compression guide chapter 6 chemical bonding planning · pdf filete lesson starter ionic vs....

Planning Guide

PACING • 45 min pp. 175 –177 TR 34 Predicting Bond Character from Electronegativity Differences

TE Lesson Starter, p. 175 g TE Reading Skill Builder Brainstorming, p. 175 bSection 1 Introduction to Chemical Bonding

PACING • 90 min pp. 178 –189Section 2 Covalent Bonding and Molecular Compounds

TR 35 Formation of a Covalent Bond TR 36 Bond Length and Stability TR 32A Lewis Structures of the 2nd Period

Elements TR 33A Electron-Dot Notation TR 34A Drawing Lewis Structures with Many

AtomsEXT Graphing Calculator Classifying Bond Type,

p. 213

TE Lesson Starter, p. 175 g TE Reading Skill Builder Interpreting Graphic Sources of

Information, p. 179 b SE Chemistry in Action Ultrasonic Toxic-Waste Destroyer,

p. 180 TE Discussion Ultrasonic Toxic-Waste Destroyer, p. 180

g

ANC Microscale Experiments Conductivity as an Indicator of Bond Type ◆ g

174A Chapter 6 Chemical Bonding

SECTIONS LABS, DEMONSTRATIONS, AND ACTIVITIES TECHNOLOGY RESOURCES

PACING • 90 min pp. 190 –194Section 3 Ionic Bonding and Ionic Compounds

TR 37 Ionic vs. Covalent Bonding TR 38 NaCl and CsCl Crystal Lattice TR 37A How to Identify a Compound as IonicEXT Chemistry In Action Nanoscale Computers,

p. 194

TE Lesson Starter Ionic vs. Covalent Bonding, p. 190 ◆ g

TE Reading Skill-Builder Comparing and Contrasting, p. 193 b

ANC Microscale Experiments Chemical Bonds ◆ g

Chemical Bonding To shorten your instruction because of time limitations, omit Section 4.

Compression Guide

CHAPTER 6

PACING • 90 minSection 4 Metallic Bonding

pp. 195 –196 TR 39 Properties of Substances with Metallic, Ionic, and Covalent Bonds

TE Lesson Starter Malleability vs. Brittleness, p. 195 ◆ g

SE Chapter Lab Types of Bonding in Solids, pp. 216 – 217 ◆ INQUIRY

ANC Datasheets for In-Text Labs

PACING • 90 minSection 5 Molecular Geometry

pp. 197 – 207 TR 42 Molecular Shape Affects Polarity TR 43 VSEPR Models of Ammonia and Water TR 44 Geometry of Hybrid Orbitals TR 45 Predicting Polarity TR 46 Hydrogen Bonding Between Water

Molecules

TE Reading Skill-Builder Interpreting Vocabulary, p. 203 b

SE Careers in Chemistry Computational Chemist, p. 204

SE Chapter Highlights, p. 208 SE Chapter Review, pp. 209 – 213 SE Math Tutor, p. 214 SE Standardized Test Prep, p. 215ANC Chapter Test A gANC Chapter Test B a OSP Test GeneratorANC Test Item ListingOSP Scoring Rubrics and Classroom Management

Checklists

CHAPTER REVIEW, ASSESSMENT, AND STANDARDIZED TEST PREPARATION

PACING • 90 minOnline and Technology ResourcesOnline and Technology Resources

Visit go.hrw.com to find a variety of online resources. To access this chapter’s extensions, enter the keyword HC6BNDXT and click the “go” button. Click Holt Online Learning for an online edition of this textbook, and other interactive resources.

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National ScienceEducation StandardsPS 2a, 2c, 3cUCP 1– 2

SE Sample Problems and Practice Problems, pp. 184, 185, 188 g

TE Additional Sample Problems, pp. 184, 217A gEXT Additional Practice Problems g SE Math Tutor Drawing Lewis Structures, p. 214

TE Alternative Assessment, p. 180 bANC Section Review * SE Section Review, p. 189ANC Quiz *

PS 2a, 2cUCP 1– 2SPSP 1, 2, 5


PS 2a, 2cUCP 1– 2

Chapter 6 Planning Guide 174B

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SKILLS DEVELOPMENT RESOURCES REVIEW AND ASSESSMENT CORRELATIONS

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Topic: Covalent BondingSciLinks Code: HC60363Topic: Ionic BondingSciLinks Code: HC60815

Topic: Metallic BondingSciLinks Code: HC60944Topic: VSEPRSciLinks Code: HC61621

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ClassroomTechnology

Chemical Bonding

Chapter OverviewSection 1 defines chemicalbonding and uses electronega-tivity values to contrast polar-covalent, nonpolar-covalent,and ionic bonding.

Section 2 covers the charac-teristics of covalent bonding,including the relationshipbetween bond length and bondstrength and the use of Lewisstructures.

Section 3 covers the charac-teristics of ionic bonding.

Section 4 covers the charac-teristics of metallic bondingand the resulting properties of metals.

Section 5 covers theories of molecular geometry, and discusses how intermolecularattraction is affected by molecular geometry.

Concept BaseStudents may need a review of the following concepts:

• boiling points, Chapter 1

• electron configurations andorbital notation, Chapter 4

• ion formation and electronegativity, Chapter 5

CHAPTER 6

174

Chemical BondingIn nature, most atoms are joined

to other atoms by chemical bonds.

C H A P T E R 6

Computer Graphic Representation of a Molecule of Adenosine Monophosphate

C H E M I C A L B O N D I N G 175

Lesson StarterHave students imagine getting onto acrowded elevator. As people squeezeinto the confined space, they come incontact with each other. Many peoplewill experience a sense of being tooclose together. Likewise, when atomsget close enough, their outer elec-trons repel each other. At the sametime, however, each atom’s outerelectrons are strongly attracted to thenuclei of the surrounding atoms. Thedegree to which these outer electronsare attracted to other atoms deter-mines the kind of chemical bondingthat occurs between the atoms. Havestudents consider how many elec-trons exist in the outer energy levelsof the atoms of various elements.

GENERAL

SECTION 1

175

SECTION 1

OBJECTIVES

Define chemical bond.

Explain why most atoms form chemical bonds.

Describe ionic and covalentbonding.

Explain why most chemicalbonding is neither purely ionic nor purely covalent.

Classify bonding type according to electronegativitydifferences.

Introduction toChemical Bonding

A toms seldom exist as independent particles in nature. The oxygenyou breathe, the water you drink, and nearly all other substances con-sists of combinations of atoms that are held together by chemical bonds.A chemical bond is a mutual electrical attraction between the nuclei andvalence electrons of different atoms that binds the atoms together.

Why are most atoms chemically bonded to each other? As indepen-dent particles, most atoms are at relatively high potential energy.Nature, however, favors arrangements in which potential energy is min-imized. This means that most atoms are less stable existing by them-selves than when they are combined. By bonding with each other, atomsdecrease in potential energy, thereby creating more stable arrange-ments of matter.

Types of Chemical Bonding

When atoms bond, their valence electrons are redistributed in ways thatmake the atoms more stable. The way in which the electrons are redis-tributed determines the type of bonding. In Chapter 5, you read thatmain-group metals tend to lose electrons to form positive ions, orcations, and nonmetals tend to gain electrons to form negative ions, oranions. Chemical bonding that results from the electrical attractionbetween cations and anions is called ionic bonding. In purely ionic bond-ing, atoms completely give up electrons to other atoms, as illustrated inFigure 1 on the next page. In contrast to atoms joined by ionic bonding,atoms joined by covalent bonding share electrons. Covalent bondingresults from the sharing of electron pairs between two atoms (seeFigure 1). In a purely covalent bond, the shared electrons are “owned”equally by the two bonded atoms.

Ionic or Covalent?Bonding between atoms of different elements is rarely purely ionic orpurely covalent. It usually falls somewhere between these two extremes,depending on how strongly the atoms of each element attract electrons.Recall that electronegativity is a measure of an atom’s ability to attractelectrons. The degree to which bonding between atoms of two elementsis ionic or covalent can be estimated by calculating the difference in

Reading Skill Builder

BRAINSTORMING Explain thedifferent types of bonding coveredin this chapter. Then have studentswork in pairs to explain what ishappening in terms of bonding.Have them use Figures 1 and17 to compare ionic and covalentbonding; Figure 3 to compare nonpolar-covalent and polar-covalent bonding; Figures 4–6to explain covalent bonding;Figures 13, 14, and 16 to explainionic bonding; and Figure 18 formetallic bonding.

the elements’ electronegativities (see Figure 2). For example, the electronegativity difference between fluorine, F, and cesium, Cs, is 4.0 −0.7 = 3.3. (See Figure 20 on page 161 for a periodic table of electroneg-ativity values.) So, according to Figure 2, cesium-fluorine bonding is ionic.Fluorine atoms, which are highly electronegative, gain valence electrons,causing the atoms to become anions. Cesium atoms, which are less elec-tronegative, lose valence electrons, causing the atoms to become cations.

Bonding between atoms with an electronegativity difference of 1.7or less has an ionic character of 50% or less.These compounds are typ-ically classified as covalent. Bonding between two atoms of the sameelement is completely covalent. Hydrogen, for example, exists innature not as isolated atoms, but as pairs of atoms held together bycovalent bonds. The hydrogen-hydrogen bond is a nonpolar-covalentbond, a covalent bond in which the bonding electrons are shared equal-ly by the bonded atoms, resulting in a balanced distribution of electri-cal charge. Bonds having 0% to 5% ionic character, corresponding toelectronegativity differences of roughly 0 to 0.3, are generally consid-ered nonpolar-covalent bonds. In bonds with significantly differentelectronegativities, the electrons are more strongly attracted by themore-electronegative atom. Such bonds are polar, meaning that theyhave an uneven distribution of charge. Covalent bonds having 5% to50% ionic character, corresponding to electronegativity differences of0.3 to 1.7, are classified as polar. A polar-covalent bond is a covalentbond in which the bonded atoms have an unequal attraction for theshared electrons.

Nonpolar- and polar-covalent bonds are compared in Figure 3, whichillustrates the electron density distribution in hydrogen-hydrogen andhydrogen-chlorine bonds. The electronegativity difference betweenchlorine and hydrogen is 3.0 − 2.1 = 0.9, indicating a polar-covalentbond. The electrons in this bond are closer to the more-electronegativechlorine atom than to the hydrogen atom, as indicated in Figure 3b.Consequently, the chlorine end of the bond has a partial negativecharge, indicated by the symbol δ−. The hydrogen end of the bond thenhas an equal partial positive charge, δ+.

C H A P T E R 6176

SECTION 1

176

SECTION 1IONIC BONDING

COVALENT BONDING

Atoms A Atoms B

Atom C Atom D

Electrons transferred from atoms A to atoms B

Electron pair sharedbetween atom C andatom D

+

+

Many atoms

Two atomsAtom C Atom D

Cation AAnion B

+ +

+++ +

+

–

–

–

––

– –

FIGURE 1 In ionic bonding,many atoms transfer electrons. Theresulting positive and negative ionscombine due to mutual electricalattraction. In covalent bonding,atoms share electron pairs to formindependent molecules.

FIGURE 2 Differences in elec-tronegativities reflect the character ofbonding between elements. The elec-tronegativity of the less-electronega-tive element is subtracted from that ofthe more-electronegative element. Thegreater the electronegativity differ-ence, the more ionic is the bonding.

Dif

fere

nce

in e

lect

ron

egat

ivit

ies

Percentag

e ion

ic character

Nonpolar-covalent

Polar-covalent

Ionic

3.3

1.7

0.3

0

100%

50%

5%

0%

Visual StrategyHave students recognize

the fundamental difference betweenionic and covalent bonding. In ionicbonding, large numbers of oppositelycharged ions come together becauseof electrical attraction. (Be sure stu-dents identify the cations and theanions.) In covalent bonding,two atoms join by sharing one ormore of their valence electrons. Theshared electrons are simultaneouslyattracted to both atomic nuclei.

Some students will notice thechanges in sizes for atoms A and Band ions A and B. Have them recallwhat they learned about atomic radiiand ionic radii in Chapter 5.

CommonMisconceptionElectronegativity difference is only ageneral guide for determining bondingtype. For example, the electronegativitydifference between boron and fluorineis 2.0. Yet scientists know throughexperimentation that boron trifluoride,BF3, is a covalently bonded compound.

FIGURE 1

GENERAL

www.scilinks.orgTopic: Covalent BondingCode: HC60363

C H E M I C A L B O N D I N G 177 177

SECTION 1FIGURE 3 Comparison of theelectron density in (a) a nonpolar,hydrogen-hydrogen bond and (b) apolar, hydrogen-chlorine bond.Because chlorine is more electroneg-ative than hydrogen, the electrondensity in the hydrogen-chlorinebond is greater around the chlorineatom.

Hydrogen nucleiHydrogennucleus

Chlorinenucleus

(a) Nonpolar-covalent bond (b) Polar-covalent bond

δ+ δ−

Go to go.hrw.com formore practice problemsthat ask you to classifybonds.

Keyword: HC6BNDX

From Figure 20 on page 161, we know that the electronegativity of sulfur is 2.5. The elec-tronegativities of hydrogen, cesium, and chlorine are 2.1, 0.7, and 3.0, respectively. In eachpair, the atom with the larger electronegativity will be the more-negative atom.

Bonding between Electronegativity More-negativesulfur and difference Bond type atomhydrogen 2.5 − 2.1 = 0.4 polar-covalent sulfurcesium 2.5 − 0.7 = 1.8 ionic sulfurchlorine 3.0 − 2.5 = 0.5 polar-covalent chlorine

SOLUTION

SAMPLE PROBLEM A

Use electronegativity differences and Figure 2 to classify bondingbetween chlorine, Cl, and the following elements: calcium, Ca; oxygen,O; and bromine, Br. Indicate the more-negative atom in each pair.

Answers in Appendix E

Use electronegativity differences and Figure 2 to classify bonding between sulfur, S, and the following elements: hydrogen, H; cesium, Cs; and chlorine, Cl. In each pair, which atom will be more negative?

PRACTICE

1. What is the main distinction between ionic andcovalent bonding?

2. How is electronegativity used in determining theionic or covalent character of the bondingbetween two elements?

3. What type of bonding would be expected betweenthe following atoms?a. Li and Fb. Cu and Sc. I and Br

4. List the three pairs of atoms referred to in the previous question in order of increasing ionic character of the bonding between them.

Critical Thinking

5. INTERPRETING CONCEPTS Compare the followingtwo pairs of atoms: Cu and Cl; I and Cl.a. Which pair would have a bond with a greater

percent ionic character?b. In which pair would Cl have the greater

negative charge?

6. INFERRING RELATIONSHIPS The isolated K atom is larger than the isolated Br atom.a. What type of bond is expected between K

and Br?b. Which ion in the compound KBr is larger?

SECTION REVIEW

Additional Sample Problem is found on page 217A.

Practice AnswersBonding between chlorine andcalciumoxygenbromine

Electronegativity difference3.0 − 1.0 = 2.03.5 − 3.0 = 0.53.0 − 2.8 = 0.2

Bond typeionicpolar-covalentnonpolar-covalent

More-negative atomchlorineoxygenchlorine

1. Ionic bonding involves the electri-cal attraction between large numbersof anions and cations. Covalentbonding involves the sharing of electron pairs between two atoms.

2. A large difference in electronega-tivity between two atoms in a bondwill result in ionic bonding. A smalldifference in electronegativitybetween two atoms will result in covalent bonding.

3. a. ionicb. polar-covalentc. polar-covalent

4. I and Br, Cu and S, Li and F

Answers are continued on page 217A.

SECTION REVIEW

ADDITIONALSAMPLEPROBLEM GENERAL

C H A P T E R 6178

SECTION 1

178

SECTION 2

Lesson StarterHave students consider what hap-pens if a soccer ball is rolled towarda ditch. The ball rolls down the nearslope of the ditch, passes the bottomof the ditch, and rolls part of the wayup the opposite slope of the ditch. Itthen rolls back down, passes the bot-tom, and rolls part of the way up thenear side of the ditch. The ball con-tinues moving in this pattern, eachtime rolling to a lesser height, until it eventually comes to rest in the bottom of the ditch. At this point,the ball’s gravitational potential ener-gy is at a minimum. The formation ofa covalent bond between two atomsis like the ball in the ditch. The atomsalternately attract and repel eachother until they reach a distance (orbond length) at which their potential energy is minimized.

Visual StrategyHave students carefully

examine the atoms involved in thesemolecules. Point out that all theatoms are nonmetals and that theoxygen and carbon atoms can eachform more than one bond. The abilityof atoms to form more than onecovalent bond allows the formationof extremely large molecules.

Did You Know?Molecules range in size from 75 pm to more than 23,000 pm.

FIGURE 4

GENERALSECTION 2 Covalent Bonding and

Molecular Compounds

M any chemical compounds, including most of the chemicals that are inliving things and are produced by living things, are composed of mol-ecules. A molecule is a neutral group of atoms that are held together bycovalent bonds. A single molecule of a chemical compound is an indi-vidual unit capable of existing on its own. It may consist of two or moreatoms of the same element, as in oxygen, or of two or more differentatoms, as in water or sugar (see Figure 4 below). A chemical compoundwhose simplest units are molecules is called a molecular compound.

The composition of a compound is given by its chemical formula. Achemical formula indicates the relative numbers of atoms of each kind ina chemical compound by using atomic symbols and numerical subscripts.The chemical formula of a molecular compound is referred to as a mo-lecular formula. A molecular formula shows the types and numbers ofatoms combined in a single molecule of a molecular compound. Themolecular formula for water, for example, is H2O, which reflects the factthat a single water molecule consists of one oxygen atom joined by sep-arate covalent bonds to two hydrogen atoms.A molecule of oxygen, O2,is an example of a diatomic molecule. A diatomic molecule is a moleculecontaining only two atoms.

OBJECTIVES

Define molecule and molecular formula.

Explain the relationshipsamong potential energy,distance between approach-ing atoms, bond length, andbond energy.

State the octet rule.

List the six basic steps used in writing Lewis structures.

