Chapter 8: Covalent Bonding - · PDF fileOLChemLab Worksheet, p. 88 Study Guide, ... ecules in...

Chapter 8 Organizer: Covalent BondingCovalent bonds form when atoms share electrons.

Section Objectives National Standards

State/Local

StandardsResources to Assess Mastery

Section 8.11. Apply the octet rule to atoms that form

covalent bonds.2. Describe the formation of single, double,

and triple covalent bonds.3. Contrast sigma and pi bonds.4. Relate the strength of a covalent bond to

its bond length and bond dissociation energy.

UCP.2, UCP.3, UCP.5; A.1; B.1, B.2, B.3, B.4, B.6

Entry-Level Assessment Focus Transparency 27Progress Monitoring Formative Assessment, p. 245 Reading Check, pp. 243, 244, 246 Section Assessment, p. 247

Section 8.21. Translate molecular formulas into binary

molecular compound names.2. Name acidic solutions.

UCP.2, UCP.3, UCP.5; B.2; G.2

Entry-Level Assessment Focus Transparency 28Progress Monitoring Formative Assessment, pp. 249, 250, 251 Reading Check, p. 249 Section Assessment, p. 252

Section 8.31. List the basic steps used to draw Lewis

structures.2. Explain why resonance occurs, and identify

resonance structures.3. Identify three exceptions to the octet rule,

and name molecules in which these exceptions occur.

UCP.2, UCP.3, UCP.5; B.2

Entry-Level Assessment Focus Transparency 29Progress Monitoring Formative Assessment, pp. 257, 258 Reading Check, p. 259 Section Assessment, p. 260

Section 8.41. Summarize the VSEPR bonding theory.2. Predict the shape of, and the bond angles

in, a molecule.3. Define hybridization.

UCP.2, UCP.3, UCP.5; A.1; B.2, B.4

Entry-Level Assessment Focus Transparency 30Progress Monitoring Formative Assessment, p. 263 Reading Check, p. 262 Section Assessment, p. 264

Section 8.51. Describe how electronegativity is used to

determine bond type.2. Compare and contrast polar and nonpolar

covalent bonds and polar and nonpolar molecules.

3. Generalize about the characteristics of covalently bonded compounds.

UCP.2, UCP.3, UCP.5; A.1, A.2; B.1, B.2, B.4, B.6; E.2; G.3

Entry-Level Assessment Focus Transparency 31Progress Monitoring Formative Assessment, p. 267 Reading Check, pp. 266, 267 Graph Check, p. 266 Section Assessment, p. 270Summative Assessment Chapter Assessment, p. 274 ExamView® Assessment Suite CD-ROM

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Suggested PacingPeriod Section 8.1 Section 8.2 Section 8.3 Section 8.4 Section 8.5 Assessment

Single 2 1 2 1 1 1

Block 1 0.5 1 0.5 0.5 0.5

Leveled Resources Lab Materials Additional Print andTechnology Resources

Science Notebook 8.1 OLFAST FILE Chapter Resources: MiniLab Worksheet, p. 86 OL Study Guide, p. 100 OLTransparencies: Section Focus Transparency 27 BL EL

Launch Lab, p. 239: paper towels, lab gloves, paper cup, graduated cylinder, food coloring, wood splint, sodium silicate solution, ethanol 15 minMiniLab, p. 242: permanent marker, 9-inch aluminum pie pan, hot plate, sugar crystals, salt crystals, paraffin 20 min

Technology: ExamView® Assessment Suite CD-ROM StudentWorks™ Plus DVD-ROM TeacherWorks™ Plus DVD-ROM Virtual Lab CD-ROM Video Lab DVD What’s CHEMISTRY Got To Do With It? DVD Interactive Classroom DVD-ROM LabManager™ CD-ROM

Assessment: Performance Assessment in the Science Classroom Challenge Problems AL Supplemental Problems BL OL Chapter Test (Scaffolded)

FAST FILE Resources: Section Focus Transparency Masters Math Skills Transparency Masters and Worksheets Teaching Transparency Masters and Worksheets

Additional Resources: Solving Problems: A Chemistry Handbook Cooperative Learning in the Science Classroom Lab and Safety Skills in the Science Classroom glencoe.com

Lab Resources: Laboratory Manual OL CBL Laboratory Manual OL Small-Scale Laboratory Manual OL Forensics Laboratory Manual OL

Science Notebook 8.2 OLFAST FILE Chapter Resources: Study Guide, p. 101 OLTransparencies: Section Focus Transparency 28 BL EL Math Skills Transparency 9 OL EL

Science Notebook 8.3 OLFAST FILE Chapter Resources: Study Guide, p. 102 OLTransparencies: Section Focus Transparency 29 BL EL Teaching Transparency 26 OL EL

Science Notebook 8.4 OLFAST FILE Chapter Resources: Study Guide, p. 103 OLTransparencies: Section Focus Transparency 30 BL EL Teaching Transparency 27 OL EL

Science Notebook 8.5 OLFAST FILE Chapter Resources: ChemLab Worksheet, p. 88 OL Study Guide, p. 104 OLTransparencies: Section Focus Transparency 31 BL EL Teaching Transparency 28 OL EL Math Skills Transparency 9 OL EL

ChemLab, p. 272: molecular model kit 45 min

BL Below Level OL On Level AL Advanced Learners EL English Learners COOP LEARN Cooperative Learning

Customize any lab with the LabManager™ CD-ROM.

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BIG IdeaSharing Electrons To introduce this chapter’s Big Idea, have students draw the Lewis structures for hydro-gen and oxygen atoms. Ask them how many additional electrons a hydrogen atom needs in order to have the same structure as helium. 1 additional electron Ask students how many electrons an atom of oxygen needs to obtain the noble gas arrangement of neon. 2 additional elec-trons Ask students how hydrogen and oxygen atoms are able to bond to form water, H 2 O. They must share elec-trons. 2 hydrogen atoms share an electron each with 1 oxygen atom.

Tie to Previous KnowledgeHave students review the following concepts before studying this chapter:Chapter 4: atomic structureChapter 5: electron configurationChapter 6: periodic trendsChapter 7: periodic properties of elementsChapter 8: ionic bonding

Use the PhotoEstablishing Relationships The art in the chapter opener shows several different ways of describing or illustrating a water molecule: a Lewis structure, a ball-and-stick model, and a space-filling model, finally showing a group of such mol-ecules in a water droplet. Ask stu-dents to describe the differences between each method of illustrating the molecule. Answers will vary, but should focus on the different purposes served by each approach.

This DVD-ROM is an editable Microsoft® PowerPoint® presentation that includes:• a premade presentation for every chapter• additional diagnostic, formative, chapter, and Standardized Test Practice questions• animations• image bank• transparencies• links to glencoe.com

Interactive Classroom

H — O

—

H

BIG Idea Covalent bonds form when atoms share electrons.

8.1 The Covalent BondMAIN Idea Atoms gain stability when they share electrons and form covalent bonds.

8.2 Naming MoleculesMAIN Idea Specific rules are used when naming binary molecular compounds, binary acids, and oxyacids.

8.3 Molecular StructuresMAIN Idea Structural formulas show the relative positions of atoms within a molecule.

8.4 Molecular ShapesMAIN Idea The VSEPR model is used to determine molecular shape.

8.5 Electronegativity and PolarityMAIN Idea A chemical bond’s character is related to each atom’s attraction for the electrons in the bond.

ChemFacts

• The spherical shape of a water drop is due to surface tension, a phenomenon caused by forces between molecules.

• Surface tension makes water act somewhat like an elastic film. Insects called water striders are able to walk on the filmlike surface of water.

• The chemical and physical properties of water make it a unique liquid.

Covalent Bonding

Lewis structure

Ball-and-stick model

Space-filling model

Spherical water droplet

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Start-Up Activities

LAUNCH Lab

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238 Chapter 8 • Covalent Bonding

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LAUNCH LabLabRUBRIC available at glencoe.com

Purpose Students will build a polymer from silicon and ethanol that can bounce when shaped into a sphere.

Safety Precautions Approve lab safety forms before work begins. Review MSDS for ethanol and sodi-um silicate with students prior to doing the lab. Ethanol is very flam-mable and its vapors are explosive—use caution and appropriate ventilation. Sodium silicate is very alkaline and is a skin irritant. Students must wear safety goggles and gloves when handling the polymer.

Disposal Wrap the excess product and wooden splints in newspaper. Place these pieces of newspaper in a box or plastic bag and seal it. Have the sealed box or bag disposed of in a landfill that takes chemical or haz-ardous waste materials.

Teaching Strategies• Have groups vary the amount of

ethanol used in making the ball, using between 5 and 15 mL. Have students observe the effects this has on the outcome and hypothe-size about the cause.

• The ball might become brittle and crumble. Ask students to determine the cause of this.

Expected Results Students should be able to form a sphere that bounces. The product might be brittle and might crumble but can be reformed by shaping it in gloved hands.

Analysis 1. The ball loses shape when it sits, but when shaped as

a ball, it bounces. As the ball dries out, it becomes brittle and crumbles.

2. Ionic compounds form crystals, dissolve in water, and have high melting points. The ball was made of two liquids at room temperature, it bounces, does not dis-solve in water, and loses its shape over time.

Inquiry Silicon has 4 valence electrons and oxygen has 6 valence electrons. To form octets, silicon must gain 4 electrons and oxygen must gain 2 electrons. To form a bond, these atoms must share electrons. Customize this lab with

the LabManager™CD-ROM.

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IonicPolar CovalentNonpolar CovalentBond Character

Start-Up ActivitiesStart-Up Activities

Bond Character Make the following Foldable to help you organize your study of the three major types of bonding.

Visit glencoe.com to: study the entire chapter online

explore

take Self-Check Quizzes

use the Personal Tutor to work Example Problems step-by-step

access Web Links for more information, projects, and activities

find the Try at Home Lab, Breaking Covalent Bonds

LLAAUUNCH NCH LabLabWhat type of compound is used to make a Super Ball?Super Balls are often made of a silicon compound called organosilicon oxide (Si(OC H 2 C H 3 ) 2 O).

Procedure 1. Read and complete the lab safety form.2. Spread several paper towels across your desk or lab

work area. Put on lab gloves. Place a paper cup on the paper towels.

3. Using a graduated cylinder, measure 20.0 mL of sodium silicate solution, and pour it into the cup. Add one drop of food coloring and 10.0 mL of ethanol to the cup. Stir the mixture clockwise with a wooden splint for 3 s.

WARNING: Keep ethanol away from flame and spark sources, as its vapors can be explosive.

4. Working over paper towels, pour the mixture onto one of your glove-covered palms. Gently squeeze out excess liquid as the mixture solidifies.

5. Roll the solid between glove-covered hands and form a ball. Drop it on the floor and observe what happens.

6. Store the ball in an airtight container. You will need to reshape the ball before using it again.

Analysis1. Describe the properties of the ball that you observed.2. Compare the properties you observed with those of

an ionic compound.

Inquiry How many electrons do silicon and oxygen atoms need to form octets? If both atoms must gain elec-trons, how can they form a bond with each other?

STEP 1 Collect two sheets of paper, and layer them about 2 cm apart vertically.

STEP 2 Fold up the bottom edges of the sheets to form three equal tabs. Crease the fold to hold the tabs in place.

STEP 3 Staple along the fold. Label the tabs asfollows: Bond Character, Nonpolar Covalent, Polar Covalent, and Ionic.

Use this Foldable with Section 8.1. Asyou read this section, summarize what you learn about bond character and how it affects the properties of compounds.

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Chapter 8 • Covalent Bonding 239

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1 FocusFocus Transparency

Before presenting the lesson, project Section Focus Transparency 27 and have students answer the accompa-nying questions. BL EL

MAIN IdeaCovalent Bonds Draw the Lewis structures for fluorine, chlorine, oxy-gen, sulfur, and bromine on the board. Ask students to locate these elements on the periodic table and identify what they have in common. They are all nonmetals and gain electrons to form negative ions. If two of these atoms form a compound, what must they do to form an octet? They must share electrons. Draw carbon’s Lewis struc-ture and show students an atomic model of carbon. On both, point out the four places carbon can form a single covalent bond. Bond a hydro-gen atom into each of the places. Ask students what is represented. The model represents the electrons shared between a central carbon atom and four ter-minal hydrogen atoms in a C H 4 molecule. OL

2 Teach

Section 8.18.1

Chemistry JournalBonding Limerick Read students the following limerick:Bond formations can be unwell(Electrons join up in a shell)For it doesn’t seem fineTo say they combineWhen we know very well they repel!Have students develop limericks, poems, or rhymes that depict some aspect of covalent bonds. OL

Quick Demo

Potential Energy Use this demo to relate potential energy to stability. Make a pendulum using a foam-rubber ball and a string or spring. Ask where the pendulum has the greatest potential energy. when the pendu-lum is pulled up to its maximum height Ask where the pendulum has the least potential energy. when it hangs straight down Explain that the lowest potential energy occurs when the pendulum is at the lowest point in its motion. Have students note that eventually a swinging pendulum comes to a stop in the lowest energy posi-tion. Relate the pendulum’s higher-energy positions to instability. OL

Section 8.18.1

The Covalent BondAtoms gain stability when they share electrons

and form covalent bonds.

Real-World Reading Link Have you ever run in a three-legged race? Each person in the race shares one of their legs with a teammate to form a single three-legged team. In some ways, a three-legged race mirrors how atoms share electrons and join together as a unit.

Why do atoms bond?Understanding the bonding in compounds is essential to developing new chemicals and technologies. To understand why new compounds form, recall what you know about elements that do not tend to form new compounds—the noble gases. You read in Chapter 6 that all noble gases have stable electron arrangements. This stable arrangement con-sists of a full outer energy level and has lower potential energy than other electron arrangements. Because of their stable configurations, noble gases seldom form compounds.

Gaining stability The stability of an atom, ion, or compound is related to its energy; that is, lower energy states are more stable. In Chapter 7, you read that metals and nonmetals gain stability by trans-ferring (gaining or losing) electrons to form ions. The resulting ions have stable noble-gas electron configurations. From the octet rule in Chapter 6, you know that atoms with a complete octet, a configuration of eight valence electrons, are stable. In this chapter, you will learn that the sharing of valence electrons is another way atoms can acquire the stable electron configuration of noble gases. The water droplets shown in Figure 8.1 consist of water molecules formed when hydrogen and oxygen atoms share electrons.

■ Figure 8.1 Each water droplet is made up of water molecules. Each water molecule is made up of two hydrogen atoms and one oxygen atom that have bonded by sharing electrons. The shapes of the drops are due to intermolecular forces acting on the water molecules.

Objectives

◗◗ Apply the octet rule to atoms that form covalent bonds.

◗◗ Describe the formation of single, double, and triple covalent bonds.

◗◗ Contrast sigma and pi bonds.◗◗ Relate the strength of a covalent

bond to its bond length and bond dissociation energy.

Review Vocabularychemical bond: the force that holds two atoms together

New Vocabularycovalent bondmoleculeLewis structuresigma bondpi bondendothermic reactionexothermic reaction

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Chemistry JournalWhy Covalent Bonds Form Have students write a paragraph explaining why a covalent bond can form between a chlorine atom and a fluorine atom. OL

Visual LearningFigure 8.2 Have students look at the figure and discuss the potential energy change as the fluorine atoms approach each other in terms of attractive forces and repulsive forces. At a certain optimal point, attractive and repulsive forces balance, and a bond forms. If the repulsive force is greater, no bond forms and the atoms stay separated. OL

■ Caption Question Fig. 8.2 A stable bond is formed at the location of maximum net attraction.

Concept DevelopmentElectron Nature It is important for students to understand that electrons are identical, regardless of which atom or energy level they are in. Students must also understand that electrons in the electron cloud are not static. They move at nearly the speed of light, yet are attracted to the positively charged nucleus. When a covalent bond forms, nuclei of both atoms contribute to the force that attracts the electrons. A single atom does not have sole ownership of the shared electrons.

ReinforcementDot Diagrams Review the electron dot diagrams used in Chapter 7. Point out that covalent bonds can be represented in a variety of ways.

Virtual Labs CD-ROMChemistry: Matter and ChangeAnimation: Covalent Bonding

GLENCOE Technology

8.1

◗

◗

◗

◗

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■ Figure 8.2 The arrows in this diagram show the net forces of attraction and repulsion acting on two fluorine atoms as they move toward each other. The overall force between two atoms is the result of electron-electron repulsion, nucleus-nucleus repulsion, and nucleus-electron attraction. At the position of maximum net attraction, a covalent bond forms.Relate How is the stability of the bond related to the forces acting on the atoms?

What is a covalent bond?You just read that atoms can share electrons to form stable electron con-figurations. How does this occur? Are there different ways in which electrons can be shared? How are the properties of these compounds different from those formed by ions? Read on to answer these questions.

Shared electrons Atoms in nonionic compounds share electrons. The chemical bond that results from sharing valence electrons is a covalent bond. A molecule is formed when two or more atoms bond covalently. In a covalent bond, the shared electrons are considered to be part of the outer energy levels of both atoms involved. Covalent bond-ing generally can occur between elements that are near each other on the periodic table. The majority of covalent bonds form between atoms of nonmetallic elements.

Covalent bond formation Diatomic molecules, such as hydrogen ( H 2 ), nitrogen ( N 2 ), oxygen ( O 2 ), fluorine ( F 2 ), chlorine (C l 2 ), bromine (B r 2 ), and iodine ( I 2 ), form when two atoms of each element share elec-trons. They exist this way because the two-atom molecules are more sta-ble than the individual atoms.

Consider fluorine, which has an electron configuration of 1 s 2 2 s 2 2 p 5 . Each fluorine atom has seven valence electrons and needs another elec-tron to form an octet. As two fluorine atoms approach each other, sever-al forces act, as shown in Figure 8.2. Two repulsive forces act on the atoms, one from each atom’s like-charged electrons and one from each atom’s like-charged protons. A force of attraction also acts, as one atom’s protons attract the other atom’s electrons. As the fluorine atoms move closer, the attraction of the protons in each nucleus for the other atom’s electrons increases until a point of maximum net attraction is achieved. At that point, the two atoms bond covalently and a molecule forms. If the two nuclei move closer, the repulsion forces increase and exceed the attractive forces.

The most stable arrangement of atoms in a covalent bond exists at some optimal distance between nuclei. At this point, the net attraction is greater than the net repulsion. Fluorine exists as a diatomic molecule because the sharing of one pair of electrons gives each fluorine atom a stable noble-gas configuration. As shown in Figure 8.3, each fluorine atom in the fluorine molecule has one pair of electrons that are cova-lently bonded (shared) and three pairs of electrons that are unbonded (not shared). Unbonded pairs are also known as lone pairs.

Fluorineatom

Fluorineatom

Fluorinemolecule

F F

F

+

F Lonepairs

Bonding pairof electronsComplete

octets

■ Figure 8.3 Two fluorine atoms share a pair of electrons to form a covalent bond. Note that the shared electron pair gives each atom a complete octet.

Force of repulsionForce of attraction

The atoms are too far apart to have noticeable attraction or repulsion.

Each nucleus attracts the other atom’s electron cloud. Repulsion occurs between nuclei and between electron clouds.

The distance is right for the attrac-tion between one atom’s protons and the other atom’s electrons to make the bond stable.

If the atoms are forced closer together, the nuclei and electrons repel each other.

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Section 8.1 • The Covalent Bond 241


See the MiniLab worksheet in your FAST FILE.

RUBRIC available at glencoe.com

Purpose Students will compare the melting points of compounds with covalent bonds and compounds with ionic bonds.

Process Skills observe and infer, classify, compare and contrast, inter-pret data

Safety Precautions Approve lab safety forms before work begins. Do not directly inhale the vapors of the substances. Make sure the area is well ventilated or students are working under a fume hood.

Disposal Carefully remove the lid and place in a waste container.

Teaching Strategies Students should work in groups to do this lab. A candle can be substituted for the Bunsen burner.

Expected Result• paraffin melts first• sugar melts second, then burns• sodium chloride does not melt

Analysis 1. The paraffin melted first. The salt crystals

did not melt. 2. paraffin, low; sugar, medium; salt

crystals, very high 3. ionic bonds: salt

covalent bonds: paraffin and sugar 4. Ionic compounds have lower melting

points than covalently bonded compounds.

