be altered.
To possibilites are ..
:N N—O : and ˙˙ ..
˙˙ O: •Comment
The three structures for N2 O are called resonance structures. The actual bonding in N 2 O is a composite of these
16. Draw resonance structures for nitric acid: H—O—N—O
O Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 80 / Chemical Bonding I: Basic Concepts EXCEPTIONS TO THE OCTET RULE
2. Describe three types of molecules that are exceptions to the octet rule.
Draw Lewis structures for molecules that do not follow the octet rule. Types of Exceptions. Lewis structures can be drawn for many compounds with the aid of the octet rule;
however, structures of some compounds do not follow the rule. The text points out three types of molecules
that are exceptions to the octet rule:
• Molecules that have an incomplete octet.
• Molecules with an odd number of electrons.
• Molecules in which the central atom has an expanded octet.
The boron halides BX3 are well-known examples of molecules with an i ncomplete octet . They are all
planar molecules in which the boron atom has only six valence electrons. The boron atom has incomplete octet.
This situation is typical in boron chemistry. ..
.. | . .
: F—B—F :
Normally when there is a shortage of electrons, we can draw a double-bonded structure as shown below for BF3 . ..
˙˙ F : ˙˙ Two common oxides of nitrogen, NO and NO2 , have o dd numbers of electrons . Since an even
number of electrons is required for complete pairing, the octet rule cannot be satisfied. Two additional oddelectron molecules that are known to exist in our atmosphere for very short periods of time are OH (hydroxyl
radical) and HO2 (hydroperoxyl radical). ..
˙˙ . . ..
H—O—O· ˙˙ ˙˙ However, experiments indicate that each B—F bond is a single bond as shown in the first dot structure.
Also, the assignment of formal charges indicates that a structure with a double bond would have adjacent formal
charges. As was discussed above, this is unfavorable. In addition, the more electronegative F atom would have a
positive charge, rather than the preferred negative charge for fluorine.
Molecules exhibiting an e xpanded octet (having more than eight valence electrons) require the presence
of nonmetal atoms from the third period or beyond in the periodic table. Second period elements never exceed the
octet rule. Third-period elements are just as likely to exceed the octet rule as they are to follow it. Where for
example, PCl3 obeys the octet rule, gaseous PCl5 has a phosphorus atom that is joined by single bonds to five Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 81
chlorine atoms. The phosphorus atom has 10 electrons in its valence shell. This is called an expanded octet.
Cl Cl P Cl Cl The central atoms in SF4 and SF 6 , and in the interhalogen compounds ClF5 , BrF 5 , and IF7 exhibit expanded
In Chapter 10, you will need to draw a number of dot structures of molecules exhibiting expanded octets.
When the central atom is from the third period or beyond, complete the octets of the surrounding atoms first,
and then complete the central atom. If extra electron pairs remain, place them on the central atom. See Example
These exceptions seem to be telling us that an atom with a completed octet is not necessary for covalent
bonding to occur. Within the Lewis framework, it is really the sharing of electron pairs that leads to the
covalent bond. A shared pair of electrons acts to attract both atoms.
EXAMPLE 9.11 Exceptions to the Octet Rule
Draw Lewis structures for:
b. NO2 (all bonds are equivalent)
c. ClF 3
•Method of Solution
a. Gallium is in Group 3A of the periodic table, a group well known for its electron deficient elements.
GaI3 has 3 + 3(7) = 24 valence electrons. ..
. . | ..
: I—Ga—I :
This structure shows 24 electrons, the correct number. Ga, with three electron pairs, is electron deficient.
b. NO2 has 17 valence electrons. With an odd number of electrons, it cannot obey the octet rule. The best we
can do is start with 18 valence electrons as in NO2 and then remove one from the nitrogen atom (because it
is the unique atom). Two contributing structures are necessary. ..
c. . .. N—O : ←→ ˙˙ .. . ..
: O—N O
˙˙ ClF 3 has 28 valence electrons. 26 electrons are required to complete the octets of the four atoms. The
remaining two electrons are placed on the central Cl atom because chlorine is in the third period, and has
vacant 3d orbitals that can hold electrons in addition to an octet. Chlorine is said to have an expanded octet. . . .. ..
: F—Cl—F :
˙˙ | ˙˙
_______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 82 / Chemical Bonding I: Basic Concepts EXERCISE
17. Write Lewis structures for the following molecules:
b . SF 4 c. NO BOND DISSOCIATION ENERGY
2. Determine the average bond energy of a bond using given enthalpies of reaction.
Use a table of bond energies to estimate ∆ H for a reaction. Average Bond Energies. Occasionally the enthalpy change ∆ H is needed for a reaction for which
enthalpy of formation data does not exist. One approach that allows an estimate of ∆ H° uses the concept of bond
energy. Consider the gas phase atomization process:
o CH4 (g) → C(g) + 4H(g) ∆ Hrxn = 1664 kJ
In this reaction, four C—H bonds are broken; therefore we can define the average C—H bond energy as oneo
fourth of ∆ Hrxn for the reaction. Hence the average C—H bond energy in CH4 is 416 kJ/mol.
