Chapter 8 Organizer: Covalent Bonding
")' )DEA
Covalent bonds form when atoms share electrons.
Section Objectives
National
Standards
State/
Local
Standards
Resources to Assess Mastery
Section 8.1
1. 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 27
Progress Monitoring
Formative Assessment, p. 245
Reading Check, pp. 243, 244, 246
Section Assessment, p. 247
Section 8.2
1. Translate molecular formulas into binary
UCP.2, UCP.3,
UCP.5; B.2;
G.2
Entry-Level Assessment
Focus Transparency 28
Progress Monitoring
Formative Assessment, pp. 249, 250, 251
Reading Check, p. 249
Section Assessment, p. 252
Section 8.3
1. 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 29
Progress Monitoring
Formative Assessment, pp. 257, 258
Reading Check, p. 259
Section Assessment, p. 260
Section 8.4
1. 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 30
Progress Monitoring
Formative Assessment, p. 263
Reading Check, p. 262
Section Assessment, p. 264
Section 8.5
1. 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 31
Progress Monitoring
Formative Assessment, p. 267
Reading Check, pp. 266, 267
Graph Check, p. 266
Section Assessment, p. 270
Summative Assessment
Chapter Assessment, p. 274
ExamView® Assessment Suite CD-ROM
molecular compound names.
2. Name acidic solutions.
238A
Suggested Pacing
Period
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
Science Notebook 8.1 OL
FAST FILE Chapter Resources:
MiniLab Worksheet, p. 86 OL
Study Guide, p. 100 OL
Transparencies:
Section Focus Transparency 27 BL EL
Customize any lab with
the LabManager™
CD-ROM.
Lab Materials
Launch Lab, p. 239: paper towels, lab gloves, paper
cup, graduated cylinder, food coloring, wood splint,
sodium silicate solution, ethanol 15 min
MiniLab, p. 242: permanent marker, 9-inch
aluminum pie pan, hot plate, sugar crystals, salt
crystals, paraffin 20 min
Science Notebook 8.2 OL
FAST FILE Chapter Resources:
Study Guide, p. 101 OL
Transparencies:
Section Focus Transparency 28 BL EL
Math Skills Transparency 9 OL EL
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
Science Notebook 8.4 OL
FAST FILE Chapter Resources:
Study Guide, p. 103 OL
Transparencies:
Section Focus Transparency 30 BL EL
Teaching Transparency 27 OL EL
BL Below Level
OL On Level
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)
Science Notebook 8.3 OL
FAST FILE Chapter Resources:
Study Guide, p. 102 OL
Transparencies:
Section Focus Transparency 29 BL EL
Teaching Transparency 26 OL EL
Science Notebook 8.5 OL
FAST FILE Chapter Resources:
ChemLab Worksheet, p. 88 OL
Study Guide, p. 104 OL
Transparencies:
Section Focus Transparency 31 BL EL
Teaching Transparency 28 OL EL
Math Skills Transparency 9 OL EL
Additional Print and
Technology Resources
Lab Resources:
Laboratory Manual OL
CBL Laboratory Manual OL
Small-Scale Laboratory Manual OL
Forensics Laboratory Manual OL
ChemLab, p. 272: molecular model kit 45 min
AL Advanced Learners
EL English Learners
COOP LEARN Cooperative Learning
238B
Covalent Bonding
BIG Idea
this chapter’s Big Idea, have students
draw the Lewis structures for hydrogen 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 electrons Ask students how hydrogen and
oxygen atoms are able to bond to
form water, H 2O. They must share electrons. 2 hydrogen atoms share an electron
each with 1 oxygen atom.
Tie to Previous
Knowledge
Have students review the following
concepts before studying this
chapter:
Chapter 4: atomic structure
Chapter 5: electron configuration
Chapter 6: periodic trends
Chapter 7: periodic properties of
elements
Chapter 8: ionic bonding
Use the Photo
Establishing 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 molecules in a water droplet. Ask students 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.
238 Chapter 8 • Covalent Bonding
Spherical water droplet
BIG Idea Covalent bonds
form when atoms share electrons.
8.1 The Covalent Bond
MAIN Idea Atoms gain stability
when they share electrons and form
covalent bonds.
8.2 Naming Molecules
MAIN Idea Specific rules are used
when naming binary molecular
compounds, binary acids, and
oxyacids.
8.3 Molecular Structures
MAIN Idea Structural formulas
show the relative positions of atoms
within a molecule.
Space-filling model
8.4 Molecular Shapes
MAIN Idea The VSEPR model is
used to determine molecular shape.
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.
Ball-and-stick model
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.
H—O
• The chemical and physical properties
of water make it a unique liquid.
Interactive
C08_CO_874637.indd
Lewis structure
—
Sharing Electrons To introduce
H
Classroom
1
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
2/2/07
2:17:34
Start-Up Activities
LAUNCH Lab
RUBRIC available at glencoe.com
Purpose Students will build a
LAUNCH Lab
Bond Character Make the
following Foldable to help you
organize your study of the three
major types of bonding.
What type of compound is used
to make a Super Ball?
Super Balls are often made of a silicon compound called
organosilicon oxide (Si(OCH 2CH 3) 2O).
Safety Precautions Approve lab
safety forms before work begins.
Review MSDS for ethanol and sodium silicate with students prior to
doing the lab. Ethanol is very flammable 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.
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.
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.
Analysis
1. 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 electrons, how can they form a bond with each other?
Analysis
1. The ball loses shape when it sits, but when shaped as
4 PMC08_CO_874637.indd
2 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 dissolve in water, and loses its shape over time.
STEP 3 Staple along
the fold. Label the tabs as
follows: Bond Character,
Nonpolar Covalent,
Polar Covalent, and Ionic.
polymer from silicon and ethanol
that can bounce when shaped into
a sphere.
Ionic
Polar Covalent
Nonpolar Covalent
Bond Character
&/,$!",%3 Use this Foldable with Section 8.1. As
you read this section, summarize what you learn about
bond character and how it affects the properties of
compounds.
Visit glencoe.com to:
υ study the entire chapter online
υ
explore
υ
take Self-Check Quizzes
υ
use the Personal Tutor to work
Example Problems step-by-step
υ
access Web Links for more information,
projects, and activities
υ
find the Try at Home Lab, Breaking
Covalent Bonds
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 2/2/07
form a 2:17:51 PM
bond, these atoms must share electrons.