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Single Covalent BondsWhen only one pair of electrons is shared, such as in a hydrogen molecule, it is a single covalent bond. The shared electron pair is often referred to as the bonding pair. For a hydrogen molecule, shown in Figure 8.4, each covalently bonded atom equally attracts the pair of shared electrons. Thus, the two shared elec-trons belong to each atom simultaneously, which gives each hydrogen atom the noble-gas configuration of helium (1 s 2 ) and lower energy. The hydrogen molecule is more stable than either hydrogen atom is by itself.

Recall from chapter 5 that electron-dot diagrams can be used to show valence electrons of atoms. In a Lewis structure, they can represent the arrangement of electrons a molecule. A line or a pair of vertical dots between the symbols of elements represents a single covalent bond in a Lewis structure. For example, a hydrogen molecule is written as H—H or H:H.

Compare Melting PointsHow can you determine the relationship between bond type and melting point? The properties of a compound depend on whether the bonds in the compound are ionic or covalent.

Procedure 1. Read and complete the lab safety form. 2. Create a data table for the experiment. 3. Using a permanent marker, draw three lines

on the inside bottom of a disposable, 9-inch aluminum pie pan to create three, equal wedges. Label the wedges, A, B, and C.

4. Set the pie pan on a hot plate. WARNING: Hot plate and metal pie pan will burn

skin—handle with care. 5. Obtain samples of the following from your

teacher and deposit them onto the labeledwedges as follows: sugar crystals ( C 12 H 22 O 11 ), A; salt crystals (NaCl) B; paraffin ( C 23 H 48 ), C.

6. Predict the order in which the compounds will melt.

7. Turn the temperature knob on the hot plate to the highest setting. You will heat the compounds for 5 min. Assign someone to time the heating of the compounds.

8. Observe the compounds during the 5-min period. Record which compounds melt and the order in which they melt.

9. After 5 min, turn off the hot plate and remove the pie pan using a hot mitt or tongs.

10. Allow the pie pan to cool,and then place it in the proper waste container.

Analysis 1. State Which solid melted first? Which solid

did not melt? 2. Apply Based on your observations and data,

describe the melting point of each solid as low, medium, high, or very high.

3. Infer Which compounds are bonded with ionic bonds? Which are bonded with covalent bonds?

4. Summarize how the type of bonding affects the melting points of compounds.

+ →

+ →H H HHH

Hydrogen atom Hydrogen atom Hydrogen molecule

■ Figure 8.4 When two hydrogen atoms share a pair of electrons, each hydrogen atom is stable because it has a full outer-energy level.




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■ Caption Question Fig. 8.5 water: two electrons from each bond with hydrogen and two pairs of lone electrons; ammonia: two electrons from each bond with hydrogen and one pair of lone electrons; methane: two electrons from each bond with hydrogen

Build a ModelLewis Structures Give students four index cards that have been cut in half to form eight pieces. Have them write the following Lewis dot diagrams on the cards: one for C, two for H, one for S, and four for F. Tell students they will use these cards to line up unpaired electrons from one atom with the unpaired electrons from other atoms to form single covalent bonds. Have students form the compounds C F 4 and H 2 S. BL EL

Reading Check A covalent bond is shown as a pair of dots or as a short line.

Differentiated InstructionBelow Level Using a piece of acetate, draw circles to represent atoms. In each circle, represent a single atom by using its electron dot diagram. Show students two F atoms, and ask what each one needs to become stable. one electron Overlap the circles so that each F atom has eight electrons, and ask students how many electrons each atom has. eight Tell students that the shared electrons form a covalent bond. Overlap two H atoms and one O atom. Ask students how many electrons O has. eight How many electrons does each H atom have? two Ask students if H is stable with two electrons. Yes, because it has the noble gas configuration of He. Ask students how many bonds are formed. two single covalent bonds BL

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Group 17 and single bonds The halogens—the group 17 elements—such as fluorine have seven valence electrons. To form an octet, one more electron is needed. Therefore, atoms of group 17 elements form single covalent bonds with atoms of other nonmetals, such as carbon. You have already read that the atoms of some group 17 elements form covalent bonds with identical atoms. For example, fluorine exists as F 2 and chlorine exists as C l 2 .

Group 16 and single bonds An atom of a group 16 element can share two electrons and can form two covalent bonds. Oxygen is a group 16 element with an electron configuration of 1 s 2 2 s 2 2 p 4 . Water is com-posed of two hydrogen atoms and one oxygen atom. Each hydrogen atom has the noble-gas configuration of helium when it shares one electron with oxygen. Oxygen, in turn, has the noble-gas configuration of neon when it shares one electron with each hydrogen atom. Figure 8.5a shows the Lewis structure for a molecule of water. Notice that the oxygen atom has two single covalent bonds and two unshared pairs of electrons.

Group 15 and single bonds Group 15 elements form three covalent bonds with atoms of nonmetals. Nitrogen is a group 15 element with the electron con-figuration of 1 s 2 2 s 2 2 p 3 . Ammonia (N H 3 ) has three single covalent bonds. Three nitrogen electrons bond with the three hydrogen atoms leaving one pair of unshared electrons on the nitrogen atom. Figure 8.5b shows the Lewis structure for an ammonia molecule. Nitrogen also forms similar compounds with atoms of group 17 elements, such as nitrogen trifluoride (N F 3 ), nitrogen trichloride (NC l 3 ), and nitrogen tribromide (NB r 3 ). Each atom of these group 17 elements and the nitrogen atom share an electron pair.

Group 14 and single bonds Atoms of group 14 elements form four covalent bonds. A methane molecule (C H 4 ) forms when one carbon atom bonds with four hydrogen atoms. Carbon, a group 14 ele-ment, has an electron configuration of 1 s 2 2 s 2 2 p 2 . With four valence electrons, carbon needs four more elec-trons for a noble gas configuration. Therefore, when carbon bonds with other atoms, it forms four bonds. Because a hydrogen atom, a group 1 element, has one valence electron, it takes four hydrogen atoms to pro-vide the four electrons needed by a carbon atom. The Lewis structure for methane is shown in Figure 8.5c. Carbon also formts single covalent bonds with other nonmetal atoms, including those in group 17.

Reading Check Describe how a Lewis structure shows a covalent bond.

■ Figure 8.5 These chemical equations show how atoms share electrons and become stable. As shown by the Lewis structure for each molecule, all atoms in each molecule achieve a full outer energy level.Describe For the central atom in each molecule, describe how the octet rule is met.

Water

Two Single Covalent Bonds

+ →2H H — O

a

O

—

H

Three Single Covalent Bonds

Ammonia

+ →3H N

H

H — N

——

b

H

Four Single Covalent Bonds

+ →C

H

H

H — C — H

——

c

4H

Methane

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Chemistry ProjectChemistry in Medicine Ask students research the use of perfluorooctylbromide (C 8 F 17 Br) as an artificial oxygen carrier in synthetic blood. Have them report on how the compound works in the body to supply oxygen to tissues. Ask them to report on the compound’s molecular structure. What types of bonds does the molecule contain? all bonds are sigma bonds (single covalent bonds) How many electron pairs do the C, F, and Br atoms in the compound share? Carbon shares four electron pairs, forming C-C bonds, C-F bonds, and C-Br bonds. Fluorine shares one electron pair and bromine shares one electron pair. OL

IN-CLASS Example

Question Hydrogen selenide, H 2 Se, is a toxic gas with properties similar to hydrogen sulfide. Draw the Lewis structure for this molecule.

AnswerSeH H

PRACTICE ProblemsHave students refer to p. 994 for complete solutions to odd-numbered problems. The complete solutions for all problems can be found in the Solutions Manual.

1. PH H

H

2. SH

H

3. ClH

4.

ClCCl

Cl

Cl

5.

HSiH

H

H 6. Using 1 and 16 to represent atoms of

groups 1 and 16, respectively, the generic structure is:

161

1

Reading Check Sigma bonds can form from the overlap of an s orbital with another s orbital, an s orbital with a p orbital, or a p orbital with another p orbital.

The sigma bond Single covalent bonds are also called sigma bonds,represented by the Greek letter sigma (σ). A sigma bond occurs when the pair of shared electrons is in an area centered between the two atoms. When two atoms share electrons, their valence atomic orbitals overlap end to end, concentrating the electrons in a bonding orbital between the two atoms. A bonding orbital is a localized region where bonding electrons will most likely be found. Sigma bonds can form when an s orbital overlaps with another s orbital or a p orbital, or two p orbitals overlap. Water ( H 2 O), ammonia (N H 3 ), and methane (C H 4 ) have sigma bonds, as shown in Figure 8.7.

Reading Check List the orbitals that can form sigma bonds in acovalent compound.

■ Figure 8.6 The frosted-looking portions of this glass were chemically etched using hydrogen fluoride (HF), a weak acid. Hydrogen fluoride reacts with silica, the major component of glass, and forms gaseous silicon tetrafluoride (Si F 4 ) and water.

VOCABULARYACADEMIC VOCABULARY

Overlapto occupy the same area in partThe two driveways overlap at the street forming a common entrance.

EXAMPLE Problem 8.1

Lewis Structure of a Molecule The pattern on the glass shown in Figure 8.6 was made by chemically etching its surface with hydrogen fluoride (HF). Draw the Lewis structure for a molecule of hydrogen fluoride.

1 Analyze the ProblemYou are given the information that hydrogen and fluorine form the molecule hydrogen fluoride. An atom of hydrogen, a group 1 element, has only one valence electron. It can bond with any nonmetal atom when they share one pair of electrons. An atom of fluorine, a group 17 element, needs one electron to complete its octet. Therefore, a single covalent bond forms when atoms of hydrogen and fluorine bond.

2 Solve for the UnknownTo draw a Lewis structure, first draw the electron-dot diagram for each of the atoms. Then, rewrite the chemical symbols and draw a line between them to show the shared pair of electrons. Finally, add dots to show the unshared electron pairs.

H F+ → H — FHydrogen

atomFluorine

atomHydrogen fluoride

molecule

3 Evaluate the AnswerEach atom in the new molecule now has a noble-gas configuration and is stable.

PRACTICE Problems Extra Practice 79 and glencoe.com

Draw the Lewis structure for each molecule.

1. PH3 4. CC l4 2. H2S 5. Si H4

3. HCl

6. Challenge Draw a generic Lewis structure for a molecule formed between atoms of Group 1 and Group 16 elements.




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■ Caption Question Fig. 8.7 The sigma bonds are formed by the overlap of a hydrogen atom s orbital and a carbon atom p orbital.

Concept DevelopmentElectron Sharing Make sure stu-dents understand that an orbital can contain only two electrons at a time. An orbital with a single electron can only share its electron with another single-electron orbital. A double covalent bond involves two orbitals sharing four electrons between two atoms. A triple covalent bond involves three orbitals sharing six electrons between two atoms.

AssessmentAssessmentSkill Divide students into

groups. Hand each group a set of cards containing electron dot dia-grams of atoms. The set should include 2 C atoms, 2 N atoms, 2 O atoms, 6 H atoms, and 4 Cl atoms. Place a hydrogen card next to a chlo-rine card. Show students that by sharing a pair of electrons each atom obtains a stable outer electron con-figuration. Have students determine the Lewis structure for as many mol-ecules as possible by combining the cards. Remind students that multiple bonds are possible with carbon, nitrogen, and oxygen atoms. OL ELCOOP LEARN


Content BackgroundExceptional Oxygen Spins of unpaired electrons in atoms or ions cause a substance to be paramagnet-ic, that is, attracted to an external magnetic field. A substance is diamagnetic, unaffected by the mag-netic field, if all of the electrons are paired. Although the Lewis structure and bond strength for diatomic oxy-gen indicates there is a double bond between the two atoms, oxygen exhibits exceptional behavior in the laboratory. It is paramagnetic. The electrons associated with the π bond have unpaired, or parallel, spins.

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H

H H

C

Methane (CH4)

H

HH

O

Water (H2O)

HH

N

Ammonia (NH3)

H

■ Figure 8.7 Sigma bonds formed in each of these molecules when the atomic orbital of each hydrogen atom overlapped end to endwith the orbital of the central atom.Interpret Identify the types of orbitals that overlap to form the sigma bonds in methane.

Multiple Covalent BondsIn some molecules, atoms have noble-gas configurations when they share more than one pair of electrons with one or more atoms. Sharing multiple pairs of electrons forms multiple covalent bonds. A double covalent bond and a triple covalent bond are examples of multiple bonds. Carbon, nitrogen, oxygen, and sulfur atoms often form multiple bonds with other nonmetals. How do you know if two atoms will form a multiple bond? In general, the number of valence electrons needed to form an octet equals the number of covalent bonds that can form.

Double bonds A double covalent bond forms when two pairs of electrons are shared between two atoms. For example, atoms of the ele-ment oxygen only exist as diatomic molecules. Each oxygen atom has six valence electrons and must obtain two additional electrons for a noble-gas configuration, as shown in Figure 8.8a. A double covalent bond forms when each oxygen atom shares two electrons; a total of two pairs of electrons are shared between the two atoms.

Triple bonds A triple covalent bond forms when three pairs of elec-trons are shared between two atoms. Diatomic nitrogen ( N 2 ) molecules contain a triple covalent bond. Each nitrogen atom shares three electron pairs, forming a triple bond with the other nitrogen atom as shown in Figure 8.8b.

The pi bond A multiple covalent bond consists of one sigma bond and at least one pi bond. A pi bond, represented by the Greek letter pi (π), forms when parallel orbitals overlap and share electrons. The shared electron pair of a pi bond occupies the space above and below the line that represents where the two atoms are joined together.

+ →NN

+ →O O O——

N —N——

O

Two shared pairsof electrons

Three shared pairsof electrons

a

b

■ Figure 8.8 Multiple covalent bonds form when two atoms share more than one pair of electrons. a. Two oxygen atoms form a double bond. b. A triple bond forms between two nitrogen atoms.

Incorporate information from this section into your Foldable.

Personal Tutor For an online tutorial onmultiple covalent bonds, visit glencoe.com.

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Differentiated InstructionBelow Level Have students place a single toothpick between two gumdrops to form a single bond. Then, have them break 1 cm off each of two toothpicks, and place them between two gumdrops to form a double bond. Have them break 2 cm off each of three toothpicks, and place them between two gumdrops to represent a triple bond. Have students identify the shortest and longest bonds. Ask students to carefully break the bonds to determine which bond is easiest and which is the most difficult to break. A triple bond is the shortest, is the most difficult to break, and has the highest bond energy. The single bond is the longest, is the easiest to break, and has the lowest bond energy. BL EL

Concept DevelopmentBond Strength An important factor in determining the chemical reactivity of a molecule is the strength of the covalent bonds holding the molecule together. When comparing two molecules with similar composition, the one with weaker bonds is usually more reactive. Have students compare the bond energies of the molecules C H 3 Cl and C H 2 C l 2 . 1567 kJ/mol, 1482 kJ/mol Ask students which molecule is more likely to react. C H 2 C l 2 OL

EnrichmentBond Energy Use the equation shown below to calculate the change in enthalpy of a chemical reaction.

∆H = ∑(bond energies of broken bonds) - ∑(bond energies of formed bonds)

Have students determine the total enthalpy change when C l 2 reacts with C H 4 to form C H 3 Cl and HCl. The enthalpy change is −104 kJ Ask stu-dents what the negative indicates about the energy of the reaction and the strength of the bonds formed. A negative enthalpy change indicates that the reaction is exothermic, and that the bonds in the product molecules are stronger than the bonds in the reactant molecules. Have them draw the Lewis structures of the molecules and predict which bonds are broken and which are formed. Bonds broken are one Cl—Cl and four C—H. Bonds formed are one H—Cl, three C—H, and one C—Cl. AL

Reading Check As the bond type increases from single covalent, to double covalent, to triple covalent, the bond length decreases.

Interactive Figure Students can interact with the sigma and pi bonding art at glencoe.com.

Nuclei

Bond length

■ Figure 8.10 Bond length is the distance from the center of one nucleus to the center of the other nucleus of two bonded atoms.

σ bond

σ bond

σ bond

σ bondσ bond

p overlap

p overlap

H H

CC

H H

π bond

Ethene

——

H H

HC C

H

■ Figure 8.9 Notice how the multiple bond between the two carbon atoms in ethene ( C 2 H 4 ) consists of a sigma bond and a pi bond. The carbon atoms are close enough that the side-by-side p orbitals overlap and forms the pi bond. This results in a doughnut-shaped cloud around the sigma bond.

Table 8.1 Covalent Bond Type and Bond Length

Molecule Bond Type Bond Length

F 2 single covalent 1.43 × 10 -10 m

O 2 double covalent 1.21 × 10 -10 m

N 2 triple covalent 1.10 × 10 -10 m

It is important to note that molecules having multiple covalent bonds contain both sigma and pi bonds. A double covalent bond, as shown in Figure 8.9, consists of one pi bond and one sigma bond. A triple covalent bond consists of two pi bonds and one sigma bond.

The Strength of Covalent BondsRecall that a covalent bond involves attractive and repulsive forces. In a molecule, nuclei and electrons attract each other, but nuclei repel other nuclei, and electrons repel other electrons. When this balance of forces is upset, a covalent bond can be broken. Because covalent bonds differ in strength, some bonds break more easily than others. Several factors influence the strength of covalent bonds.

Bond length The strength of a covalent bond depends on the dis-tance between the bonded nuclei. The distance between the two bonded nuclei at the position of maximum attraction is called bond length, as shown in Figure 8.10. It is determined by the sizes of the two bonding atoms and how many electron pairs they share. Bond lengths for mole-cules of fluorine ( F 2 ), oxygen ( O 2 ), and nitrogen ( N 2 ) are listed in Table 8.1. Notice that as the number of shared electron pairs increases, the bond length decreases.

Bond length and bond strength are also related: the shorter the bond length, the stronger the bond. Therefore, a single bond, such as that in F 2 , is weaker than a double bond, such as that in O 2 . Likewise, the double bond in O 2 is weaker than the triple bond in N 2 .

Reading Check Relate covalent bond type to bond length.

Interactive Figure To see an animation of sigma and pi bonding, visit glencoe.com.

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Self-Check Quiz glencoe.com

3 AssessCheck for UnderstandingAsk students to draw the Lewis structure for the molecule H 2 S. OL

SH

H

Reteach Draw the structural formulas for C 2 H 6 , C 2 H 4 , and C 2 H 2 . Ask students to identify all sigma and pi bonds. Using a table of bond energies and bond length, compare the bond lengths and bond energies of C—C, C=C, and C≡C. Remind students that as the number of bonds between two atoms increases, the bond length becomes shorter and the bond itself becomes stronger. BL

ExtensionHave interested students research trinitrotoluene (TNT) and determine the available bond energy if all bonds in a molecule of this compound were broken. AL

Section 8.18.1 Assessment 7. Most covalent bonds form between nonmetallic elements. 8. Atoms share valence electrons; the shared electrons complete the octet of

each atom. 9. Student Lewis structures should show the sharing of a single pair of

electrons, two pairs of electrons, and three pairs of electrons, respectively, for single, double, and triple covalent bonds.

10. Valence electrons are involved in both types of bonds. In covalent bonds, atoms share electrons, whereas in ionic bonds, electrons are transferred between atoms.

11. A sigma bond is a single covalent bond formed from the direct overlap of orbitals. A pi bond is the parallel overlap of p orbitals.

12. Student graphs should show that as bond length decreases, the bond dissociation energy increases.

13. a. C—H: less energy than C≡C b. C—H: less energy than C=C

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Section 8.18.1 Assessment

Table 8.2 Bond-Dissociation Energy

Molecule Bond-Dissociation Energy

F 2 159 kJ/mol

O 2 498 kJ/mol

N 2 945 kJ/mol

Section Summary◗ ◗ Covalent bonds form when atoms

share one or more pairs of electrons.

◗ ◗ Sharing one pair, two pairs, and three pairs of electrons forms single, double, and triple covalent bonds, respectively.

◗ ◗ Orbitals overlap directly in sigma bonds. Parallel orbitals overlap in pi bonds. A single covalent bond is a sigma bond but multiple covalent bonds are made of both sigma and pi bonds.

◗ ◗ Bond length is measured nucleus-to-nucleus. Bond-dissociation energy is needed to break a covalent bond.