Explain how to determineLewis structures for moleculescontaining single bonds,multiple bonds, or both.

Explain why scientists use resonance structures to represent some molecules.

(a) Water molecule,H2O

(c) Sucrose molecule,C12H22O11

(b) Oxygen molecule,O2

FIGURE 4 The models for (a) water, (b) oxygen, and (c) sucrose, or tablesugar, represent a few examples of the many molecular compounds in andaround us. Atoms within molecules may form one or more covalent bonds.

Formation of a Covalent Bond

As you read in Section 1, nature favors chemical bonding because mostatoms have lower potential energy when they are bonded to otheratoms than they have as they are independent particles. In the case ofcovalent bond formation, this idea is illustrated by a simple example, theformation of a hydrogen-hydrogen bond.

Picture two isolated hydrogen atoms separated by a distance largeenough to prevent them from influencing each other. At this distance,the overall potential energy of the atoms is arbitrarily set at zero, asshown in part (a) of Figure 5.

Now consider what happens if the hydrogen atoms approach eachother. Each atom has a nucleus containing a single positively chargedproton.The nucleus of each atom is surrounded by a negatively chargedelectron in a spherical 1s orbital. As the atoms near each other, theircharged particles begin to interact. As shown in Figure 6, the approach-ing nuclei and electrons are attracted to each other, which correspondsto a decrease in the total potential energy of the atoms. At the sametime, the two nuclei repel each other and the two electrons repel eachother, which results in an increase in potential energy.

The relative strength of attraction and repulsion between thecharged particles depends on the distance separating the atoms. Whenthe atoms first “sense” each other, the electron-proton attraction isstronger than the electron-electron and proton-proton repulsions. Thus,the atoms are drawn to each other and their potential energy is lowered,as shown in part (b) of Figure 5.

The attractive force continues to dominate and the total potentialenergy continues to decrease until, eventually, a distance is reached atwhich the repulsion between the like charges equals the attraction ofthe opposite charges. This is shown in part (c) of Figure 5. At this point,which is represented by the bottom of the valley in the curve, potentialenergy is at a minimum and a stable hydrogen molecule forms. A closerapproach of the atoms, shown in part (d) of Figure 5, results in a sharprise in potential energy as repulsion becomes increasingly greater thanattraction.


SECTION 2

Visual StrategyAsk students why the

potential energy of the atoms in-creases as the atoms get very closeto or far from each other. Have students consider the attractionbetween the electrons and the nuclei,as well as the repulsion between thetwo nuclei and between the twoelectrons.

FIGURE 5

GENERAL

Distance between hydrogen nuclei (pm)

Pote

nti

al e

ner

gy

(kJ/

mo

l)

−436

0

75

75 pm

(a)(b)

(c)

(d)

+

FIGURE 5 Potential energychanges during the formation of ahydrogen-hydrogen bond. (a) Theseparated hydrogen atoms do notaffect each other. (b) Potentialenergy decreases as the atoms aredrawn together by attractive forces.(c) Potential energy is at a minimumwhen attractive forces are balancedby repulsive forces. (d) Potentialenergy increases when repulsionbetween like charges outweighsattraction between opposite charges.

FIGURE 6 The arrows indicatethe attractive and repulsive forcesbetween the electrons (shown aselectron clouds) and nuclei of twohydrogen atoms. Attraction betweenparticles corresponds to a decreasein potential energy of the atoms,while repulsion corresponds to an increase.

–

–

++

Both nuclei repeleach other, as do both

electron clouds.

The nucleus of one atomattracts the electron cloud of

the other atom, and vice versa.


INTERPRETING GRAPHIC SOURCES OF INFORMATION Have stu-dents read the selections on theFormation of a Covalent Bond andCharacteristics of a Covalent Bondsilently. Lead a discussion about thefeatures of the graph in Figure 5that correspond to the changes inpotential energy described in thetext. You may wish to highlightthese features on an overheadtransparency as you discuss them.Have students make notes on theirown copies of the graph.

C H A P T E R 6180

SECTION 1

180

SECTION 2

Chemistry in ActionClass DiscussionHave students assess the benefitsand drawbacks of ProfessorHoffmann’s device. One possible benefit includes reducing the amountof toxic wastes in the environment.Two possible drawbacks are that thismethod requires considerable energyand would not remove all toxins fromwaste.

Alternative AssessmentHave students research the existingmethods used to dispose of toxicwastes in your region.

Answers1. The use of ultrasonic waves

allows organic pollutants (andsome inorganic wastes) to reactand be broken down into non-harmful substances without having to create bulk high-temperature or high-pressureconditions. Therefore, this proce-dure is safer to use and morefriendly to the environment thanare more common methods, suchas combustion (as in incinera-tors). In addition, this processhas the potential to be cheaperthan combustion methods.

2. The ultrasound waves generatebubbles in the water, called cavi-tation. A second ultrasoundbreaks these bubbles. The inter-nal pressure of these bubbles isextremely high and causes thetemperature of the water rightaround the bubbles to reachabout 5000°C. Hence, the hotspots in the water are localized,and the temperature of the bulkwater remains low.

GENERAL

Ultrasonic Toxic-Waste DestroyerPaints, pesticides, solvents, and sulfides are just a few componentsof the 3 million tons of toxic waste that flow out of U.S. factoriesevery year. Some of this waste ends up in groundwater and contaminates our streams and drinking water.

Eliminating hazardous waste is a constant challenge. Unfortu-nately, today’s disposal methodsoften damage the environment asmuch as they help it. Incineratorsburning certain waste, for example,produce dioxins, one of the mostdangerous class of toxins known to man.

Finding new methods to destroytoxic waste is a puzzle. MichaelHoffmann, a professor of environ-mental chemistry at the CaliforniaInstitute of Technology, thinks thatpart of the solution lies in sound-wave technology.

According to Hoffmann, cavitationis the key to eliminating certainchemical wastes from polluted water.Cavitation occurs when the pressurein water is made to fluctuate fromslightly above to slightly below nor-mal, causing bubbles. The bubblesare unstable and collapse, creatingtiny areas of extremely high pressureand heat. The pressure inside a col-lapsing bubble can be 1000 timesgreater than normal, and the tem-perature reaches about 5000°C—just a bit cooler than the surface ofthe sun. These conditions are harsh

enough to combust most toxic-wastecompounds in the water, breakingthem down into harmlesscomponents.

Hoffmann has employed a devicethat uses ultrasound—sound wavesat frequencies just above the rangeof human hearing—to create cavita-tion in polluted water. As waterflows between two panels that gen-erate ultrasound at different fre-quencies, the ultrasonic wavesgenerated by one panel form cavita-tion bubbles. An instant later, theultrasound produced by the otherpanel collapses the bubbles. Theintense pressure and heat generatedbreak down toxic compounds intoinnocuous substances, such as car-bon dioxide, chloride ions, andhydrogen ions.

“With ultrasound,” saysHoffmann, “we can harness fre-quencies . . . of about 16 kilohertzup to 1 megahertz, and different . . .compounds are destroyed morereadily at one frequency versusanother . . . applying a particularfrequency range, we can destroy a very broad range of chemicalcompounds.”

The device destroys simple toxinsin a few minutes and other toxins in several hours. To be destroyedcompletely, some compounds mustform intermediate chemicals firstand then be treated again. To besure the waste is totally removed,scientists use sophisticated tracking

methods to trace what happens to every single molecule of the toxin.

The ultrasound toxic-wastedestroyer treats about 10% of alltypes of waste, eliminating bothorganic and inorganic compounds,such as hydrogen cyanide, TNT, andmany pesticides. While the devicecannot destroy complex mixtures of compounds, such as those foundin raw sewage, it does have manyadvantages over current technolo-gies. Aside from having no harmfulenvironmental side effects, ultrasonicwaste destruction is cheaper and simpler than the process of combustion.

1. How does Dr. Hoffmann’s ultra-sound device benefit society?

2. Briefly explain why the bulk tem-perature of the water remainslow (at room temperature).

Questions

www.scilinks.orgTopic: UltrasoundCode: HC61576

Characteristics of the Covalent Bond

In Figure 5, the bottom of the valley in the curve represents the balancebetween attraction and repulsion in a stable covalent bond. At thispoint, the electrons of each hydrogen atom of the hydrogen moleculeare shared between the nuclei. As shown below in Figure 7, the mole-cule’s electrons can be pictured as occupying overlapping orbitals, mov-ing about freely in either orbital.

The bonded atoms vibrate a bit, but as long as their potential energyremains close to the minimum, they are covalently bonded to each other.The distance between two bonded atoms at their minimum potentialenergy, that is, the average distance between two bonded atoms, is thebond length. The bond length of a hydrogen-hydrogen bond is 75 pm.

In forming a covalent bond, the hydrogen atoms release energy asthey change from isolated individual atoms to parts of a molecule. Theamount of energy released equals the difference between the potentialenergy at the zero level (separated atoms) and that at the bottom of thevalley (bonded atoms) in Figure 5. The same amount of energy must beadded to separate the bonded atoms. Bond energy is the energy requiredto break a chemical bond and form neutral isolated atoms. Scientistsusually report bond energies in kilojoules per mole (kJ/mol), whichindicates the energy required to break one mole of bonds in isolatedmolecules. For example, 436 kJ of energy is needed to break the hydrogen-hydrogen bonds in one mole of hydrogen molecules and form two molesof separated hydrogen atoms.

The energy relationships described here for the formation of ahydrogen-hydrogen bond apply generally to all covalent bonds.However, bond lengths and bond energies vary with the types of atomsthat have combined. Even the energy of a bond between the same twotypes of atoms varies somewhat, depending on what other bonds theatoms have formed. These facts should be considered when examiningthe data in Table 1 on the next page. The first three columns in the tablelist bonds, bond lengths, and bond energies of atoms in specific diatom-ic molecules. The last three columns give average values of specifiedbonds in many different compounds.


SECTION 2

Visual StrategyThe electron of an

isolated hydrogen atom exists mostlywithin a predictable radius of thenucleus. Have students describe howthe location of the electrons differs ina hydrogen molecule. (The electronsspend more time in between thehydrogen nuclei, resulting in anincreased electron density there.)

Did You Know?A microwave oven works by bathingfood with radiation that has a frequency of about 3 GHz. This radia-tion rotates the water molecules,generating energy as heat and there-by cooking the food.

Be sure that students understand thatlonger bonds tend to be weaker. Bondstrength is reflected by the amount ofenergy needed to break a bond, orthe bond energy.

Teaching Tip

FIGURE 7

Electron clouds

Nuclei

Hydrogen atoms

Region of orbitaloverlap

Hydrogen molecule

FIGURE 7 The orbitals of thehydrogen atoms in a hydrogen mol-ecule overlap, allowing each electronto feel the attraction of both nuclei.The result is an increase in electrondensity between the nuclei.

✔

CHAPTER CONNECTION

Students may want to review electron configurations and orbitalnotation, discussed in Chapter 4.

All individual hydrogen atoms contain a single, unpaired electron in a1s atomic orbital.When two hydrogen atoms form a molecule, they shareelectrons in a covalent bond. As Figure 8 shows, sharing electrons allowseach atom to have the stable electron configuration of helium, 1s2. Thistendency for atoms to achieve noble-gas configurations by bonding cova-lently extends beyond the simple case of a hydrogen molecule.

The Octet Rule

Unlike other atoms, the noble-gas atoms exist independently in nature.They possess a minimum of energy existing on their own because of thespecial stability of their electron configurations. This stability resultsfrom the fact that, with the exception of helium and its two electrons ina completely filled outer shell, the noble-gas atoms’ outer s and porbitals are completely filled by a total of eight electrons. Other main-group atoms can effectively fill their outermost s and p orbitals withelectrons by sharing electrons through covalent bonding. Such bond

Hydrogenatoms

Bonding electron pair in overlapping orbitals

H ↑1s

H ↓1s

Hydrogenmolecule

H ↑1s

H ↓1s

C H A P T E R 6182

SECTION 1

182

SECTION 2

Table 1 Students can see a clearrelationship between bond lengthand bond strength by plotting agraph with bond length on the x-axisand bond energy on the y-axis. Evenwithout the graph, they may see thatthe bond length decreases as thestrength of the bond—as represent-ed by bond energy—increases.

Also, students should understandthat the bonds, bond lengths, andbond energies in the first threecolumns of the table correspond to specific diatomic molecules. Thebond lengths and bond energies listed in the fifth and sixth columns,on the other hand, are average val-ues for the bonds specified in thefourth column. The bonds listed inthe fourth column exist in many different molecules, and their lengthsand energies vary from molecule tomolecule.

GENERALTABLE STRATEGY Average bond Average bond Average bond Average bondBond length (pm) energy (kJ/mol) Bond length (pm) energy (kJ/mol)

HIH 75 436 CIC 154 346

FIF 142 159 CIN 147 305

ClICl 199 243 CIO 143 358

BrIBr 229 193 CIH 109 418

III 266 151 CICl 177 327

HIF 92 569 CIBr 194 285

HICl 127 432 NIN 145 163

HIBr 141 366 NIH 101 386

HII 161 299 OIH 96 459

TABLE 1 Bond Lengths and Bond Energies for Selected Covalent Bonds

FIGURE 8 By sharing electrons inoverlapping orbitals, each hydrogenatom in a hydrogen molecule experi-ences the effect of a stable 1s2 configuration.

formation follows the octet rule: Chemical compounds tend to form sothat each atom, by gaining, losing, or sharing electrons, has an octet ofelectrons in its highest occupied energy level.

Let’s examine how the bonding in a fluorine molecule illustrates theoctet rule. An independent fluorine atom has seven electrons in its high-est energy level ([He]2s22p5). Like hydrogen atoms, fluorine atomsbond covalently with each other to form diatomic molecules, F2. Whentwo fluorine atoms bond, each atom shares one of its valence electronswith its partner. The shared electron pair effectively fills each atom’soutermost energy level with an octet of electrons, as illustrated inFigure 9a. Figure 9b shows another example of the octet rule, in whichthe chlorine atom in a molecule of hydrogen chloride, HCl, achieves anoutermost octet by sharing an electron pair with an atom of hydrogen.

Exceptions to the Octet RuleMost main-group elements tend to form covalent bonds according tothe octet rule. However, there are exceptions. As you have seen, hydro-gen forms bonds in which it is surrounded by only two electrons. Boron,B, has just three valence electrons ([He]2s22p1). Because electron pairsare shared in covalent bonds, boron tends to form bonds in which it issurrounded by six electrons. In boron trifluoride, BF3, for example, theboron atom is surrounded by its own three valence electrons plus onefrom each of the three fluorine atoms bonded to it. Other elements canbe surrounded by more than eight electrons when they combine withthe highly electronegative elements fluorine, oxygen, and chlorine. Inthese cases of expanded valence, bonding involves electrons in d orbitalsas well as in s and p orbitals. Examples of compounds that have anexpanded valence include PF5 and SF6, as shown in Table 5.


SECTION 2

Fluorine molecule

Bondingelectronpair inoverlappingorbitals

F ↑↓1s 2s

2p

↑↓ ↑↓ ↑↓ ↑

F ↑↓1s 2s

2p

↑↓ ↑↓ ↑↓ ↓

Fluorine atoms

F ↑↓1s 2s

2p

↑↓ ↑↓ ↑↓ ↑

F ↑↓1s 2s

2p

↑↓ ↑↓ ↑↓ ↓

(a)

Bonding electron pairin overlapping orbitals

Hydrogen and chlorine atoms

H

(b)

↑

Cl ↑↓1s

1s

2s2p

↑↓3s↑↓↑↓ ↑↓ ↑↓

3p

↑↓ ↑↓ ↓

Hydrogen chloride molecule

H ↑

Cl ↑↓1s 2s

2p

1s

↑↓3s↑↓↑↓ ↑↓ ↑↓

3p

↑↓ ↑↓ ↓

FIGURE 9 (a) By sharingvalence electrons in overlappingorbitals, each atom in a fluorinemolecule feels the effect ofneon’s stable configuration,[He]2s22p6. (b) In a hydrogenchloride molecule, the hydrogenatom effectively fills its 1sorbital with two electrons, whilethe chlorine atom experiencesthe stability of an outermostoctet of electrons.

CHAPTER CONNECTION

Boron trifluoride, BF3, mentioned on this page as an exception to theoctet rule, will easily attract an atomor molecule that has an unbondedpair of electrons, such as ammonia,NH3. This kind of reaction yields aproduct that is more stable becausean octet is formed around boron.The ability of BF3 to attract a pair of electrons makes it a Lewis acid,which is described in further detail inChapter 14. This type of acid is namedafter Gilbert Lewis, who also proposedthe octet rule and the electron-dotsymbols—called Lewis structures—introduced on the next page.

Lewis Structures

Electron-dot notation can also be used to represent molecules. Forexample, a hydrogen molecule, H2, is represented by combining thenotations of two individual hydrogen atoms, as follows.

HH

Electron-Dot Notation

Covalent bond formation usually involves only theelectrons in an atom’s outermost energy levels, or theatom’s valence electrons. To keep track of these elec-trons, it is helpful to use electron-dot notation.Electron-dot notation is an electron-configuration no-tation in which only the valence electrons of an atomof a particular element are shown, indicated by dotsplaced around the element’s symbol. The inner-shellelectrons are not shown. For example, the electron-dot notation for a fluorine atom (electron configura-tion [He]2s22p5) may be written as follows.

In general, an element’s number of valence electronscan be determined by adding the superscripts of theelement’s noble-gas notation. In this book, theelectron-dot notations for elements with 1–8 valenceelectrons are written as shown in Figure 10.

F

C H A P T E R 6184

SECTION 1

184

SECTION 2

Number of valence electrons

Electron-dotnotation Example

1

2

3

4

5

6

7

8

X Na

X Mg

X B

X C

X N

X O

X F

X Ne

FIGURE 10 To write an element’s electron-dot notation, determine the element’s number of valence electrons. Then place a corresponding number of dotsaround the element’s symbol, as shown.

a. A hydrogen atom has only one occupied energy level, the n = 1 level, which contains asingle electron. Therefore, the electron-dot notation for hydrogen is written as follows.

b. The group notation for nitrogen’s family of elements is ns2np3, which indicates that nitro-gen has five valence electrons. Therefore, the electron-dot notation for nitrogen is writ-ten as follows.