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 hazardous 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 hypothesize 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.
Customize this lab with
the LabManager™
CD-ROM.
Chapter 8 • Covalent Bonding
239
Section 8.1
1 Focus
Focus Transparency
Before presenting the lesson, project
Section Focus Transparency 27 and
have students answer the accompanying questions. BL EL
MAIN Idea
Covalent Bonds Draw the Lewis
structures for fluorine, chlorine, oxygen, 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 structure and show students an atomic
model of carbon. On both, point out
the four places carbon can form a
single covalent bond. Bond a hydrogen atom into each of the places. Ask
students what is represented. The
model represents the electrons shared
between a central carbon atom and four terminal hydrogen atoms in a CH 4 molecule. OL
2 Teach
Section 8.1
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 Vocabulary
chemical bond: the force
that holds two atoms together
New Vocabulary
covalent bond
molecule
Lewis structure
sigma bond
pi bond
endothermic reaction
exothermic reaction
The Covalent Bond
-!). )DEA Atoms 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 consists 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 transferring (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.
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 pendulum 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 position. Relate the pendulum’s
higher-energy positions to
instability. OL
240 Chapter 8 • Covalent Bonding
Chemistry Journal
1
Bonding Limerick
Read students the following limerick:
C08_S1_874637.indd
Bond formations can be unwell
(Electrons join up in a shell)
For it doesn’t seem fine
To say they combine
When we know very well they repel!
Have students develop limericks, poems, or rhymes that depict some aspect
of covalent bonds. OL
2/2/07
9:40:16
Visual Learning
Force of repulsion
Force 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.
You just read that atoms can share electrons to form stable electron configurations. 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 bonding 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 (Cl 2), bromine
(Br 2), and iodine (I 2), form when two atoms of each element share electrons. They exist this way because the two-atom molecules are more stable than the individual atoms.
Consider fluorine, which has an electron configuration of 1s 22s 22p 5.
Each fluorine atom has seven valence electrons and needs another electron to form an octet. As two fluorine atoms approach each other, several 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 covalently bonded (shared) and three pairs of electrons that are unbonded
(not shared). Unbonded pairs are also known as lone pairs.
6 AMC08_S1_874637.indd
Chemistry Journal
2
Why
Covalent Bonds Form Have
students write a paragraph explaining why
a covalent bond can form between a
chlorine atom and a fluorine atom. OL
If the atoms are forced
closer together, the
nuclei and electrons
repel each other.
The distance is right for the attraction between one atom’s protons
and the other atom’s electrons to
make the bond stable.
What is a covalent bond?
Figure 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
■ 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?
■
Concept Development
Electron 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.
■ 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.
F
+
Fluorine
atom
Complete
octets
Reinforcement
Dot Diagrams Review the
electron dot diagrams used in
Chapter 7. Point out that covalent
bonds can be represented in a variety
of ways.
F
Fluorine
atom
F F
Caption Question Fig. 8.2
A stable bond is formed at the location of
maximum net attraction.
Bonding pair
of electrons
Lone
pairs
Fluorine
molecule
GLENCOE Technology
Virtual Labs CD-ROM
2/2/07
9:40:26 AM
Chemistry: Matter and Change
Animation: Covalent Bonding
Section 8.1 • The Covalent Bond
241
Compare Melting Points
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, interpret 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.
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 labeled
wedges as follows: sugar crystals (C 12H 22O 11),
A; salt crystals (NaCl) B; paraffin (C 23H 48), C.
6. Predict the order in which the compounds
will melt.
Expected Result
• paraffin melts first
• sugar melts second, then burns
• sodium chloride does not melt
Chapter 8 • Covalent Bonding
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.
When 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 electrons belong to each atom simultaneously, which gives each
hydrogen atom the noble-gas configuration of helium (1s 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.
should work in groups to do this lab.
A candle can be substituted for the
Bunsen burner.
242
Analysis
Single Covalent Bonds
Teaching Strategies Students
Customize this lab with
the LabManager™
CD-ROM.
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.
How can you determine the relationship
between bond type and melting point? The
Disposal Carefully remove the lid
and place in a waste container.
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.
7. Turn the temperature knob on the hot plate
■ 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.
H
Hydrogen atom
C08_S1_874637.indd
3
→
+
+
H
Hydrogen atom
→
HH
Hydrogen molecule
2/2/07
9:40:27
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 Cl 2.
Group 14 and single bonds Atoms of group
14 elements form four covalent bonds. A methane
molecule (CH 4) forms when one carbon atom bonds
with four hydrogen atoms. Carbon, a group 14 element, has an electron configuration of 1s 22s 22p 2. With
four valence electrons, carbon needs four more electrons 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 provide 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.
7 AMC08_S1_874637.indd
■
a
+
O
→ H—O
—
2H
H
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 1s 22s 22p 4. Water is composed 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.
Build a Model
Two Single Covalent Bonds
Lewis Structures Give students
Ammonia
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 CF 4 and H 2S.
BL EL
b
+
N
→ H—N
—
3H
—
H
H
Three Single Covalent Bonds
Methane
c
Reading Check A covalent bond is
shown as a pair of dots or as a short line.
4H
+
—
H
C
→ H—C—H
—
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 configuration of 1s 22s 22p 3. Ammonia (NH 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 (NF 3),
nitrogen trichloride (NCl 3), and nitrogen tribromide
(NBr 3). Each atom of these group 17 elements and the
nitrogen atom share an electron pair.
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
Water
H
Four Single Covalent Bonds
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.
■
Reading Check Describe how a Lewis structure
shows a covalent bond.
Differentiated Instruction
4
Below
Level Using a piece of acetate, draw circles to represent atoms. In each circle,
2/2/07
9:40:30 AM
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
Section 8.1 • The Covalent Bond
243
EXAMPLE Problem 8.1
IN-CLASS Example
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.
Question Hydrogen selenide,
H 2Se, is a toxic gas with properties
similar to hydrogen sulfide. Draw
the Lewis structure for this molecule.
Answer
H
Se
Have 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.
P
■ 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 (SiF 4)
and water.
3. H
Cl
Solve for the Unknown
To 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
H
3
+
F
Fluorine
atom
→
H—F
Hydrogen fluoride
molecule
Evaluate the Answer
Each atom in the new molecule now has a noble-gas
configuration and is stable.