7. Identify the type of atom that generally forms covalent bonds.

8. Describe how the octet rule applies to covalent bonds.

9. Illustrate the formation of single, double, and triple covalent bonds using Lewis structures.

10. Compare and contrast ionic bonds and covalent bonds.

11. Contrast sigma bonds and pi bonds.

12. Apply Create a graph using the bond-dissociation energy data in Table 8.2 and the bond-length data in Table 8.1. Describe the relationship between bond length and bond-dissociation energy.

13. Predict the relative bond-dissociation energies needed to break the bonds in the structures below.

a. H — C C — H——— b.

————

H H

HH

C C——

Bonds and energy An energy change occurs when a bond between atoms in a molecule forms or breaks. Energy is released when a bond forms, but energy must be added to break a bond. The amount of energy required to break a specific covalent bond is called bond-dissociation energy and is always a positive value. The bond-dissociation energies for the covalent bonds in molecules of fluorine, oxygen, and nitrogen are listed in Table 8.2.

Bond-dissociation energy also indicates the strength of a chemical bond because of the inverse relationship between bond energy and bond length. As indicated in Table 8.1 and Table 8.2, the smaller bond length, the greater the bond-dissociation energy. The sum of the bond-dissociation energy values for all of the bonds in a molecule is the amount of chemical potential energy in a molecule of that compound.

The total energy change of a chemical reaction is determined from the energy of the bonds broken and formed. An endothermic reaction occurs when a greater amount of energy is required to break the exist-ing bonds in the reactants than is released when the new bonds form in the products. An exothermic reaction occurs when more energy is released during product bond formation than is required to break bonds in the reactants. See Figure 8.11.

■ Figure 8.11 Breaking the C–C bonds in charcoal and the O–O bonds in the oxygen in air requires an input of energy. Energy is released as heat and light when bonds form producing C O 2 . Thus, the burning of charcoal is an exothermic reaction.

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Section 8.28.21 Focus

Focus TransparencyBefore presenting the lesson, project Section Focus Transparency 28 and have students answer the accompa-nying questions. BL EL

MAIN IdeaNaming Molecules Ask students how they would identify or name their grandmother’s sister in terms of their familial relationship. Their grandmother’s sister is their great aunt. Ask them to explain how they determined that name. They should describe a systematic way of naming relatives. Mother’s or father’s brothers are uncles. Mother’s or father’s sisters are aunts. Explain that molecules are named using a systematic approach. OL BL

2 TeachVisual LearningTable 8.3 Review oxidation numbers, naming ionic compounds, and writing formula units. Unlike ionic compounds, the number of each atom that makes up a molecule is identified by a prefix. Write several molecular formulas on the board and have students use Table 8.3 to practice naming molecules. OL

Interactive Table Students can interact with the table at glencoe.com.

DemonstrationSafety Precautions

Disposal Dispose of the solid in a landfill approved to receive chemical waste.ProcedurePlace a large can in the sink. Hold the magnesium ribbon with tongs and light it. Hold the burning ribbon in the can. If the room lights are turned off, students can safely

observe the light. WARNING: Wear goggles when burning magnesium. Do not look directly at the burning metal. In a darkened room, place a small piece of roll sulfur in a def-lagrating spoon and briefly heat it in a burner flame. The sulfur will ignite and burn with a low blue flame. WARNING: Use a fume hood; S O 2 vapor is toxic.

Forming BondsPurposeto demonstrate the formation of ionic and covalent bondsMaterialsMagnesium ribbon (5 cm); roll sulfur (2 g); tongs; large metal can; deflagrating spoon; laboratory burner

Section 8.28.2

Objectives

◗◗ Translate molecular formulas into binary molecular compound names.

◗◗ Name acidic solutions.

Review Vocabularyoxyanion: a polyatomic ion in which an element (usually a nonmetal) is bonded to one or more oxygen atoms

New Vocabularyoxyacid

Naming MoleculesMAIN Idea Specific rules are used when naming binary molecular compounds, binary acids, and oxyacids.

Real-World Reading Link You probably know that your mother’s mother is your grandmother, and that your grandmother’s sister is your great-aunt. But what do you call your grandmother’s brother’s daughter? Naming molecules requires a set of rules, just as naming family relationships requires rules.

Naming Binary Molecular CompoundsMany molecular compounds have common names, but they also have scientific names that reveal their composition. To write the formulas and names of molecules, you will use processes similar to those described in Chapter 7 for ionic compounds.

Start with a binary molecular compound. Note that a binary molecu-lar compound is composed only of two nonmetal atoms—not metal atoms or ions. An example is dinitrogen monoxide ( N 2 O), a gaseous anesthetic that is more commonly known as nitrous oxide or laughing gas. The naming of nitrous oxide is explained in the following rules. 1. The first element in the formula is always named first, using the

entire element name. N is the symbol for nitrogen.

2. The second element in the formula is named using its root and adding the suffix -ide. O is the symbol for oxygen so the second word is oxide.

3. Prefixes are used to indicate the number of atoms of each element that are present in the compound. Table 8.3 lists the most common prefixes used. There are two atoms of nitrogen and one atom of oxygen, so the

first word is dinitrogen and second word is monoxide. There are exceptions to using the prefixes shown in Table 8.3. The

first element in the compound name never uses the mono- prefix. For example, CO is carbon monoxide, not monocarbon monoxide. Also, if using a prefix results in two consecutive vowels, one of the vowels is usually dropped to avoid an awkward pronunciation. For example, notice that the oxygen atom in CO is called monoxide, not monooxide.

Table 8.3 Prefixes in Covalent Compounds

Number of Atoms Prefix Number of Atoms Prefix

1 mono- 6 hexa-

2 di- 7 hepta-

3 tri- 8 octa-

4 tetra- 9 nona-

5 penta- 10 deca-

Interactive Table Explore naming covalent compounds at glencoe.com.




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IN-CLASS Example

Question PC l 5 is a colorless to pale, yellow solid with a pungent odor mainly used as a gentle chlori-nating agent in manufacturing dyes and pharmaceuticals. Name this compound.

Answer The first element, P, is phospho-rus. The second element, Cl, is chlorine. Add the suffix -ide to the root of chlorine to make chloride. Now modify the name to indicate the number of atoms present in a molecule. There are no subscripts given for phosphorus, indi-cating that there is only one atom of this ele-ment present. Chloride has the subscript 5, the prefix for which is penta-. Hence, the com-pound is called phosphorus pentachloride.


14. carbon dioxide 15. sulfur dioxide 16. nitrogen trifluoride 17. carbon tetrachloride 18. A s 2 O 3

ReinforcementMolecule Mobile Have students make a mobile that can be used to name molecules. Display the mobile in the classroom. BL EL

Reading Check nitrogen tri-hydride, dinitrogen tetrahydride, nitrogen oxide

ResultsMagnesium burns with a bright, white-hot flame. Sulfur burns with a low blue flame. The stability of MgO is evidenced by the large amount of heat and light produced when it is formed, compared to the heat and light pro-duced in the formation of S O 2 . Thus, the ionic bonds in MgO are stronger (more stable) than the covalent bonds in S O 2. .

AnalysisAsk these questions: 1. In the demonstration, with what element do

magnesium and sulfur react? oxygen 2. Write balanced chemical equations for the

chemical reactions. 2Mg(s) + O 2 (g) → 2MgO(s); S(s) + O 2 (g) → S O 2 (g)

AssessmentAssessmentKnowledge If Mg forms an

ionic bond, and S forms a covalent bond, which bond type is the stron-ger bond? Why? The ionic bond is stron-ger because more energy is released during its formation. OL

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Common names for some molecular compounds Have you ever enjoyed an icy, cold glass of dihydrogen monoxide on a hot day? You probably have but you most likely called it by its common name, water. Recall from Chapter 7 that many ionic compounds have common names in addition to their scientific ones. For example, baking soda is sodium hydrogen carbonate and common table salt is sodium chloride.

Many binary molecular compounds, such as nitrous oxide and water, were discovered and given common names long before the present-day naming system was developed. Other binary covalent compounds that are generally known by their common names rather than their scientific names are ammonia (N H 3 ), hydrazine ( N 4 H 4 ), and nitric oxide (NO).

Reading Check Apply What are the scientific names for ammonia, hydrazine, and nitric oxide?

EXAMPLE Problem 8.2

Naming Binary Molecular Compounds Name the compound P 2 O 5 , which is used as a drying and dehydrating agent.

1 Analyze the ProblemYou are given the formula for a compound. The formula contains the elements and the number of atoms of each element in one molecule of the compound. Because only two different elements are present and both are nonmetals, the compound can be named using the rules for naming binary molecular compounds.

2 Solve for the UnknownFirst, name the elements involved in the compound.

phosphorus The first element, represented by P, is phosphorus.

oxide The second element, represented by O, is oxygen. Add the suffix –ide to the root of oxygen, ox-.

phosphorus oxide Combine the names.

Now modify the names to indicate the number of atoms present in a molecule.

diphosphorus pentoxide From the formula P2O5, you know that two phosphorus atoms and five oxygen atoms make up a molecule of the compound. From Table 8.3, you know that di- is the prefix for two and penta- is the prefix for five. The a in penta- is not used because oxide begins with a vowel.

3 Evaluate the AnswerThe name diphosphorus pentoxide shows that a molecule of the compound contains two phosphorus atoms and five oxygen atoms, which agrees with the compound’s chemical formula, P2O5.

PRACTICE Problems Extra Practice Page 979 and glencoe.com

Name each of the binary covalent compounds listed below.

14. CO2

15. SO2

16. NF3

17. CCl418. Challenge What is the formula for diarsenic trioxide?

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Section 8.2 • Naming Molecules 249


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AssessmentAssessmentKnowledge Show students

a chart that contains molecular formulas and have students name each molecule or acid. Use a second chart that contains names of both molecules and acids. Have students determine the correct molecular formula for each substance. OL

Differentiated InstructionAdvanced Learners Have interested students research and write a report on the chemical and physical properties, sources, and uses of compounds that contain only nitrogen and oxygen. These compounds are nitrogen monoxide, nitrogen dioxide, dinitrogen monoxide, dinitrogen trioxide, dinitrogen tetroxide, and dinitrogen pentoxide. AL

Chemistry ProjectMolecular Compounds Using a 2 × 2 grid, have students determine all possible molecules formed when either carbon or hydrogen reacts with either oxygen or fluorine. Have them use the oxidation states of 4+ for carbon, 1+ for hydrogen, 2- for oxygen, and 1- for fluorine. Write the correct formula and name for each compound. OL

Quick Demo

Acid Formation Add 2 drops of bromothymol blue indicator to 50.0 mL of distilled water in a flask. Place a straw in the water and have a student blow through the straw. The indicator will turn from blue to yellow because the carbon dioxide exhaled from the student reacts with the water forming an acidic solution.C O 2 + H 2 O → H 2 C O 3 Write the formula for the acid formed ( H 2 C O 3 ) on the board, and ask students if it is a binary acid or an oxyacid. oxyacid Ask students to name the acid. carbonic acid OL

Table 8.4 Naming Oxyacids

Compound Oxyanion Acid Suffix Acid Name

HCl O 3 chlorate -ic chloric acid

HCl O 2 chlorite -ous chlorous acid

HN O 3 nitrate -ic nitric acid

HN O 2 nitrite -ous nitrous acid

Naming AcidsWater solutions of some molecules are acidic and are named as acids. Acids are important compounds with specific properties and will be discussed at length in Chapter 18. If a compound produces hydrogen ions ( H + ) in solution, it is an acid. For example, HCl produces H + in solution and is an acid. Two common types of acids exist—binary acids and oxyacids.

Naming binary acids A binary acid contains hydrogen and one other element. The naming of the common binary acid known as hydrochloric acid is explained in the following rules. 1. The first word has the prefix hydro- to name the hydrogen part of

the compound. The rest of the first word consists of a form of the root of the second element plus the suffix -ic. HCl (hydrogen and

chlorine) becomes hydrochloric.

2. The second word is always acid. Thus, HCl in a water solution is

called hydrochloric acid.

Although the term binary indicates exactly two elements, a few acids that contain more than two elements are named according to the rules for naming binary acids. If no oxygen is present in the formula for the acidic compound, the acid is named in the same way as a binary acid, except that the root of the second part of the name is the root of the polyatomic ion that the acid contains. For example, HCN, which is composed of hydrogen and the cyanide ion, is called hydrocyanic acid in solution.

Naming oxyacids An acid that contains both a hydrogen atom and an oxyanion is referred to as an oxyacid. Recall from Chapter 7 that an oxyanion is a polyatomic ion containing one or more oxygen atoms. The following rules explain the naming of nitric acid (HN O 3 ), an oxyacid. 1. First, identify the oxyanion present. The first word of an oxyacid’s

name consists of the root of the oxyanion and the prefix per- or hypo- if it is part of the name, and a suffix. If the oxyanion’s name ends with the suffix -ate, replace it with the suffix -ic. If the name of the oxyanion ends with the suffix -ite, replace it with the suffix -ous. NO3, the nitrate ion, becomes nitric.

2. The second word of the name is always acid. HNO3 (hydrogen and

the nitrate ion) becomes nitric acid.

Table 8.4 shows how the names of several oxyacids follow these rules. Notice that the hydrogen in an oxyacid is not part of the name.





19. hydroiodic acid 20. chloric acid 21. chlorous acid 22. sulfuric acid 23. hydrosulfuric acid 24. HI O 4 25. AgCl 26. H 2 O 27. Cl F 3 28. P 2 O 3 29. S 2 F 10

30. H 2 C O 3

Interactive Table Students can interact with the table at glencoe.com.

AssessmentAssessmentKnowledge Have students

write molecular formulas, names of molecules, names of acids, or formu-las of acids on the front of note cards. Have another student write the correct formula or name on the back of each note card. Use these note cards for review. EL

Chemistry JournalForm Compounds Give students the following chart with the answers left blank. Have them write the formula and name the molecule that forms in their journal. Remind students that usu-ally the least electronegative atom is written first in a compound. OL

Br F ClC CB r 4

carbon tetrabromideC F 4

carbon tetrafluorideCC l 4

carbon tetrachlorideP PB r 3

phosphorus tribromideP F 3

phosphorus trifluoridePC l 3

phosphorus trichlorideS SB r 2

sulfur dibromideS F 2

sulfur difluorideSC l 2

sulfur dichloride

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You have learned that naming covalent compounds follows different sets of rules depending on the composition of the compound. Table 8.5 summarizes the formulas and names of several covalent compounds. Note that an acid, whether a binary acid or an oxyacid, can have a com-mon name in addition to its compound name.

Writing Formulas from NamesThe name of a molecular compound reveals its composition and is important in communicating the nature of the compound. Given the name of any binary molecule, you should be able to write the correct chemical formula. The prefixes used in a name indicate the exact num-ber of each atom present in the molecule and determine the subscripts used in the formula. If you are having trouble writing formulas from the names for binary compounds, you might want to review the naming rules listed on pages at the beginning of this section.

The formula for an acid can also be derived from the name. It is helpful to remember that all binary acids contain hydrogen and one other element. For oxyacids—acids containing oxyanions—you will need to know the names of the common oxyanions. If you need to review oxyanion names, see Table 7.9 in the previous chapter.


Name the following acids. Assume each compound is dissolved in water.

19. HI 20. HCl O3 21. HCl O2 22. H2SO4 23. H2S

24. Challenge What is the formula for periodic acid?


Give the formula for each compound.

25. silver chloride

26. dihydrogen oxide

27. chlorine trifluoride

28. diphosphorus trioxide

29. strontium acetate

30. Challenge What is the formula for carbonic acid?

Table 8.5 Formulas and Names of Some Covalent Compounds

Formula Common Name Molecular Compound Name

H 2 O water dihydrogen monoxide

N H 3 ammonia nitrogen trihydride

N 2 H 4 hydrazine dinitrogen tetrahydride

HCl muriatic acid hydrochloric acid

C 9 H 8 O 4 aspirin 2-(acetyloxy)benzoic acid

Interactive Table Explore naming covalent compounds glencoe.com.

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Section 8.2 • Naming Molecules 251


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■ Caption Question Fig. 8.12H 2 S O 3 is an oxyacid; HBr is a binary acid.

3 AssessCheck for UnderstandingShow students a 4 × 4 grid containing eight molecules and their corresponding names, randomly arranged. Cover each square and play a game by matching the formula of the molecule with its correct name. BL EL

ReteachMany common air pollutants form acids in water. Have students create tables that have the following column heads: Formula of Pollutant, Name of Molecule, Formula of Acid, and Name of Acid. For each of the following pollutants, provide information for one column and have students complete the table: S O 2 (sulfur dioxide, H 2 S O 3 , sulfurous acid), C O 2 (carbon dioxide, H 2 C O 3 , carbonic acid), S O 3 (sulfur trioxide, H 2 S O 4 , sulfuric acid), N O 2 (nitrogen dioxide, HN O 3 , nitric acid). OL

ExtensionHave students write answers and questions for a game using the categories Molecular Formulas, Names of Molecules, Names of Acids, and Formulas for Acids. OL

Section 8.28.2 Assessment 31. name the first element in the formula first, name the second element using

its root plus the suffix –ide, add prefixes to indicate the number of atoms of each element present

32. a molecule composed of only two nonmetal elements 33. A binary acid contains hydrogen and one other element. An oxyacid

contains hydrogen, another element, and oxygen. 34. There are two atoms of nitrogen; use the prefix di– with the name nitrogen.

There are four atoms of oxygen, so use the prefix tetra– + the root of oxygen + the ending –ide. The name is dinitrogen tetroxide.

35. S 2 O 3 , HI O 3 , N 2 O, HF 36. a. N 2 O 3 b. NO c. HCl d. HCl O 3 e. H 2 S O 4 f. H 2 S O 3


Look at the formula ofthe molecule.

No(NO2)

Examples:HBr, H2SO3,

and N02

No(HBr)

Name the first element in themolecule. Use a prefix if the numberof atoms is greater than one. Toname the second element, indicate the number present by using a prefix +root of second element + -ide.

Name as an acid.Is there an oxygenpresent in the compound?

Hydro + root of second element + -ic, then acid.

Root of oxyanion present + -icif the anion ends in -ate, or + -ousif the anion ends in -ite, then acid.

Does the compoundform an acidicaqueous solution?

Yes(H2SO3 and HBr)

Yes(H2SO3)

NO2 is nitrogen dioxide.

H2SO3 is sulfurous acid.HBr (aq) is hydrobromic acid.

■ Figure 8.12 Use this flowchart to name molecular compounds when their formulas are known.Apply Which compound above is an oxyacid? Which is a binary acid?

The flowchart in Figure 8.12 can help you determine the name of a molecular covalent compound. To use the chart, start at the top and work downward by reading the text contained in the colored boxes and applying it to the formula of the compound you wish to name.

Section Summary◗ ◗ Names of covalent molecular

compounds include prefixes for the number of each atom present. The final letter of the prefix is dropped if the element name begins with a vowel.

◗ ◗ Molecules that produce H + in solu-tion are acids. Binary acids contain hydrogen and one other element. Oxyacids contain hydrogen and an oxyanion.

31. MAIN Idea Summarize the rules for naming binary molecular compounds.

32. Define a binary molecular compound.

33. Describe the difference between a binary acid and an oxyacid.

34. Apply Using the system of rules for naming binary molecular compounds, describe how you would name the molecule N 2 O 4 .

35. Apply Write the molecular formula for each of these compounds: iodic acid, disulfur trioxide, dinitrogen monoxide, and hydrofluoric acid.

36. State the molecular formula for each compound listed below. a. dinitrogen trioxide d. chloric acid b. nitrogen monoxide e. sulfuric acid c. hydrochloric acid f. sulfurous acid

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MAIN IdeaMolecular Structures Give the students a set of snap-together toy blocks and ask them to make a pyramid. Ask them if there was a set pattern they followed. The shape of the blocks determined how they could build the pyramid. Tell the students that molecular shapes are determined by the way atoms share electrons. OL

2 TeachQuick Demo

Octet Models Place a circle with the electron dot diagram of a sulfur atom and two circles with the electron dot diagram of oxygen on a felt board. Ask students how these atoms could be arranged to satisfy the octet rule. Students will use trial-and-error methods. Tell students that a systematic approach can be used to determine the Lewis structure of molecules, including those that show exceptions to the octet rule. OL EL

Visual LearningBuild Models Using a molecular model kit or marshmallows and toothpicks, have stu-dents build models of N H 4 + , C H 2 O, BC l 3, and HF. Have students create a chart showing the ball-and-stick model, Lewis structure, and structural formula for each substance. OL EL

■ Caption Question Fig. 8.13 All models show the number and type of each atom; Lewis structures, structural formulas, ball-and-stick models, and space-filling mod-els show geometric shape; Lewis structures show distribution of valence electrons among bonding pairs and lone pairs. Space-filling models show the relative sizes of the atoms.