N

H

SOLUTION

SAMPLE PROBLEM B

a. Write the electron-dot notation for hydrogen.b. Write the electron-dot notation for nitrogen.

For more help, go to the Math Tutor at the end of this chapter.

B-1 Write the electron-dot notationfor the following elements:a. phosphorusb. siliconc. sulfurd. chlorinee. xenon

Ans. a.

b.

c.

d.

e. Xe

Cl

S

Si

P


The pair of dots represents the shared electron pair of the hydrogen-hydrogen covalent bond. For a molecule of fluorine, F2, the electron-dotnotations of two fluorine atoms are combined.

Here also the pair of dots between the two symbols represents theshared pair of a covalent bond. In addition, each fluorine atom is sur-rounded by three pairs of electrons that are not shared in bonds. Anunshared pair, also called a lone pair, is a pair of electrons that is notinvolved in bonding and that belongs exclusively to one atom.

The pair of dots representing a shared pair of electrons in a covalentbond is often replaced by a long dash. According to this convention,hydrogen and fluorine molecules are represented as follows.

These representations are all Lewis structures, formulas in whichatomic symbols represent nuclei and inner-shell electrons, dot-pairs ordashes between two atomic symbols represent electron pairs in covalentbonds, and dots adjacent to only one atomic symbol represent unsharedelectrons. It is common to write Lewis structures that show only theelectrons that are shared, using dashes to represent the bonds. A struc-tural formula indicates the kind, number, arrangement, and bonds butnot the unshared pairs of the atoms in a molecule. For example, FIF andHICl are structural formulas.

The Lewis structures (and therefore the structural formulas) formany molecules can be drawn if one knows the composition of themolecule and which atoms are bonded to each other. The followingsample problem illustrates the basic steps for writing Lewis structures.The molecule described in this problem contains bonds with singleshared electron pairs. A single covalent bond, or a single bond, is acovalent bond in which one pair of electrons is shared between twoatoms.

FIFHIH

F F


SECTION 2

1. Determine the type and number of atoms in the molecule.The formula shows one carbon atom, one iodine atom, and three hydrogen atoms.

2. Write the electron-dot notation for each type of atom in the molecule.Carbon is from Group 14 and has four valence electrons. Iodine is from Group 17 andhas seven valence electrons. Hydrogen has one valence electron.

C I H

SOLUTION

SAMPLE PROBLEM C

Draw the Lewis structure of iodomethane, CH3I.


CHAPTER CONNECTION

Structural formulas can be used todiscern the difference between twoisomers—compounds with the sameformula but a different arrangementof atoms. Isomers are discussed inChapter 22.


Some students might derive Lewisstructures more easily by represent-ing the valence electrons from eachatom in the molecule or ion with adifferent symbol. Emphasize, how-ever, that this is only an intermediaterepresentation, and that the othersymbols should then be converted todots. For example,

FF FFor��

��

��

�

HH HHor�

Teaching Tip


✔

Multiple Covalent Bonds

Atoms of some elements, especially carbon, nitrogen, and oxygen, canshare more than one electron pair. A double covalent bond, or simply adouble bond, is a covalent bond in which two pairs of electrons areshared between two atoms. A double bond is shown either by two side-by-side pairs of dots or by two parallel dashes. All four electrons in adouble bond “belong” to both atoms. In ethene, C2H4, for example, twoelectron pairs are simultaneously shared by two carbon atoms.

orH

H

H

HCJC

H

H

H

HC C

C H A P T E R 6186

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186

SECTION 2

Go to go.hrw.com formore practice problemsthat ask you to drawLewis structures.

Keyword: HC6BNDX

3. Determine the total number of valence electrons available in the atoms to be combined.

C 1 � 4e– = 4e–

I 1 � 7e– = 7e–

3H 3 � 1e– = 3e–

14e–

4. Arrange the atoms to form a skeleton structure for the molecule. If carbon is present, it isthe central atom. Otherwise, the least-electronegative atom is central (except for hydrogen,which is never central). Then connect the atoms by electron-pair bonds.

5. Add unshared pairs of electrons to each nonmetal atom (except hydrogen) such that eachis surrounded by eight electrons.

or

6. Count the electrons in the structure to be sure that the number of valence electrons usedequals the number available. Be sure the central atom and other atoms besides hydrogenhave an octet.There are eight electrons in the four covalent bonds and six electrons in the threeunshared pairs, giving the correct total of 14 valence electrons.

HICII

H

H

C IHH

H

C IHH

H

1. Draw the Lewis structure of ammonia, NH3.

2. Draw the Lewis structure for hydrogen sulfide, H2S.

3. Draw the Lewis structure for silane, SiH4.

4. Draw the Lewis structure for phosphorus trifluoride, PF3.

Answers in Appendix EPRACTICE

CHAPTER CONNECTION

The geometries of hydrocarbon dou-ble and triple bonds are discussed inChapter 22.

Practice Answers1. or

2. or

3. or

4. or FIPIF

F

PFF

F

HISiIH

H

H

SiHH

HH

HISIHS HH

HINIH

H

N HHH

A triple covalent bond, or simply a triple bond, is a covalent bond inwhich three pairs of electrons are shared between two atoms. For exam-ple, elemental nitrogen, N2, like hydrogen and the halogens, normallyexists as diatomic molecules. In this case, however, each nitrogen atom,which has five valence electrons, acquires three electrons to completean octet by sharing three pairs of electrons with its partner. This is illus-trated in the Lewis structure and the formula structure for N2, as shownbelow.

or

Figure 11 represents nitrogen’s triple bond through orbital notation.Like the single bonds in hydrogen and halogen molecules, the triplebond in nitrogen molecules is nonpolar.

Carbon forms a number of compounds containing triple bonds. Forexample, the compound ethyne, C2H2, contains a carbon-carbon triplebond.

or

Double and triple bonds are referred to as multiple bonds, or multi-ple covalent bonds. Double bonds in general have greater bond ener-gies and are shorter than single bonds. Triple bonds are even strongerand shorter. Table 2 compares average bond lengths and bond energiesfor some single, double, and triple bonds.

In writing Lewis structures for molecules that contain carbon, nitro-gen, or oxygen, one must remember that multiple bonds between pairsof these atoms are possible. (A hydrogen atom, on the other hand, hasonly one electron and therefore always forms a single covalent bond.)The need for a multiple bond becomes obvious if there are not enoughvalence electrons to complete octets by adding unshared pairs. SampleProblem D on the next page shows how to deal with this situation.

HICKCIHCC HH

NKNN N


SECTION 2

Table 2 Once again, have students observe the relationshipbetween bond length and bond energy. In general, stronger bonds(those with greater bond energies) are shorter bonds.

The CJO bond energy in carbon diox-ide is 799 kJ/mol, which is muchstronger than the average CJO bondenergy. However, it is still weaker thana CKO triple bond.

Teaching Tip

GENERALTABLE STRATEGY

Nitrogen molecule

N ↑↓1s 2s

2p

↑↓ ↓ ↓ ↓

N ↑↓1s 2s

2p

↑↓ ↑ ↑ ↑

FIGURE 11 In a molecule of nitro-gen, N2, each nitrogen atom is sur-rounded by six shared electrons plusone unshared pair of electrons. Thus,each nitrogen atom follows the octetrule in forming a triple covalent bond.

Average bond Average bond Average bond Average bondBond length (pm) energy (kJ/mol) Bond length (pm) energy (kJ/mol)

CIC 154 346 CIO 143 358

CJC 134 612 CJO 120 732

CKC 120 835 CKO 113 1072

CIN 147 305 NIN 145 163

CJN 132 615 NJN 125 418

CKN 116 887 NKN 110 945

TABLE 2 Bond Lengths and Bond Energies for Single and Multiple Covalent Bonds

✔

C H A P T E R 6188

SECTION 1

188

SECTION 2


Did You Know?Formaldehyde, the molecule men-tioned in Sample Problem D, isbelieved to cause cancer. Because of its bacterial killing characteristic,it was at one time used to preserveanimals in biology labs. Now, how-ever, safer chemicals are used.

Practice Answers1.

2. HICKN

OJCJO


Go to go.hrw.com formore practice problemsthat ask you to drawLewis structures.

Keyword: HC6BNDX

1. Determine the number of atoms of each element present in the molecule.The formula shows one carbon atom, two hydrogen atoms, and one oxygen atom.

2. Write the electron-dot notation for each type of atom.Carbon is from Group 14 and has four valence electrons. Oxygen, which is in Group 16,has six valence electrons. Hydrogen has only one electron.

3. Determine the total number of valence electrons available in the atoms to be combined.

C 1 � 4e– = 4e–

O 1 � 6e– = 6e–

2H 2 � 1e– = 2e–

12e–

4. Arrange the atoms to form a skeleton structure for the molecule, and connect the atoms by electron-pair bonds.

5. Add unshared pairs of electrons to each nonmetal atom (except hydrogen) such that eachis surrounded by eight electrons.

6a. Count the electrons in the Lewis structure to be sure that the number of valence electronsused equals the number available.The structure above has six electrons in covalent bonds and eight electrons in four lone pairs, for a total of 14 electrons. The structure has two valence electrons too many.

6b. If too many electrons have been used, subtract one or more lone pairs until the total numberof valence electrons is correct. Then move one or more lone electron pairs to existing bondsbetween non-hydrogen atoms until the outer shells of all atoms are completely filled.Subtract the lone pair of electrons from the carbon atom. Then move one lone pair ofelectrons from the oxygen to the bond between carbon and oxygen to form a double bond.

or

There are eight electrons in covalent bonds and four electrons in lone pairs, for a total of 12 valence electrons.

HICJO

H

C OHH

C OHH

C OHH

C O H

SOLUTION

SAMPLE PROBLEM D

1. Draw the Lewis structure for carbon dioxide, CO2.

2. Draw the Lewis structure for hydrogen cyanide, which containsone hydrogen atom, one carbon atom, and one nitrogen atom.


Draw the Lewis structure for methanal, CH2O, which is also known as formaldehyde.

PRACTICE


Resonance Structures

Some molecules and ions cannot be represented adequately by a singleLewis structure. One such molecule is ozone, O3, which can be repre-sented by either of the following Lewis structures.

or

Notice that each structure indicates that the ozone molecule has twotypes of OIO bonds, one single and one double. Chemists once specu-lated that ozone split its time existing as one of these two structures, con-stantly alternating, or “resonating,” from one to the other. Experiments,however, revealed that the oxygen-oxygen bonds in ozone are identical.Therefore, scientists now say that ozone has a single structure that is theaverage of these two structures. Together the structures are referred toas resonance structures or resonance hybrids. Resonance refers to bond-ing in molecules or ions that cannot be correctly represented by a singleLewis structure. To indicate resonance, a double-headed arrow is placedbetween a molecule’s resonance structures.

Covalent-Network Bonding

All the covalent compounds that you have read about so far are mole-cular. They consist of many identical molecules held together by forcesacting between the molecules. (You will read more about these inter-molecular forces in Section 5.) There are many covalently bonded com-pounds that do not contain individual molecules, but instead can bepictured as continuous, three-dimensional networks of bonded atoms.You will read more about covalently bonded networks in Chapter 7.

OJOIO OIOJO←⎯→

OIOJOOJOIO


SECTION 2

Lewis structures assume that electronsare found only between the two atomsin a bond. However, sometimes elec-trons are delocalized, which meansthat they are distributed over three ormore atoms. In such cases, Lewis struc-tures do not clearly model the locationof the electrons in the bond unless res-onance structures are used.

Did You Know?Diamond, the hardest materialknown, is actually made of many car-bon atoms that are bonded togetherwith single bonds. A diamond crystalis an example of a covalent network.

1. a. the distance at which two covalently bonded atoms minimizetheir potential energyb. the energy required to break achemical bond and form neutral,isolated atoms (in the gas phase)

2. Chemical compounds tend to formso that each atom gains, loses, orshares electrons until an octet of elec-trons exists in its highest occupiedenergy level.

3. a. oneb. twoc. three

4. a.

b.

c.

d.

e.

5. Draw the Lewis structures. HNNHhas a double N-N bond, and H2NNH2has a single N-N bond. Therefore,HNNH has the stronger N-N bond.

OF F

SiClCl

ClCl

H C C Cl

CHH

HBr

I Br

SECTION REVIEW

Teaching Tip

1. Define the following:

a. bond length b. bond energy

2. State the octet rule.

3. How many pairs of electrons are shared in thefollowing types of covalent bonds?

a. a single bond

b. a double bond

c. a triple bond

4. Draw the Lewis structures for the followingmolecules:

a. IBr d. SiCl4b. CH3Br e. OF2

c. C2HCl

Critical Thinking

5. APPLYING MODELS Compare the moleculesH2NNH2 and HNNH. Which molecule has thestronger N-N bond?

SECTION REVIEW

✔

C H A P T E R 6190

SECTION 1

190

SECTION 3

Lesson StarterUse plastic, interlocking buildingblocks to demonstrate the differencebetween ionic and covalent bonding.Assemble a number of blocks so thatall sides of the object created areflush. This is a model of a molecule—the atoms are bonded to create anindividual unit capable of existing onits own. Assemble a number of blocksso that pieces of the various blocksare jutting out of the sides of theobject created. This represents a smallportion of the ionic arrangementwithin an ionic compound. The piecesof blocks jutting out represent cationsand anions, which attract more ions(more blocks) to create a large arrayof packed ions. The result is a hugeaggregate of ions that have minimizedtheir potential energy through electri-cal attraction (as opposed to electronsharing).

CommonMisconceptionRemind students that the chemicalformulas for ionic compounds repre-sent the simplest formula of the compound, as opposed to the for-mulas for molecules, which repres-ent separate molecules that are eacha discrete group of atoms.

GENERALSECTION 3 Ionic Bonding and

Ionic Compounds

M ost of the rocks and minerals that make up Earth’s crust consist ofpositive and negative ions held together by ionic bonding. A familiarexample of an ionically bonded compound is sodium chloride, or com-mon table salt, which is found in nature as rock salt. A sodium ion, Na+,has a charge of 1+. A chloride ion, Cl−, has a charge of 1−. There is anelectrical force of attraction between oppositely charged ions. In sodi-um chloride, these ions combine in a one-to-one ratio—Na+Cl−—sothat each positive charge is balanced by a negative charge. The chemi-cal formula for sodium chloride is usually written simply as NaCl.

An ionic compound is composed of positive and negative ions that arecombined so that the numbers of positive and negative charges are equal.Most ionic compounds exist as crystalline solids (see Figure 12). A crys-tal of any ionic compound is a three-dimensional network of positiveand negative ions mutually attracted to one another. As a result, in con-trast to a molecular compound, an ionic compound is not composed ofindependent, neutral units that can be isolated and examined. Thechemical formula of an ionic compound merely represents the simplestratio of the compound’s combined ions that gives electrical neutrality.

The chemical formula of an ionic compound shows the ratio of the ions present in a sample of any size. A formula unit is the simplestcollection of atoms from which an ionic compound’s formula can beestablished. For example, one formula unit of sodium chloride, NaCl, isone sodium cation plus one chloride anion. (In the naming of amonatomic anion, the ending of the element's name is replaced with-ide. See Chapter 7 for more details.)

The ratio of ions in a formula unit depends on the charges of the ionscombined. For example, to achieve electrical neutrality in the ionic com-pound calcium fluoride, two fluoride anions, F−, each with a charge of1−, must balance the 2+ charge of each calcium cation, Ca2+. Therefore,the formula of calcium fluoride is CaF2.

Formation of Ionic Compounds

Electron-dot notation can be used to demonstrate the changes thattake place in ionic bonding. Ionic compounds do not ordinarily form bythe combination of isolated ions, but consider for a moment a sodium

OBJECTIVES

Compare and contrast achemical formula for a molecular compound withone for an ionic compound.

Discuss the arrangements of ions in crystals.

Define lattice energy andexplain its significance.

List and compare the distinc-tive properties of ionic andmolecular compounds.

Write the Lewis structure for a polyatomic ion given theidentity of the atoms com-bined and other appropriateinformation.

FIGURE 12 Like most ionic com-pounds, sodium chloride is a crys-talline solid.

atom and a chlorine atom approaching each other. The two atoms areneutral and have one and seven valence electrons, respectively.

Sodium atom Chlorine atom

We have already seen that atoms of sodium and the other alkali met-als readily lose one electron to form cations. And we have seen thatatoms of chlorine and the other halogens readily gain one electron toform anions. The combination of sodium and chlorine atoms to produceone formula unit of sodium chloride can thus be represented as follows.

+ ⎯→ +

Sodium atom Chlorine atom Sodium cation Chloride anion

The transfer of an electron from the sodium atom to the chlorine atomtransforms each atom into an ion with a noble-gas configuration. In thecombination of calcium with fluorine, two fluorine atoms are needed toaccept the two valence electrons given up by one calcium atom.

+ + ⎯→ + +

Calcium atom Fluorine atoms Calcium cation Fluoride anions

Characteristics of Ionic BondingRecall that nature favors arrangements in which potential energy isminimized. In an ionic crystal, ions minimize their potential energy bycombining in an orderly arrangement known as a crystal lattice (seeFigure 13). The attractive forces at work within an ionic crystal includethose between oppositely charged ions and those between the nucleiand electrons of adjacent ions. The repulsive forces include thosebetween like-charged ions and those between electrons of adjacentions. The distances between ions and their arrangement in a crystal rep-resent a balance among all these forces. Sodium chloride’s crystal struc-ture is shown in Figure 14 below.

F −F −Ca2+FFCa

Cl −Na+ClNa

ClNa


SECTION 3

Visual StrategyEmphasize that an

ion in an ionic crystal is surroundedby ions of opposite charge. It isimpossible to say that an ion“belongs” more to one of its neighbors than it does to another.The models of sodium chloride inFigure 14 below and Figure 15on the next page help illustrate this.