S
H
PRACTICE Problems
Cl
Draw the Lewis structure for each molecule.
1.
2.
3.
6.
Cl
4.
2
Hydrogen
atom
H
2. H
Analyze the Problem
You 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.
H
PRACTICE Problems
1. H
1
C
Cl
Extra Practice Page 979 and glencoe.com
PH 3
4. CCl 4
H 2S
5. SiH 4
HCl
Challenge Draw a generic Lewis structure for a molecule formed
between atoms of Group 1 and Group 16 elements.
Cl
5.
H
H
Si
H
H
6. Using 1 and 16 to represent atoms of
groups 1 and 16, respectively, the generic
structure is:
1
16
VOCABULARY
ACADEMIC VOCABULARY
Overlap
to occupy the same area in part
The two driveways overlap at the street
forming a common entrance.
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 2O), ammonia (NH 3), and methane (CH 4)
have sigma bonds, as shown in Figure 8.7.
Reading Check List the orbitals that can form sigma bonds in a
covalent compound.
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.
244 Chapter 8 • Covalent Bonding
Chemistry Project
5
Chemistry in
Medicine Ask students research the use of perfluorooctylbromide
C08_S1_874637.indd
(C 8F 17Br) 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
2/2/07
9:40:32
■
H
O
N
H
H
H
Water (H2O)
C
H
H
Ammonia (NH3)
Multiple Covalent Bonds
In 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 element oxygen only exist as diatomic molecules. Each oxygen atom has six
valence electrons and must obtain two additional electrons for a noblegas 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 electrons 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
H
H
Concept Development
H
Methane (CH4)
■ Figure 8.7 Sigma bonds formed in each
of these molecules when the atomic orbital of
each hydrogen atom overlapped end to end
with the orbital of the central atom.
Interpret Identify the types of orbitals
that overlap to form the sigma bonds in
methane.
Assessment
&/,$!",%3
Incorporate information
from this section into
your Foldable.
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.
2 AMC08_S1_874637.indd
a
O
+
O
→
O—O
Electron Sharing Make sure students 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.
Skill Divide students into
Figure 8.8b.
Two shared pairs
of electrons
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.
■ 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.
groups. Hand each group a set of
cards containing electron dot diagrams 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 chlorine card. Show students that by
sharing a pair of electrons each atom
obtains a stable outer electron configuration. Have students determine
the Lewis structure for as many molecules as possible by combining the
cards. Remind students that multiple
bonds are possible with carbon,
nitrogen, and oxygen atoms. OL EL
COOP LEARN
&/,$!",%3
RUBRIC available at glencoe.com
b
6
N
+
N
→
N—
—N
Three shared pairs
of electrons
Content Background
Personal Tutor For an online tutorial on
multiple covalent bonds, visit glencoe.com.
2/2/07
Exceptional Oxygen Spins of
unpaired electrons in atoms or ions
cause a substance to be paramagnetic, that is, attracted to an external
magnetic field. A substance is
diamagnetic, unaffected by the mag9:40:38 AM
netic field, if all of the electrons are
paired. Although the Lewis structure
and bond strength for diatomic oxygen 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.
Section 8.1 • The Covalent Bond
245
Concept Development
Bond 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
CH 3Cl and CH 2Cl 2. 1567 kJ/mol, 1482
kJ/mol Ask students which molecule
is more likely to react. CH 2Cl 2 OL
■ Figure 8.9 Notice how the multiple
bond between the two carbon atoms in
ethene (C 2H 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 doughnutshaped cloud around the sigma bond.
p overlap
σ bond
σ bond
H
H
C
σ bond
σ bond
C
H
H
p overlap
Interactive Figure To see an animation of
sigma and pi bonding, visit glencoe.com.
σ bond
H
C—C
H
H
H
π bond
Ethene
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.
Enrichment
Bond Energy Use the equation
The Strength of Covalent Bonds
shown below to calculate the change
in enthalpy of a chemical reaction.
Recall 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.
∆H = ∑(bond energies of broken
bonds) - ∑(bond energies of
formed bonds)
Have students determine the total
enthalpy change when Cl 2 reacts
with CH 4 to form CH 3Cl and HCl.
The enthalpy change is −104 kJ Ask students 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
Bond length The strength of a covalent bond depends on the distance 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 molecules 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.
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.
■
Nuclei
Reading Check As the bond type
increases from single covalent, to double
covalent, to triple covalent, the bond
length decreases.
Bond length
Table 8.1
Covalent Bond Type and
Bond Length
Molecule
Bond Type
Bond Length
F2
single covalent
1.43 × 10 -10 m
O2
double covalent
1.21 × 10 -10 m
N2
triple covalent
1.10 × 10 -10 m
Differentiated Instruction
Interactive Figure Students can
interact with the sigma and pi bonding
art at glencoe.com.
246 Chapter 8 • Covalent Bonding
7
Below Level
Have students place a single toothpick between two gumdrops to form
C08_S1_874637.indd
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
2/2/07
9:40:41
Table 8.2
3 Assess
Bond-Dissociation Energy
Molecule
Bond-Dissociation Energy
Check for Understanding
F2
159 kJ/mol
O2
498 kJ/mol
Ask students to draw the Lewis
structure for the molecule H 2S. OL
N2
945 kJ/mol
H
S
H
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 bonddissociation 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 existing 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.
Section 8.1
Reteach
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 CO 2.
Thus, the burning of charcoal is an
exothermic reaction.
■
Extension
Have interested students research
trinitrotoluene (TNT) and determine
the available bond energy if all bonds
in a molecule of this compound were
broken. AL
Assessment
Section Summary
7.
◗ Covalent bonds form when atoms
share one or more pairs of electrons.
8. Describe how the octet rule applies to covalent bonds.
◗ 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.
Identify the type of atom that generally forms 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.
13. Predict the relative bond-dissociation energies needed to break the bonds in the
structures below.
b. H
H
— —
a. H — C —
— C—H
— —
Section 8.1
-!). )DEA
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.
◗ Bond length is measured nucleus-tonucleus. Bond-dissociation energy is
needed to break a covalent bond.
1 AMC08_S1_874637.indd
Draw the structural formulas for
C 2H 6, C 2H 4, and C 2H 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
C —C
H
H
Assessment
8
2/2/07
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.