Section 8.38.3

Molecular StructuresMAIN Idea Structural formulas show the relative positions of atoms within a molecule.

Real-World Reading Link As a child, you might have played with plastic building blocks that connected only in certain ways. If so, you probably noticed that the shape of the object you built depended on the limited ways the blocks interconnected. Building molecules out of atoms works in a similar way.

Structural FormulasIn Chapter 7, you learned about the structure of ionic compounds—substances formed from ionic bonds. The covalent molecules you have read about in this chapter have structures that are different from those of ionic compounds. In studying the molecular structures of covalent compounds, models are used as representations of the molecule.

The molecular formula, which shows the element symbols and numerical subscripts, tells you the type and number of each atom in a molecule. As shown in Figure 8.13, there are several different models that can be used to represent a molecule. Note that in the ball-and-stick and space-filling molecular models, atoms of each specific element are represented by spheres of a representative color, as shown in Table R-1 on page 968. These colors are used for identifying the atoms if the chemical symbol of the element is not present.

One of the most useful molecular models is the structural formula, which uses letter symbols and bonds to show relative positions of atoms. You can predict the structural formula for many molecules by drawing the Lewis structure. You have already seen some simple examples of Lewis structures, but more involved structures are needed to help you determine the shapes of molecules.

PH3

H — P — H

H

Molecular formula

Structural formula

H — P — H

HLewis structure

Space-fillingmolecular model

Ball-and-stickmolecular model

■ Figure 8.13 All of these models can be used to show the relative locations of atoms and electrons in the phosphorus trihydride (phosphine) molecule.Compare and contrast the types of information contained in each model.

Objectives

◗◗ List the basic steps used to draw Lewis structures.

◗◗ Explain why resonance occurs, and identify resonance structures.

◗◗ Identify three exceptions to the octet rule, and name molecules in which these exceptions occur.

Review Vocabularyionic bond: the electrostatic force that holds oppositely charged particles together in an ionic compound

New Vocabularystructural formularesonancecoordinate covalent bond

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Section 8.3 • Molecular Structures 253


Apply ChemistryDyes Organic dyes are organic molecules that absorb select wave-lengths of visible light and reflect all other wavelengths. These dyes are used to produce colorful fabrics. They are also used in recordable compact discs. Recordable CDs have a layer of colored, organic dye sand-wiched between a clear rigid backing and a reflective surface. A laser is used to burn data onto the CD. The bonds in the organic dye molecule absorb the laser light and change the structure of the molecule, producing an opaque substance. This change in structure is irreversible; therefore, data can be written only once on any part of the compact disc.

Concept DevelopmentLewis Structure Using the systematic approach, have students draw Lewis structures for several molecules. Remind them that carbon atoms can form chains with other carbon atoms. The carbon atoms will bond together with a single, double, or triple bond, depending on the available electron pairs. Have students draw the Lewis structure for ethane ( C 2 H 6 ), ethene ( C 2 H 4 ), and ethyne ( C 2 H 2 ), using the systematic technique. OL

ReinforcementRepresenting Molecules Point out that both Lewis structures and structural formulas are used to show the arrangement and ratio of atoms in a compound. Lewis structures can be used to represent both ionic and covalent compounds. Structural formulas are used only to represent molecules. Structural formulas use a line to represent a shared pair of electrons. However, unshared, lone, pairs of electrons are still represented by electron dots.

Chemistry ProjectGilbert Lewis Have students research Gilbert Newton Lewis and make posters depicting his original Lewis structures. Students should share the posters with the class. EL

Virtual Labs CD-ROMChemistry: Matter and ChangeVideo: Lewis Structures

GLENCOE Technology

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Lewis structures Although it is fairly easy to draw Lewis structures for most compounds formed by nonmetals, it is a good idea to follow a regular procedure. Whenever you need to draw a Lewis structure, follow the steps outlined in this Problem-Solving Strategy.

Problem-Solving StrategyDrawing Lewis Structures

1. Predict the location of certain atoms.

The atom that has the least attraction for shared electrons will be the central atom in the molecule. This element is usually the one closer to the left side of the periodic table. The central atom is located in the center of the molecule; all other atoms become terminal atoms.

Hydrogen is always a terminal, or end, atom. Because it can share only one pair of electrons, hydrogen can be connected to only one other atom.

2. Determine the number of electrons available for bonding.

This number is equal to the total number of valence electrons in the atoms that make up the molecule.

3. Determine the number of bonding pairs.

To do this, divide the number of electrons available for bonding by two.

4. Place the bonding pairs.

Place one bonding pair (single bond) between the central atom and each of the terminal atoms.

5. Determine the number of bonding pairs remaining.

To do this, subtract the number of pairs used in Step 4 from the total number of bonding pairs determined in Step 3. These remaining pairs include lone pairs as well as pairs used in double and triple bonds. Place lone pairs around each terminal atom (except H atoms) bonded to the central atom to satisfy the octet rule. Any remaining pairs will be assigned to the central atom.

6. Determine whether the central atom satisfies the octet rule.

Is the central atom surrounded by four electron pairs? If not, it does not satisfy the octet rule. To satisfy the octet rule, convert one or two of the lone pairs on the terminal atoms into a double bond or a triple bond between the terminal atom and the central atom. These pairs are still associated with the terminal atom as well as with the central atom. Remember that carbon, nitrogen, oxygen, and sulfur often form double and triple bonds.

Apply the StrategyStudy Example Problems 8.3 through 8.5 to see how the steps in the Problem-Solving Strategy are applied.

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IN-CLASS Example

Question Silicon tetrahydride is a gas at room temperature that under-goes spontaneous combustion in air. What is the Lewis structure for Si H 4 ?

Answer

1 Si atom × 4 valence electrons ___ Si atom

+ 4 H atoms × 1 valence electron ___ H atom

= 8 total electrons ___ 2

= 4 bonding pairs

HSH

H

H Use all bonding pairs. Silicon has an octet

and hydrogen only shares one pair. This Lewis structure satisfies the octet rule.

Content BackgroundSpace Molecules In dust clouds throughout the universe, dust parti-cles provide a surface for chemical reactions to take place. Atoms join together to create larger molecules. As these molecules tumble through space, they change energy and give off radiation in the form of radio waves. In May 2000, scientists dis-covered the simple sugar molecule ribose floating in the dust cloud called Sagittarius B2. Astronomers have identified more than 120 space molecules.


37.

B

H

HH

38.

FNF

F

Chemistry JournalLewis Structures Have students predict the Lewis structures for the following common molecules: carbon monoxide (CO), carbon dioxide (C O 2 ), sulfur dioxide (S O 2 ), sulfur trioxide (S O 3 ), hydrogen cyanide (HCN), dinitrogen monoxide ( N 2 O), nitrogen dioxide (N O 2 ), and nitrogen monoxide (NO). OL

EXAMPLE Problem 8.3

Lewis Structure for a Covalent Compound with Single Bonds Ammonia is a raw material used in the manufacture of many materials, including fertilizers, cleaning products, and explosives. Draw the Lewis structure for ammonia (N H 3 ).

1 Analyze the ProblemAmmonia molecules consist of one nitrogen atom and three hydrogen atoms. Because hydrogen must be a terminal atom, nitrogen is the central atom.

2 Solve for the UnknownFind the total number of valence electrons available for bonding.

1 N atom × 5 valence electrons__1 N atom

+ 3 H atoms × 1 valence electron__1 H atom

= 8 valence electrons

There are 8 valence electrons available for bonding.

8 electrons__2 electrons/pair

= 4 pairs Determine the total number of bonding pairs. To do this, divide the number of available electrons by two.

Four pairs of electrons are available for bonding.

H

H —N —H

—

Place a bonding pair (a single bond) between the central nitrogen atom and each terminal hydrogen atom.

Determine the number of bonding pairs remaining.

4 pairs total - 3 pairs used Subtract the number of pairs used in these bonds from the total number of pairs of electrons available.

= 1 pair available

The remaining pair—a lone pair—must be added to either the terminal atoms or the central atom. Because hydrogen atoms can have only one bond, they have no lone pairs.

H

H —N —H

—

Place the remaining lone pair on the central nitrogen atom.

3 Evaluate the AnswerEach hydrogen atom shares one pair of electrons, as required, and the central nitrogen atom shares three pairs of electrons and has one lone pair, providing a stable octet.


37. Draw the Lewis structure for B H3.

38. Challenge A nitrogen trifluoride molecule contains numerous lone pairs. Draw its Lewis structure.

Dimensional Analysis page 956

Math Handbook




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IN-CLASS Example

Question Selenium trioxide is a yellowish-white crystalline powder that decomposes on heating and is a very toxic compound. What is the Lewis structure for Se O 3 ?

Answer

1 Se atom × 6 valence electrons ___ Se atom

+

3 O atoms × 6 valence electrons ___ O atom

= 24 valence electrons ___ 2

= 12 electron pairs

OSeO

O

All bonded atoms have an octet.


39. C CH

H

H

H

40. C SS

Apply ChemistryRipening Fruits and vegetables naturally produce ethylene gas ( C 2 H 4 ) as they ripen. For some fruits and vegetables, the ripening process is accelerated when they are exposed to ethylene. The effect of ethylene on fruits and vegetables is dependent on the sensitivity of the item. Tell stu-dents that they can accelerate the ripening of fruit at home by placing unripe fruit in a closed paper bag. The trapped ethylene gas accelerates the ripening process.

Differentiated InstructionAdvanced Learners Have students make posters that list the steps of the systematic approach as they determine the Lewis structure of one of the following molecules or ions: O H - , P F 6- , B F 3 , Cl O 3- , S F 3 N, or C H 2 Cl 2 . AL

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EXAMPLE Problem 8.4

Lewis Structure for a Covalent Compound with Multiple Bonds Carbon dioxide is a product of all cellular respiration. Draw the Lewis structure for carbon dioxide (C O 2 ).

1 Analyze the ProblemThe carbon dioxide molecule consists of one carbon atom and two oxygen atoms. Because carbon has less attraction for shared electrons, carbon is the central atom, and the two oxygen atoms are terminal.


1 C atom × 4 valence electrons__1C atom

+ 2 O atoms × 6 valence electrons__1O atom


There are 16 valence electrons available for bonding.


= 8 pairs Determine the total number of bonding pairs by dividing the number of available electrons by two.

Eight pairs of electrons are available for bonding.

O — C — OPlace a bonding pair (a single bond) between the central carbon atom and each terminal oxygen atom.

Determine the number of bonding pairs remaining. Subtract the number of pairs used in these bonds from the total number of pairs of electrons available.

8 pairs total - 2 pairs used Subtract the number of pairs used in these bonds from the total number of pairs of electrons available.

= 6 pairs available

O — C —O Add three lone pairs to each terminal oxygen atom.

Determine the number of bonding pairs remaining.

6 pairs available - 6 pairs used Subtract the lone pairs from the pairs available.= 0 pairs available

Examine the incomplete structure above (showing the placement of the lone pairs). Note that the carbon atom does not have an octet and that there are no more electron pairs available. To give the carbon atom an octet, the molecule must form double bonds.

O C O—— —— Use a lone pair from each O atom to form a double bond with the C atom.

3 Evaluate the AnswerBoth carbon and oxygen now have an octet, which satisfies the octet rule.


39. Draw the Lewis structure for ethylene, C2H4.

40. Challenge A molecule of carbon disulfide contains both lone pairs and multiple-covalent bonds. Draw its Lewis structure.

Personal Tutor For an online tutorial on greatest common factors, visit glencoe.com.

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AssessmentAssessmentSkill On a note card, write

the chemical formula for a molecule or polyatomic ion. Ask students to draw the Lewis structure on the other side of the card. OL

Concept DevelopmentModel Resonance Choose three students to represent atoms in the polyatomic ion N O 2-. Arrange the three students in a sequence with the nitrogen atom as the central atom. One bond is a double bond, and the other bond is a single bond. Ask students to determine the resonance structures. OL EL

IN-CLASS Example

Question What is the correct Lewis structure for the polyatomic ion Br O 4-?

Answer

–

Br O

O

O

O


41. +

HNH

H

H

42. –

Cl O

O

O

O

1. Write the molecular formula for theobromine. C 7 H 8 N 4 O 2

2. Identify the number of sigma and pi bondspresent in this molecule. 22 sigma bonds and 4 pi bonds

theobromine

HN N

NN

O

O

EnrichmentChemistry of Chocolate The Mayans and Aztecs consumed chocolate, as do many people throughout the world today. Cocoa and choco-late products contain theobromine, which is chemically related to caffeine, though lacking caffeine’s stimulant effect. OL

Lewis structures for polyatomic ions Although the unit acts as an ion, the atoms within a polyatomic ion are covalently bonded. The procedure for drawing Lewis structures for polyatomic ions is similar to drawing them for covalent compounds. The main difference is in find-ing the total number of electrons available for bonding. Compared to the number of valence electrons present in the atoms that make up the ion, more electrons are present if the ion is negatively charged and fewer are present if the ion is positive. To find the total number of electrons available for bonding, first find the number available in the atoms pres-ent in the ion. Then, subtract the ion charge if the ion is positive, and add the ion charge if the ion is negative.

EXAMPLE Problem 8.5

Lewis Structure for a Polyatomic Ion Draw the correct Lewis structure for the polyatomic ion phosphate (P O 4 3- ).

1 Analyze the ProblemYou are given that the phosphate ion consists of one phosphorus atom and four oxygen atoms and has a charge of 3-. Because phosphorus has less attraction for shared electrons than oxygen, phosphorus is the central atom and the four oxygen atoms are terminal atoms.


1 P atom × 5 valence electrons__P atom

+ 4 O atoms × 6 valence electrons__O atom

+ 3 electrons from the negative charge = 32 valence electrons


= 16 pair Determine the total number of bonding pairs.

O

O — P —O

—

O

— Draw single bonds from each terminal oxygen atom to the central phosphorus atom.

16 pairs total - 4 pairs used Subtract the number of pairs used from the total number of pairs of electrons available.

= 12 pairs available

Add three lone pairs to each terminal oxygen atom. 12 pairs available - 12 lone pairs used = 0

O — P —O

O3-

O

——

Subtracting the lone pairs used from the pairs available verifies that there are no electron pairs available for the phosphorus atom. The Lewis structure for the phosphate ion is shown.

3 Evaluate the AnswerAll of the atoms have an octet, and the group has a net charge of 3-.


41. Draw the Lewis structure for the N H4+ ion.

42. Challenge The Cl O4- ion contains numerous lone pairs.

Draw its Lewis structure.

Real-World Chemistry Phosphorus and Nitrogen

Algal blooms Phosphorus and nitrogen are nutrients required for algae growth. Both can enter lakes and streams from discharges of sewage and industrial waste, and in fertilizer runoff. If these substances build up in a body of water, a rapid growth of algae, known as an algal bloom, can occur, forming a thick layer of green slime over the water’s surface. When the algae use up the supply of nutrients, they die and decompose. This process reduces the amount of dissolved oxygen in the water that is available to other aquatic organisms.




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Content BackgroundExceptions to the Octet Rule Many compounds form that do not have eight electrons around each atom. Compounds that contain beryllium and boron as the central atom are often deficient in electrons. Expanded octets, those having more than eight electrons, occur only around a central nonmetallic atom from period 3 or higher because they have d orbitals available.

AssessmentAssessmentKnowledge Phosgene is

a colorless, highly toxic gas used in warfare. The formula for this molecule is CC l 2 O. Have students draw two resonance structures for this molecule. OL


43. –N

O

O

–N

O

O

44. S

O

O

S

O

O

45.

OO O

OO O

46. 2–S

O

O O

2–S

O

O O

2–S

O

O O

Differentiated InstructionBelow Level Using index cards that have been cut in half, have students write the elec-tron dot diagram for C on one card and the electron dot diagram for O on three cards. Ask students to use the cards to form the polyatomic C O 3 2- ion. Tell students to indi-cate the ion’s charge by placing large square brackets around the structure and writing a 2- superscript outside the brackets. Student structures should show a central carbon atom, three ter-

minal oxygen atoms, two single bonds, and one double bond. Ask students if the atoms could be arranged another way and have octets for all atoms. Two other resonance structures are possible. The location of the double bond determines the reso-nance structure. BL EL

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■ Figure 8.14 The nitrate ion ( NO 3 - ) exhibits resonance. a. These resonance structures differ only in the location of the dou-ble bond. The locations of the nitrogen and oxy-gen atoms stay the same. b. The actual nitrate ion is like an average of the three resonance structures in a. The dotted lines indicate possi-ble locations of the double bond.

O

O

ON

O

NO

-

O

ON

O

-

O

- ON

-O

O

a b

Exceptions to the Octet RuleGenerally, atoms attain an octet when they bond with other atoms. Some molecules and ions, however, do not obey the octet rule. There are several reasons for these exceptions.

Odd number of valence electrons First, a small group of mole-cules might have an odd number of valence electrons and be unable to form an octet around each atom. For example, N O 2 has five valence electrons from nitrogen and 12 from oxygen, totaling 17, which cannot form an exact number of electron pairs. See Figure 8.15. Cl O 2 and NO are other examples of molecules with odd numbers of valence electrons.

VOCABULARYSCIENCE USAGE V. COMMON USAGE

ResonanceScience usage: a phenomenon related to the stability of a molecule; a large vibration in a mechanical system caused by a small periodic stimulusThe new molecule had several resonance structures.

Common usage: a quality of richness or varietyThe sound of the orchestra had resonance.

NO O

Incomplete octet

■ Figure 8.15 The central nitrogen atom in this NO 2 molecule does not satisfy the octet rule; the nitrogen atom has only seven electrons in its outer energy level.

Resonance StructuresUsing the same sequence of atoms, it is possible to have more than one correct Lewis structure when a molecule or polyatomic ion has both a double bond and a single bond. Consider the polyatomic ion nitrate (N O 3 - ), shown in Figure 8.14a. Three equivalent structures can be used to represent the nitrate ion.

Resonance is a condition that occurs when more than one valid Lewis structure can be written for a molecule or ion. The two or more correct Lewis structures that represent a single molecule or ion are referred to as resonance structures. Resonance structures differ only in the position of the electron pairs, never the atom positions. The location of the lone pairs and bonding pairs differs in resonance structures. The molecule O 3 and the polyatomic ions N O 3 - , N O 2 - , S O 3 2- , and C O 3 2- commonly form resonance structures.

It is important to note that each molecule or ion that undergoes resonance behaves as if it has only one structure. Refer to Figure 8.14b. Experimentally measured bond lengths show that the bonds are identi-cal to each other. They are shorter than single bonds but longer than double bonds. The actual bond length is an average of the bonds in the resonance structures.


Draw the Lewis resonance structures for the following molecules.

43. NO2- 44. SO2 45. O3

46. Challenge Draw the Lewis resonance structure for the ion S O32- .

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Differentiated InstructionBelow Level Use the following analogy to clarify coordinate covalent bonds to students. You borrow a CD from a friend. While using the CD, you treat the CD as belonging to you, yet at the same time your friend counts that CD as part of his or her CD collection. BL

■ Caption Question Fig. 8.16 Yes, each atom involved in the bond has a com-plete octet.