FIGURE 13

GENERAL

Cl−Na+

Cl−

Na+

(a) (b)

+

++

++

++++

+

+ +

+

−−

−

− −−

−−−−

−−

−

−

FIGURE 13 The ions in an ioniccompound lower their potentialenergy by forming an orderly, three-dimensional array in which the positive and negative charges arebalanced. The electrical forces of attraction between oppositelycharged ions extend over long dis-tances, causing a large decrease in potential energy.

FIGURE 14 Two models of thecrystal structure of sodium chlorideare shown. (a) To illustrate the ions’actual arrangement, the sodium andchloride ions are shown with theirelectron clouds just touching.(b) In an expanded view, the distancesbetween ions have been exaggeratedin order to clarify the positioning ofthe ions in the structure.

www.scilinks.orgTopic: Ionic BondingCode: HC60815

Figure 15 shows the crystal structure of sodium chloride in greaterdetail. Within the arrangement, each sodium cation is surrounded by sixchloride anions. At the same time, each chloride anion is surrounded bysix sodium cations. Attraction between the adjacent oppositely chargedions is much stronger than repulsion by other ions of the same charge,which are farther away.

The three-dimensional arrangements of ions and the strengths ofattraction between them vary with the sizes and charges of the ions andthe numbers of ions of different charges. For example, in calcium fluo-ride, there are two anions for each cation. Each calcium cation is sur-rounded by eight fluoride anions. At the same time, each fluoride ion issurrounded by four calcium cations, as shown in Figure 16.

To compare bond strengths in ionic compounds, chemists comparethe amounts of energy released when separated ions in a gas cometogether to form a crystalline solid. Lattice energy is the energy releasedwhen one mole of an ionic crystalline compound is formed from gaseousions. Lattice energy values for a few common ionic compounds areshown in Table 3. The negative energy values indicate that energy isreleased when the crystals are formed.

C H A P T E R 6192

SECTION 1

192

SECTION 3

Visual StrategyHave students look

at the portion of sodium chloride’scrystal arrangement from the per-spective of the chloride ion, which issurrounded by six sodium ions. Thenhave them consider the perspectiveof a sodium ion, which is surroundedby six chloride ions. Ask students todeduce the ratio of sodium ions tochloride ions in sodium chloride (one-to-one).

Students are often confused aboutthe difference in signs of bond ener-gies (Table 2) and lattice energies(Table 3). The lattice energy describesthe energies associated with the for-mation of bonds, whereas the bondenergy describes the energies associ-ated with the breaking of bonds. Inany case, the bonded atoms are morestable than the separate ones. Thelarger the magnitude of the latticeenergy or the bond energy, the morestable the bonding.

Teaching Tip

FIGURE 15 Na+

Cl–

(a) (b)

Calcium ion, Ca2+

Fluoride ion, F−

FIGURE 15 The figure shows theions that most closely surround a chloride anion and a sodium cationwithin the crystal structure of NaCl.The structure is composed such that (a) six Na+ ions surround eachCl− ion. At the same time, (b) six Cl−

ions surround each Na+ ion (whichcannot be seen but whose location isindicated by the dashed outline).

FIGURE 16 In the crystal structure of calcium fluoride, CaF2,each calcium cation is surroundedby eight fluoride anions and eachfluoride ion is surrounded by fourcalcium cations. This is the closestpossible packing of the ions inwhich the positive and negativecharges are balanced.

✔

A Comparison of Ionic and Molecular Compounds

The force that holds ions together in ionic compounds is a very strong over-all attraction between positive and negative charges. In a molecular com-pound, the covalent bonds of the atoms making up each molecule are alsostrong. But the forces of attraction between molecules are much weakerthan the forces among formula units in ionic bonding.This difference in thestrength of attraction between the basic units of molecular and ionic com-pounds gives rise to different properties in the two types of compounds.

The melting point, boiling point, and hardness of a compound dependon how strongly its basic units are attracted to each other. Because theforces of attraction between individual molecules are not very strong,many molecular compounds melt at low temperatures. In fact, many mol-ecular compounds are already completely gaseous at room temperature.In contrast, the ions in ionic compounds are held together by strongattractive forces, so ionic compounds generally have higher melting andboiling points than do molecular compounds.

Ionic compounds are hard but brittle. Why? In an ionic crystal, evena slight shift of one row of ions relative to another causes a large buildupof repulsive forces, as shown in Figure 17. These forces make it difficultfor one layer to move relative to another, causing ionic compounds tobe hard. If one layer is moved, however, the repulsive forces make thelayers part completely, causing ionic compounds to be brittle.

In the solid state, the ions cannot move, so the compounds are notelectrical conductors. In the molten state, ionic compounds are electri-cal conductors because the ions can move freely to carry electrical cur-rent. Many ionic compounds can dissolve in water. When they dissolve,their ions separate from each other and become surrounded by watermolecules. These ions are free to move through the solution, so suchsolutions are electrical conductors. Other ionic compounds do not dis-solve in water, however, because the attractions between the water mol-ecules and the ions cannot overcome the attractions between the ions.


SECTION 3

Lattice energyCompound (kJ/mol)

NaCl −787.5

NaBr −751.4

CaF2 −2634.7

LiCl −861.3

LiF −1032

MgO −3760

KCl −715

TABLE 3 Lattice Energies of Some CommonIonic Compounds

FIGURE 17 (a) The attractionbetween positive and negative ionsin a crystalline ionic compoundcauses layers of ions to resist motion.(b) When struck with sufficientforce, the layers shift so that ions ofthe same charge approach eachother, causing repulsion. As a result,the crystal shatters along the planes.

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

Strong repulsion

Attraction

Attraction

Forcedislocates crystal plane

(a) (b)

Lattice energies are listed as negativevalues to indicate that energy is givenoff when the gaseous ions cometogether to form the ionic crystal.

Table 3 The lattice energy must beovercome in order for an ionic com-pound to dissolve in water.Have students form a hypothesisabout which two ionic compoundslisted in the table are the least solu-ble in water (MgO and CaF2).

GENERALTABLE STRATEGY

Teaching Tip✔


COMPARING AND CONTRASTING Have studentsuse the section “A Comparison ofIonic and Molecular Compounds”as a starting point for a chart com-paring and contrasting covalent andionic bonding. Have students rereadSections 1, 2, and 3 to help themcomplete their charts. Suggest thatthey incorporate the boldfacedterms in these sections as well asany other relevant information.

Polyatomic Ions

Certain atoms bond covalently with each other to form a group ofatoms that has both molecular and ionic characteristics. A chargedgroup of covalently bonded atoms is known as a polyatomic ion.Polyatomic ions combine with ions of opposite charge to form ioniccompounds. The charge of a polyatomic ion results from an excess ofelectrons (negative charge) or a shortage of electrons (positivecharge). For example, an ammonium ion, a common positively chargedpolyatomic ion, contains one nitrogen atom and four hydrogen atomsand has a single positive charge. Its formula is NH4

+, sometimes writtenas [NH4]+ to show that the group of atoms as a whole has a charge of1+. The seven protons in the nitrogen atom plus the four protons in thefour hydrogen atoms give the ammonium ion a total positive charge of11+. An independent nitrogen atom has seven electrons, and four inde-pendent hydrogen atoms have a total of four electrons. When theseatoms combine to form an ammonium ion, one of their electrons is lost,giving the polyatomic ion a total negative charge of 10−.

Lewis structures for the ammonium ion and some common negativepolyatomic ions—the nitrate, sulfate, and phosphate ions—are shownbelow. To find the Lewis structure for a polyatomic ion, follow the stepsof Sample Problem D, with the following exception. If the ion is nega-tively charged, add to the total number of valence electrons a numberof electrons corresponding to the ion’s negative charge. If the ion is pos-itively charged, subtract from the total number of valence electrons anumber of electrons corresponding to the ion’s positive charge.

Ammonium ion Nitrate ion Sulfate ion Phosphate ion

P OOO

O3−

S OOO

O2−

N OO

O−

N HHH

H+

C H A P T E R 6194

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194

SECTION 3

CommonMisconceptionMany students forget that althoughpolyatomic ions are involved in ionicbonding, the ions themselves are heldtogether by covalent bonding. Forexample, the atoms in the ammoniumion are held together by covalentbonding, but ammonium ions andchloride ions are held together byionic bonding in the compoundammonium chloride. We call ammoni-um chloride an ionic compound.

1. Two possible examples are sodiumchloride, NaCl, and magnesium chlo-ride, MgCl2.

2. a.

b.

3. Ionic compounds consist of posi-tive and negative ions bound to-gether by electrical attraction.Molecular compounds are groups ofatoms held together by covalentbonding, or the sharing of electrons.

4. Compound B is probably a molecu-lar substance; compound A is probablyionic. The intermolecular attractionsthat hold molecules together areweaker than ionic attraction, resultingin lower melting and boiling points inmolecular substances.

5. Rb2S, K2S, Li2S

(CaI2)Ca2+

+ I – + I

–

⎯→Ca + I + I

Li+ + Cl

–(LiCl)

Li + Cl ⎯→

SECTION REVIEW

1. Give two examples of an ionic compound.

2. Use electron-dot notation to demonstrate the formation of ionic compounds involving thefollowing:

a. Li and Cl

b. Ca and I

3. Distinguish between ionic and molecular compounds in terms of the basic units that each is composed of.

4. Compound B has lower melting and boiling points than compound A. At the same tempera-ture, compound B vaporizes faster than compoundA. If one of these compounds is ionic and theother is molecular, which would you expect to bemolecular? ionic? Explain your reasoning.

Critical Thinking

5. ANALYZING DATA The melting points for the com-pounds Li2S, Rb2S, and K2S are 900°C, 530°C, and840°C, respectively. List these three compounds inorder of increasing lattice energy.

SECTION REVIEW

Chemistry in ActionGo to go.hrw.com for a full-lengtharticle on nanoscale computers.

Keyword: HC6BNDX

SECTION 4

OBJECTIVES

Describe the electron-seamodel of metallic bonding,and explain why metals aregood electrical conductors.

Explain why metal surfacesare shiny.

Explain why metals are mal-leable and ductile but ionic-crystalline compounds are not.

Metallic Bonding

C hemical bonding is different in metals than it is in ionic, molecular, orcovalent-network compounds. This difference is reflected in the uniqueproperties of metals. They are excellent electrical conductors in the solidstate—much better conductors than even molten ionic compounds. Thisproperty is due to the highly mobile valence electrons of the atoms thatmake up a metal. Such mobility is not possible in molecular compounds, inwhich valence electrons are localized in electron-pair bonds between neu-tral atoms. Nor is it possible in solid ionic compounds, in which electronsare bound to individual ions that are held in place in crystal structures.

The Metallic-Bond Model

The highest energy levels of most metal atoms are occupied by very fewelectrons. In s-block metals, for example, one or two valence electronsoccupy the outermost orbital, and all three outermost p orbitals, whichcan hold a total of six electrons, are vacant. In addition to completelyvacant outer p orbitals, d-block metals also possess many vacant dorbitals in the energy level just below their highest energy level.

Within a metal, the vacant orbitals in the atoms’ outer energy levelsoverlap. This overlapping of orbitals allows the outer electrons of theatoms to roam freely throughout the entire metal. The electrons aredelocalized, which means that they do not belong to any one atom butmove freely about the metal’s network of empty atomic orbitals. Thesemobile electrons form a sea of electrons around the metal atoms, whichare packed together in a crystal lattice (see Figure 18). The chemicalbonding that results from the attraction between metal atoms and the sur-rounding sea of electrons is called metallic bonding.

Metallic PropertiesThe freedom of electrons to move in a network of metal atoms accountsfor the high electrical and thermal conductivity characteristic of all met-als. In addition, metals are both strong absorbers and reflectors of light.Because they contain many orbitals separated by extremely small energydifferences, metals can absorb a wide range of light frequencies. Thisabsorption of light results in the excitation of the metal atoms’ electronsto higher energy levels. However, in metals the electrons immediately fallback down to lower levels, emitting energy in the form of light at a fre-quency similar to the absorbed frequency. This re-radiated (or reflected)light is responsible for the metallic appearance or luster of metal surfaces.


SECTION 4

Lesson StarterPlace a sample of sodium chloride ormagnesium chloride and a sample ofmossy zinc on the table top. Hit bothwith a hammer. (Use eye protection,and make sure that students are at asufficient distance from the samplesto avoid the possibility of salt frag-ments getting in their eyes. You maywant to put the salt in a plastic bagbefore you hit it.) The salt is brittleand shatters. The zinc will just flattenbecause it is malleable. The malleabil-ity of metal is a property that resultsfrom metallic bonding.

GENERAL

CHAPTER CONNECTION

The luster of metals is caused by theemission of photons when excitedelectrons return to the ground state.This principle was discussed inChapter 4.

Because electrons are delocalizedand mobile, they can move from theregion around one nucleus to theregion around another. This allows forthe conductivity of heat and electricity.

The band theory of conductivity ispresented with Group 14 of theElements Handbook.

USING THE HANDBOOK

Teaching Tip✔

FIGURE 18 The model shows a portion of the crystal structure ofsolid sodium. The atoms are arrangedso that each sodium atom is sur-rounded by eight other sodiumatoms. The atoms are relatively fixedin position, while the electrons arefree to move throughout the crystal,forming an electron sea.

www.scilinks.orgTopic: Metallic BondingCode: HC60944

Most metals are also easy to form into desired shapes. Two importantproperties related to this characteristic are malleability and ductility.Malleability is the ability of a substance to be hammered or beaten intothin sheets. Ductility is the ability of a substance to be drawn, pulled, orextruded through a small opening to produce a wire. The malleability andductility of metals are possible because metallic bonding is the same inall directions throughout the solid. When struck, one plane of atoms in ametal can slide past another without encountering resistance or break-ing bonds. By contrast, recall from Section 3 that shifting the layers of anionic crystal causes the bonds to break and the crystal to shatter.

Metallic Bond StrengthMetallic bond strength varies with the nuclear charge of the metalatoms and the number of electrons in the metal’s electron sea. Both ofthese factors are reflected in a metal’s enthalpy of vaporization. Theamount of energy as heat required to vaporize the metal is a measure ofthe strength of the bonds that hold the metal together. The enthalpy ofvaporization is defined as the amount of energy absorbed as heat whena specified amount of a substance vaporizes at constant pressure. Someenthalpies of vaporization for metals are given in Table 4.

C H A P T E R 6196

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SECTION 4

AnalogyMetallically bonded atoms are like large objects embedded in the material used for packing boxes,which represents the electron sea.In the same way that the objects can move freely relative to eachother within the packing material,the nuclei of metals can move freelyrelative to one another in the elec-tron “packing material”; malleabilityand ductility result.

1. Metallic bonding is generallyviewed as the result of the mutualsharing of many electrons by manyatoms. Each atom contributes itsvalence electrons to a region sur-rounding the atoms. These electronsare then free to move about themostly vacant outer orbitals of all themetal atoms. The mobile electrons areoften referred to as an electron sea.

2. In general, the higher the enthalpyof vaporization, the stronger themetallic bonding.


SECTION REVIEW

Period Element

Second Li Be147 297

Third Na Mg Al97 128 294

Fourth K Ca Sc77 155 333

Fifth Rb Sr Y76 137 365

Sixth Cs Ba La64 140 402

TABLE 4 Enthalpies of Vaporization of Some Metals (kJ/mol)

1. Describe the electron-sea model of metallicbonding.

2. What is the relationship between metallic bondstrength and enthalpy of vaporization?

3. Explain why most metals are malleable and ductilebut ionic crystals are not.

Critical Thinking

4. ORGANIZING IDEAS Explain why metals are goodelectrical conductors.

SECTION REVIEW

FIGURE 19 Unlike ionic crystalline compounds, most metalsare malleable. This property allowsiron, for example, to be shaped intouseful tools.

• Developmentally Delayed • Learning Disabled

Have students make a poster, claymodel, or other representation of theproperties of substances with metal-lic, ionic, and covalent bonds. Askstudents to label their representationwith the name of the bond type andat least three examples of each typeof bond. Students may display andexplain their representations to smallor large groups of other students.

StrategiesStrategiesINCLUSIONINCLUSION

SECTION 5

OBJECTIVES

Explain VSEPR theory.

Predict the shapes of molecules or polyatomic ions using VSEPR theory.

Explain how the shapes ofmolecules are accountedfor by hybridization theory.

Describe dipole-dipole forces, hydrogen bonding,induced dipoles, and Londondispersion forces and theireffects on properties such asboiling and melting points.

Explain what determines molecular polarity.

Molecular Geometry

T he properties of molecules depend not only on the bonding ofatoms but also on molecular geometry—the three-dimensionalarrangement of a molecule’s atoms in space. The polarity of eachbond, along with the geometry of the molecule, determines molecularpolarity, or the uneven distribution of molecular charge. As you willread, molecular polarity strongly influences the forces that act betweenmolecules in liquids and solids.

A chemical formula reveals little information about a molecule’sgeometry. After performing many tests designed to reveal the shapes ofvarious molecules, chemists developed two different, equally successfultheories to explain certain aspects of their findings. One theory accountsfor molecular bond angles. The other is used to describe the orbitals thatcontain the valence electrons of a molecule’s atoms.

VSEPR Theory

As shown in Figure 20, diatomic molecules, like those of hydrogen, H2,and hydrogen chloride, HCl, must be linear because they consist of onlytwo atoms. To predict the geometries of more-complicated molecules,one must consider the locations of all electron pairs surrounding thebonded atoms. This is the basis of VSEPR theory.

The abbreviation VSEPR stands for “valence-shell, electron-pairrepulsion,” referring to the repulsion between pairs of valence electronsof the atoms in a molecule. VSEPR theory states that repulsion betweenthe sets of valence-level electrons surrounding an atom causes these setsto be oriented as far apart as possible. How does the assumption thatelectrons in molecules repel each other account for molecular shapes?For now let us consider only molecules with no unshared valence elec-tron pairs on the central atom.

Let’s examine the simple molecule BeF2. The beryllium atom formsa covalent bond with each fluorine atom and does not follow the octetrule. It is surrounded by only the two electron pairs that it shares withthe fluorine atoms.