Self-Check Quiz glencoe.com
9:40:49 AM
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
Section 8.1 • The Covalent Bond
247
Section 8.2
1 Focus
Focus Transparency
Before presenting the lesson, project
Section Focus Transparency 28 and
have students answer the accompanying questions. BL EL
MAIN Idea
Section 8.2
Objectives
Naming Molecules
◗ Translate molecular formulas into
binary molecular compound names.
◗ Name acidic solutions.
MAIN Idea Specific rules are used when naming binary molecular
compounds, binary acids, and oxyacids.
Review Vocabulary
Real-World Reading Link You probably know that your mother’s mother
oxyanion: a polyatomic ion in which
an element (usually a nonmetal) is
bonded to one or more oxygen atoms
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.
New Vocabulary
Naming Binary Molecular Compounds
oxyacid
Many 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 molecular compound is composed only of two nonmetal atoms—not metal
atoms or ions. An example is dinitrogen monoxide (N 2O), 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
Naming 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
first word is dinitrogen and second word is monoxide.
2 Teach
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.
Visual Learning
Table 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
Table 8.3
Interactive Table Students
can interact with the table at
glencoe.com.
Prefixes in Covalent Compounds
Interactive Table Explore
naming covalent compounds
at glencoe.com.
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-
Demonstration
Forming Bonds
C08_S2_874637.indd
1
Safety
Precautions
to demonstrate the formation of
ionic and covalent bonds
Disposal Dispose of the solid in a landfill
approved to receive chemical waste.
Materials
Procedure
Purpose
Magnesium ribbon (5 cm); roll
sulfur (2 g); tongs; large metal can;
deflagrating spoon; laboratory
burner
248 Chapter 8 • Covalent Bonding
Place 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 deflagrating 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;
SO 2 vapor is toxic.
2/2/07
9:48:42
EXAMPLE Problem 8.2
IN-CLASS Example
Naming Binary Molecular Compounds Name the compound P 2O 5, which is
used as a drying and dehydrating agent.
1
Question PCl 5 is a colorless to
pale, yellow solid with a pungent
odor mainly used as a gentle chlorinating agent in manufacturing dyes
and pharmaceuticals. Name this
compound.
Analyze the Problem
You 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 Unknown
First, 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.
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, indicating that there is only one atom of this element present. Chloride has the subscript 5,
the prefix for which is penta-. Hence, the compound is called phosphorus pentachloride.
Now modify the names to indicate the number of atoms present in a molecule.
diphosphorus pentoxide
3
From the formula P 2O 5, 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.
Evaluate the Answer
The 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, P 2O 5.
PRACTICE Problems
PRACTICE Problems
Have 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.
Extra Practice Page 979 and glencoe.com
Name each of the binary covalent compounds listed below.
14.
15.
16.
17.
18.
CO 2
SO 2
NF 3
CCl 4
Challenge What is the formula for diarsenic trioxide?
14.
15.
16.
17.
18.
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 (NH 3), hydrazine (N 4H 4), and nitric oxide (NO).
Reinforcement
Molecule Mobile Have students
make a mobile that can be used to
name molecules. Display the mobile
in the classroom. BL EL
Reading Check nitrogen trihydride, dinitrogen tetrahydride,
nitrogen oxide
Reading Check Apply What are the scientific names for ammonia,
hydrazine, and nitric oxide?
2
Results
Analysis
Magnesium 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 produced in the formation of SO 2. Thus, the ionic
bonds in MgO are stronger (more stable) than
the covalent bonds in SO 2..
Ask 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) → SO 2(g)
2 AMC08_S2_874637.indd
carbon dioxide
sulfur dioxide
nitrogen trifluoride
carbon tetrachloride
As 2O 3
2/2/07
9:48:47 AM
Assessment
Knowledge If Mg forms an
ionic bond, and S forms a covalent
bond, which bond type is the stronger bond? Why? The ionic bond is stronger because more energy is released during its
formation. OL
Section 8.2 • Naming Molecules 249
Naming Acids
Assessment
Water 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.
Knowledge 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
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
Quick Demo
chlorine) becomes hydrochloric.
2. The second word is always acid. Thus, HCl in a water solution is
called hydrochloric acid.
Acid Formation Add 2 drops
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.
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.
CO 2 + H 2O → H 2CO 3
Write the formula for the acid
formed (H 2CO 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
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 (HNO 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.
NO 3, the nitrate ion, becomes nitric.
2. The second word of the name is always acid. HNO 3 (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.
Table 8.4 Naming Oxyacids
Compound
HClO 3
chlorate
Acid Suffix
-ic
Acid Name
chloric acid
HClO 2
chlorite
-ous
chlorous acid
HNO 3
nitrate
-ic
nitric acid
HNO 2
nitrite
-ous
nitrous acid
Differentiated Instruction
Chemistry Project
3
Advanced Learners
Have interested
Molecular Compounds Using a 2 × 2
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
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
C08_S2_874637.indd
250 Chapter 8 • Covalent Bonding
Oxyanion
2/2/07
9:48:49
Table 8.5
Interactive Table Explore
naming covalent compounds
glencoe.com.
Formulas and Names of
Some Covalent Compounds
Common Name
Formula
Have 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.
Molecular Compound Name
H 2O
water
dihydrogen monoxide
NH 3
ammonia
nitrogen trihydride
N 2H 4
hydrazine
dinitrogen tetrahydride
HCl
muriatic acid
hydrochloric acid
C 9H 8O 4
aspirin
2-(acetyloxy)benzoic acid
PRACTICE Problems
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
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 common name in addition to its compound name.
PRACTICE Problems
Extra Practice Page 979 and glencoe.com
Name the following acids. Assume each compound is dissolved in water.
19. HI
20. HClO 3
21. HClO 2
22. H 2SO 4
24. Challenge What is the formula for periodic acid?
23. H 2S
Writing Formulas from Names
Interactive Table Students
can interact with the table at
glencoe.com.
The 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 number 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.
PRACTICE Problems
hydroiodic acid
chloric acid
chlorous acid
sulfuric acid
hydrosulfuric acid
HIO 4
AgCl
H 2O
ClF 3
P 2O 3
S 2F 10
H 2CO 3
Assessment
Knowledge Have students
write molecular formulas, names of
molecules, names of acids, or formulas 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
Extra Practice Page 979 and glencoe.com
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?