Visual LearningExpanded Octets Have students locate the nonmetals on the periodic table located in period 3 or higher. These elements can have expanded octets. Explain that phosphorus, a group 15 nonmetal with 5 valence electrons, can share all of its elec-trons. This expands its octet to s p 3 d, a configuration capable of holding 10 electrons. Using marshmallows to represent atoms and toothpicks to represent single covalent bonds, have students make models of P F 5 , SC l 6 , and Cl F 5 . AL

Reading Check the molecule has an odd number of valence electrons; the molecule is stable with less than eight valence electrons; the molecule is stable with more than eight valence electrons (expanded octet)

3 AssessCheck for UnderstandingWrite a molecular formula on the board. Have students draw the Lewis structure for the molecule. OL

ReteachReview the steps used in drawing Lewis structures using the molecules N 2 O and Be F 2 . Have students deter-mine whether these molecules are exceptions to the octet rule or if they exhibit resonance. OL

ExtensionHave students research the com-pounds formed from the noble gas xenon. Have them draw the Lewis structure that represents each mole-cule formed. OL

→ H — B — N — H

H

—

H

—

H

—

H

—

H

—

H

—

H

—

H

—The boron atom has no electrons to share, whereas the nitrogen atom has two electrons to share.

The nitrogen atom sharesboth electrons to form thecoordinate covalent bond.

H — B + N — H

■ Figure 8.16 In this reaction between boron trihydride (B H 3 ) and ammonia (N H 3 ), the nitrogen atom donates both electrons that are shared by boron and ammonia, forming a coordinate covalent bond.Interpret Does the coordinate covalent bond in the product molecule satisfy the octet rule?

+ ClCl

Cl

Cl

ClP

Cl

P

Cl

Cl ClCl

Expanded octet

■ Figure 8.17 Prior to the reaction of PC l 3 and C l 2 , every reactant atom follows the octet rule. After the reaction, the product, PC l 5 , has an expanded octet containing ten electrons.

Suboctets and coordinate covalent bonds Another exception to the octet rule is due to a few compounds that form suboctets—stable configurations with fewer than eight electrons present around an atom. This group is relatively rare, and B H 3 is an example. Boron, a group 3 nonmetal, forms three covalent bonds with other nonmetallic atoms.

H

H — B — H

—

The boron atom shares only six electrons, to few to form an octet. Such compounds tend to be reactive and can share an entire pair of electrons donated by another atom.

A coordinate covalent bond forms when one atom donates both of the electrons to be shared with an atom or ion that needs two electrons to form a stable electron arrangement with lower potential energy. Refer to Figure 8.16. Atoms or ions with lone pairs often form coordinate covalent bonds with atoms or ions that need two more electrons.

Expanded octets The third group of compounds that does not fol-low the octet rule has central atoms that contain more than eight valence electrons. This electron arrangement is referred to as an expanded octet. An expanded octet can be explained by considering the d orbital that occurs in the energy levels of elements in period three or higher. An example of an expanded octet, shown in Figure 8.17, is the bond forma-tion in the molecule PC l 5 . Five bonds are formed with ten electrons shared in one s orbital, three p orbitals, and one d orbital. Another example is the molecule S F 6 , which has six bonds sharing 12 electrons in an s orbital, three p orbitals, and two d orbitals. When you draw the Lewis structure for these compounds, extra lone pairs are added to the central atom or more than four bonding atoms are present in the molecule.

Reading Check Summarize three reasons why some molecules do not conform to the octet rule.





IN-CLASS Example

Question Selenium hexafluoride is a toxic gas used as an electrical insu-lator. What is the Lewis structure for this compound?

Answer Each fluorine has an octet and the central Se atom has an expanded octet.


47.

F

Cl

F

F

48.

PCl Cl

ClCl

Cl

49.

SF F

FF

F

F

Section 8.38.3 Assessment 50. types and number of atoms; rough molecular shape 51. determine central atom and terminal atoms; determine number of bonding

electrons; determine bonding pairs; connect terminal atoms to the central atom with single bonds; determine remaining number of bonding pairs; apply octet rule and form double or triple bonds if needed

52. expanded octet, PC l 5 ; odd number of valence electrons, Cl O 2 ; less than an octet, B H 3

53. No, a molecule or polyatomic ion must have both a single bond and a double bond in order to display resonance.

54. N O orN N ON

55.

NC[ ]– AsF F

FF

F

F

FSiF

F

F

– –

C O

O

O

H

Section 8.38.3 AssessmentSection Summary◗ ◗ Different models can be used to

represent molecules.

◗ ◗ Resonance occurs when more than one valid Lewis structure exists for the same molecule.

◗ ◗ Exceptions to the octet rule occur in some molecules.

50. MAIN Idea Describe the information contained in a structural formula.

51. State the steps used to draw Lewis structures.

52. Summarize exceptions to the octet rule by correctly pairing these molecules and phrases: odd number of valence electrons, PC l 5 , Cl O 2 , B H 3 , expanded octet, less than an octet.

53. Evaluate A classmate states that a binary compound having only sigma bonds displays resonance. Could the classmate’s statement be true?

54. Draw the resonance structures for the dinitrogen oxide ( N 2 O) molecule.

55. Draw the Lewis structures for C N - , Si F 4 , HC O 3 - , and, As F 6 - .

EXAMPLE Problem 8.6

Lewis Structure: Exception to the Octet Rule Xenon is a noble gas that will form a few compounds with nonmetals that strongly attract electrons. Draw the correct Lewis structure for xenon tetrafluoride (Xe F 4 ).

1 Analyze the ProblemYou are given that a molecule of xenon tetrafluoride consists of one xenon atom and four fluorine atoms. Xenon has less attraction for electrons, so it is the central atom.

2 Solve for the UnknownFirst, find the total number of valence electrons.

1 Xe atom × 8 valence electrons__1Xe atom

+ 4 F atoms × 7 valence electrons__1F atom



= 18 pairs Determine the total number of bonding pairs.

XeF F

FF

Use four bonding pairs to bond the four F atoms to the central Xe atom.

18 pairs available - 4 pairs used = 14 pairs available Determine the number of remaining pairs.

14 pairs - 4 F atoms ×3 pairs_1F atom

= 2 pairs unused Add three pairs to each F atom to obtain an octet. Determine how many pairs remain.

XeF F

F

F

FF

Place the two remaining pairs on the central Xe atom.

3 Evaluate the AnswerThis structure gives xenon 12 total electrons—an expanded octet—for a total of six bond positions. Xenon compounds, such as the Xe F4 shown here, are toxic because they are highly reactive.


Draw the expanded octet Lewis structure for each molecule.

47. ClF3 48. PC l549. Challenge Draw the Lewis structure for the molecule formed when six fluorine atoms

and one sulfur atom bond covalently.

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MAIN IdeaMolecular Shape Bring four inflated balloons to class and tell students the balloons represent bonding pairs of electrons. Hold the ends of two balloons together in the center and ask students to predict the shape of the molecule. linear Hold three balloons together in the center and ask students to predict the shape of the molecule. trigonal planar (a triangle) Finally, hold all four balloons together in the center and ask for the shape. tetrahedral OL

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Virtual Lab

CD-ROM Molecular Geometry

Quick Demo

Model Electron Sharing Using small balloons to repre-sent shared pairs of electrons and larger balloons to represent unshared pairs of electrons, make models to represent HCl, H 2 O, N H 3 , and C H 4 . Use the fol-lowing balloon combinations for the models: HCl, one small and three large; H 2 O, two small and two large; N H 3 , three small and one large; C H 4 , four small. Tie the balloons together and ask students to notice the shape of the resultant molecules. EL BL

Section 8.48.4

Molecular ShapesMAIN Idea The VSEPR model is used to determine molecular shape.

Real-World Reading Link Have you ever rubbed two balloons in your hairto create a static electric charge on them? If you brought the balloons together, their like charges would cause them to repel each other. Molecular shapes are also affected by the forces of electric repulsion.

VSEPR ModelThe shape of a molecule determines many of its physical and chemical properties. Often, shapes of reactant molecules determine whether or not they can get close enough to react. Electron densities created by the overlap of the orbitals of shared electrons determine molecular shape. Theories have been developed to explain the overlap of bonding orbitals and can be used to predict the shape of the molecule.

The molecular geometry, or shape, of a molecule can be determined once a Lewis structure is drawn. The model used to determine the molecular shape is referred to as the Valence Shell Electron Pair Repulsion model, or VSEPR model. This model is based on an arrange-ment that minimizes the repulsion of shared and unshared electron pairs around the central atom.

Bond angle To understand the VSEPR model better, imagine bal-loons that are inflated to similar sizes and tied together, as shown in Figure 8.18. Each balloon represents an electron-dense region. The repulsive force of this electron-dense region keeps other electrons from entering this space. When a set of balloons is connected at a central point, which represents a central atom, the balloons naturally form a shape that minimizes interactions between the balloons.

The electron pairs in a molecule repel one another in a similar way. These forces cause the atoms in a molecule to be positioned at fixed angles relative to one another. The angle formed by two terminal atoms and the central atom is a bond angle. Bond angles predicted by VSEPR are supported by experimental evidence.

Unshared pairs of electrons are also important in determining the shape of the molecule. These electrons occupy a slightly larger orbital than shared electrons. Therefore, shared bonding orbitals are pushed together by unshared pairs.

■ Figure 8.18 Electron pairs in a mole-cule are located as far apart as they can be, just as these balloons are arranged. Two pairs form a linear shape. Three pairs form a trigonal planar shape. Four pairs form a tetrahedral shape.

Objectives

◗◗ Summarize the VSEPR bonding theory.

◗◗ Predict the shape of, and the bond angles in, a molecule.

◗◗ Define hybridization.

Review Vocabularyatomic orbital: the region around an atom’s nucleus that defines an electron’s probable location

New VocabularyVSEPR modelhybridization

Linear TetrahedralTrigonal planar

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Section 8.4 • Molecular Shapes 261


Math in ChemistryGeometry Using a molecular model set, construct molecules, beginning with H 2 . Ask students why a hydrogen molecule is considered linear. A straight line can be drawn through the two atoms. Next, construct Be H 2 , and ask students to determine its shape. linear Construct B H 3 , and have students notice that all four atoms lie in the same plane. Cut a triangle from transparency film, place it on top of the molecule, and have students determine the bond angle. 120° The molecular shape is referred to as trigonal planar. Construct a C H 4 molecule, and have students cut four equilateral triangles and assemble them over the molecular model, forming a geometric model. Ask students to describe the shape. tetrahedron Construct N H 3 and have students assemble a geometric shape to encompass the molecule. Tell stu-dents the shape is known as trigonal pyramidal. Students should notice that there are no interior points in this molecular shape, and each atom aligns with a vertex. Construct an H 2 O molecule, and ask students to determine the shape. bent Point out that this molecular shape is planar, and the atoms form an obtuse angle. OL EL

ExtensionComplex Geometry Repeat the activity above with a PC l 5 molecule and the S F 6 molecule. Have students determine the geometric shapes of the molecules. trigonal bipyramidal, octa-hedral AL EL

■ Caption Question Fig. 8.19 four

Reading Check one

Interactive Figure Students can interact with molecular shapes at glencoe.com.

Differentiated InstructionAdvanced Learners Have students research the molecular orbital theory. This theory describes electrons in molecules by using molecular orbitals instead of atomic orbitals. Students should use diagrams and report their findings to the class. AL

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VOCABULARYWORD ORIGIN

Trigonal planarcomes from the Latin wordstrigonum, which means triangular, and plan-, which means flat

Connection Biology The shape of food molecules is important to our sense of taste. The surface of your tongue is covered with taste buds, each of which contains from 50 to 100 taste receptor cells. Taste receptor cells can detect five distinct tastes—sweet, bitter, salty, sour, and umami (the taste of MSG, monosodium glutamate)—but each receptor cell responds best to only one taste.

The shapes of food molecules are determined by their chemical structures. When a molecule enters a taste bud, it must have the correct shape for the nerve in each receptor cell to respond and send a message to the brain. The brain then interprets the message as a certain taste. When such molecules bind to sweet receptors, they are sensed as sweet. The greater the number of food molecules that fit a sweet receptor cell, the sweeter the food tastes. Sugars and artificial sweeteners are not the only sweet molecules. Some proteins found in fruits are also sweet mol-ecules. Some common molecular shapes are illustrated in Table 8.5.

HybridizationA hybrid occurs when two things are combined and the result has char-acteristics of both. For example, a hybrid automobile uses both gas and electricity as energy sources. During chemical bonding, different atomic orbitals undergo hybridization. To understand this, consider the bond-ing involved in the methane molecule (C H 4 ). The carbon atom has four valence electrons with the electron configuration[He]2 s 2 2 p 2 . You might expect the two unpaired p electrons to bond with other atoms and the 2s electrons to remain an unshared pair. However, carbon atoms under-go hybridization, a process in which atomic orbitals mix and form new, identical hybrid orbitals.

The hybrid orbitals in a carbon atom are shown in Figure 8.19. Note that each hybrid orbital contains one electron that it can share with another atom. The hydrid orbital is called an s p 3 orbital because the four hybrid orbitals form from one s orbital and three p orbitals. Carbon is the most common element that undergoes hybridization.

The number of atomic orbitals that mix and form the hybrid orbital equals the total number of pairs of electrons, as shown in Table 8.5. In addition, the number of hybrid orbitals formed equals the number of atomic orbitals mixed. For example, AlC l 3 has a total of three pairs of electrons and VSEPR predicts a trigonal planar molecular shape. This shape results when one s and two p orbitals on the central atom, Al, mix and form three identical s p 2 hybrid orbitals.

Lone pairs also occupy hybrid orbitals. Compare the hybrid orbitals of BeC l 2 and H 2 O in Table 8.6. Both compounds contain three atoms. Why does an H 2 O molecule contain s p 3 orbitals? There are two lone pairs on the central oxygen atom in H 2 O. Therefore, there must be four hybrid orbitals—two for bonding and two for the lone pairs.

Recall from Section 8.1 that multiple covalent bonds consist of one sigma bond and one or more pi bonds. Only the two electrons in the sigma bond occupy hybrid orbitals such as sp and s p 2 . The remaining unhybridized p orbitals overlap to form pi bonds. It is important to note that single, double, and triple covalent bonds contain only one hybrid orbital. Thus, C O 2 , with two double bonds, forms sp hybrid orbitals.

Reading Check State the number of electrons that are available for bonding in a hybrid s p 3 orbital.

Energy

p2

s2

→ →

sp3

→ → → →

→

→

Carbon

H

HH

C

CH4

sp3

sp3

sp3

sp3H

■ Figure 8.19 A carbon atom’s 2s and 2p electrons occupy the hybrid s p 3 orbitals. Notice that the hybrid orbitals have an intermediate amount of potential energy when compared with the energy of the original s and p orbitals. According to VSEPR theory, a tetrahedral shape mini-mizes repulsion between the hybrid orbit-als in a C H 4 molecule.Identify How many faces does the tetrahedral shape formed by the sp 3 orbitals have?

Interactive Figure To see an animation of molecular shapes, visit glencoe.com.

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Interactive Table Students can interact with the molecular shapes at glencoe.com.

AssessmentAssessmentPerformance Using the

VSEPR model, have students build models that have these shapes: linear, trigonal planer, trigonal bipyramidal, and octahedral. OL

3 AssessCheck for UnderstandingHave students use the Lewis struc-ture for As H 3 to predict the shape of the molecule, the bond angle, and the hybrid orbitals of arsenic. OL

ReteachHave students use balls of modeling clay to represent nuclei and nonva-lence electrons, colored toothpicks to represent bonded electrons, and noncolored toothpicks to represent lone pairs. Construct C H 4 , and have students determine the hybrid orbit-als of carbon and the shape of the molecule. s p 3 , tetrahedral Repeat for N H 3 . s p 3 , trigonal pyramidal Continue with H 2 O. s p 3 , bent OL

ExtensionAsk students to explain how a car-bon atom has four bonding sites when its electron configuration of [He]2s22p2 suggests it should only have two bonding sites. The 2s22p2

orbitals undergo hybridization, forming four s p 3 bonding orbitals.

AssessmentAssessmentPerformance Have students

write the formula of a simple mole-cule on a sheet of paper. Have another student draw a model of the molecule, predict the geometry, the hybrid, and the bond angle. OL

Chemistry ProjectSulforaphane The compound sulforaphane is an ingredient in broccoli and has been discovered to be a potent anticancer agent. Have students research sulforaphane ( C 5 H 11 OS) and draw models to represent this molecule. AL

Chemistry JournalMolecular Shape Have students predict the molecular shape and bond angle of the common molecules C O 2 , N O 2 , C H 2 O, CC l 4 , H 2 S, and HF. OL

S ti 8 4 M l l Sh 263

Table 8.6 Molecular Shapes

MoleculeTotal Pairs

Shared Pairs

Lone Pairs

Hybrid Orbitals

Molecular Shape*

BeC I 2 2 2 0 sp

Linear

180°

AIC I 3 3 3 0 s p 2

Trigonal planar

120°

C H 4 4 4 0 s p 3

Tetrahedral

109.5°

P H 3 4 3 1 s p 3

Trigonal pyramidal107.3°

H 2 O 4 2 2 s p 3

Bent104.5°

NbB r 5 5 5 0 s p 3 d

Trigonal bipyramidal

120°

90°

S F 6 6 6 0 s p 3 d 2

Octahedral

90°

90°

Interactive Table Explore mol-ecular shapes at glencoe.com.

The BeC l 2 molecule contains only two pairs of electrons shared with the central Be atom. These bonding electrons have the maximum separation, a bond angle of 180°, and the molecular shape is linear.

When the central atom in a molecule has four pairs of bonding electrons, as C H 4 does, the shape is tetrahedral. The bond angles are 109.5°.

P H 3 has three single covalent bonds and one lone pair. The lone pair takes up a greater amount of space than the shared pairs. There is stronger repulsion between the lone pair and the bonding pairs than between two bonding pairs. The resulting geometry is trigonal pyramidal, with 107.3°

bond angles.

The three bonding electron pairs in AlC l 3 have maximum separation in a trigonal planar shape with 120° bond angles.

*Balls represent atoms, sticks represent bonds, and lobes represent lone pairs of electrons.

Water has two covalent bonds and two lone pairs. Repulsion between the lone pairs causes the angle to be 104.5°, less than both tetrahedral and trigonal pyramid. As a result, water molecules have a bent shape.

The NbB r 5 molecule has five pairs of bonding electrons. The trigonal bipyramidal shape minimizes the repulsion of these shared electron pairs.

As with NbB r 5 , S F 6 has no unshared electron pairs on the central atom. However, six shared pairs arranged about the central atom result in an octahedral shape.


Section 8.4 • Molecular Shapes 263


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Section 8.48.4 Assessment 61. VSEPR theory determines molecular geometry based on the repulsive

nature of electron pairs around a central atom. 62. The bond angle is the angle formed by any two terminal atoms and the

central atom. 63. A lone pair occupies more space than a shared electron pair, thus, the

presence of a lone pair pushes the bonding pairs closer together. 64. The orbital containing a lone electron pair occupies more space than a

shared electron pair. 65. s p 3 and 109°

66. P F 3 is trigonal pyramidal with s p 3 hybrid orbitals. P F 5 is trigonal bipyramidal with s p 3 d hybrid orbitals. Shape is determined by the type of hybrid orbital.

67. Refer to the Solutions Manual for Lewis structures.C S 2 : linear, 180°, sp C H 2 O: trigonal planar, 120°, s p 2 H 2 Se: bent, 104.5°, s p 3 C Cl 2 F 2 : tetrahedryl, 109°, s p 3 N Cl 3 : trigonal pyramidal, 107°, s p 3

IN-CLASS Example

Question Sulfur dichloride is a cherry-red liquid at room tempera-ture. What is the shape of sulfur dichloride? Determine the bond angle, and identify the type of hybrid.

Answer All atoms have an octet. The shape is bent, with a bond angle of 104.5°, and it is an s p 3 hybrid.


56. trigonal planar, 120°, s p 2 57. bent, 104.5°, s p 3 58. linear, 180°, sp 59. tetrahedral, 109°, s p 3 60. tetrahedral, 109°, s p 3

Section 8.48.4 AssessmentSection Summary◗ ◗ VSEPR model theory states that elec-

tron pairs repel each other and deter-mine both the shape of and bond angles in a molecule.

◗ ◗ Hybridization explains the observed shapes of molecules by the presence of equivalent hybrid orbitals.

61. MAIN Idea Summarize the VSEPR bonding theory.

62. Define the term bond angle.

63. Describe how the presence of a lone electron pair affects the spacing of shared bonding orbitals.

64. Compare the size of an orbital that has a shared electron pair with one that has a lone pair.

65. Identify the type of hybrid orbitals present and bond angles for a molecule with a tetrahedral shape.

66. Compare the molecular shapes and hybrid orbitals of P F 3 and P F 5 molecules . Explain why their shapes differ.

67. List in a table, the Lewis structure, molecular shape, bond angle, and hybrid orbitals for molecules of C S 2 , C H 2 O, H 2 0Se, CC l 2 F 2 , and NC l 3 .