According to VSEPR theory, the shared pairs will be as far away fromeach other as possible.As shown in Figure 21a on the next page, the dis-tance between electron pairs is maximized if the bonds to fluorine are

Be FF


SECTION 5

When ionic compounds dissolve in water, their ions are hydrated bywater molecules as they dissociate,or separate from one another.

Visual StrategyStudents should realize

that in the ball-and-stick representa-tions in this textbook, all atoms aredrawn the same size. Relative atomicand ionic sizes are shown only inspace-filling models.

FIGURE 20

GENERAL

Teaching Tip

(a) Hydrogen, H2

(b) Hydrogen chloride, HCl

FIGURE 20 Ball-and-stick modelsillustrate the linearity of diatomicmolecules. (a) A hydrogen moleculeis represented by two identical balls(the hydrogen atoms) joined by asolid bar (the covalent bond). (b) Ahydrogen chloride molecule is com-posed of dissimilar atoms, but it isstill linear.

✔

on opposite sides of the beryllium atom, 180° apart. Thus, allthree atoms lie on a straight line. The molecule is linear.

If we represent the central atom in a molecule by the let-ter A and we represent the atoms bonded to the centralatom by the letter B, then according to VSEPR theory, BeF2is an example of an AB2 molecule, which is linear. Can youdetermine what an AB3 molecule looks like? The three A—B bonds stay farthest apart by pointing to the corners ofan equilateral triangle, giving 120° angles between thebonds. This trigonal-planar geometry is shown in Figure 21bfor the AB3 molecule boron trifluoride, BF3.

The central atoms in AB4 molecules follow the octet ruleby sharing four electron pairs with B atoms. The distancebetween electron pairs is maximized if each A—B bondpoints to one of four corners of a tetrahedron. This geome-try is shown in Figure 21c for the AB4 molecule methane,CH4. The same figure shows that in a tetrahedral molecule,each of the bond angles formed by the A atom and any twoof the B atoms is equal to 109.5°.

The shapes of various molecules are summarized inTable 5. B can represent a single type of atom, a group ofidentical atoms, or a group of different atoms on the samemolecule. The shape of the molecule will still be based onthe forms given in the table. However, different sizes of Bgroups distort the bond angles, making some bond angleslarger or smaller than those given in the table.

C H A P T E R 6198

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198

SECTION 5

Visual StrategyGive students four nails,

clay, and a protractor to duplicate themodels shown. Have them arrangetwo of the nails so that the nailstouch each other and the nail headsare as far apart as possible. They canhold the nails in place with the clay.Then have them measure the anglebetween the nail heads. Repeat using three and four nails.

The central atom of a molecule orpolyatomic ion usually is the onewith the lowest electronegativity. Themore electronegative atoms are gen-erally around the outside of a mole-cule or polyatomic ion, so they shareelectrons with fewer other atoms.Note, however, that H is never a central atom, because it can bond to only one other atom.

Exceptions of the type shown inSample Problem E (where the centralatom has fewer than eight electrons)are common for Group 2 and Group 13elements. Group 2 elements in cova-lent compounds commonly have onlyfour valence electrons (two bonds);BeF2 in Figure 21a is an example ofsuch an exception. Group 13 elementsin covalent compounds commonlyhave only six valence electrons (threebonds); BF3 in Figure 21b is an exam-ple of such an exception.

Teaching Tip

Teaching Tip

FIGURE 21

(b) Boron trifluoride, BF3

(c) Methane, CH4

(a) Beryllium fluoride, BeF2

AB2

AB3

AB4

180°

120°

109.5°

FIGURE 21 Ball-and-stick models show the shapes of (a) AB2, (b) AB3, and (c) AB4molecules according to VSEPR theory.

First write the Lewis structure for BCl3. Boron is in Group 13 and has three valenceelectrons.

Chlorine is in Group 17, so each chlorine atom has seven valence electrons.

The total number of available valence electrons is therefore 24e− (3e− from boron and 21e−

from chlorine). The following Lewis structure uses all 24e−.

This molecule is an exception to the octet rule because in this case B forms only threebonds. Boron trichloride is an AB3 type of molecule. Therefore, according to VSEPR theo-ry, it should have trigonal-planar geometry.

Cl BCl

Cl

Cl

B

SOLUTION

SAMPLE PROBLEM E

Use VSEPR theory to predict the molecular geometry of boron trichloride, BCl3.

✔

✔


SECTION 5

Go to go.hrw.com formore practice problemsthat ask you to predictmolecular geometry.

Keyword: HC6BNDX

1. Use VSEPR theory to predict the molecular geometry of the following molecules:

a. HI b. CBr4 c. CH2Cl2

Answers in Appendix EPRACTICE

E-1 Use VSEPR theory to predict themolecular geometry of the followingmolecules:

a. CCl4b. HCNc. SiBr4

Ans. a. tetrahedralb. linearc. tetrahedral

Practice Answers1. a. linear

b. tetrahedralc. tetrahedral

Visual StrategyInform students that a

pair of unshared electrons will takeup more space than a pair of elec-trons in a bond will. As a result, pairsof unshared electrons play a signifi-cant role in determining the geome-try of a molecule, despite the factthat only atoms are taken intoaccount when we visualize the shapeof a molecule.

FIGURE 22


VSEPR Theory and Unshared Electron PairsAmmonia, NH3, and water, H2O, are examples of molecules inwhich the central atom has both shared and unshared electron pairs(see Table 5 on the next page for their Lewis structures). How doesVSEPR theory account for the geometries of these molecules?

The Lewis structure of ammonia shows that in addition to thethree electron pairs it shares with the three hydrogen atoms, thecentral nitrogen atom has one unshared pair of electrons.

VSEPR theory postulates that the lone pair occupies space aroundthe nitrogen atom just as the bonding pairs do. Thus, as in an AB4molecule, the electron pairs maximize their separation by assumingthe four corners of a tetrahedron. Lone pairs do occupy space, but ourdescription of the observed shape of a molecule refers to the positionsof atoms only. Consequently, as shown in Figure 22a, the moleculargeometry of an ammonia molecule is that of a pyramid with a trian-gular base.The general VSEPR formula for molecules such as ammo-nia is AB3E, where E represents the unshared electron pair.

A water molecule has two unshared electron pairs. It is an AB2E2molecule. Here, the oxygen atom is at the center of a tetrahedron,with two corners occupied by hydrogen atoms and two by theunshared pairs (Figure 22b). Again, VSEPR theory states that thelone pairs occupy space around the central atom but that the actualshape of the molecule is determined by the positions of the atomsonly. In the case of water, this results in a “bent,”or angular, molecule.

N HHH

H

(a) Ammonia, NH3

107°

(b) Water, H2O

105°

N

H

H

O

H

H

FIGURE 22 The locations ofbonds and unshared electrons areshown for molecules of (a) ammoniaand (b) water. Although unsharedelectrons occupy space around thecentral atoms, the shapes of the mole-cules depend only on the position of the molecules’ atoms, as clearlyshown by the ball-and-stick models.

www.scilinks.orgTopic: VSEPR TheoryCode: HC61621

In Figure 22b, note that the bond angles in ammonia and water aresomewhat less than the 109.5° bond angles of a perfectly tetrahedralmolecule. These angles are smaller because the unshared electron pairsrepel electrons more strongly than do bonding electron pairs.

Table 5 also includes an example of an AB2E type molecule. Thistype of molecule results when a central atom forms two bonds andretains one unshared electron pair.

Finally, in VSEPR theory, double and triple bonds are treated in thesame way as single bonds. And polyatomic ions are treated similarly tomolecules. (Remember to consider all of the electron pairs present inany ion or molecule.) Thus, Lewis structures and Table 5 can be usedtogether to predict the shapes of polyatomic ions as well as moleculeswith double or triple bonds.

C H A P T E R 6200

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SECTION 5

Table 5 You may return to the useof nails and clay to illustrate the mol-ecular shapes in this table. Once stu-dents can visualize how the electronpairs repel one another, the three-dimensional geometry is morestraightforward.

It might be useful to point out thatAB2E structures are similar to AB3structures, but with an unshared elec-tron pair in place of one of thebonds. Similarly, AB3E and AB2E2structures are similar to AB4 struc-tures, but with one and two unsharedelectron pairs, respectively, in place ofbonds. This will help students to rec-ognize the hybridizations later in thechapter. sp2 for AB3 and AB2E; sp3

for AB4, AB3E, and AB2E2. The mole-cule types AB5 and AB6 do not followthe octet rule. These two types ofstructures involve expanded valenceshells and sp3d and sp3d 2 orbitals.

Teaching Tip

TABLE STRATEGY

Atoms Lonebonded to pairs of Type of Formula Lewis

Molecular shape central atom electrons molecule example structure

Linear 2 0 AB2 BeF2

3 0 AB3 BF3

Bent or Angular 2 1 AB2E ONF

Tetrahedral 4 0 AB4 CH4

3 1 AB3E NH3

Bent or Angular 2 2 AB2E2 H2O

5 0 AB5 PCl5

Octahedral 6 0 AB6 SF6 S

F

F F

F

F

F

ClIP

Cl

Cl Cl

Cl

Trigonal-bipyramidal

OHH

H HH

NTrigonal-pyramidal

HICIH

H

H

N

FO

FB

F

FTrigonal-planar

FIBeIF

TABLE 5 VSEPR Theory and Molecular Geometry

90o

120o

90o

90o

✔

Hybridization

VSEPR theory is useful for explaining the shapes of molecules. However,it does not reveal the relationship between a molecule’s geometry and theorbitals occupied by its bonding electrons. To explain how the orbitals ofan atom become rearranged when the atom forms covalent bonds, a dif-ferent model is used. This model is called hybridization, which is the mix-ing of two or more atomic orbitals of similar energies on the same atom toproduce new hybrid atomic orbitals of equal energies.

Methane, CH4, provides a good example of how hybridization is usedto explain the geometry of molecular orbitals. The orbital notation for acarbon atom shows that it has four valence electrons, two in the 2sorbital and two in 2p orbitals.

C ↑↓1s 2s

2p

↑↓ ↑ ↑


SECTION 5

F-1 Use VSEPR theory to predict theshape of the each of the followingmolecules and ions:

a. AsF5b. SeF6c. CF4d. NO3

–

Ans. a. trigonal-bipyramidalb. octahedralc. tetrahedrald. trigonal-planar

Practice Answers1. a. bent or angular

b. trigonal-pyramidal

Neither Lewis structures nor VSEPRtheory can explain why the twounpaired valence electrons in carbonresult in four bonds of equal energy.Hybridization theory explains that thetwo electrons in carbon’s 2s orbitalare also used for bonding, allowingall four valence electrons around car-bon to form covalent bonds.

Teaching Tip


Go to go.hrw.com formore practice problemsthat ask you to useVSEPR theory.

Keyword: HC6BNDX

a. The Lewis structure of carbon dioxide shows two carbon-oxygen double bonds and nounshared electron pairs on the carbon atom. To simplify the molecule’s Lewis structure,we represent the covalent bonds with lines instead of dots.

This is an AB2 molecule, which is linear.

b. The Lewis structure of a chlorate ion shows three oxygen atoms and an unshared pair ofelectrons surrounding a central chlorine atom. Again, lines are used to represent thecovalent bonds.

The chlorate ion is an AB3E type. It has trigonal-pyramidal geometry, with the three oxy-gen atoms at the base of the pyramid and the chlorine atom at the top.

O

Cl−

OO

OJCJO

SOLUTION

SAMPLE PROBLEM F

1. Use VSEPR theory to predict the molecular geometries of themolecules whose Lewis structures are given below.

a. b.Cl

ClIPIClFISIF


a. Use VSEPR theory to predict the shape of a molecule of carbon dioxide, CO2.b. Use VSEPR theory to predict the shape of a chlorate ion, ClO3

–.

PRACTICE

✔

We know from experiments that a methane molecule has tetrahedralgeometry. How does carbon form four equivalent, tetrahedrallyarranged covalent bonds by orbital overlap with four other atoms?

Two of carbon’s valence electrons occupy the 2s orbital, and twooccupy the 2p orbitals. Recall that the 2s orbital and the 2p orbitals havedifferent shapes. To achieve four equivalent bonds, carbon’s 2s andthree 2p orbitals hybridize to form four new, identical orbitals called sp3

orbitals. The superscript 3 indicates that three p orbitals were includedin the hybridization; the superscript 1 on the s is understood. The sp3

orbitals all have the same energy, which is greater than that of the 2sorbital but less than that of the 2p orbitals, as shown in Figure 23.

Hybrid orbitals are orbitals of equal energy produced by the combina-tion of two or more orbitals on the same atom. The number of hybridorbitals produced equals the number of orbitals that have combined.Bonding with carbon sp3 orbitals is illustrated in Figure 24a for a mol-ecule of methane.

Hybridization also explains the bonding and geometry of manymolecules formed by Group 15 and 16 elements. The sp3 hybridizationof a nitrogen atom ([He]2s22p3) yields four hybrid orbitals—oneorbital containing a pair of electrons and three orbitals that each con-tain an unpaired electron. Each unpaired electron is capable of form-ing a single bond, as shown for ammonia in Figure 24b. Similarly, twoof the four sp3 hybrid orbitals on an oxygen atom ([He]2s22p4) areoccupied by two electron pairs and two are occupied by unpaired elec-trons. Each unpaired electron can form a single bond, as shown forwater in Figure 24c.

Carbon’s orbitalsbefore hybridization

Carbon’s orbitals aftersp3 hybridization

Hybridization

Ene

rgy 2p

↑ ↑

sp3

↑ ↑ ↑ ↑

2s↑↓

1s↑↓

1s↑↓

C H A P T E R 6202

SECTION 1

202

SECTION 5

H

O

H

H

N

H

H

(a) Methane, CH4 (c) Water, H2O(b) Ammonia, NH3

H

C

H

H

H

CHAPTER CONNECTION

Be sure students are familiar withorbital notation, which was intro-duced in Chapter 4.

CommonMisconceptionThe labeling of hybrid orbitals is different from that used for electronconfigurations. The superscripts inhybrid orbitals refer to the number of orbitals of that particular type that are involved in the orbital blending. The superscripts in electronconfigurations refer to the number of electrons in that type of orbital.

FIGURE 23 The sp3 hybridizationof carbon’s outer orbitals combinesone s and three p orbitals to formfour sp3 hybrid orbitals. Wheneverhybridization occurs, the resultinghybrid orbitals are at an energy levelbetween the levels of the orbitalsthat have combined.

FIGURE 24 Bonds formed by the overlap of the 1s orbitals ofhydrogen atoms and the sp3 orbitalsof (a) carbon, (b) nitrogen, and (c) oxygen. For the sake of clarity,only the hybrid orbitals of the cen-tral atoms are shown.

The linear geometry of molecules such as beryllium fluoride, BeF2,(see Table 5) is made possible by hybridization involving the s orbitaland one available empty p orbital to yield sp hybrid orbitals. Thetrigonal-planar geometry of molecules such as boron fluoride, BF3, ismade possible by hybridization involving the s orbital, one singly occu-pied p orbital, and one empty p orbital to yield sp2 hybrid orbitals. Thegeometries of sp, sp2, and sp3 hybrid orbitals are summarized in Table 6.

Intermolecular Forces

As a liquid is heated, the kinetic energy of its particles increases. At theboiling point, the energy is sufficient to overcome the force of attractionbetween the liquid’s particles. The particles pull away from each otherand enter the gas phase. Boiling point is therefore a good measure ofthe force of attraction between particles of a liquid. The higher the boil-ing point, the stronger the forces between particles.

The forces of attraction between molecules are known as intermolec-ular forces. Intermolecular forces vary in strength but are generally weaker than bonds that join atoms in molecules, ions in ionic com-pounds, or metal atoms in solid metals. Compare the boiling points of themetals and ionic compounds in Table 7 on the next page with those ofthe molecular substances listed. Note that the values for ionic com-pounds and metals are much higher than those for molecular substances.


SECTION 5

The number of orbitals involved inhybridization determines the geometryof a molecule. There must be spacearound the atom for every hybridorbital; each orbital will be as far apartas possible from every other orbital.

Teaching Tip✔


INTERPRETINGVOCABULARY Point out themeanings of the prefixes inter- andintra-.

inter-: between or among;together

intra-: within; insideExplain that molecules are affect-ed by both intermolecular forcesand intramolecular forces.Intermolecular forces are thosebetween molecules. The attractionbetween the positive part of onepolar molecule and the negativepart of another are intermolecularforces. Intramolecular forces arethose within the molecule. A cova-lent bond within a molecule is anintramolecular force. Ask studentsto list other examples of inter- andintramolecular forces. Also havestudents brainstorm a list of otherwords that use these prefixes andtheir definitions.

Atomic Type of Number oforbitals hybridization hybrid orbitals Geometry

s, p sp 2

s, p, p sp2 3

s, p, p, p sp3 4

TABLE 6 Geometry of Hybrid Orbitals

Linear

Trigonal-planar

Tetrahedral

180°

120°

109.5°

Molecular Polarity and Dipole-Dipole ForcesThe strongest intermolecular forces exist between polar molecules.Polar molecules act as tiny dipoles because of their uneven charge dis-tribution. A dipole is created by equal but opposite charges that are sep-arated by a short distance. The direction of a dipole is from the dipole’spositive pole to its negative pole. A dipole is represented by an arrowwith a head pointing toward the negative pole and a crossed tail situ-ated at the positive pole. The dipole created by a hydrogen chloridemolecule, which has its negative end at the more electronegative chlo-rine atom, is indicated as follows.

The negative region in one polar molecule attracts the positiveregion in adjacent molecules, and so on throughout a liquid or solid.Theforces of attraction between polar molecules are known as dipole-dipoleforces. These forces are short-range forces, acting only between nearbymolecules.The effect of dipole-dipole forces is reflected, for example, bythe significant difference between the boiling points of iodine chloride,IICl, and bromine, BrIBr. The boiling point of polar iodine chloride is97°C, whereas that of nonpolar bromine is only 59°C. The dipole-dipoleforces responsible for the relatively high boiling point of ICl are illus-trated schematically in Figure 25.