C08_S2_874637.indd
9 AM
Chemistry Journal
4
2/7/07
Form 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 usually the least electronegative atom is written first in a compound.
OL
Br
11:52:18 AM
F
Cl
C
CBr 4
CF 4
CCl 4
carbon tetrabromide
carbon tetrafluoride
carbon tetrachloride
P
PBr 3
PF 3
PCl 3
phosphorus tribromide phosphorus trifluoride phosphorus trichloride
S
SBr 2
SF 2
SCl 2
sulfur dibromide
sulfur difluoride
sulfur dichloride
Section 8.2 • Naming Molecules 251
■
Caption Question Fig. 8.12
Look at the
formula of
the molecule.
H 2SO 3 is an oxyacid; HBr is a binary acid.
3 Assess
Does the compound
form an acidic
aqueous solution?
Check for Understanding
Show 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
Name the first element in the
molecule. Use a prefix if the number
of atoms is greater than one. To
name the second element, indicate the
number present by using a prefix +
root of second element + -ide.
Name as an acid.
Is there an oxygen
present in the
compound?
No
(HBr)
Many 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: SO 2
(sulfur dioxide, H 2SO 3, sulfurous
acid), CO 2 (carbon dioxide, H 2CO 3,
carbonic acid), SO 3 (sulfur trioxide,
H 2SO 4, sulfuric acid), NO 2 (nitrogen
dioxide, HNO 3, nitric acid). OL
Yes
(H2SO3)
Hydro + root of second
element + -ic, then acid.
NO2 is nitrogen dioxide.
Root of oxyanion present + -ic
if the anion ends in -ate, or + -ous
if the anion ends in -ite, then acid.
HBr (aq) is hydrobromic acid.
H2SO3 is sulfurous 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 8.2
Extension
Assessment
Section Summary
31.
◗ 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.
32. Define a binary molecular compound.
◗ Molecules that produce H + in solution are acids. Binary acids contain
hydrogen and one other element.
Oxyacids contain hydrogen and an
oxyanion.
Section 8.2
No
(NO2)
Yes
(H2SO3 and HBr)
Reteach
Have students write answers and
questions for a game using the
categories Molecular Formulas,
Names of Molecules, Names of
Acids, and Formulas for Acids. OL
Examples:
HBr, H2SO3,
and N02
MAIN Idea
Summarize the rules for naming binary molecular compounds.
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 2O 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
Assessment
C08_S2_874637.indd
5
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.
252 Chapter 8 • Covalent Bonding
2/2/07
35. S 2O 3, HIO 3, N 2O, HF
36. a. N 2O 3
b. NO
c. HCl
d. HClO 3
e. H 2SO 4
f. H 2SO 3
Self-Check Quiz glencoe.com
9:48:55
Section 8.3
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 Vocabulary
ionic bond: the electrostatic force
that holds oppositely charged particles
together in an ionic compound
New Vocabulary
structural formula
resonance
coordinate covalent bond
Section 8.3
1 Focus
Molecular Structures
Focus Transparency
MAIN 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 Formulas
In 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.
Before presenting the lesson, project
Section Focus Transparency 29 and
have students answer the accompanying questions. BL EL
MAIN Idea
Molecular 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 Teach
Quick 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-anderror 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
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.
H—P—H
PH3
Molecular formula
H
Lewis structure
Space-filling
molecular model
H—P—H
H
Structural formula
Ball-and-stick
molecular model
Visual Learning
Build Models Using a molecular model kit
or 1marshmallows and toothpicks, have students build models of NH 4 +, CH 2O, BCl 3, and
HF. Have students create a chart showing the
ball-and-stick model, Lewis structure, and
structural formula for each substance. OL EL
C08_S3_874637.indd
2/2/07
■ 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 mod9:59:52 AMels 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 8.3 • Molecular Structures 253
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.
Apply Chemistry
Dyes Organic dyes are organic
molecules that absorb select wavelengths 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 sandwiched 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.
Problem-Solving Strategy
Drawing 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.
Concept Development
To do this, divide the number of electrons available for bonding by
two.
Lewis 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 2H 6), ethene
(C 2H 4), and ethyne (C 2H 2), using the
systematic technique. OL
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.
Reinforcement
Representing 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.
GLENCOE Technology
Virtual Labs CD-ROM
Chemistry: Matter and Change
Video: Lewis Structures
254 Chapter 8 • Covalent Bonding
Apply the Strategy
Study Example Problems 8.3 through 8.5 to see how the steps in
the Problem-Solving Strategy are applied.
Chemistry Project
C08_S3_874637.indd
2
Gilbert Lewis Have students research
Gilbert Newton Lewis and make posters
depicting his original Lewis structures.
Students should share the posters with
the class. EL
2/2/07
9:59:56
EXAMPLE Problem 8.3
IN-CLASS Example
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 (NH 3).
1
2
Analyze the Problem
Math Handbook
Question Silicon tetrahydride is a
gas at room temperature that undergoes spontaneous combustion in air.
What is the Lewis structure for SiH 4?
Ammonia molecules consist of one nitrogen
atom and three hydrogen atoms. Because
hydrogen must be a terminal atom, nitrogen
is the central atom.
Dimensional Analysis
page 956
Answer
4 valence electrons
1 Si atom × ___
Si atom
1 valence electron
+ 4 H atoms × ___
H atom
8___
total electrons
=
= 4 bonding pairs
2
H
Solve for the Unknown
Find the total number of valence electrons available for bonding.
5 valence electrons
1 valence electron
1 N atom × __ + 3 H atoms × __
1 N atom
1 H atom
= 8 valence electrons
There are 8 valence electrons available for bonding.
8 electrons
__
= 4 pairs
2 electrons/pair
—
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.
Content Background
4 pairs total - 3 pairs used
= 1 pair available
Space Molecules In dust clouds
throughout the universe, dust particles 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 discovered the simple sugar molecule
ribose floating in the dust cloud
called Sagittarius B2. Astronomers
have identified more than 120 space
molecules.
Subtract the number of pairs used in
these bonds from the total number of
pairs of electrons 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—N—H
H
3
H
H
Use all bonding pairs. Silicon has an octet
and hydrogen only shares one pair. This
Lewis structure satisfies the octet rule.
Four pairs of electrons are available for bonding.