The actual bond energies of the individual C—H bonds in CH4 are not the same as the average value. Even
so, the use of average bond energies makes it possible to estimate the enthalpy changes of certain reactions.
Some of the bond energies in Table 9.4 of the textbook are listed here in Table 9.4.
Table 9.4 Bond Energies of Some Common Bonds
Bond Energy (kJ/mol)
We can estimate ∆ H for the combustion of methane as follows:
CH4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(g)
CH4 + 2O 2 ∑BE (reactants) C + 4H + 4O – ∑BE (products) CO2 + 2H2 O
First we break 4 C—H bonds and 2 O
O bonds, where ∑BE (reactants) is the sum of the bond energies of
bonds broken. Then let the atoms recombine to form the 2 C
O and 4 O—H bonds of the products. ∑BE Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 83
(products) is the sum of the bond energies of all bonds formed from the atoms. Keep in mind that breaking
bonds is an endothermic process, and making new bonds is an exothermic process. Therefore
∆ H = ∑BE (reactants) – ∑BE (products)
∆ H = [4 BE (C—H) + 2 BE (O O)] – [2 BE (C O) + 4 BE (O—H)] Inserting values from Table 9.5 gives:
∆ H = [4(414) + 2(499)] – [2(802) + 4(460)]
= 2654 kJ – 3444 kJ
∆ H = –790 kJ
If we compare this value with the standard enthalpy of combustion of methane, we get
CH4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) o ∆ Hrxn = –890 kJ At first it seems there is a large discrepancy of 100 kJ. However, in our calculation, H2 O is present as a gas!
But ∆ H° refers to the standard state of H2 O as a liquid. Therefore, the 100 kJ difference is largely due to the ∆ H
of vaporization of two moles of H 2 O, which is 81.4 kJ. Two important conclusions can be drawn here;
(1) Bond energies are used to estimate enthalpies of reaction of gas phase reactions, and (2) enthalpies of reaction
calculated from average bond energies are only approximate values.
EXAMPLE 9.12 U se of Bond Energies to Estimate ∆ Hrxn
Estimate ∆ Hrxn for the reaction:
Cl 2 (g) + I 2 (g) → 2ICl(g)
using bond energies given in Table 9.4 (text), and given that BE (I—Cl) = 210 kJ/mol.
•Method of Solution
∆ H = ∑BE (reactants) – ∑BE (products)
∑BE (reactants) = BE (Cl—Cl) + BE (I—I)
= 243 + 151 = 394 kJ
∑ BE (products) = 2BE (I—Cl) = 2(210) = 420 kJ
∆ H = 394 kJ – 420 kJ
= –26 kJ
18. Given the N N and H—H bond energies in Table 9.5 and standard enthalpy of reaction: 1
N 2 (g) + 2 H 2 (g) → NH3 (g)
2 o ∆ Hrxn = –46.3 kJ/mol calculate the average N—H bond energy in ammonia. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 84 / Chemical Bonding I: Basic Concepts
19. Given the bond energies in Table 9.5, calculate the enthalpy of formation for ammonia, NH 3 .
N 2 (g) + 2 H 2 (g) → NH3 (g)
2 ∆ Hf = ? _______________________________________________________________________________ CONCEPTUAL QUESTIONS
1. Explain the difference between electron affinity and electronegativity. 2. Do formal charges represent actual charges? What do they mean? 3. Do resonance structures actually exist? Describe why the concept of resonance is needed. PRACTICE TEST
1. Arrange the following ionic compounds in order of increasing lattice energy: RbI, MgO, CaBr2 .
2. Which compound in each pair has the higher melting point?
a. RbI or KI
b. MgCl2 or MgBr2
3. Draw Lewis dot structures for
b. CS 2
e. CO 2–
f. P2 Cl 4 g. SO4 d. SO2
h. PO4 4. Draw resonance structures for SO3 .
5. Which of the following elements is the most electronegative?
a. Ca b. P c. As d. K
6. Which one of the following is the most polar bond?
a. B—C b. B—N c. B—S d. C—C
7. Which bond has the the greatest percent ionic character?
a. CO b. SO c. NaI d. NaBr
8. Which molecule has a Lewis structure that does not obey the octet rule?
a. NO b. PF3 c . CS 2 d. HCN e. BF4
9. Which molecule has covalent bonding between atoms?
a. NaF b. K2 O c. ICl d. SrI2 e. none of these
10. Indicate the molecule in each pair that does not follow the octet rule.
a. PCl5 or AsCl3 b. BCl 3 or CCl4 c. NH3 or CH 3 d. SbCl 5 or NCl 3 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 85
11. Consider the following Lewis structures for sulfate ion. Which is the more reasonable structure in terms of
formal changes? .. a. : O: ..