⎯→HICl

C H A P T E R 6204

SECTION 1

204

SECTION 5

Remind students that boiling point isa measure of the force of attractionbetween molecules in covalent com-pounds, between ions in ionic com-pounds, and between atoms in metals.

Table 7 Have students use the boil-ing points on this table to assess therelative strengths of the forces ofattraction at work in molecular, ionic,and metallic substances.

CommonMisconceptionSome might infer from the table on this page that covalent bondinginvolves the weakest force of attrac-tion. In fact, covalent bonding involvesthe strongest force of attraction—asevidenced by the fact that boiling adiamond is more difficult than boilingany metal. The attraction that mustbe overcome to boil a molecularcompound is the intermolecularattraction, not covalent bonding.

TABLE STRATEGY

Teaching Tip Bonding type Substance bp (1 atm, °C)

Nonpolar-covalent H2 −253(molecular) O2 −183

Cl2 −34Br2 59CH4 −164CCl4 77C6H6 80

Polar-covalent PH3 −88(molecular) NH3 −33

H2S −61H2O 100HF 20HCl −85ICl 97

Ionic NaCl 1413MgF2 2239

Metallic Cu 2567Fe 2750W 5660

TABLE 7 Boiling Points and Bonding Types

✔

Computational ChemistComputational chemistry is the study ofmolecules, their properties, and the inter-action between molecules using mathe-matical equations that are based on thelaws of quantum mechanics and whichdescribe the motion of the electrons.Today, widely-available software pack-ages exist that allow chemists to solvethese equations for molecules understudy.

Computational chemists combinetheir expertise in mathematics, theircomputer skills, and their love of chem-istry. Computational chemistry has beenused in industry to aid in the discoveryof new pharmaceutical drugs and ofnew catalysts. Computational chemistsare employed in all areas of chemistryand work closely with experimentalchemists.

The polarity of diatomic molecules such as ICl is determined by justone bond. For molecules containing more than two atoms, molecularpolarity depends on both the polarity and the orientation of each bond.A molecule of water, for example, has two hydrogen-oxygen bonds in which the more-electronegative oxygen atom is the negative pole of each bond. Because the molecule is bent, the polarities of these two bonds combine to make the molecule highly polar, as shown inFigure 26. An ammonia molecule is also highly polar because the dipolesof the three nitrogen-hydrogen bonds are additive, combining to createa net molecular dipole. In some molecules, individual bond dipoles can-cel one another, causing the resulting molecular polarity to be zero.Carbon dioxide and carbon tetrachloride are molecules of this type.

A polar molecule can induce a dipole in a nonpolar molecule by tem-porarily attracting its electrons. The result is a short-range intermolecu-lar force that is somewhat weaker than the dipole-dipole force. Theforce of an induced dipole accounts for the solubility of nonpolar O2 inwater. The positive pole of a water molecule attracts the outer electrons


SECTION 5

Visual StrategyEmphasize to students

that a large difference in electroneg-ativities between the atoms in a mol-ecule does not necessarily mean thatthe molecule will be polar. Dipoleswithin a molecule can cancel eachother out, resulting in a nonpolarmolecule, as in carbon tetrachloride.Polarity is a result of both the exis-tence of polar covalent bonds and thenature of the molecule’s geometry.

FIGURE 26

GENERAL

� +

� –

(a) Water, H2O Ammonia, NH3

� +

� –

Key:

I

Cl

Dipole-dipoleforces� +

� + � +

� +� –� –

� – � –

Carbon dioxide, CO2(no molecular dipole)

Carbon tetrachloride, CCl4(no molecular dipole)

(b)

FIGURE 25 Ball-and-stick modelsillustrate the dipole-dipole forcesbetween molecules of iodine chlo-ride, ICl. In each molecule, the high-ly electronegative chlorine atom hasa partial negative charge, leavingeach iodine atom with a partial posi-tive charge. Consequently, the nega-tive and positive ends of neighboringmolecules attract each other.

FIGURE 26 (a) The bond polari-ties in a water or an ammonia mole-cule are additive, causing themolecule as a whole to be polar.(b) In molecules of carbon tetrachlo-ride and carbon dioxide, the bondpolarities extend equally and sym-metrically in different directions,canceling each other’s effect andcausing each molecule as a whole to be nonpolar.

of an adjacent oxygen molecule. The oxygen molecule, then, has aninduced negative pole on the side toward the water molecule and an induced positive pole on the opposite side. The result is an attractionto the water molecule, as shown in Figure 27.

Hydrogen BondingSome hydrogen-containing compounds, such as hydrogen fluoride(HF), water (H2O), and ammonia (NH3), have unusually high boilingpoints. This is explained by the presence of a particularly strong type ofdipole-dipole force. In compounds containing HIF, HIO, or HINbonds, the large electronegativity differences between hydrogen atomsand fluorine, oxygen, or nitrogen atoms make the bonds connectingthem highly polar. This gives the hydrogen atom a positive charge thatis almost half as large as that of a proton. Moreover, the small size of thehydrogen atom allows the atom to come very close to an unshared pairof electrons on an adjacent molecule. The intermolecular force in whicha hydrogen atom that is bonded to a highly electronegative atom isattracted to an unshared pair of electrons of an electronegative atom in anearby molecule is known as hydrogen bonding.

Hydrogen bonds are usually represented by dotted lines connectingthe hydrogen-bonded hydrogen to the unshared electron pair of theelectronegative atom to which it is attracted, as illustrated for water inFigure 28. The effect of hydrogen bonding can be seen by comparingthe boiling points in Table 7. Look at phosphine, PH3, compared withhydrogen-bonded ammonia, NH3. How does hydrogen sulfide, H2S,compare with hydrogen-bonded water, H2O?

C H A P T E R 6206

SECTION 1

206

SECTION 5

CommonMisconceptionMany students think of hydrogenbonding as a separate type of inter-molecular force. In fact, it is a partic-ularly strong dipole-dipole force.

Did You Know?Hydrogen bonds are responsible forholding DNA in its helical shape.

Answer to In-TextQuestionH2S boils at a much lower tempera-ture than does H2O. This is explainedby the fact that oxygen is so electro-negative that hydrogen bondingoccurs between H2O molecules.

Water, H2O Oxygen, O2

� +

� +

� –

� +

� +

� +

� –

� –

FIGURE 27 Dipole-induceddipole interaction. The positive poleof a water molecule causes a tempo-rary change in the electron distribu-tion of an oxygen molecule. Thenegative pole induced in the oxygenmolecule is then attracted to thepositive pole of the water molecule.

FIGURE 28 Space-filling models illustrate hydrogen bondingbetween water molecules. The dotted lines indicate the attractionbetween electronegative oxygenatoms and electropositive hydrogenatoms of neighboring molecules.

www.scilinks.orgTopic: Hydrogen BondingCode: HC60777

London Dispersion ForcesEven noble-gas atoms and molecules that are nonpolar experience aweak intermolecular attraction. In any atom or molecule—polar ornonpolar—the electrons are in continuous motion. As a result, at anyinstant the electron distribution may be slightly uneven. The momen-tary, uneven charge creates a positive pole in one part of the atom ormolecule and a negative pole in another. This temporary dipole canthen induce a dipole in an adjacent atom or molecule. The two are heldtogether for an instant by the weak attraction between the temporarydipoles, as illustrated in Figure 29. The intermolecular attractions result-ing from the constant motion of electrons and the creation of instanta-neous dipoles are called London dispersion forces, after Fritz London,who first proposed their existence in 1930.

London forces act between all atoms and molecules. But they are theonly intermolecular forces acting among noble-gas atoms and nonpolarmolecules. This fact is reflected in the low boiling points of the noblegases and nonpolar molecular compounds listed in Table 7. BecauseLondon forces are dependent on the motion of electrons, their strengthincreases with the number of electrons in the interacting atoms or mol-ecules. In other words, London forces increase with increasing atomic ormolar mass. This trend can be seen by comparing the boiling points ofthe gases helium, He, and argon, Ar; hydrogen, H2, and oxygen, O2; andchlorine, Cl2, and bromine, Br2.


SECTION 5

CommonMisconceptionSome students might think that amolecule experiences either dipole-dipole forces or London dispersionforces with another molecule. In fact,all molecules experience London dis-persion forces. In polar molecules,London dispersion forces are insignif-icant relative to dipole-dipole forces.

1. VSEPR theory and hybridizationtheory

2. a.

b.

c.

3. Factors include the number ofbonds formed by each atom in themolecule, the number of lone pairs of electrons on the atoms, and thehybridization of some of the atoms’orbitals.

4. One s orbital and three p orbitalshave combined to form four hybridorbitals of equal energy.

5. Hydrogen bonding is responsiblefor water’s high boiling point. In gen-eral, boiling point is a measure of theamount of energy required to over-come intermolecular attractions.Hydrogen bonding, a particularlystrong dipole-dipole force, causes apowerful attraction between watermolecules, which results in a highboiling point.


trigonal-planar

BClCl

Cl

tetrahedral

CII

II

SO

bent or angular

O

SECTION REVIEW

FIGURE 29 When an instanta-neous, temporary dipole develops in a helium atom, it induces a dipolein a neighboring atom.

Momentary dipolein one helium atom

Dipole induced inneighboring atom

Weakattractive

force

+2 +2

–

–+2

–

–+2

–

–– –

1. What two theories can be used to predict molecu-lar geometry?

2. Draw the Lewis structure and predict the molecu-lar geometry of the following molecules:

a. SO2 b. CI4 c. BCl33. What factors affect the geometry of a molecule?

4. Explain what is meant by sp3 hybridization.

5. What type of intermolecular force contributes tothe high boiling point of water? Explain.

Critical Thinking

6. INFERRING RELATIONSHIPS What experimentalproperty directly correlates with the strength ofthe intermolecular forces? Briefly explain youranswer.

SECTION REVIEW

C H A P T E R 6208208

1. A chemical bond is a linkbetween atoms resulting fromthe mutual attraction of theirnuclei and electrons.

2. The three major types of chemi-cal bonding are ionic, covalent,and metallic. In ionic bonding,large numbers of oppositelycharged ions join because ofmutual electrical attraction. Incovalent bonding, atoms join bysharing electron pairs. In metallicbonding, atoms join through anattraction to a sea of valenceelectrons.

3. In general, the greater the elec-tronegativity difference betweentwo atoms is, the more ionic thebond between them is.

4. a. Polar refers to bonds that havean uneven distribution of charge.b. In a polar-covalent bond thereis an unequal attraction for theelectron pair, resulting in one ofthe bonded atoms possessing apartial negative charge and theother atom possessing a partialpositive charge. In a nonpolar-covalent bond the electron pair is shared equally by the bondedatoms.

5. In general, atoms will form achemical bond if their potentialenergy is lowered by doing so.

6. See page 217A.

7. K and Cl, K and Br, Si and Cl, Sand O, H and I, C and H, Se and S

8. See page 217A.

9. Bonding is stronger between theions in sodium chloride becauseits lattice energy is greater (morenegative). Greater lattice energyindicates stronger ionic bonding.

REVIEW ANSWERS

CHAPTER REVIEW

C H A P T E R H I G H L I G H T S

• Most atoms are chemically bonded to other atoms. The threemajor types of chemical bonding are ionic, covalent, and metallic.

• In general, atoms of metals bond ionically with atoms of non-metals, atoms of metals bond metallically with each other, andatoms of nonmetals bond covalently with each other.

chemical bondionic bondingcovalent bondingnonpolar-covalent bondpolarpolar-covalent bond

Vocabulary

Introduction to Chemical Bonding

• Atoms in molecules are joined by covalent bonds. In a cova-lent bond, two atoms share one or more pairs of electrons.

• The octet rule states that many chemical compounds tend toform bonds so that each atom shares or has eight electrons inits highest occupied energy level.

• Bonding within many molecules and ions can be indicated by aLewis structure. Molecules or ions that cannot be correctlyrepresented by a single Lewis structure are represented by res-onance structures.

moleculemolecular compoundchemical formulamolecular formulabond energyelectron-dot notation

Lewis structurestructural formulasingle bondmultiple bondresonance

Vocabulary

Covalent Bonding and Molecular Compounds

ionic compoundformula unitlattice energypolyatomic ion

Vocabulary • An ionic compound is a three-dimensional network of positiveand negative ions mutually attracted to one another.

• Ionic compounds tend to be harder and more brittle and tohave higher boiling points than materials containing only cova-lently bonded atoms.

Ionic Bonding and Ionic Compounds

• The “electron sea” formed in metallic bonding gives metals theirproperties of high electrical and thermal conductivity, malleabili-ty, ductility, and luster.

Metallic Bonding

metallic bondingmalleabilityductility

Vocabulary

• VSEPR theory is used to predict the shapes of molecules basedon the fact that electron pairs strongly repel each other.

• Hybridization theory is used to predict the shapes of moleculesbased on the fact that orbitals within an atom can mix to formorbitals of equal energy.

• Intermolecular forces include dipole-dipole forces and Londondispersion forces. Hydrogen bonding is a special case of dipole-dipole forces.

Molecular Geometry

VSEPR theoryhybridizationhybrid orbitalsdipolehydrogen bondingLondon dispersion forces

Vocabulary

C H A P T E R R E V I E W


CHAPTER REVIEW

10. A molecule is a neutral group oftwo or more atoms—usuallynonmetals—held together bycovalent bonds.

11. a. The distance at which poten-tial energy is at a minimum, bondlength, is the point at which thereis a balance between attractionand repulsion between atoms ina covalent bond.b. In general, higher bond ener-gies correspond to shorter bondlengths.

12. The electrons in a covalent bondoccupy overlapping orbitals; eachelectron is free to occupy eitherof the orbitals, but both are morelikely to be in the space betweenthe nuclei of the bonded atoms.

13. a pair of electrons that is notinvolved in bonding but insteadbelongs exclusively to one atom

14. A noble-gas configuration corre-sponds to a set of outer s and porbitals that is completely filledwith a total of eight electrons.These eight outer electrons arereferred to as an octet. Atomsthat possess a noble-gas configu-ration are very stable becausethe potential energy of their elec-trons is relatively low. The octetrule states that elements willgain, lose, or share electrons toform a noble-gas configuration.

16. The least-electronegative atom isusually the central atom (exceptfor hydrogen, which is never cen-tral). If carbon is present, howev-er, it is usually the central atom,regardless of what other atomsare in the molecule.

17. A single bond involves one pairof electrons, as in the bondbetween hydrogen and oxygen inwater, H2O. A double bond

15. a. 1b. 7c. 2

d. 6e. 3

f. 5g. 4

Introduction to ChemicalBondingSECTION 1 REVIEW

1. What is a chemical bond?2. Identify and define the three major types

of chemical bonding.3. What is the relationship between electronegativ-

ity and the ionic character of a chemical bond?4. a. What is the meaning of the term polar, as

applied to chemical bonding?b. Distinguish between polar-covalent and

nonpolar-covalent bonds.5. In general, what determines whether atoms will

form chemical bonds?

PRACTICE PROBLEMS

6. Determine the electronegativity difference,the probable bond type, and the more-electronegative atom with respect to bondsformed between the following pairs of atoms.(Hint: See Sample Problem A.)a. H and Ib. S and Oc. K and Brd. Si and Cle. K and Clf. Se and Sg. C and H

7. List the bonding pairs described in item 6 inorder of increasing covalent character.

8. Use orbital notation to illustrate the bonding in each of the following molecules:a. chlorine, Cl2b. oxygen, O2c. hydrogen fluoride, HF

9. The lattice energy of sodium chloride, NaCl, is −787.5 kJ/mol. The lattice energy of potassiumchloride, KCl, is −715 kJ/mol. In which com-pound is the bonding between ions stronger?Why?

Covalent Bonding and MolecularCompoundsSECTION 2 REVIEW

10. What is a molecule?11. a. What determines bond length?

b. In general, how are bond energies and bondlengths related?

12. Describe the general location of the electrons ina covalent bond.

13. As applied to covalent bonding, what is meantby an unshared or lone pair of electrons?

14. Describe the octet rule in terms of noble-gasconfigurations and potential energy.

15. Determine the number of valence electrons inan atom of each of the following elements:a. Hb. Fc. Mgd. Oe. Alf. Ng. C

16. When drawing Lewis structures, which atom isusually the central atom?

17. Distinguish between single, double, and triplecovalent bonds by defining each and providingan illustration of each type.

18. In writing Lewis structures, how is the need formultiple bonds generally determined?

PRACTICE PROBLEMS

19. Use electron-dot notation to illustrate the num-ber of valence electrons present in one atom ofeach of the following elements. (Hint: SeeSample Problem B.)a. Lib. Cac. Cld. Oe. Cf. Pg. Alh. S

210

CHAPTER REVIEW CHAPTER REVIEW

20. Use electron-dot structures to demonstrate theformation of ionic compounds involving the fol-lowing elements:a. Na and Sb. Ca and Oc. Al and S

21. Draw Lewis structures for each of the followingmolecules. (Hint: See Sample Problem D.)a. contains one C and four F atomsb. contains two H and one Se atomc. contains one N and three I atomsd. contains one Si and four Br atomse. contains one C, one Cl, and three H atoms

22. Determine the type of hybrid orbitals formedby the boron atom in a molecule of boron fluo-ride, BF3.

23. Draw Lewis structures for each of the followingmolecules. Show resonance structures, if theyexist.a. O2b. N2c. COd. SO2

24. Draw Lewis structures for each of the followingpolyatomic ions. Show resonance structures, ifthey exist.a. OH−

b. H3C2O2−

c. BrO3−

Ionic Bonding and IonicCompoundsSECTION 3 REVIEW

25. a. What is an ionic compound?b. In what form do most ionic compounds

occur?26. a. What is a formula unit?

b. What are the components of one formulaunit of CaF2?

27. a. What is lattice energy?b. In general, what is the relationship between

lattice energy and the strength of ionic bonding?

28. a. In general, how do ionic and molecular com-pounds compare in terms of melting points,boiling points, and ease of vaporization?

b. What accounts for the observed differencesin the properties of ionic and molecular compounds?

c. Cite three physical properties of ionic compounds.