H—N—H
S
H
Determine the total number of
bonding pairs. To do this, divide the
number of available electrons by two.
Place the remaining lone pair on the
central nitrogen atom.
Evaluate the Answer
Each 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.
PRACTICE Problems
Extra Practice Page 980 and glencoe.com
37. Draw the Lewis structure for BH 3.
38. Challenge A nitrogen trifluoride molecule contains numerous lone
pairs. Draw its Lewis structure.
6 AMC08_S3_874637.indd
PRACTICE Problems
Have 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.
Chemistry Journal
3
Lewis
Structures Have students predict
the Lewis structures for the following
common molecules: carbon monoxide
(CO), carbon dioxide (CO 2), sulfur dioxide
(SO 2), sulfur trioxide (SO 3), hydrogen
cyanide (HCN), dinitrogen monoxide
(N 2O), nitrogen dioxide (NO 2), and
nitrogen monoxide (NO). OL
2/2/07
9:59:57 AM
H
37.
B
H
38.
H
F
F
N
F
Section 8.3 • Molecular Structures
255
EXAMPLE Problem 8.4
IN-CLASS Example
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 (CO 2).
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 SeO 3?
1
The 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.
2
Answer
6___
valence electrons
+
Se atom
6 valence electrons
3 O atoms × ___
O atom
24 valence electrons
___
=
2
= 12 electron pairs
Analyze the Problem
Solve for the Unknown
Find the total number of valence electrons available for bonding.
1 Se atom ×
4 valence electrons
6 valence electrons
1 C atom × __ + 2 O atoms × __
1C atom
There are 16 valence electrons available for bonding.
16 electrons
__
= 8 pairs
2 electrons/pair
Personal Tutor For an online tutorial on
greatest common factors, visit glencoe.com.
Se
O
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
= 6 pairs available
PRACTICE Problems
O—C—O
Have 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.
39.
H
40. S
C
C
C
Add three lone pairs to each terminal
oxygen atom.
6 pairs available - 6 pairs used
= 0 pairs available
Subtract the lone pairs from the
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.
H
—C —
—O
O—
S
3
Use a lone pair from each O atom to
form a double bond with the C atom.
Evaluate the Answer
Both carbon and oxygen now have an octet, which satisfies the
octet rule.
Apply Chemistry
256 Chapter 8 • Covalent Bonding
Subtract the number of pairs used in
these bonds from the total number of
pairs of electrons available.
Determine the number of bonding pairs remaining.
H
Ripening Fruits and vegetables
naturally produce ethylene gas
(C 2H 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 students 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.
Place a bonding pair (a single bond)
between the central carbon atom and
each terminal oxygen atom.
O—C—O
All bonded atoms have an octet.
H
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
O
1O atom
= 16 valence electrons
PRACTICE Problems
Extra Practice Page 980 and glencoe.com
39. Draw the Lewis structure for ethylene, C 2H 4.
40. Challenge A molecule of carbon disulfide contains both lone pairs
and multiple-covalent bonds. Draw its Lewis structure.
Differentiated Instruction
C08_S3_874637.indd
4
Advanced 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: OH -, PF 6-, BF 3, ClO 3-,
SF 3N, or CH 2Cl 2. AL
2/2/07
9:59:59
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 finding 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 present in the ion. Then, subtract the ion charge if the ion is positive, and
add the ion charge if the ion is negative.
Assessment
Skill 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 Development
Model Resonance Choose three
students to represent atoms in the
polyatomic ion NO 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
EXAMPLE Problem 8.5
Lewis Structure for a Polyatomic Ion Draw the correct Lewis
structure for the polyatomic ion phosphate (PO 4 3-).
1
Analyze the Problem
You 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.
2
Solve for the Unknown
Real-World Chemistry
Find the total number of valence electrons available for bonding.
Phosphorus and Nitrogen
5 valence electrons
6 valence electrons
1 P atom × __ + 4 O atoms × __
P atom
Question What is the correct
Lewis structure for the polyatomic
ion BrO 4 -?
O atom
+ 3 electrons from the negative charge = 32 valence electrons
32 electrons
__
= 16 pair
2 electrons/pair
—
O
Determine the total number of
bonding pairs.
Answer
Draw single bonds from each
terminal oxygen atom to the
central phosphorus atom.
—
O—P—O
O
Subtract the number of pairs
used from the total number of
pairs of electrons available.
—
—
O—P—O
O
3
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.
All of the atoms have an octet, and the group has a net charge of 3-.
Extra Practice Page 980 and glencoe.com
41. Draw the Lewis structure for the NH 4 + ion.
42. Challenge The ClO 4 - ion contains numerous lone pairs.
Draw its Lewis structure.
Enrichment
Chemistry of Chocolate The Mayans and
5
Aztecs
consumed chocolate, as do many people
throughout the world today. Cocoa and chocolate products contain theobromine, which is
chemically related to caffeine, though lacking
caffeine’s stimulant effect. OL
9 AMC08_S3_874637.indd
O
Algal blooms Phosphorus and
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.
Evaluate the Answer
PRACTICE Problems
Br
O
Add three lone pairs to each terminal oxygen atom.
12 pairs available - 12 lone pairs used = 0
3-
–
O
O
16 pairs total - 4 pairs used
= 12 pairs available
O
IN-CLASS Example
PRACTICE Problems
Have 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.
H
1. Write the molecular formula for
theobromine. C 7H 8N 4O 2
42.
2. Identify the number of sigma and2/2/07
pi bonds
10:00:00 AM
present in this molecule. 22 sigma bonds and
O
4 pi bonds
N
H
H
–
O
Cl
O
O
O
N
HN
O
+
H
41.
N
N
theobromine
Section 8.3 • Molecular Structures
257
Content Background
Exceptions 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.
■ Figure 8.14 The nitrate ion (NO 3 -)
exhibits resonance. a. These resonance
structures differ only in the location of the double bond. The locations of the nitrogen and oxygen atoms stay the same. b. The actual nitrate
ion is like an average of the three resonance
structures in a. The dotted lines indicate possible locations of the double bond.
Have 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.
VOCABULARY
SCIENCE USAGE V. COMMON USAGE
Resonance
Science usage: a phenomenon related
to the stability of a molecule; a large
vibration in a mechanical system
caused by a small periodic stimulus
The new molecule had several
resonance structures.