EXAMPLE Problem 8.7

Find the Shape of a Molecule Phosphorus trihydride, a colorless gas, is produced when organic materials, such as fish flesh, rot. What is the shape of a phosphorus trihydride molecule? Identify the bond angle size and hybrid orbitals.

1 Analyze the ProblemYou are given the information that a phosphorus trihydride molecule has three, terminal hydrogen atoms bonded to a central phosphorus atom.

2 Solve for the UnknownFind the total number of valence electrons and the number of electron pairs.

1 P atom × 5 valence electrons__1P atom

+ 3 H atoms × 1 valence electron__1F atom



= 4 pairs Determine the total number of bonding pairs.

→H H

HLewis structure

H — P — H

H

Molecular shape

PDraw the Lewis structure, using one pair of electrons to bond each H atom to the central P atom and assigning the lone pair to the P atom.

The molecular shape is trigonal pyramidal with a 107° bond angle and s p3 hybrid orbitals.

3 Evaluate the AnswerAll electron pairs are used and each atom has a stable electron configuration.


Determine the molecular shape, bond angle, and hybrid orbitals for each molecule.

56. BF3 58. BeF2

57. OCl2 59. CF4

60. Challenge For a N H4+ ion, identify its molecular shape, bond angle, and hybrid orbitals.

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1 FocusFocus Transparency

Before presenting the lesson, project Section Focus Transparency 31 and have students answer the accompa-nying questions. BL EL

MAIN IdeaElectronegativity and Polarity Have two students help with a quick demo. Ask both students to pull on a rope with equal strength. Tell the class the rope represents a shared pair of electrons. Ask the class what this represents when atoms share electrons. equal sharing of electrons Ask one student to pull harder than the other student. The second student should be pulled toward the first stu-dent. Ask the class what this repre-sents when atoms share electrons. unequal sharing of electrons Have stu-dents identify which atoms have a greater tendency to gain electrons. the ones with the greatest pull on the elec-trons Ask students what type of bond is represented if the electron is com-pletely pulled away from one atom. an ionic bond OL

2 Teach

Section 8.58.5

Video Lab

DVD Formation of Compounds

Quick Demo

Polar Water Rub a rubber rod with wool or silk to induce a charge and bring it near a stream of water from an opened buret. Have students observe the water move toward the charged rod. Repeat the procedure using paint thinner. nothing happens Have stu-dents try to explain the results. When polarity is discussed in this section, refer to this Quick Demo and point out that water is a polar molecule, whereas paint thinner is nonpolar. Partial charges on water molecules are attracted to the rubber rod, which has a negative charge. OL

Section 8.58.5

Objectives

◗◗ Describe how electronegativity is used to determine bond type.

◗◗ Compare and contrast polar and nonpolar covalent bonds and polar and nonpolar molecules.

◗◗ Generalize about the characteristics of covalently bonded compounds.

Review Vocabularyelectronegativity: the relative ability of an atom to attract electrons in a chemical bond

New Vocabularypolar covalent bond

Electronegativity and PolarityMAIN Idea A chemical bond’s character is related to each atom’s attraction for the electrons in the bond.

Real-World Reading Link The stronger you are, the more easily you can do pull-ups. Just as people have different abilities for doing pull-ups, atoms in chemical bonds have different abilities to attract (pull) electrons.

Electron Affinity, Electronegativity, and Bond CharacterThe type of bond formed during a reaction is related to each atom’s attraction for electrons. Electron affinity is a measure of the tendency of an atom to accept an electron. Excluding noble gases, electron affinity increases with increasing atomic number within a period and decreases with increasing atomic number within a group. The scale of electroneg-ativities—shown in Figure 8.20—allows chemists to evaluate the elec-tron affinity of specific atoms in a compound. Recall from Chapter 6 that electronegativity indicates the relative ability of an atom to attract electrons in a chemical bond. Note that electronegativity values were assigned, whereas electron affinity values were measured.

Electronegativity The version of the periodic table of the elements shown in Figure 8.20 lists electronegativity values. Note that fluorine has the greatest electronegativity value (3.98), while francium has the least (0.7). Because noble gases do not generally form compounds, indi-vidual electronegativity values for helium, neon, and argon are not list-ed. However, larger noble gases, such as xenon, sometimes bond with highly electronegative atoms, such as fluorine.

Electronegativity Values for Selected Elements

MetalMetalloidNonmetal

78

Pt2.2

79

Au2.4

80

Hg1.9

81

Tl1.8

82

Pb1.8

83

Bi1.9

28

Ni1.91

29

Cu1.90

30

Zn1.65

31

Ga1.81

46

Pd2.20

47

Ag1.93

48

Cd1.69

49

In1.78

50

Sn1.96

13

Al1.61

1

H2.20

3

Li0.98

11

Na0.93

19

K0.82

37

Rb0.82

55

Cs0.7987

Fr0.7

88

Ra0.9

89

Ac1.1

56

Ba0.89

57

La1.10

72

Hf1.3

73

Ta1.5

74

W1.7

75

Re1.9

76

Os2.2

77

Ir2.2

38

Sr0.95

39

Y1.22

40

Zr1.33

41

Nb1.6

42

Mo2.16

20

Ca1.00

21

Sc1.36

22

Ti1.54

23

V1.63

24

Cr1.66

43

Tc2.10

44

Ru2.2

25

Mn1.55

26

Fe1.83

27

Co1.88

45

Rh2.28

12

Mg1.31

4

Be1.57

85

At2.2

32

Ge2.01

33

As2.18

34

Se2.55

35

Br2.96

51

Sb2.05

52

Te2.1

53

I2.66

14

Si1.90

15

P2.19

16

S2.58

17

Cl3.16

5

B2.04

6

C2.55

7

N3.04

8

O3.44

9

F3.98

84

Po2.0

Figure 8.20 Electronegativity values are derived by comparing an atom’s attrac-tion for shared electrons to that of a fluo-rine’s atom attraction for shared electrons. Note that the electronegativity values for the lanthanide and actinide series, which are not shown, range from 1.12 to 1.7.

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Section 8.5 • Electronegativity and Polarity 265


■ In-Text Questions What percent ionic character is a bond between two atoms that have an elec-tronegativity diff erence of 2.00? About 60% Where would LiBr be plotted on the graph? just to the left of NaBr on the curve, at an electronegativity diff erence of 1.98

Reading Check 0%

Graph Check approximately 74%

Differentiated InstructionBelow Level Have a simulated tug-of-war with seven people on one side and one on the other to represent transfer of electrons. Have two equally sized people represent a nonpolar molecule. Finally, have one large person and one small person represent a polar bond. BL EL

IdentifyMisconceptions

A common misconception is that the bond between two atoms is purely covalent or purely ionic.Uncover the Misconception Have students write analogies to represent nonpolar covalent bonds, polar covalent bonds, and ionic bonds.Demonstrate the Concept Have students imagine they are eating at a restaurant with another person and each orders a different type of sandwich to be shared. If they share equally, what would they do? They would cut the sandwich in half, and each would get a half. Ask students what unequal sharing would be. The shared portions are not equal. If one person eats his or her entire sandwich and then takes some of the other person’s sandwich, what bond type does this represent? ionic Assess New Knowledge Using the graph of electro-negativities and percent ionic character in Figure 8.21, have stu-dents determine the bond type of the several pairs of atoms. Ask them to also identify which atom is more likely to have the electron. H and Br polar covalent, Br; C and O polar covalent, O; Li and F ionic, F; O and O nonpolar covalent, the electrons are shared equally. OL

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Table 8.7 EN Differenceand Bond Character

Electronegativity Difference Bond Character

> 1.7 mostly ionic

0.4 - 1.7 polar covalent

< 0.4 mostly covalent

0 nonpolar covalent

Bond character A chemical bond between atoms of different ele-ments is never completely ionic or covalent. The character of a bond depends on how strongly each of the bonded atoms attracts electrons. As shown in Table 8.7, the character and type of a chemical bond can be predicted using the electronegativity difference of the elements that bond. Electrons in bonds between identical atoms have an electronega-tivity difference of zero—meaning that the electrons are equally shared between the two atoms. This type of bond is considered nonpolar cova-lent, or a pure covalent bond. On the other hand, because different ele-ments have different electronegativities, the electron pairs in a covalent bond between different atoms are not shared equally. Unequal sharing results in a polar covalent bond. When there is a large difference in the electronegativity between bonded atoms, an electron is transferred from one atom to the other, which results in bonding that is primarily ionic.

Bonding is not often clearly ionic or covalent. An electronegativity difference of 1.70 is considered 50 percent covalent and 50 percent ionic. As the difference in electronegativity increases, the bond becomes more ionic in character. Generally, ionic bonds form when the electro-negativity difference is greater than 1.70. However, this cutoff is some-times inconsistent with experimental observations of two nonmetals bonding together. Figure 8.21 summarizes the range of chemical bond-ing between two atoms. What percent ionic character is a bond between two atoms that have an electronegativity difference of 2.00? Where would LiBr be plotted on the graph?

Reading Check Analyze What is the percent ionic character of a pure covalent bond?

0 1.0 3.02.0

Perc

ent

ioni

c ch

arac

ter

75

50

25

0

Electronegativity difference

Electronegativity and Bond Character

HClAlP

Ionic

Covalent

HF

NaBrMgO

CaO

N2

■ Figure 8.21 This graph shows that the difference in electronegativity between bonding atoms determines the percent ionic character of the bond. Above 50% ionic character, bonds are mostly ionic.

Graph CheckDetermine the percent ionic charcter of calcium oxide.

C08_S5_874637.indd 2 2/2/07 10:16:27 AM



Chemistry JournalBond Type Using the electronegativity difference, have students determine if the bonds in the following molecules are polar covalent, nonpolar covalent, or ionic: H 2 S, C H 4 , N H 3 , KCl, N 2 , CaO, C O 2 , HCl. KCl and CaO are ionic, N 2 is nonpolar covalent, and the rest are polar covalent. OL

Quick Demo

Polar Molecules Place two or three stirring bars with non-stick coating on a flat sur-face. The stirring bars should be spaced away from each other and have random orientation. Bring a magnet near the stirring bars. The stirring bars will align themselves with the magnetic field. Tell students that a polar molecule behaves much the same way in an electric field because it has a positive pole and a negative pole.

Chemistry ProjectChromatography Have student groups research the separation technique known as paper chromatography. Ask them to describe how polarity differences are used to separate a mixture of substances. Students should outline a lab activity to show how chromatography works and per-form the lab for the class. OL

AssessmentAssessmentPerformance Show

students molecular models of common molecules. Have them determine the bond type present and then determine the type of molecule based on its symmetry. The mole-cules S O 2 , C S 2 , C H 3 Cl, C H 2 O, HCl, and H 2 S can be used. OL

Interactive Figure Students can interact with the bond-type art at glencoe.com.

Reading Check Water has polar covalent molecules that align with the electric field of the charged balloon.

δ⁻δ⁺

H — Cl

Electronegativity Cl = 3.16Electronegativity H = 2.20Difference = 0.96

■ Figure 8.22 Chlorine’s electronegativity is higher than that of hydrogen. Therefore, in a molecule containing hydrogen and chlorine, the shared pair of electrons is with the chlorine atom more often than it is with the hydrogen atom. Symbols are used to indicate the partial charge at each end of the molecule from this unequal sharing of electrons.

Polar Covalent BondsAs you just learned, polar covalent bonds form because not all atoms that share electrons attract them equally. A polar covalent bond is simi-lar to a tug-of-war in which the two teams are not of equal strength. Although both sides share the rope, the stronger team pulls more of the rope toward its side. When a polar bond forms, the shared electron pair or pairs are pulled toward one of the atoms. Thus, the electrons spend more time around that atom than the other atom. This results in partial charges at the ends of the bond.

The Greek letter delta (δ) is used to represent a partial charge. In a polar covalent bond, δ

- represent a partial negative charge and δ + rep-

resents a partial positive charge. As shown in Figure 8.22, δ -

and δ + can be added to a molecular model to indicate the polarity of the cova-lent bond. The more-electronegative atom is at the partially negative end, while the less-electronegative atom is at the partially positive end. The resulting polar bond often is referred to as a dipole (two poles).

Molecular polarity Covalently bonded molecules are either polar or nonpolar; which type depends on the location and nature of the covalent bonds in the molecule. A distinguishing feature of nonpolar molecules is that they are not attracted by an electric field. Polar mole-cules, however, are attracted by an electric field. Because polar mole-cules are dipoles with partially charged ends, they have an uneven electron density. This results in the tendency of polar molecules to align with an electric field.

Polarity and molecular shape You can learn why some mole-cules are polar and some are not by comparing water ( H 2 O) and carbon tetrachloride (CC l 4 ) molecules. Both molecules have polar covalent bonds. According to the data in Figure 8.20, the electronegativity difference between a hydrogen atom and a oxygen atom is 1.24. The electronegativity difference between a chlorine atom and a carbon atom is 0.61. Although these electronegativity differences vary, a H—O bond and a C—Cl bond are considered to be polar covalent.

δ + δ - δ + δ -

H—O C—Cl

According to their molecular formulas, both molecules have more than one polar covalent bond. However, only the water molecule is polar.

Reading Check Apply Why does a statically charged balloon cause a slow stream of water from a faucet to bend when placed next to it?

Careers In chemistry

Flavor Chemist A flavor chemist, or flavorist, must know how chemi-cals react and change in different conditions. A degree in chemistry is an asset, but is not required. Most flavorists work for companies that supply flavors to the food and bever-age industries. A certified flavorist trains for five years in a flavor labo-ratory, passes an oral examination, and then works under supervision for another two years. For more information on chemistry careers, visit glencoe.com.

Interactive Figure To see an animation of bond types, visit glencoe.com.




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■ Caption Question Fig. 8.24 No, detergents must be used because the polar water molecules cannot dissolve the nonpolar oil molecules.

Quick Demo

Polarity Made Apparent In four labeled beakers, put 100 mL of methanol, ethanol, 2-propanol (isopropyl alcohol) and acetone. To each beaker, add a few crystals of Reichardt’s dye and stir. Reichardt’s dye is a compound that indicates the polarity of solvents by color. It should be green-blue in methanol, blue-violet in ethanol, red-violet in 2-propanol, and orange in acetone. As the polarity of the solvent increases, the light absorbed by the dye changes. Yellow is most polar and green-blue is the least polar.

Chemistry JournalLike Dissolves Like Add 20 mL of carbon tetrachloride (C Cl 4 )and 20 mL of water to a beaker. The liquids form layers and do not mix. C Cl 4 is more dense than water and forms the lower layer. In a second beaker, add 20 ml of C Cl 4 only. In a third beaker, add 20 mL of water only. Add a few crystals of iodine and and a few crystals of potassium permanganate to the two beakers. The iodine dissolves in the nonpolar solvent C Cl 4 . The potassium permanganate dissolved in the

polar water. Water is polar and attracts the ions present in the ionic compound potassi-um permanganate. Have students draw a water molecule and identify the portion of the molecule that attracts the potassium ion and the portion that attracts the permanga-nate ion. The central oxygen atom has a partial nega-tive charge and attracts the potassium ion. The two hydrogen atoms create a partial positive region that attracts the permanganate ion. OL

Data Analysis lab

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The shape of a H 2 O molecule, as determined by VSEPR, is bent because the central oxygen atom has lone pairs of electrons, as shown in Figure 8.23a. Because the polar H—O bonds are asymmetric in a water molecule, the molecule has a definite positive end and a definite nega-tive end. Thus, it is polar.

A CC l 4 molecule is tetrahedral, and therefore, symmetrical, as shown in Figure 8.23b. The electric charge measured at any distance from its center is identical to the charge measured at the same distance to the opposite side. The average center of the negative charge is located on the chlorine atom. The positive center is also located on the carbon atom. Because the partial charges are balanced, CC l 4 is a nonpolar mol-ecule. Note that symmetric molecules are usually nonpolar, and mole-cules that are asymmetric are polar as long as the bond type is polar.

Is the molecule of ammonia (N H 3 ), shown in Figure 8.23c, polar? It has a central nitrogen atom and three terminal hydrogen atoms. Its shape is a trigonal pyramidal because of the lone pair of electrons present on the nitrogen atom. Using Figure 8.20, you can find that the electronegativity difference of hydrogen and nitrogen is 0.84 making each N—H bond polar covalent. The charge distribution is unequal because the molecule is asymmetric. Thus, the molecule is polar.

Solubility of polar molecules The physical property known as solubility is the ability of a substance to dissolve in another substance. The bond type and the shape of the molecules present determine solu-bility. Polar molecules and ionic compounds are usually soluble in polar substances, but nonpolar molecules dissolve only in nonpolar substances, as shown in Figure 8.24. Solubility is discussed in detail in Chapter 14.

■ Figure 8.24 Symmetric covalent molecules, such as oil and most petroleum products, are nonpolar. Asymmetric molecules, such as water, are usually polar. As shown in this photo, polar and nonpolar sub-stances usually do not mix.Infer Will water alone clean oil from a fabric?

■ Figure 8.23 A molecule’s shape deter-mines its polarity.

H H

N

HNH3

δ⁻

δ⁺

δ⁺

δ⁺Hδ⁺

Oδ⁻

H δ⁺

H2O

Cl

Cl

δ⁻

δ⁻

δ⁻

δ⁻

Cl

ClC δ⁺

CCl4

a b c

The asymmetric shape of an ammonia molecule results in an unequal charge distribution and the molecule is polar.

The symmetry of a CC l 4molecule results in an equal distribution of charge, and the molecule is nonpolar.

The bent shape of a water molecule makes it polar.

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Cultural DiversityMedical Research The use of herbal remedies has been well documented in China over the past 2500 years. Western scientists and pharmaceutical companies are looking at covalent compounds contained in these remedies as sources of possible drugs for the treatment of cancer and AIDS. The National Cancer Institute recently compiled data that catalogs the medicinal effects of Chinese herbs. In 1992, the National Institute of Health created an Office of Alternative Medicine to evaluate a variety of medical therapies, including five related to Chinese herbal medicine. Several U.S. pharmaceutical companies have formed over the past few years to develop drugs from Chinese plants.

The ChemLab located at the end of the chapter can be used at this point in the lesson.

Data-Analysis LabData-Analysis Lab

About the Lab• Phenol and benzoic acid have sim-

ilar structures as shown below.

HO

Phenol

HO

OBenzoic acid

• For additional details about origi-nal experiment see: Orna, Mary V. Chemistry, Color, and Art. Journal of Chemical Education. 78, no. 10 (2001) 1305

• Note that both the peak intensities and the time elapsed before each peaks occurs (for phenol and ben-zoic acid) become more equal as the concentration of water in the mobile base decreases.

Think Critically 1. Phenol has lower retention

time relative to benzoic acid because it has less attraction to the water.

2. Phenol is in excess because it has the largest peak. The area under the peak corresponds to the amount of the component present in the mixture.

3. Benzoic acid is the more polar compound because it shows a greater attraction to water; phenol is the less polar compound.

4. Answers and reasoning will vary. A 50% methanol/50% water mobile- phase solvent mixture extracts more of the mixture components (phenol and benzoic acid) than the other mixtures, has good separation between the two peak intensities, and requires less time.

Data Analysis labData Analysis labBased on Real Data*

Interpret DataHow does the polarity of the mobile phase affect chromatograms? Chromatography is a technique in which a moving phase transports and separates the components of a mixture. A chromatograph is created by recording the intensity of each component carried in the mov-ing phase versus time. The peak intensities on the chromatograph indicate the amount of each component present in the mixture. High-performance liquid chromatography, or HPLC, is used by analytical chemists to separate mixtures of solutes. During HPLC, components that are strongly attracted to the extracting sol-vent are retained longer by the moving phase and tend to appear early on a chromatograph. Several scientists performed HPLC using a meth-anol-water mixture as the extracting solvent to separate a phenol-benzoic acid mixture. Their results are shown in the graph to the right.

Think Critically1. Explain the different retention times shown

on the chromatograms. 2. Infer from the graph the component, phenol

or benzoic acid, that is in excess. Explain your answer.