29. a. What is a polyatomic ion?b. Give two examples of polyatomic ions.c. In what form do such ions often occur

in nature?

Metallic BondingSECTION 4 REVIEW

30. a. How do the properties of metals differ fromthose of both ionic and molecular compounds?

b. What specific property of metals accounts for their unusual electrical conductivity?

31. What properties of metals contribute to theirtendency to form metallic bonds?

32. a. What is metallic bonding?b. How can the strength of metallic bonding

be measured?

Molecular GeometrySECTION 5 REVIEW

33. a. How is the VSEPR theory used to classifymolecules?

b. What molecular geometry would be expectedfor F2 and HF?

34. According to the VSEPR theory, what moleculargeometries are associated with the followingtypes of molecules?a. AB2b. AB3c. AB4d. AB5e. AB6

35. Describe the role of each of the following in pre-dicting molecular geometries:a. unshared electron pairsb. double bonds

C H A P T E R 6210

involves two electron pairs, as inthe bond between the carbonand oxygen atoms in a moleculeof carbon dioxide, CO2. A triplebond involves three electronpairs, as in the bond betweencarbon and nitrogen in hydrogencyanide, HCN.

18. A multiple bond is needed whenthere are not enough valenceelectrons to complete octets byadding unshared pairs.

19–21. See page 217B.

22. sp2 hybrid orbitals

23–24. See page 217B.

25. a. An ionic compound is com-posed of cations and anions suchthat the total positive and nega-tive charges are equal.b. Most ionic compounds occurnaturally as crystalline solids.

26. a. the simplest collection of atomsfrom which an ionic compound’sformula can be establishedb. one calcium ion, Ca2+, andtwo fluoride ions, F−

27. a. the energy released when onemole of an ionic crystalline com-pound is formed from gaseous ionsb. The greater the lattice energy,the stronger the ionic bonding.

28. a. Ionic compounds have highermelting and boiling points thanmolecular compounds do, andthey do not vaporize at roomtemperature.b. The differences in the proper-ties of ionic and molecular com-pounds are generally a result ofdifferences in how strongly thecompound’s basic units are heldtogether.c. hardness, brittleness, electricalconductivity in the molten state

29. a. a charged group of covalentlybonded atomsb. Some common examples ofpolyatomic anions include the

CHAPTER REVIEW


nitrate ion, NO3−, the ammonium

ion, NH4+, the sulfate ion, SO4

2−,and the phosphate ion, PO4

3−.c. Polyatomic ions combine withions of opposite charge to formionic compounds.

30. a. Metals are better heat conduc-tors than ionic or molecular com-pounds. In the solid state, metalsare more easily deformed and arebetter electrical conductors thansolid ionic or molecular com-pounds. Unlike ionic and molecu-lar compounds, metals are alsoshiny in appearance.b. Metals are good electricalconductors because of the pres-ence of highly mobile electronswithin the bonding networks oftheir atoms.

31. Most contain sparsely populatedoutermost orbitals, they have lowionization energies, and theyhave low electronegativities.

32. a. Metallic bonding results fromthe attraction between metalatoms and a sea of surroundingelectrons.b. A metal’s enthalpy of vapor-ization is a measure of thestrength of the metal’s bonding.

33. a. According to VSEPR theory,the shapes of molecules are clas-sified based on the number ofbonding electron pairs and lonepairs that surround a molecule’scentral atom.b. Both molecules are linear.

34. a. linearb. trigonal-planarc. tetrahedrald. trigonal-bipyramidale. octahedral

35. a. Unshared electron pairs occu-py space as bonded electrons do,but they are not part of the visu-alized molecular geometry.b. Double (and triple) bonds aretreated the same as single bonds.

CHAPTER REVIEW

36. a. What are hybrid orbitals?b. What determines the number of hybrid

orbitals produced by the hybridization of an atom?

37. a. What are intermolecular forces?b. In general, how do these forces compare in

strength with those in ionic and metallic bonding?

c. What types of molecules have the strongestintermolecular forces?

38. What is the relationship between electronegativi-ty and the polarity of a chemical bond?

39. a. What are dipole-dipole forces?b. What determines the polarity of a molecule?

40. a. What is meant by an induced dipole?b. What is the everyday importance of this type

of intermolecular force?41. a. What is hydrogen bonding?

b. What accounts for its extraordinary strength?42. What are London dispersion forces?

PRACTICE PROBLEMS

43. According to the VSEPR theory, what moleculargeometries are associated with the followingtypes of molecules?a. AB3Eb. AB2E2c. AB2E

44. Use hybridization to explain the bonding inmethane, CH4.

45. For each of the following polar molecules, indi-cate the direction of the resulting dipole:a. HIFb. HIClc. HIBrd. HII

46. Determine whether each of the following bondswould be polar or nonpolar:a. HIHb. HIOc. HIFd. BrIBre. HIClf. HIN

47. On the basis of individual bond polarity and ori-entation, determine whether each of the follow-ing molecules would be polar or nonpolar:a. H2Ob. I2c. CF4d. NH3e. CO2

48. Draw a Lewis structure for each of the followingmolecules, and then use the VSEPR theory topredict the molecular geometry of each:a. SCl2b. PI3c. Cl2Od. NH2Cle. SiCl3Brf. ONCl

49. Draw a Lewis structure for each of the followingpolyatomic ions, and then use VSEPR theory todetermine the geometry of each:a. NO3

−

b. NH4+

c. SO42−

d. ClO2−

50. Arrange the following pairs from strongest toweakest attraction:a. polar molecule and polar moleculeb. nonpolar molecule and nonpolar moleculec. polar molecule and iond. ion and ion

51. Determine the geometry of the following molecules:a. CCl4b. BeCl2c. PH3

52. What types of atoms tend to form the followingtypes of bonding?a. ionicb. covalentc. metallic

MIXED REVIEW

212

36. a. Hybrid orbitals are identicallyshaped orbitals of equal energythat are produced by the mixingof two or more atomic orbitals ofsimilar, but not identical, ener-gies on the same atom.b. The number of hybrid orbitalsproduced is always equal to thenumber of atomic orbitals thathave combined.

37. a. Intermolecular forces are theforces of attraction between molecules.b. Intermolecular forces areweaker than the forces involvedin ionic and metallic bonding.c. The strongest intermolecularforces occur between polar molecules.

38. The larger the difference in elec-tronegativities of the atoms inthe bond, the greater the polarityof that bond.

39. a. Dipole-dipole forces are theforces of attraction betweenpolar molecules.b. The overall polarity of a mole-cule is determined by the polarityof the molecule’s individualbonds as well as the orientationof the bonds with respect to oneanother.

40. a. An induced dipole is aninstantaneous dipole that is pro-duced in a nonpolar moleculewhen the molecule’s electronsare momentarily attracted by apolar molecule.b. Induced dipoles account forthe solubility of nonpolar com-pounds, such as oxygen, in polarcompounds, such as water.

41. a. Hydrogen bonding is a particu-larly strong dipole-dipole forcethat occurs among moleculescontaining hydrogen atoms andand highly electronegative atoms,such as those of N,O, and F.b. Because of the large elec-tronegativity difference between

CHAPTER REVIEW CHAPTER REVIEW

53. What happens to the energy level and stabilityof two bonded atoms when they are separatedand become individual atoms?

54. Draw the three resonance structures for sulfurtrioxide, SO3.

55. a. How do ionic and covalent bonding differ?b. How does an ionic compound differ from a

molecular compound?c. How does an ionic compound differ from a

metal?56. Write the electron-dot notation for each of the

following elements:a. He d. Pb. Cl e. Bc. O

57. Write the structural formula for methanol,CH3OH.

58. How many K+ and S2− ions would be in one for-mula unit of the ionic compound formed bythese ions?

59. Explain metallic bonding in terms of the sparse-ly populated outermost orbitals of metal atoms.

60. Explain the role of molecular geometry indetermining molecular polarity.

61. How does the energy level of a hybrid orbitalcompare with the energy levels of the orbitals itwas formed from?

62. Aluminum’s enthalpy of vaporization is284 kJ/mol. Beryllium’s enthalpy of vaporiza-tion is 224 kJ/mol. In which element is thebonding stronger between atoms?

63. Determine the electronegativity difference,the probable bonding type, and the more-electronegative atom for each of the followingpairs of atoms:a. Zn and O c. S and Clb. Br and I

64. Draw the Lewis structure for each of the following molecules:a. PCl3 c. CH3NH2b. CCl2F2

65. Write the Lewis structure for BeCl2. (Hint:Beryllium atoms do not follow the octet rule.)

66. Draw a Lewis structure for each of the followingpolyatomic ions and determine their geometries:a. NO2

− c. NH4+

b. NO3−

67. Why do most atoms tend to chemically bondedto other atoms?

68. Inferring Relationships The length of a bondvaries depending on the type of bond formed.Predict and compare the lengths of the carbon-carbon bonds in the following molecules.Explain your answer. (Hint: See Table 2.)

C2H6 C2H4 C2H269. Why does F generally form covalent bonds with

great polarity?70. Explain what is wrong with the following Lewis

structures, and then correct each one.

a.

b.

c.

71. Ionic compounds tend to have higher boilingpoints than covalent substances do. Both ammo-nia, NH3, and methane, CH4, are covalent com-pounds, yet the boiling point of ammonia is130°C higher than that of methane. What mightaccount for this large difference?

72. Figure 18 shows a model for a body-centeredcubic crystal. Review the Properties tables for allof the metals in the Elements Handbook. Whatmetals exist in body-centered cubic structures?

USING THE HANDBOOK

NClCl

Cl

HICJOIH

O

SHIHI

H

HICICIH

H

H

H

HICJCIH HICKCIH

H H

CRITICAL THINKING

C H A P T E R 6212

CHAPTER REVIEW


H and F, N, or O, an atom of hydrogen has a positive chargethat is almost half as large as thatof a proton. This, coupled with thesmall size of the hydrogen atom,results in a very strong dipole-dipole attraction.

42. London dispersion forces areintermolecular forces resultingfrom the creation of induceddipoles.

43. a. trigonal-pyramidalb. bent or angularc. bent or angular

44. The carbon atom contains fourvalence electrons, two in the 2s and two in the 2p orbitals.Hybridization of the 2s orbital andthe three 2p orbitals creates foursp3 hybrid orbitals, each of whichcan bond with a hydrogen atomto form four covalent bonds.

45. The direction of the dipole istoward

46.

47.

48–49. See page 217B.

50. d, c, a, b

51. a. tetrahedralb. linearc. trigonal-pyramidal

52. a. metals and nonmetalsb. nonmetals onlyc. metals only

53. The energies of the atomsincrease and the atoms becomeless stable.

Answers are continued on pages217B–C.

a. polarb. nonpolarc. nonpolar

d. polare. nonpolar

a. nonpolarb. polarc. polar

d. nonpolare. polarf. polar

a. Fb. Cl

c. Brd. I

CHAPTER REVIEW

73. Group 14 of the Elements Handbook contains adiscussion of semiconductors and the band the-ory of metals. How does this model explain theelectrical conductivity of metals?

74. Prepare a report on the work of Linus Pauling.a. Discuss his work on the nature of the chemi-

cal bond.b. Linus Pauling was an advocate of the use of

vitamin C as a preventative for colds.Evaluate Pauling’s claims. Determine if thereis any scientific evidence that indicateswhether vitamin C helps prevent colds.

75. Covalently bonded solids, such as silicon, an ele-ment used in computer components, are harderthan pure metals. Research theories that explainthe hardness of covalently bonded solids andtheir usefulness in the computer industry.Present your findings to the class.

76. Natural rubber consists of long chains of carbonand hydrogen atoms covalently bonded togeth-er. When Goodyear accidentally dropped a mix-ture of sulfur and rubber on a hot stove, theenergy from the stove joined these chainstogether to make vulcanized rubber. Vulcan wasthe Roman god of fire. The carbon-hydrogenchains in vulcanized rubber are held together by two sulfur atoms that form covalent bondsbetween the chains. These covalent bonds arecommonly called disulfide bridges. Exploreother molecules that have such disulfidebridges. Present your findings to the class.

RESEARCH & WRITING

77. Searching for the perfect artificial sweetener—great taste with no Calories—has been the focusof chemical research for some time. Moleculessuch as sucralose, aspartamine, and saccharineowe their sweetness to their size and shape. Onetheory holds that any sweetener must havethree sites that fit into the proper taste buds onthe tongue. This theory is appropriately knownas the “triangle theory.” Research artificialsweeteners to develop a model to show how thetriangle theory operates.

78. Devise a set of criteria that will allow you toclassify the following substances as ionic or non-ionic: CaCO3, Cu, H2O, NaBr, and C (graphite).Show your criteria to your instructor.

79. Performance Assessment Identify 10 commonsubstances in and around your home, and indi-cate whether you would expect these substancesto contain ionic, covalent, or metallic bonds.

ALTERNATIVE ASSESSMENT

Graphing CalculatorClassifying Bond Type

Go to go.hrw.com for a graphing calculatorexercise that asks you to classify bondingtype based on electronegativities of atoms.

Keyword: HC6BNDX

214

1. a. ··Si···· ··

b. ··Sr···· ··

2. a. HH··S

····

··

b. ··O····

··H··C

····

··O····

···H

ANSWERS

CHAPTER REVIEW

Math Tutor DRAWING LEWIS STRUCTURES

1. Draw the electron dot notations for a siliconatom and a strontium atom.

2. Draw Lewis structures for hydrogen sulfide,H2S, and for formic acid, HCO2H.

SAMPLE

PRACTICE PROBLEMS

Drawing Lewis dot structures can help you understand how valence electrons partici-pate in bonding. Dots are placed around the symbol of an element to represent theelement’s valence electrons. For example, carbon has four valence electrons, and itsLewis dot structure is usually written as ··C

······. An atom of fluorine has seven valence

electrons. Fluorine’s Lewis dot structure can be written as ··F····

··. When Lewis structuresfor covalently bonded atoms are written, the dots may be placed as needed to showthe electrons shared in each bond. Most atoms bond in a way that gives them a stableoctet of s and p electrons in the highest energy level. So, whenever possible, dotsshould be arranged in a way that represents a stable octet around each atom.

Draw the Lewis dot structure for a molecule of sulfur dichloride, SCl2.First, write the electron dot notation for each atom.

··Cl····

·· ··Cl····

·· ··S····

··

Next, determine the total number of valence electrons in the atoms.

S 1 × 6e− = 6e−

2Cl 2 × 7e− = 14e−

Total e− = 20e−

Arrange the atoms to form a skeleton structure for the molecule, and place electron pairsbetween atoms to represent covalent bonds. You can predict the arrangement of atoms byfiguring out how many covalent bonds each atom must form in order to achieve a stableoctet. Each chlorine atom, which has 7 valence electrons, must form a single covalent bond.Sulfur, which has 6 valence electrons, must form two covalent bonds.The only possible struc-ture is ClISICl.

Finally, insert dots representing the remaining electrons (16 in this case), in order to giveeach atom an octet.

··Cl····

··S····

··Cl····

··

C H A P T E R 6214

Problem Solving S-

• Hydrogen is an exception to the octet rule because hydrogen has only one electron and becomes stable with two electrons.

• Some elements, such as boron, can bond without achieving an octet because theyhave three or fewer electrons to share.


To give students practice under morerealistic testing conditions, give them60 minutes to answer all of the ques-tions in this Standardized TestPreparation.

1. A 2. C

3. B 4. D

5. A 6. D

7. A 8. B

9. C

10. Hybridization explains how theorbitals of an atom becomerearranged when the atom formscovalent bonds.

11. Ionic crystals are brittle becauseshifting of the layers of ionsresults in large repulsive forcesthat cause the layers to partcompletely.

12. Naphthalene has the strongerintermolecular forces even thoughit is nonpolar, because its boilingpoint is higher than that of aceticacid. Boiling point is directly corre-lated to strength of intermolecularforces; the stronger the intermo-lecular forces, the more energyneeded to break all the intermolec-ular forces, and therefore the high-er the boiling point. Naphthalene isso large that its dispersion forcesare greater than the sum of thedispersion forces and hydrogenbonding in acetic acid.

Answers are continued on page 217C.

TEST ANSWERS

CHAPTER REVIEW

Answer the following items on a separate piece of paper.

1.A chemical bond results from the mutual attrac-tion of the nuclei forA. electrons.B. neutrons.C. protons.D. dipoles.

2.A polar covalent bond is likely to form betweentwo atoms thatA. are similar in electronegativity.B. are of similar size.C. differ in electronegativity.D. have the same number of electrons.

3.The Lewis structure of HCN containsA. one double bond and one single bond.B. one triple bond and one single bond.C. two single bonds.D. two double bonds.

4.According to VSEPR theory, the moleculargeometry for CH3

+ isA. tetrahedral.B. trigonal-pyramidal.C. bent or angular.D. None of the above

5.Which molecule contains a double bond?A. COCl2B. C2H6C. CF4D. SF2

6.Which molecule is polar?A. CCl4B. CO2C. SO3D. none of these

7.What is the hybridization of the carbon atomsin C2H2?A. spB. sp2

C. sp3

D. The carbon atoms do not hybridize in C2H2.

8.Which of the following compounds is predictedto have the highest boiling point?A. HClB. CH3COOH (Note: The two oxygen atoms

bond to the carbon.)C. Cl2D. SO2

9.An unknown substance is an excellent electricalconductor in the solid state and is malleable.What type of chemical bonding does this sub-stance exhibit?A. ionic bondingB. molecular bondingC. metallic bondingD. cannot determine from the information given

SHORT ANSWER

10.What does the hybridization model helpexplain?

11.Explain why ionic crystals are brittle.

EXTENDED RESPONSE

12.Naphthalene, C10H8, is a nonpolar molecule and has a boiling point of 218°C. Acetic acid,CH3CO2H, is a polar molecule and has a boilingpoint of 118°C. Which substance has thestronger intermolecular forces? Briefly explainyour answer.

13.Describe and explain the potential energychanges that occur during the formation of acovalent bond.

When several questions refer tothe same graph, table, drawing, or passage, answerthe questions you are sure of first.