Common usage: a quality of
richness or variety
The sound of the orchestra had
resonance.
N –
■
S
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.
Incomplete octet
S
O
O
N
O
O
O
Using 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
(NO 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 NO 3 -, NO 2 -, SO 3 2-, and CO 3 2commonly 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 identical 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.
Extra Practice Page 980 and glencoe.com
O
O
46. O
S
O
2–
O
2–
O
S
O
Odd number of valence electrons First, a small group of molecules might have an odd number of valence electrons and be unable to
form an octet around each atom. For example, NO 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. ClO 2 and NO
are other examples of molecules with odd numbers of valence electrons.
Differentiated Instruction
O
S
N
Generally, 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.
O
O
O
O
Exceptions to the Octet Rule
O
O
O
O
43. NO 2 44. SO 2
45. O 3
46. Challenge Draw the Lewis resonance structure for the ion SO 3 2-.
O
45. O
O
-
Draw the Lewis resonance structures for the following molecules.
O
O
N
-
PRACTICE Problems
O
44. O
-
O
O
Resonance Structures
43. O N –
O
b
N
O
N
Assessment
PRACTICE Problems
-
O
O
Knowledge Phosgene is
a colorless, highly toxic gas used
in warfare. The formula for this
molecule is CCl 2O. Have students
draw two resonance structures for
this molecule. OL
a
O
2–
O
258 Chapter 8 • Covalent Bonding
6
Below Level
Using index cards that have
C08_S3_874637.indd
been cut in half, have students write the electron 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 CO 3 2- ion. Tell students to indicate 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 resonance structure. BL EL
2/2/07
10:00:20
—
H
—
→
—
—
—
H
N—H
H
—
—
H—B +
H
H
—
H
H—B—N—H
H
H
The nitrogen atom shares
both electrons to form the
coordinate covalent bond.
The boron atom has no electrons
to share, whereas the nitrogen
atom has two electrons to share.
■ Caption Question Fig. 8.16 Yes,
each atom involved in the bond has a complete octet.
■ Figure 8.16 In this reaction between
boron trihydride (BH 3) and ammonia (NH 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?
Visual Learning
Expanded 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 electrons. This expands its octet to sp 3d,
a configuration capable of holding
10 electrons. Using marshmallows
to represent atoms and toothpicks to
represent single covalent bonds, have
students make models of PF 5, SCl 6,
and ClF 5. AL
—
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 BH 3 is an example.
Boron, a group 3 nonmetal, forms three covalent bonds with other
nonmetallic atoms.
H—B—H
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.
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)
Expanded octets The third group of compounds that does not follow 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 formation in the molecule PCl 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 SF 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.
3 Assess
Check for Understanding
Write a molecular formula on the
board. Have students draw the Lewis
structure for the molecule. OL
Reading Check Summarize three reasons why some molecules do
not conform to the octet rule.
Reteach
Figure 8.17 Prior to the reaction of PCl 3 and Cl 2, every reactant atom follows
the octet rule. After the reaction, the product, PCl 5, has an expanded octet containing
ten electrons.
■
0 AMC08_S3_874637.indd
Cl
+
P
Cl
Cl
Cl
Cl
Cl
Cl
Cl
P
Review the steps used in drawing
Lewis structures using the molecules
N 2O and BeF 2. Have students determine whether these molecules are
exceptions to the octet rule or if they
exhibit resonance. OL
Cl
Cl
Expanded octet
Extension
Differentiated Instruction
7
Below
Level Use the following analogy
2/2/07
Have students research the compounds formed from the noble gas
xenon. Have them draw the Lewis
structure that represents each mole10:00:22 AM
cule formed. OL
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
Section 8.3 • Molecular Structures 259
EXAMPLE Problem 8.6
IN-CLASS Example
Question Selenium hexafluoride is
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 (XeF 4).
a toxic gas used as an electrical insulator. What is the Lewis structure for
this compound?
1
Answer Each fluorine has an octet and
2
Analyze the Problem
You 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.
Solve for the Unknown
First, find the total number of valence electrons.
the central Se atom has an expanded octet.
8 valence electrons
7 valence electrons
1 Xe atom × __ + 4 F atoms × __ = 36 valence electrons
1Xe atom
PRACTICE Problems
Determine the number of remaining pairs.
3 pairs
14 pairs - 4 F atoms × _ = 2 pairs unused
Add three pairs to each F atom to obtain an octet.
Determine how many pairs remain.
1F atom
F
F
F
3
F
Cl
Cl
Place the two remaining pairs on the central Xe atom.
F
Evaluate the Answer
This structure gives xenon 12 total electrons—an expanded octet—for a total of six bond positions.
Xenon compounds, such as the XeF 4 shown here, are toxic because they are highly reactive.
F
PRACTICE Problems
Cl
Extra Practice Page 980 and glencoe.com
Draw the expanded octet Lewis structure for each molecule.
P
49.
F
F
18 pairs available - 4 pairs used = 14 pairs available
F
Cl
Use four bonding pairs to bond the four F atoms to
the central Xe atom.
Xe
Xe
Cl
48.
Determine the total number of bonding pairs.
F
F
Have 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.
47.
1F atom
36 electrons
__
= 18 pairs
2 electrons/pair
47. ClF 3
48. PCl 5
49. Challenge Draw the Lewis structure for the molecule formed when six fluorine atoms
and one sulfur atom bond covalently.
Cl
F
F
F
Section 8.3
S
F
F
F
Assessment
Section Summary
50.
◗ Different models can be used to
represent molecules.
51. State the steps used to draw Lewis structures.
◗ Resonance occurs when more than
one valid Lewis structure exists for
the same molecule.
◗ Exceptions to the octet rule occur in
some molecules.
MAIN Idea
Describe the information contained in a structural formula.
52. Summarize exceptions to the octet rule by correctly pairing these molecules
and phrases: odd number of valence electrons, PCl 5, ClO 2, BH 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 2O) molecule.
55. Draw the Lewis structures for CN -, SiF 4, HCO 3 -, and, AsF 6 -.
Section 8.3
Assessment
C08_S3_874637.indd
8
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, PCl 5; odd number of valence electrons, ClO 2; less than an
octet, BH 3
53. No, a molecule or polyatomic ion must have both a single bond and a
double bond in order to display resonance.