Data and Observations

240

Inte

nsit

y

60

40

20

0

–20

Time (s)480 720 960 1200 1440

Chromatograms of Phenol and Benzoic Acid inDifferent Compositions of Mobile Phase Solvent

75% methanol/25% water mobile phase



Benzoicacid

Benzoicacid

Benzoicacid

Phenol

PhenolPhenol

*Data obtained from: Joseph, Seema M. and Palasota, John A. 2001. The combined effects of pH and percent methanol on the HPLC separation of benzoic acid and phenol. Journal of Chemical Education 78:1381.

3. Infer which component of the mixture has more polar molecules.

4. Determine the most effective composition of the mobile phase (of those tested) for sepa-rating phenol from benzoic acid. Explain.

Properties of Covalent CompoundsTable salt, an ionic solid, and table sugar, a covalent solid, are similar in appearance. However, these compounds behave differently when heated. Salt does not melt, but sugar melts at a relatively low temperature. Does the type of bonding in a compound affect its properties?

Intermolecular forces Differences in properties are a result of differences in attractive forces. In a covalent compound, the covalent bonds between atoms in molecules are strong, but the attraction forces between molecules are relatively weak. These weak attraction forces are known as intermolecular forces, or van der Waals forces, which are dis-cussed in Chapter 12. Intermolecular forces vary in strength but are weaker than the bonds that join atoms in a molecule or ions in an ionic compound.

There are different types of intermolecular forces. Between nonpolar molecules, the force is weak and is called a dispersion force, or induced dipole. The force between oppositely charged ends of two polar mole-cules is called a dipole-dipole force. The more polar the molecule, the stronger the dipole-dipole force. The third force, a hydrogen bond, is especially strong. It forms between the hydrogen end of one dipole and a fluorine, oxygen, or nitrogen atom on another dipole.





Section 8.58.5 Assessment 68. The greater the EN difference, the greater the ionic nature of the bond. 69. A polar covalent bond has unequal sharing of electrons. The electrons are

pulled toward one of the atoms, generating partial charges on the ends. 70. It has a greater electron density on one side of the molecule. 71. The solid state of a molecule is crystalline. A molecular solid is a

nonconductor, is soft, and has a low melting point. 72. If the difference is zero, the bond is considered nonpolar covalent; if

between 0.4 and 1.7, polar covalent; if greater than 1.7, ionic. 73. brittle, nonconductors of heat and electricity, extremely hard

74. a. mostly covalent b. mostly covalent c. polar covalent 75. S Cl 2 , polar; C S 2 , nonpiolar; C F 4 , nonpolar, 76. polar; because it is asymmetric 77.

SF F

FF

F

F

polar nonpolar

FSF

F

F

3 AssessCheck for UnderstandingAsk students to compare the charac-teristics of ionic solids to those of covalent and network solids. OL

ReteachHave students make a chart and dis-cuss the properties of ionic, metallic, covalent, and network solids. Point out that a high melting point is a result of a three-dimensional attrac-tion of an atom or ion for electrons. This occurs in network solids, metal-lic solids, and ionic solids. Covalent solids are held together by intermo-lecular attractions and have lower melting points. OL

ExtensionHave students investigate intermo-lecular bonds. Students should use drawings to represent each type of van der Waals force and give an example. Display students’ work. OL

AssessmentAssessmentKnowledge Write several

chemical formulas on the board. Have students predict the general physical properties of each com-pound based on its bonding. OL


Forces and properties The properties of covalent molecular com-pounds are related to the relatively weak intermolecular forces holding the molecules together. These weak forces result in the relatively low melting and boiling points of molecular substances compared with those of ionic substances. That is why, when heated moderately, sugar melts but salt does not. Weak intermolecular forces also explain why many molecular substances exist as gases or vaporize readily at room temperature. Oxygen ( O 2 ), carbon dioxide (C O 2 ), and hydrogen sulfide ( H 2 S) are examples of covalent gases. Because the hardness of a sub-stance depends on the intermolecular forces between individual mole-cules, many covalent molecules are relatively soft solids. Paraffin, found in candles and other products, is a common example of a covalent solid.

In the solid phase, molecules align to form a crystal lattice. This molecular lattice is similar to that of an ionic solid, but with less attrac-tion between particles. The structure of the lattice is affected by molecu-lar shape and the type of intermolecular force. Most molecular information has been determined by studying molecular solids.

Covalent Network SolidsThere are some solids, often called covalent network solids, that are composed only of atoms interconnected by a network of covalent bonds. Quartz and diamond are two common examples of network solids. In contrast to molecular solids, network solids are typically brittle, non-conductors of heat or electricity, and extremely hard. Analyzing the structure of a diamond explains some of its properties. In a diamond, each carbon atom is bonded to four other carbon atoms. This tetrahe-dral arrangement, which is shown in Figure 8.25, forms a strongly bonded crystal system that is extremely hard and has a very high melt-ing point.

68. MAIN Idea Summarize how electronegativity difference is related to bond character.

69. Describe a polar covalent bond.

70. Describe a polar molecule.

71. List three properties of a covalent compound in the solid phase.

72. Categorize bond types using electronegativity difference.

73. Generalize Describe the general characteristics of covalent network solids.

74. Predict the type of bond that will form between the following pair of atoms: a. H and S b. C and H c. Na and S.

75. Identify each molecule as polar or nonpolar: SC l2 , C S2 , and C F4 .

76. Determine whether a compound made of hydrogen and sulfur atoms is polar or nonpolar.

77. Draw the Lewis structures for the molecules S F 4 and S F 6 . Analyze each structure to determine whether the molecule is polar or nonpolar.

Section Summary◗ ◗ The electronegativity difference deter-

mines the character of a bond between atoms.

◗ ◗ Polar bonds occur when electrons are not shared equally forming a dipole.

◗ ◗ The spatial arrangement of polar bonds in a molecule determines the overall polarity of a molecule.

◗ ◗ Molecules attract each other by weak intermolecular forces. In a covalent network solid, each atom is covalently bonded to many other atoms.

■ Figure 8.25 Network solids are often used in cutting tools because of their extreme hardness. Here, a diamond-tipped saw blade cuts through stone.

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PurposeStudents will learn how van der Waals forces provide the sticking power for geckos’ toes.

BackgroundGecko toes look like suction cups, but they do not function that way. Suction was ruled out as an explana-tion for how they stick to surfaces by testing gecko feet in a vacuum. Other ideas, such as static cling or water-based attraction, have also been ruled out by experiment.

Teaching StrategiesThe diagram has five numbered text blocks that explain the source of the gecko’s phenomenal sticking power.• Number 1 shows a gecko’s toe

magnified, showing rows contain-ing hundreds of thousands of tiny hairs, or setae.

• Numbers 2 and 3 explain how the setae branch out into even smaller structures called spatulae. The angle at which setae meet the sur-face is important factor in deter-mining the holding power.

• Number 4 explains how large sur-face area and van der Waals forces act to create a significant gripping strength. Van der Waals forces are intermolecular forces; that is, they act between molecules.

• Number 5 explains how a gecko lets go by curling its toes. Curling the toes reduces the number of setae making contact with the sur-face at the optimal angle. This results in decreased van der Waals forces and reduced gripping strength.Chemistry


Invent Some other applications that have been dis-cussed include less painful adhesive bandages, super-grip athletic equipment, and sticky tools for nanosurgery.

ChemistryInvent Using their knowledge of how geckos stick to surfaces, scientists are developing applications for geckolike materials. Some possible applications include mini-robots that climb walls and tape that sticks even under water. What uses for a new sticky geckolike material can you think of? For more on gecko-tech, visit glencoe.com.

Sticky Feet: How Geckos GripFor a gecko, hanging from a wall or a ceiling is no great feat. The key to a gecko’s amazing grip is found on each of its toes. Researchers have determined that a gecko’s grip depends on the sticking power of atoms themselves.

Temporary dipole Temporary dipole

Surface Spatula

Attractionδ+δ- δ+δ-

Gecko toe The bottom of a gecko’s toe is covered with millions of tiny hairs, called setae, arranged in rows.

1

Spatulae Setae are complex structures. The end of each seta has microscopic branches called spatulae

2

Surface area Each seta has a relatively enormous surface area because of its vast number of spatulae.

3

Sticking Van der Waals forces form between a surface and a gecko’s spatulae. When multiplied by the spatulae‘s vast surface areas, the sum of the weak van der Waals forces is more than enough to balance the pull of gravity and hold a gecko in place.

4

Letting go A gecko simply curls its toes when it wants to move. This reduces the amount of surface contact and the van der Waals forces, and a gecko loses its grip.

5

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How It Works 271

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between the two resonance structures by swapping the position of a lone pair and a covalent bond.

OOO

OOO

Customize this lab with the LabManager™CD-ROM.

See the ChemLab worksheet in your FAST FILE.


PreparationTime Allotment one class period

Process Skills applying concepts, comparing and contrasting, formu-lating models

Safety Precautions Approve lab safety forms before work begins.

Procedure• Prominently display the color code

key used to identify atoms of each element.

• Review the molecular geometry examples in Table 8.6 with students prior to beginning the lab, or ask them to review the table on their own before beginning.

Analyze and Conclude 1. increasing flexibility: triple, double, sin-

gle; increasing strength: single, double, triple

2. The H 2 O molecule has two bonds and two lone pairs around the central atom. The lone pairs take up space around the central atom and repel the bonding elec-trons, causing the bent shape. The C O 2 molecule has two double bonds with no lone pairs. The bonding electrons repel to form the linear shape, which maximizes the distance between electron densities.

3. The S O 3 molecule undergoes resonance. It has a central S and three terminal O atoms. One of the terminal O atoms forms a double bond. Three resonance structures exist, one for each possible location of the double bond.

4. The following molecules are polar: H 2 O, P H 3 , HCN, and CO. All others are nonpolar.

Inquiry ExtensionStudents should assemble two models, each with a central oxygen atom jointed to two ter-minal oxygen atoms. One terminal atom is joined by a single bond, the other by a double bond. The location of these two bonds deter-mines the two resonance structures. The Lewis structures should show that you can convert

MODEL MOLECULAR SHAPES

INQUIRY EXTENSIONModel Use a molecular model kit to build the two resonance structures of ozone ( O 3 ). Then, use Lewis structures to explain how you can convert between the two resonance structures by interchanging a lone pair for a covalent bond.

Background: Covalent bonding occurs when atoms share valence electrons. In the Valence Shell Electron Pair Repulsion (VSEPR) theory, the way in which valence electrons of bonding atoms are positioned is the basis for predicting a molecule’s shape. This method of visualizing shape is also based on the molecule’s Lewis structure.

Question: How do the Lewis structure and the positions of valence electrons affect the shape of the covalent compound?

Materialsmolecular model kit

Safety Precautions

Procedure 1. Read and complete the lab safety form. 2. Create a table to record your data. 3. Note and record the color used to represent each of

the following atoms in the molecular model kit: hydrogen (H), oxygen (O), phosphorus (P), carbon (C), fluorine (F), sulfur (S), and nitrogen (N).

4. Draw the Lewis structures of the H 2 , O 2 , and N 2molecules.

5. Obtain two hydrogen atoms and one connector from the molecular model kit, and assemble a hydrogen ( H 2 ) molecule. Observe that your model represents a single-bonded diatomic hydrogen molecule.

6. Obtain two oxygen atoms and two connectors from the molecular model kit, and assemble an oxygen ( O 2 ) molecule. Observe that your model represents a double-bonded diatomic oxygen molecule.

7. Obtain two nitrogen atoms and three connectors from the molecular model kit, and assemble one nitrogen ( N 2 ) molecule. Observe that your model represents a triple-bonded diatomic nitrogen molecule.

8. Recognize that diatomic molecules such as those formed in this lab are always linear. Diatomic mole-cules are made up of only two atoms and two points (atoms) can only be connected by a straight line.

9. Draw the Lewis structure of water ( H 2 O), and construct its molecule.

10. Classify the shape of the H 2 O molecule using infor-mation in Table 8.6.

11. Repeat Steps 9 and 10 for the P H 3 , C F 4 , C O 2 , S O 3 , HCN, and CO molecules.

Analyze and Conclude 1. Think Critically Based on the molecular models

you built and observed in this lab, rank single, dou-ble, and triple bonds in order of increasing flexibility and increasing strength.

2. Observe and Infer Explain why H 2 O and C O 2molecules have different shapes.

3. Analyze and Conclude One of the molecules from this lab undergoes resonance. Identify the molecule that has three resonance structures, draw the struc-tures, and explain why resonance occurs.

4. Recognize Cause and Effect Use the electronega-tivity difference to determine the polarity of the molecules in Steps 9–11. Based on their calculated bond polarities and the models constructed in this lab, determine the molecular polarity of each structure.

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272 ChemLab

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Chapter 8 • Study Guide 273Vocabulary Puzzlemaker glencoe.com

Study Guide

Using the VocabularyTo reinforce chapter vocabulary, have students write a sentence using each term. OL EL

Review Strategies• Have students name and write

formulas and Lewis structures for molecular compounds and acids. OL

• Have students identify the hybrid orbitals present in a molecule.

• Discuss resonance hybrids and exceptions to the octet rule. OL

• Have students determine molecu-lar shapes and bond angles. OL

• Have students use electronegativity difference to determine bond type and polarity. OL

Students can visit glencoe.com to:• study the entire chapter online• access Web links for more

information, projects, and activities• review content online with

the Interactive Tutor and take Self-Check Quizzes

• take Chapter Tests and Standardized Test Practice

• use Study to Go to download content onto a PDA

Vocabulary Puzzlemaker

For additional practice with vocabulary, have students access the Vocabulary Puzzlemaker online at glencoe.com.

Use the ExamView® Assessment SuiteCD-ROM to:• create multiple versions of tests• create modified tests with one

mouse click• edit existing questions and add

your own questions• build tests aligned with state

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• change English tests to Spanish with one mouse click

• track students’ progress using the Teacher Management System

What’s CHEMISTRYGot To Do With It?

DVD Th e Plastics

Download quizzes, key terms, and flash cards from glencoe.com.

BIG Idea Covalent bonds form when atoms share electrons.

Section 8.1Section 8.1 The Covalent Bond

MAIN Idea Atoms gain stability when they share electrons and form covalent bonds.

Vocabulary• covalent bond (p. 241)• endothermic reaction (p. 247)• exothermic reaction (p. 247)• Lewis structure (p. 242)• molecule (p. 241)• pi bond (p. 245)• sigma bond (p. 244)

Key Concepts• Covalent bonds form when atoms share one or more pairs of electrons.• Sharing one pair, two pairs, and three pairs of electrons forms single, double, and

triple covalent bonds, respectively.• Orbitals overlap directly in sigma bonds. Parallel orbitals overlap in pi bonds. A

single covalent bond is a sigma bond but multiple covalent bonds are made of both sigma and pi bonds.

• Bond length is measured nucleus-to-nucleus. Bond dissociation energy is needed to break a covalent bond.

Section 8.2Section 8.2 Naming Molecules

MAIN Idea Specific rules are used when naming binary molecular compounds, binary acids, and oxyacids.

Vocabulary• oxyacid (p. 250)

Key Concepts• Names of covalent molecular compounds include prefixes for the number of each

atom present. The final letter of the prefix is dropped if the element name begins with a vowel.

• Molecules that produce H + in solution are acids. Binary acids contain hydrogen and one other element. Oxyacids contain hydrogen and an oxyanion.

Section 8.3Section 8.3 Molecular Structures

MAIN Idea Structural formulas show the relative positions of atoms within a molecule.

Vocabulary• coordinate covalent bond (p. 259)• resonance (p. 258)• structural formula (p. 253)

Key Concepts• Different models can be used to represent molecules.• Resonance occurs when more than one valid Lewis structure exists for the same

molecule.• Exceptions to the octet rule occur in some molecules.

Section 8.4Section 8.4 Molecular Shapes

MAIN Idea The VSEPR model is used to determine molecular shape.

Vocabulary• hybridization (p. 262)• VSEPR model (p. 261)

Key Concepts• VSEPR model theory states that electron pairs repel each other and determine both

the shape of and bond angles in a molecule.• Hybridization explains the observed shapes of molecules by the presence of

equivalent hybrid orbitals.

Section 8.5Section 8.5 Electronegativity and Polarity

MAIN Idea A chemical bond’s character is related to each atom’s attraction for the electrons in the bond.

Vocabulary• polar covalent bond (p. 266)

Key Concepts• The electronegativity difference determines the character of a bond between atoms.• Polar bonds occur when electrons are not shared equally forming a dipole.• The spatial arrangement of polar bonds in a molecule determines the overall

polarity of a molecule.• Molecules attract each other by weak intermolecular forces. In a covalent network

solid, each atom is covalently bonded to many other atoms.

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Chapter Test glencoe.com274 Chapter 8 • Assessment

Assessment

Section 8.1

Mastering Concepts 78. Atoms lose, gain, or share electrons to

achieve an octet. Covalent bonding occurs when atoms share electrons to achieve an octet.

79. The nucleus of one atom attracts the electrons of the other atom, and they share one or more pairs of electrons.

80. Molecules bond covalently. 81. As the atoms approach, the net force of

attraction increases. At a certain optimal distance between atoms, the net attrac-tive force is maximized. If the atoms move closer than the optimal distance, repulsive force exceeds attractive force.

82. single covalent bond: sigma bond; dou-ble bond: a sigma bond and a pi bond; triple bond: one sigma and two pi bonds

Mastering Problems 83. N: 5, 3; As: 5, 3; Br: 7, 1; Se: 6, 2 84. a. single bonds: sigma bonds; double

bond: one sigma bond and one pi bond

b. single bonds: sigma bonds; triple bond: one sigma and two pi bonds

85. The triple bond in CO is the shortest and the strongest.

86. The triple bond in C ≡ N − is shorter and stronger.

87. a, c, b

Section 8.2

Mastering Concepts 88. Answers should agree with Figure 8.12

on page 252. 89. when it releases H + in water solution 90. Sulfur hexafluoride:1 S atom bonded

with 6 F atoms. Disulfur tetrafluoride:2 S atoms bonded with 4 F atoms.

91. The name silicon indicates 1 Si atom. The prefix di- means 2 and oxide indicates oxygen. The correct formula is Si O 2 .

Mastering Problems 92. HCl O 2 : chlorous acid; H 3 P O 4 : phosphoric

acid; H 2 Se: hydroselenic acid; HCl O 3 : chloric acid

93. a. nitrogen trifluoride b. nitrogen monoxide c. sulfur trioxide

d. silicon tetrafluoride 94. a. selenium dioxide c. dinitrogen tetrafluoride b. selenium trioxide d. tetrasulfur tetranitride 95. a. S F 2 b. Si Cl 4 c. C F 4 d. H 2 S O 3 96. a. Si O 2 b. HBr O 2 c. Cl F 3 d. HBr

Section 8.3

Mastering Concepts 97. the number of valence electrons for each atom 98. Arsenic has five bonding positions with a total of

10 shared electrons. This is greater that the eight electrons that occupy an octet.

99. B H 3 only has six electrons and does not have an electron arrangement with a low amount of poten-tial energy. It will share a lone pair with another molecule to form this electron arrangement.

100. Antimony has five valence electrons, one lone pair, and three positions where it can share one electron with a chlorine atom. This will form SbC l 3 . Antimony can also expand its octet and bond with all five valence electrons, forming Sb Cl 5 .

Mastering Problems 101. Refer to the Solutions Manual for structures. 102. Refer to the Solutions Manual for structures.

Section 8.1Mastering Concepts

78. What is the octet rule, and how is it used in covalent bonding?

79. Describe the formation of a covalent bond. 80. Describe the bonding in molecules. 81. Describe the forces, both attractive and repulsive, that

occur as two atoms move closer together. 82. How could you predict the presence of a sigma or pi

bond in a molecule?

Mastering Problems 83. Give the number of valence electrons in N, As, Br, and

Se. Predict the number of covalent bonds needed for each of these elements to satisfy the octet rule.

84. Locate the sigma and pi bonds in each of the molecules shown below.

a.

H — C — H

O— —

b. H — C C — H———

85. In the molecules CO, C O 2 , and C H 2 O, which C—O bond is the shortest? Which C—O bond is the strongest?

86. Consider the carbon-nitrogen bonds shown below:

H

C — N—

H—

H

H —

—H

—andC N-≡

Which bond is shorter? Which is stronger?