Standardized Test Prep

C H A P T E R 6216216

RECOMMENDED TIMETeacher preparation: 40 minutesClass time: 45–50 minutes

RATINGSTEACHER PREPARATION 3STUDENT SETUP 2CONCEPT LEVEL 2CLEANUP 2

MATERIALSChemicals:• copper wire• aluminum shot• silicon dioxide (sand)• sodium chloride, NaCl• sucrose• calcium chloride, CaCl2 (unknown

solid)

Equipment (per group):• beakers, 50 mL (6)• Bunsen burner• deionized water• evaporating dishes (6)• graduated cylinder, 10 mL• LED conductivity tester• spatula• test tubes, small, with solid rubber

stoppers (6)• test-tube rack• tongs• wire gauze, support stand, iron

ring, and clay triangle

SOLUTION/MATERIALSPREPARATION1. Check the LED electrodes to makesure they are performing correctly.

CHAPTER LAB

Types of Bonding in Solids

BACKGROUNDThe purpose of this experiment is to relate certainproperties of solids to the type of bonding the solidshave. These observable properties depend on thetype of bonding that holds the molecules, atoms, orions together in each solid. Depending on the typeof bonding, solids may be described as ionic, molecu-lar, metallic, or covalent network solids. The proper-ties to be studied are relative melting point, solubilityin aqueous solution, and electrical conductivity.

SAFETY

For review of safety, please see the Safety in theChemistry Laboratory in the front of your book.

PREPARATION1. Make a data table in which to record the results

of melting, water solubility, solid conductivity,aqueous solution conductivity, and type of bond-ing of each substance tested.

PROCEDURE1. Place 1 g samples of each substance into sepa-

rate evaporating dishes.

2. Touch the electrodes of the conductivity testerto each solid. After each test, rinse with distilledwater and carefully dry the electrodes. Notewhich substances conducted electricity.

3. Place one evaporating dish on a triangle, andheat with a Bunsen burner. As soon as a solidmelts, remove the flame.

4. Repeat this procedure for every substance. Donot heat any substance for more than 5 min.There may be some substances that will notmelt.

OBJECTIVES

• Observe the physical properties of differentsolids.

• Relate knowledge of these properties to thetype of bonding in each solid.

• Identify the type of bonding in an unknownsolid.

MATERIALS

• beakers, 50 mL (6)

• Bunsen burner

• copper wire

• deionized water

• evaporating dishes or crucibles (6)

• graduated cylinder, 10 mL

• aluminum shot

• LED conductivity tester

• silicon dioxide (sand)

• sodium chloride (NaCl)

• spatula

• sucrose

• test tubes, small, with solid rubberstoppers (6)

• test-tube rack

• tongs

• unknown substance

• wire gauze, support stand, iron ring, andclay triangle

TEACHER‘S NOTES

216216

EASY HARD

1 2 3 4

CHAPTER LAB INQUIRYL A B?

217 217

13-1 TEACHER‘S NOTES

5. Note which substances melted and how long thesubstances took to melt.

6. Place five test tubes in the test-tube rack. Place0.5 g of each solid into its own individual testtube. Add 5 mL of deionized water to each testtube. Stopper and shake each test tube in anattempt to dissolve the solid.

7. Note which substances dissolved in the water.

8. Place the solutions or mixtures into separate50 mL beakers, and immerse the electrodes ofthe conductivity tester. Rinse the electrodes withthe solvent (deionized water) before and aftereach test. Note which substances conduct electricity.

CLEANUP AND DISPOSAL9. Dispose of solids and solutions in containers

designated by your teacher.

10. Clean all equipment and return it to its properplace.

11. Wash your hands thoroughly aftercleaning up your area and equipment.

ANALYSIS AND INTERPRETATION1. Analyzing Methods: Why did you rinse the elec-

trodes before each conductivity test?

2. Analyzing Methods: Why did you use deionizedwater in making the solutions?

3. Organizing Data: List the results that each typeof bonding should show.

2. Note: Among molecules, there can be discrepancies depending onwhether the molecule is polar, non-polar, acidic, or basic. This experimentwas designed to test polar com-pounds that are neither acidic norbasic. A weak electrolyte could alsobe used. Different results will occur ifother types of molecules are used.

REQUIRED PRECAUTIONS• Apron and goggles should be worn

at all times.• Students should be sure to keep

hands and papers away from theBunsen burner flame.

TECHNIQUES TODEMONSTRATEDemonstrate how to use the LEDconductivity tester.

SAMPLE DATASee page 217D for Data Table.

PRE-LAB DISCUSSIONDiscuss the various types of bonding,and have the students predict whatthe physical properties should be.

DISPOSALSet out disposal containers for stu-dents. Thoroughly clean all labwareafter use.

ANALYSIS ANDINTERPRETATION—ANSWERS1. The electrodes need to be washedwith deionized water before eachconductivity test to ensure that noneof the previous material is clinging tothe electrodes and thereby giving afalse conductivity reading.2. Deionized water is used to makethe solutions instead of tap waterbecause tap water contains ions,which could give a false positive con-ductivity test result.

Answers are continued on page 217D.

CONCLUSIONS1. Inferring Conclusions: What type of bonding

describes each substance? Explain your reasoning.

2. Inferring Conclusions: Comparing the proper-ties of your unknown solid with the propertiesof the known solids, determine the type of bond-ing present in your unknown solid.

EXTENSIONS1. Evaluating Methods: Is it possible, for a specific

type of bonding, for these properties to varyfrom what was observed in this experiment? Ifso, give an example of such a variance.

2. Applying Conclusions: Think about diamond.What would you predict to be the results of thisexperiment performed on diamond, and whatwould you predict the bond type to be?

TEACHER‘S NOTES


Answers from page 2086.

8.a. Cl

⎯→

Cl

b. O

⎯→

O

c. F

H ↓

1s

↑↑↓

2p

↑↓↑↓2s

↑↓1s

↓↓

2p

↑↓↑↓2s

↑↓1s

↓↓

2p

↑↓↑↓2s

↑↓1s

↑↑

2p

↑↓↑↓2s

↑↓1s

↑↑

2p

↑↓↑↓2s

↑↓1s

↓↑↓

3p

↑↓↑↓3s

↑↓↑↓

2p

↑↓↑↓2s

↑↓1s

↑↑↓

3p

↑↓↑↓3s

↑↓↑↓

2p

↑↓↑↓2s

↑↓1s

↓↑↓

3p

↑↓↑↓3s

↑↓↑↓

2p

↑↓↑↓2s

↑↓1s

↑↑↓

3p

↑↓↑↓3s

↑↓↑↓

2p

↑↓↑↓2s

↑↓1s

REVIEW ANSWERSAns.a. c.

b. d.

Additional Sample Problem from page 188

D-1 Determine the Lewis structure for each of the following molecules:a. O2 c. C2H2b. C2H4

Ans.

a. c.

b.

Continued from page 1963. Metallic bonding is not directional but is uniformthroughout the solid. One group of atoms can slide pastanother group of atoms amid the electron sea withoutbreaking any attractions. Ionic crystals are held to-gether by strong electrical attraction between opposite-ly charged ions. Moving one plane of ions relative toanother requires breaking this attraction.

4. A metal is a good electrical conductor because itsvalence electrons are delocalized and move freely amongthe metal’s network of empty valence orbitals. Therefore,the electrons can move from one end of the metal to theother end; that is, the metal conducts electricity.

Continued from page 2076. The boiling point of a molecule is directly correlatedto the strength of the intermolecular forces. The boilingpoint is the temperature at which a molecule goes fromthe liquid state to the gaseous state. In the gaseousstate, all the intermolecular forces are essentially bro-ken. The higher the boiling point, the more energyneeded to break the intermolecular forces, the strongerthe intermolecular forces.

SECTION REVIEW

SECTION REVIEW

CH

HCH

H

CH HCO O

ClHCH HH

H

CH HOH

H

OH H

217A

CONTINUATION OF ANSWERS AND TEACHER’S NOTES


A-1 Ask students to complete the following chart:

Ans. a. polar-covalent; C c. polar-covalent; Ob. nonpolar-covalent; d. ionic; Cl

same electronegativity e. ionic; S

Continued from page 1775. electronegativities: Cu = 1.9, Cl = 3.0, I = 2.5 differences in electronegativity: CuCl = 1.1 and ICl = 0.5a. CuCl has the greater percent ionic character.b. The chlorine in CuCl has greater negative partialcharge than the chlorine in ICl. There is a greater differ-ence in electronegativities between Cu and Cl; there-fore, there is a greater separation of charge between Cuand Cl than between I and Cl.

6. a. Electronegativity for Br = 2.8 and for K = 0.8. Thedifference between the electronegativities is 2.0; there-fore, an ionic bond is expected between K and Br.b. Br− is larger than K+. Because an ionic bond is expect-ed between the two, Br atom becomes Br− and K atombecomes K+. According to the ionic radii in Figure 19,Br− is larger than K+.

SECTION REVIEW


Electro-Elements negativity Bond More-negativebonded difference type atom

a. C and H 0.4b. C and S 0.0c. O and H 1.4d. Na and Cl 2.1e. Cs and S 1.8


C-1 Draw the Lewis structure for each of the followingcompounds:a. H2O c. CH3OHb. CH4 d. HCl



a. H and I 0.4 polar- Icovalent

b. S and O 1.0 polar- Ocovalent

c. K and Br 2.0 ionic Brd. Si and Cl 1.2 polar- Cl

covalente. K and Cl 2.2 ionic Clf. Se and S 0.1 nonpolar- S

covalentg. C and H 0.4 polar- C

covalent

⎫⎬⎭

⎫⎬⎭

⎫⎬⎭

⎫⎬⎭

⎫⎬⎭

⎫⎬⎭

⎫⎬⎭

⎫⎬⎭⎫⎬⎭

⎫⎬⎭

⎫⎬⎭

⎫⎬⎭

⎫⎬⎭


49. a. c.

b. d.

54.

55. a. In ionic bonding, valence electrons of the atomsof the less-electronegative element are donatedentirely to the atoms of the more-electronegativeelement. In covalent bonding, valence electrons areshared between the bonded atoms.b. A molecular compound consists of individual unitscapable of existing on their own. An ionic compoundconsists of an arrangement of a large number ofions. There is no discrete, independent particle in anionic compound.c. An ionic compound is held together by electricalattraction between ions. A metal is held together bythe sharing by atoms of a sea of mobile valenceelectrons.

56. a. c. e.

b. d.

57.

58. two potassium cations, K+, and one sulfide anion, S2−

59. Vacant orbitals in the outer energy levels of themetal’s atoms overlap and are occupied by elec-trons from adjacent atoms.

60. Although a particular bond in a molecule may bepolar, it is the arrangement of all the polar bonds inspace—or molecular geometry—that determineswhether a molecule is polar.

61. It lies between the energy levels of the orbitalsfrom which it was made.

62. aluminum

HICIOIH

H

H

PCl

BOHe

SOO

O

←⎯→ SOO

O

SOO

O

←⎯→

O_

OCl

bent or angularH NH +

HH

tetrahedral

OO 2_

OOS

tetrahedral

O N_

OO

trigonal-planar

Continued from page 210

19. a. d. g.

b. e. h.

c. f.

20. a.

b.

c.

21. a. d.

b. e.

c.

23. a. b. c.

d.

24. a.

b.

c.


48. a. d.

b. e.

c. f.bent or angular

ClNOCl O Cl

bent or angular

ClCl ClSi

Brtetrahedral

P II

I

trigonal-pyramidal

ClH N H

trigonal-pyramidal

Cl ClSbent or angular

O OOBr

_

CHH

CO

H_

O CHH

C

_

O

O

H←⎯→

HO_

O OS O OS←⎯→

C ON NO O

NI II

H

HC ClH

SeH H

SiBr BrBr

BrCF FF

F

(Al2S3)Al3+

+ Al3+

+ S 2–

+ S 2–

+ S 2–

⎯→Al + Al + S + S + S

(CaO)⎯→ Ca2+

+ O 2–

Ca + O

Na + Na + S Na+

+ Na+

+ S 2–

(Na2S)⎯→

Li O Al

Ca C S

Cl P

217B

c. The top chlorine atoms contain too many elec-trons. Correct structure:

71. Because of its shape (tetrahedral), methane is anonpolar molecule. As a result, CH4 molecules arenot strongly attracted to one another. In contrast,ammonia is a polar molecule. As a result, NH3 mol-ecules are attracted to one another. Therefore, ittakes more energy to separate ammonia moleculesand change ammonia to a gas than to separatemethane molecules.

72. Lithium, sodium, potassium, rubidium, cesium, bari-um, and radium are listed as bcc.

73. The electrons in metals require very little energy toenter the conduction band, which is a set of over-lapping orbitals. Thus, electrons can move freelythrough a metal with a small applied voltage.

Refer to the One-Stop Planner CD-ROM for appropriatescoring rubrics for items 74–77. Look for these points ineach report:

74. a. Students should mention Dr. Pauling’s role inestablishing the electronegativity scale and in dis-covering the hydrogen bond. For this work,Dr. Pauling received a Nobel Prize.b. Some scientists criticize Pauling for being anadvocate of something as trendy as vitamin supple-ments. Be sure students examine real scientificdata, rather than political responses, to make their decision.

75. Students should discuss some properties of metalloids.

76. Answers may vary. Proteins are another type ofsubstance that can have disulfide bridges.

77. Answers may vary.

78. Students might use either physical or chemical prop-erties to distinguish between these substances. Ionicsubstances will be solid at room temperature, elimi-nating water. They will not conduct electricity, elimi-nating Cu. And they will be composed of more thanone type of atom, eliminating C. Ionic solids are brit-tle and have high melting points.

79. Students’ answers may vary greatly. Some possibleresponses include plastics (covalent), paper (cova-lent), wood (covalent), dishes (ionic or covalent),pots (metallic), and cloth (covalent).

ClINICl

Cl

217C


63.

64. a. c.

b.

65.

66. a. c.

b.

67. Most atoms are bonded to other atoms in naturebecause bonding lowers their potential energy.

68. The carbon-carbon triple bond in C2H2 is thestrongest of the bonds. Therefore, it is the shortestbecause stronger bonds are shorter bonds. The car-bon-carbon single bond in C2H6 is the weakest ofthe bonds, so it is the longest. From Table 2, stu-dents should surmise that the lengths of the single,double, and triple carbon-carbon bonds are about154 pm, 134 pm, and 120 pm, respectively.

69. F has the highest electronegativity and thereforehas the greatest attraction for the electrons itshares in a covalent bond.

70. a. Hydrogen is never a central atom. The secondhydrogen atom has two single bonds and can haveonly one single bond. The total number of valenceelectrons is wrong. Correct structure:

b. The octet rule is not followed for the carbonatom or for the central oxygen atom. Correct struc-ture:

HICIOIH

O

SIH

H

trigonal-planar

NO OO

_tetrahedral

NHH

HH +N

bent or angular

O O_

Cl Be Cl

Cl

FC ClF

H

H HC N HH

Cl ClClP


a. Zn and O 1.9 ionic Ob. Br and I 0.3 polar- Br

covalentc. S and Cl 0.5 polar- Cl

covalent


217D

conduct electricity in solution or in the solid state. Thelow melting point is due to the presence of weak inter-molecular forces that are easily broken when going fromsolid to liquid. The lack of conductivity is due to the lackof ions in the solution. Copper and aluminum are metal-lic because they have very high melting points and con-duct electricity in the solid state. This conductivity is dueto the easily mobile valence electrons in the solid. SiO2is a covalent network because it has a very high meltingpoint and does not conduct electricity. Its lack of con-ductivity is due to the strength of the covalent bonds(and lack of ions). It has a very high melting pointbecause the solid contains only covalent bonds and hasno weak intermolecular forces.2. Students should identify the unknown as ionic andsimilar in character to NaCl.

Continued from page 21513. There is a repulsive region at small bond distances, a

minimum well at an intermediate bond distance, andan asymptotic region at large bond distances. Therepulsive region is due to the repulsion between thetwo positive nuclei at distances where the nuclei areclose to each other. The asymptotic region is wherethe atoms are far apart and therefore do not interactwith each other. The potential energy decreases asthe atoms come closer together and are attracted toeach other. The minimum well is the region where theattractive forces are balanced by the repulsive forces.This distance is the equilibrium bond length.


3. Ionic bonding shows high melting, solubility in water,no conductivity as a solid, but conductivity as an aque-ous solution. Molecular bonding shows low melting,solubility in water, no conductivity as a solid or in aque-ous solution. Metallic bonding shows high meltingpoint, no solubility in water, conductivity as a solid, butno conductivity in aqueous solution. Covalent networkbonding shows high melting point, no solubility inwater, no conductivity as solid or in aqueous solution.

CONCLUSIONS—ANSWERS1. Sodium chloride is ionic because it has a high meltingpoint, and conducts electricity when dissolved in waterbut not in the solid phase. This conductivity in solution iscaused by ionic solids dissolving and forming ions insolution, which conduct the electricity. Sucrose is molec-ular because it has a low melting point and does not

CHAPTER LAB

TEST ANSWERS

DATA TABLE

AluminumNaCl Sucrose Copper wire shot SiO2 CaCl2

Melting does not melt melts easily does not does not does not does notmelt melt melt melt or

just meltsWater solubility soluble soluble insoluble insoluble insoluble solubleSolid conductivity no no yes yes no noAqueous solution yes (strong) no (or weak) no no no yes(or mixture)conductivityType of bonding ionic molecular metallic metallic covalent ionic

network

EXTENSIONS—ANSWERS1. Discrepancies may arise because of differences in themolecular type of bonding. If the molecule is nonpolar,then it will not be soluble in aqueous solution. If themolecule is acidic or basic (forms ions in solution), thenit will show some electrical conductivity in solution.However, this conductivity will not be as great as in anionic solution.2. Diamond would not melt, would not be soluble inaqueous solution, and would not conduct electricity asa solid or mixture. Hence, diamond must have covalent-network bonding.

compression guide chapter 6 chemical bonding planning · pdf filete lesson starter ionic vs....

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