.. S 2– .. b. : O: ..
. S : .. :
O 2– ..
.. : .. :
O 12. Assign formal changes to the atoms in the Lewis structures shown. a. ..
O: N b. ..
˙˙ c. .. ..
[: O—Cl :]–
˙˙ ˙˙ 13. Write a mathematical expression that would allow you to calculate the lattice energy of CaCl2 using Hess's
14. Calculate the N
kJ/mol. N bond energy, given that the standard enthalpy of formation of atomic nitrogen is 473 15. Estimate the enthalpy change for the following reaction
N2 (g) + O 2 (g) → 2NO(g)
Hint: NO has a double bond. BE (N O) = 630 kJ ANSWERS
1. a. ·Mg· 2. a. K + b. :Se:
˙ .. .
c. ·Al· .. .. d. :Br·
˙˙ e. :Xe:
˙˙ .. b . :S: 2– ˙˙ .. c. :N:3– d. : I :– ˙˙ ˙˙ 3. MgO(s)→ Mg 2+ (g) + O 2– (g)
7. Back e. Sr2+
∆ H = lattice energy a. CaCl2
a. 1 b. 2 c. 2 Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 86 / Chemical Bonding I: Basic Concepts
.8. + H | .. a. H —N—H
H ˙˙ : Cl :
.. | ..
b. : Cl—N—Cl :
c. : F—C—F :
˙˙ | ˙˙
: Cl :
: Cl : ˙˙
13. ˙˙ P
O—O < S—O < N—O < Na—O
polar covalent, ionic, ionic, nonpolar covalent 14. a. The formal charges in NH3 are: The N atom: 0.
The H atoms: 0.
The S atom: 0.
The O atoms: 0. b. The formal charges in SO 2 are:
15. a. O –
C—O .. +
b. F 2–
F .. .. 16. H—O—N O
:O: ←→ .. H—O—N—O :
: O: ˙˙
17. a. ..
: Br—Al—Br :
˙˙ | ˙˙
: Br :
˙˙ b. .. | . .
: F—S—F :
˙ ˙ | ˙˙
˙˙ c. ..
O ˙˙ 18. 391 kJ
19. –50.7 kJ Conceptual Questions
3. Back Electron affinity is a property of a gaseous atom, while electronegativity is a property of an atom within a
molecule. Both have to do with attracting electrons but they are very different properties. Electron affinity is
related to the energy released by an uncombined atom in the gaseous state when it adds an additional
electron. Electronegativity is a measure of the ability of an atom within a molecule to attract an electron
pair in a bond it has formed.
Formal charges do not represent actual charges of the atoms in a molecule. However they do give an
indication of the sign (+ or –), if not the magnitude, of the charge and we use formal charges to choose
from several possible Lewis structures.
When a molecule cannot be accurately represented by a single Lewis structure we resort to resonance
structures. These structures have the same placement of atoms but different positions of electron pairs. The
electron pairs do not actually move from position to position within the molecule. Therefore none of the
resonance structures actually exist. Rather the real molecule is said to be a composite of the resonance
structures. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding I: Basic Concepts / 1 87
2. RbI > CaBr2 > MgO
b. MgCl 2 3. a. : I—Br : b. S .. .. .. .. C ˙˙ ˙˙ S ˙˙ ˙˙ .. ..
˙˙ O Cl
c . Cl Cl P Cl Cl .. .. . .
S—O: ←→ d. O ˙˙ ˙˙ O: e. : C : O: ..
O S ˙˙
: Cl : : Cl :
: P—P :
: Cl : : Cl :
˙˙ f. .. g. .. 2– ..
.. O ˙˙
11. P 3– ..
.. : : .. :
.. : O: ..
.. : O:
4. .. h. :O: | ..
S—O: ←→ ˙˙ .. || ..
:O—S—O: ←→ :O—S
O ˙˙ b.
b. BCl 3 c. CH 3
d. SbCl 5
b. Because fewer formal charges are present. Particularly the formal charge on sulfur is zero compared to
+2 in a. _
.. 12. a. :N +
N O: ˙˙ _
b. [:O—H ]–
c. [:O—Cl :]–
o 13. Lattice energy = –∆ Hf (CaCl 2 ) + ∆ Hsub (Ca) + first ionization energy (Ca) + second ionization energy(Ca)
+ bond dissociation energy (Cl2 ) + 2 (–electron affinity (Cl))
14. 946 kJ
15. 180 kJ
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