87. Rank each of the molecules below in order of the short-est to the longest sulfur-oxygen bond length.

a. S O 2 b. S O 3 2— c. S O 4 2—


88. Explain how molecular compounds are named. 89. When is a molecular compound named as an acid? 90. Explain the difference between sulfur hexafluoride and

disulfur tetrafluoride. 91. Watches The quartz crystals used in watches are made

of silicon dioxide. Explain how you use the name to determine the formula for silicon dioxide.

Mastering Problems 92. Complete Table 8.8.

Table 8.8 Acid Names

Formula Name

HCI O 2

H 3 P O 4

H 2 Se

HCI O 3

93. Name each molecule. a. N F 3 c. S O 3 b. NO d. Si F 4 94. Name each molecule. a. Se O 2 c. N 2 F 4 b. Se O 3 d. S 4 N 4 95. Write the formula for each molecule. a. sulfur difluoride c. carbon tetrafluoride b. silicon tetrachloride d. sulfurous acid 96. Write the formula for each molecule. a. silicon dioxide c. chlorine trifluoride b. bromous acid d. hydrobromic acid


97. What must you know in order to draw the Lewis struc-ture for a molecule?

98. Doping Agent Material scientists are studying the prop-erties of polymer plastics doped with As F 5 . Explain why the compound As F 5 is an exception to the octet rule.

99. Reducing Agent Boron trihydride (B H 3 ) is used as reducing agent in organic chemistry. Explain why B H 3 often forms coordinate covalent bonds with other molecules.

100. Antimony and chlorine can form antimony trichloride or antimony pentachloride. Explain how these two ele-ments can form two different compounds.

Mastering Problems101. Draw three resonance structures for the polyatomic

ion C O 3 2- . 102. Draw the Lewis structures for these molecules, each

of which has a central atom that does not obey theoctet rule.

a. PC l 5 c. Cl F 5 b. B F 3 d. Be H 2

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Chapter 8 • Assessment 275Chapter Test glencoe.com

Mastering Problems 103. Refer to the Solutions Manual for

structures. 104. Refer to the Solutions Manual for

structures. 105. P and Se because they are period 3 and

higher and have a d sublevel available

Section 8.4

Mastering Concepts 106. the repulsive nature of electron pairs

around a central atom 107. four 108. a. linear, 180° b. linear, 180° c. trigonal planar, 120° d. tetrahedral, 109° 109. The theory of hybridization explains the

shapes of molecules by the formation of identical hybrid orbitals from the atomic orbitals of the atoms in the mol-ecule; five identical s p 3 d orbitals

Mastering Problems 110. Xe F 4 : s p 3 d 2 ; Te F 4 : s p 3 d; Kr F 2 : s p 3 d;

O F 2 : s p 3 111. a. linear b. tetrahedral 112. a. bent, 104.5°, s p 3 b. trigonal pyramidal, 107°, s p 3 c. bent, 104.5°, s p 3 d. trigonal planar, 120°, s p 2

Section 8.5

Mastering Concepts 113. It increases left to right in a period and

decreases top to bottom in a group. 114. A nonpolar molecule has a symmetric

distribution of charge, while a polar molecule has a concentration of elec-trons on one side of the molecule.

115. Electrons in a polar bond are closer to the more electronegative atom because of unequal sharing. Those in a nonpolar bond are shared equally.

116. A covalent molecular solid is soft and has a low melting point because of weak intermolecular forces. A covalent network solid has a high melting point and is very hard because of the strength of the network of covalent bonds.

Mastering Problems 117. The following are circled: a. O b. F c. Cl 118. a. C b. N c. S d. O 119. Si—O 120. in order of increasing polarity: c, a, b, e, d 121. equal distribution of charge in a symmetrical

molecule 122. a. polar, asymmetrical b. nonpolar, symmetrical c. polar, asymmetrical d. nonpolar, symmetrical 123. S F 2 and S F 4 are polar. S F 6 is nonpolar.

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103. Draw two resonance structures for the polyatomic ion HC O 2 — .

104. Draw the Lewis structure for a molecule of each of these compounds and ions.

a. H 2 S c. S O 2 b. B F 4 — d. SeC l 2 105. Which elements in the list below are capable of forming

molecules in which one of its atoms has an expanded octet? Explain your answer.

a. B d. O b. C e. Se c. P


106. What is the basis of the VSEPR model? 107. What is the maximum number of hybrid orbitals a

carbon atom can form? 108. What is the molecular shape of each molecule? Estimate

the bond angle for each molecule, assuming that thereis not a lone pair.

a. A B—

b. A B A——

c.

A

A B A——

—

d.

A

A B A

A

——

——

109. Parent Compound PC l 5 is used as a parent compound to form many other compounds. Explain the theory of hybridization and determine the number of hybrid orbitals present in a molecule of PC l 5 .

Mastering Problems 110. Complete Table 8.9 by identifying the expected hybrid

on the central atom. You might find drawing the mole-cule’s Lewis structure helpful.

Table 8.9 Structures

Formula Hybrid Orbital Lewis Structure

Xe F 4

Te F 4

Kr F 2

O F 2

111. Predict the molecular shape of each molecule. a. COS b. C F 2 C l 2 112. For each molecule listed below, predict its molecular

shape and bond angle, and identify the hybrid orbitals. Drawing the Lewis structure might help you.

a. SC l 2 c. HOF b. N H 2 Cl d. B F 3


113. Describe electronegativity trends in the periodic table. 114. Explain the difference between nonpolar molecules and

polar molecules. 115. Compare the location of bonding electrons in a polar

covalent bond with those in a nonpolar covalent bond. Explain your answer.

116. What is the difference between a covalent molecular solid and a covalent network solid? Do their physical properties differ? Explain your answer.

Mastering Problems 117. For each pair, indicate the more polar bond by circling

the negative end of its dipole. a. C—S, C—O b. C—F, C—N c. P—H, P—Cl 118. For each of the bonds listed, tell which atom is more

negatively charged. a. C—H c. C—S b. C—N d. C—O 119. Predict which bond is the most polar. a. C—O c. C—Cl b. Si—O d. C—Br 120. Rank the bonds according to increasing polarity. a. C—H d. O—H b. N—H e. Cl—H c. Si—H 121. Refrigerant The refrigerant known as freon-14 is an

ozone-damaging compound with the formula C F 4 . Why is the C F 4 molecule nonpolar even though it contains polar bonds?

122. Determine if these molecules and ion are polar. Explain your answers.

a. H 3 O + c. H 2 S b. PC l 5 d. C F 4 123. Use Lewis structures to predict the molecular polarities

for sulfur difluoride, sulfur tetrafluoride, and sulfur hexafluoride.

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Chapter Test glencoe.com276 Chapter 8 • Assessment

Mixed Review 124. a. ClO c. P Cl 5 b. H 3 As O 4 d. H 2 S 125. a. phosphorus trichloride b. dichlorine heptoxide c. tetraphosphorus hexoxide d. nitrogen monoxide 126. a. SeF

F

b. ClO

O

–

c. PO O

O

3–

d.

PCl

Cl

Cl

O

e.

FGeF

F

F

127. The polar molecules are C H 3 Cl, ClF, and NC l 3 because each molecule is asym-metric and the charge is not distributed evenly.

128. in order of least to greatest bond character: e, d, a, c, b

129. ClF

FF

s p 3 d 130. Single covalent: 2 shared electrons, any

halogen or group 17 element Double covalent: 4 shared electrons,

group 16 elements Triple covalent: 6 shared electrons,

group 15 elements

Think Critically 131. Concept maps will vary. 132. The name arsenic(III) oxide states that

arsenic has an oxidation number of 3+ and oxide is 2–. The correct formu-la is A s 2 O 3 . The name diarsenic trioxide states that there are two atoms of arsenic and three atoms of oxygen. The correct formula is A s 2 O 3. Even

though they are named differently, they both repre-sent the same formula.

133. Column 2: the electrostatic attraction of a positive ion for a negative ion; the sharing of electrons between two atoms; the attraction of a positive ion for delocalized electrons; atoms covalently bonded to many other atoms in a crystal lattice. Column 3: hard, rigid, brittle, crystalline, high melting point, nonconductor in the solid state; soft, low melting point, nonconductor in the solid state; a crystal that conducts heat and electricity, malleable, ductile, high melting point; crystal is hard, rigid, brittle, nonconductor. Sample answers for Column 4: NaCl, C O 2 , Ag, diamond.

134. Sigma bonds are the N—H bonds and the C—N bonds as well as one of the C—O bonds. The other C—O bond is a pi bond. The lone pairs are located on both N atoms.

135. a. polar b. nonpolar c. polar 136. The first carbon (bonded to three H atoms and one

C atom) atom is a s p 3 hybrid because it has 4 bond-ing positions. The second carbon atom (bonded to one C atom and one N atom) is an sp hybrid because it has two bonding positions.

Mixed Review 124. Write the formula for each molecule. a. chlorine monoxide b. arsenic acid c. phosphorus pentachloride d. hydrosulfuric acid 125. Name each molecule. a. PC l 3 b. C l 2 O 7 c. P 4 O 6

d. NO 126. Draw the Lewis structure for each molecule or ion. a. Se F 2 b. Cl O 2 -

c. P O 3 3— d. POC l 3 e. Ge F 4

127. Determine which of the molecules are polar. Explain your answers.

a. C H 3 Cl b. ClF c. NC l 3 d. B F 3

e. C S 2 128. Arrange the bonds in order of least to greatest polar

character. a. C—O b. Si—O c. Ge—O d. C—Cl e. C—Br 129. Rocket Fuel In the 1950s, the reaction of hydrazine

with chlorine trifluoride (Cl F 3 ) was used as a rocket fuel. Draw the Lewis structure for Cl F 3 and identify the hybrid orbitals.

130. Complete Table 8.10, which shows the number of elec-trons shared in a single covalent bond, a double covalent bond, and a triple covalent bond. Identify the group of atoms that will form each of these bonds.

Table 8.10 Shared Pairs

Bond TypeNumber of Shared

ElectronsAtoms that Form

the Bond

Single covalent

Double covalent

Triplecovalent

Think Critically

131. Organize Design a concept map that explains how VSEPR model theory, hybridization theory, and molecu-lar shape are related.

132. Compare and contrast the two covalent compounds identified by the names arsenic(III) oxide and diarsenic trioxide.

133. Make and Use Tables Complete Table 8.11, using what you learned in Chapters 7 and 8.

Table 8.11 Properties and Bonding

SolidBond

DescriptionCharacteristic

of SolidExample

Ionic

Covalent molecular

Metallic

Covalent network

134. Apply Urea, whose structure is shown below, is a com-pound used in manufacturing plastics and fertilizers. Identify the sigma bond, pi bonds, and lone pairs pres-ent in a molecule of urea.

NH

H

H

HC

O

N

— —

135. Analyze For each of the characteristics listed below, identify the polarity of a molecule with that characteristic.

a. solid at room temperature b. gas at room temperature c. attracted to an electric current 136. Apply The structural formula for acetonitrile, C H 3 CN,

is shown below.

H

HH C ———C N

Examine the structure of the acetonitrile molecule. Determine the number of carbon atoms in the molecule, identify the hybrid present in each carbon atom, and explain your reasoning.

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Challenge Problem

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Chapter 8 • Assessment 277Chapter Test glencoe.com

Challenge Problem 137. a. H HC C

HH

C

O

O

b. C 2 H 2 = 1671 kJ/mol; HCOOH = 1986 kJ/mol

Cumulative Review 138. 2.34 g/mL; density 139. a. CaC O 3 b. KCl O 3 c. Ag C 2 H 3 O 2 d. CuS O 4 e. (N H 4 ) 3 P O 4 140. a. sodium iodide b. iron(III) nitrat c. strontium hydroxide d. cobalt(II) chloride e. magnesium bromate

Additional Assessment

Chemistry


141.

C

H

H

H O HO

H

H

C

Answers will vary. Students might note that the presence of –OH groups make ethylene glycol miscible in water and contribute to its relatively high boiling point and relatively low freezing point.

142. Answers should include a discussion of the nonpolar end of a detergent molecule and the polar end of the same molecule, which allow it to attract both water and oil.

144. A, s p 2 ; B, s p 3 ; C, s p 2 145. C 8 H 5 N O 4 2-

Document-Based QuestionsData obtained from: Fleming, Declan., 2002. The Chemiluminescence of Luminol, Exemplarchem, Royal Society of Chemistry.

143. C 8 H 7 O 2 N 3

C N

NC C

C H

HH

H

C

C

H

HH

C

C

O

ON

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Additional AssessmentChallenge ProblemChallenge Problem 137. Examine the bond-dissociation energies for the various

bonds listed in Table 8.12.

Table 8.12 Bond-Dissociation Energies

BondBond-Dissociation

Energy (kJ/mol)Bond

Bond-Dissociation Energy (kJ/mol)

C—C 348 O—H 467

C=C 614 C—N 305

C≡C 839 O=O 498

N—N 163 C—H 416

N=N 418 C—O 358

N≡N 945 C=O 745

a. Draw the correct Lewis structures for C 2 H 2 and HCOOH.

b. Determine the amount of energy needed to break apart each of these molecules.

Cumulative Review 138. Table 8.13 lists a liquid’s mass and volume data. Create a

line graph of this data with the volume on the x-axis and the mass on the y-axis. Calculate the slope of the line. What information does the slope give you? (Chapter 2)

Table 8.13 Mass v. Volume

Volume Mass

4.1 mL 9.36 g

6.0 mL 14.04 g

8.0 mL 18.72 g

10.0 mL 23.40 g

139. Write the correct chemical formula for each compound. (Chapter 7)

a. calcium carbonate b. potassium chlorate c. silver acetate d. copper(II) sulfate e. ammonium phosphate 140. Write the correct chemical name for each compound.

(Chapter 7) a. NaI b. Fe(N O 3 ) 3

c. Sr(OH ) 2 d. CoC l 2 e. Mg(Br O 3 ) 2

Chemistry

141. Antifreeze Research ethylene glycol, an antifreeze-coolant, to learn its chemical formula. Draw its Lewis structure and identify the sigma and pi bonds.

142. Detergents Choose a laundry detergent to research and write an essay about its chemical composition. Explain how it removes oil and grease from fabrics.

Document-Based QuestionsLuminol Crime-scene investigators often use the covalent compound luminol to find blood evidence. The reaction between luminol, certain chemicals, and hemoglobin, a pro-tein in blood, produces light. Figure 8.26 shows a ball-and-stick model of luminol.

Data obtained from: Fleming, Declan., 2002. The Chemiluminescence of Luminol, Exemplarchem, Royal Society of Chemistry.

a

b

c

■ Figure 8.26

143. Determine the molecular formula for luminol and draw its Lewis structure.

144. Indicate the hybrid present on the atoms labeled A, B, and C in Figure 8.26.

CC

CH

H

C

C

CC

C

H O

NH2 O

O-

APA ion

O-

■ Figure 8.27

145. When luminol comes in contact with the iron ion in hemoglobin, it reacts to produce N a 2 APA, water, nitrogen, and light energy. Given the structural for-mula of the APA ion in Figure 8.27, write the chemi-cal formula for the polyatomic APA ion.

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278 Chapter 8 • Assessment Standardized Test Practice glencoe.com

Standardized Test PracticeMultiple Choice 1. D 2. A 3. C 4. D 5. B 6. C 7. D 8. D 9. A 10. B

Standardized Test PracticeCumulative

Multiple Choice

1. The common name of Si I 4 is tetraiodosilane. What is its molecular compound name?A. silane tetraiodideB. silane tetraiodineC. silicon iodideD. silicon tetraiodide

2. Which compound contains at least one pi bond?A. C O 2 B. CHC l 3 C. As I 3 D. Be F 2

Use the graph below to answer Questions 3 and 4.

Elec

tron

egat

ivit

y

Atomic number

5

4

3

2

1

01 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

3. What is the electronegativity of the element with atomic number 14?A. 1.5B. 1.9C. 2.0D. 2.2

4. An ionic bond would form between which pairs of elements?A. atomic number 3 and atomic number 4B. atomic number 7 and atomic number 8C. atomic number 4 and atomic number 18D. atomic number 8 and atomic number 12

5. Which is the Lewis structure for silicon disulfide?A. S SSi

B. SS Si SS

C. S SSi

D. SiS S

6. The central selenium atom in selenium hexafluo-ride forms an expanded octet. How many electron pairs surround the central Se atom?A. 4 C. 6B. 5 D. 7

Use the table below to answer Questions 7 and 8.

Bond Dissociation Energies at 298 K

Bond kJ/mol Bond kJ/mol

Cl–Cl 242 N≡N 945

C–C 345 O–H 467

C–H 416 C–O 358

C–N 305 C=O 745

H–I 299 O=O 498

H–N 391

7. Which diatomic gas has the shortest bond between its two atoms?A. HI C. C l 2 B. O 2 D. N 2

8. Approximately how much energy will it take to break all the bonds present in the molecule below?

O

— —

HHN

H

C

——

—

OCC

HH

H

H—

A. 3024 kJ/molB. 4318 kJ/molC. 4621 kJ/molD. 5011 kJ/mol

9. Which compound does NOT have a bent molecular shape?A. Be H 2 C. H 2 OB. H 2 S D. Se H 2

10. Which compound is nonpolar?A. H 2 S C. Si H 3 ClB. CC l 4 D. As H 3

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Standardized Test Practice glencoe.com Chapter 8 • Assessment 279

Short Answer 11. Nitric acid (HN O 3 ) and nitrous acid

(NH O 2 ). The -ic suffix indicates the larger number of oxygen atoms; the -ous suffix indicates the lower number of oxygen atoms.

12. 580 nm 13. 5.2 × 10 14 Hz

Extended Response 14. The lab partner is incorrect because he

has taken a simple average of the masses of each isotope. An average atomic mass must take into account the percent abundance of each isotope and find a weighted average. The correct answer (28.09 amu) is found as follows: (27.98 amu × 92.21%) + (28.98 amu × 4.70%) + (29.97 amu × 3.09%).

SAT Subject Test: Chemistry 15. B 16. A 17. D 18. D 19. A

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14.X

Short Answer SAT Subject Test: Chemistry

11. Oxyacids contain hydrogen and an oxyanion. There are two different oxyacids that contain hydrogen, nitrogen, and oxygen. Identify these two oxyacids. How can they be distinguished on the basis of their names and formulas?

Use the atomic emission spectrum below to answer Questions 12 and 13.

400 500

Nanometers

600 700

12. Estimate the wavelength of the photons being emitted by this element.

13. Find the frequency of the photons being emitted by this element.

Extended Response

Use the table below to answer Question 14.

Percent Abundance of Silicon Isotopes

Isotope Mass Percent Abundance

28 Si 27.98 amu 92.21 %

29 Si 28.98 amu 4.70 %

30 Si 29.97 amu 3.09 %

14. Your lab partner calculates the average atomic mass of these three silicon isotopes. His average atomic mass value is 28.98 amu. Explain why your lab partner is incorrect, and show how to calculate the correct average atomic mass.

Use the list of separation techniques below to answer Questions 15 to 17.

A. filtration D. chromatographyB. distillation E. sublimationC. crystallization

15. Which technique separates components of a mixture with different boiling points?

16. Which technique separates components of a mixture based on the size of its particles?

17. Which technique is based on the stronger attraction some components have for the stationary phase compared to the mobile phase?

Use the table below to answer Questions 18 to 19.

Electron-Dot Structures

Group 1 2 13 14 15 16 17 18

Diagram Li Be B C N O F Ne

18. Based on the Lewis structures shown, which elements will combine in a 2:3 ratio?A. lithium and carbonB. beryllium and fluorineC. beryllium and nitrogenD. boron and oxygenE. boron and carbon

19. How many electrons will beryllium have in its outer energy level after it forms an ion to become chemi-cally stable?A. 0 D. 6B. 2 E. 8C. 4

NEED EXTRA HELP?

If You Missed Question . . . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Review Section . . . 8.2 8.1 8.5 8.5 8.3 8.3 8.1 8.1 8.4 8.5 8.2 5.2 5.2 4.3 3.3 3.3 3.3 7.2 7.2

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