260 Chapter 8 • Covalent Bonding
2/2/07
54. N N
O or N N
O
[ C N ]–
–
O
F
55.
F
F
F
Si
F
F
H
C
O
–
F
As
O
F
F
F
Self-Check Quiz glencoe.com
10:00:23
Section 8.4
Objectives
◗ Summarize the VSEPR
bonding theory.
◗ Predict the shape of, and the bond
angles in, a molecule.
◗ Define hybridization.
Review Vocabulary
atomic orbital: the region around
an atom’s nucleus that defines an
electron’s probable location
Section 8.4
1 Focus
Molecular Shapes
Focus Transparency
MAIN Idea The VSEPR model is used to determine molecular
shape.
Real-World Reading Link Have you ever rubbed two balloons in your hair
to 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 Model
The 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 arrangement that minimizes the repulsion of shared and unshared electron
pairs around the central atom.
New Vocabulary
VSEPR model
hybridization
Before presenting the lesson, project
Section Focus Transparency 30 and
have students answer the accompanying questions. BL EL
Bond angle To understand the VSEPR model better, imagine balloons 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.
MAIN Idea
Molecular 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
2 Teach
Quick Demo
Model Electron Sharing
Using small balloons to represent shared pairs of electrons
and larger balloons to represent
unshared pairs of electrons,
make models to represent HCl,
H 2O, NH 3, and CH 4. Use the following balloon combinations for
the models: HCl, one small and
three large; H 2O, two small and
two large; NH 3, three small and
one large; CH 4, four small. Tie
the balloons together and ask
students to notice the shape of
the resultant molecules. EL BL
Figure 8.18 Electron pairs in a molecule 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.
■
Linear
Trigonal planar
Tetrahedral
Virtual Lab
C08_S4_874637.indd
1
2/2/07
10:08:37 AM
CD-ROM Molecular Geometry
Section 8.4 • Molecular Shapes 261
Math in Chemistry
VOCABULARY
WORD ORIGIN
Trigonal planar
comes from the Latin words
trigonum, which means triangular,
and plan-, which means flat
Hybridization
■ Figure 8.19 A carbon atom’s 2s
and 2p electrons occupy the hybrid sp 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 minimizes repulsion between the hybrid orbitals in a CH 4 molecule.
Identify How many faces does the
tetrahedral shape formed by the sp 3
orbitals have?
H
sp3
H
Caption Question Fig. 8.19 four
H
→
→
→
→
→
■
H
sp3
→
activity above with a PCl 5 molecule
and the SF 6 molecule. Have students
determine the geometric shapes of
the molecules. trigonal bipyramidal, octahedral AL EL
p2
→
Complex Geometry Repeat the
sp3
sp3
CH4
Energy
Extension
C
sp3
→
Geometry 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 BeH 2,
and ask students to determine its
shape. linear Construct BH 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
CH 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 NH 3 and have
students assemble a geometric shape
to encompass the molecule. Tell students 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 2O 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
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 molecules. Some common molecular shapes are illustrated in Table 8.5.
s2
Carbon
Interactive Figure To see an animation
of molecular shapes, visit glencoe.com.
A hybrid occurs when two things are combined and the result has characteristics 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 bonding involved in the methane molecule (CH 4). The carbon atom has four
valence electrons with the electron configuration[He]2s 22p 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 undergo 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 sp 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, AlCl 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 sp 2 hybrid orbitals.
Lone pairs also occupy hybrid orbitals. Compare the hybrid orbitals
of BeCl 2 and H 2O in Table 8.6. Both compounds contain three atoms.
Why does an H 2O molecule contain sp 3 orbitals? There are two lone
pairs on the central oxygen atom in H 2O. 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 sp 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, CO 2, with two double bonds, forms sp hybrid orbitals.
Reading Check State the number of electrons that are available for
bonding in a hybrid sp 3 orbital.
Reading Check one
Differentiated Instruction
C08_S4_874637.indd
Interactive Figure Students can
interact with molecular shapes at
glencoe.com.
262 Chapter 8 • Covalent Bonding
2
Advanced 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
2/2/07
10:09:07
Table 8.6 Molecular Shapes
Molecule
Total
Pairs
Shared
Pairs
Lone
Pairs
Hybrid
Orbitals
2
2
0
sp
Interactive Table Explore molecular shapes at glencoe.com.
The BeCl 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.
180°
BeCI 2
Linear
3
AICI 3
3
0
Interactive Table Students can
interact with the molecular shapes at
glencoe.com.
Molecular Shape*
The three bonding electron pairs in
AlCl 3 have maximum separation in a
trigonal planar shape with 120° bond
angles.
120°
sp 2
Assessment
Performance Using the
VSEPR model, have students build
models that have these shapes: linear,
trigonal planer, trigonal bipyramidal,
and octahedral. OL
Trigonal planar
109.5°
4
CH 4
4
0
sp 3
Tetrahedral
4
PH 3
3
1
PH 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.
sp 3
107.3°
Trigonal pyramidal
4
H 2O
2
2
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.
sp 3
104.5°
Bent
90°
5
NbBr 5
5
0
sp 3d
120°
Trigonal bipyramidal
6
SF 6
6
0
90°
sp 3d 2
90°
When the central atom in a molecule
has four pairs of bonding electrons,
as CH 4 does, the shape is tetrahedral.
The bond angles are 109.5°.
The NbBr 5 molecule has five pairs
of bonding electrons. The trigonal
bipyramidal shape minimizes the
repulsion of these shared electron
pairs.
3 Assess
Check for Understanding
Have students use the Lewis structure for AsH 3 to predict the shape of
the molecule, the bond angle, and
the hybrid orbitals of arsenic. OL
Reteach
Have students use balls of modeling
clay to represent nuclei and nonvalence electrons, colored toothpicks to
represent bonded electrons, and
noncolored toothpicks to represent
lone pairs. Construct CH 4, and have
students determine the hybrid orbitals of carbon and the shape of the
molecule. sp 3, tetrahedral Repeat for
NH 3. sp 3, trigonal pyramidal Continue
with H 2O. sp 3, bent OL
Extension
As with NbBr 5, SF 6 has no unshared
electron pairs on the central atom.
However, six shared pairs arranged
about the central atom result in an
octahedral shape.
Octahedral
Ask students to explain how a carbon atom has four bonding sites
when its electron configuration of
[He]2s 22p 2 suggests it should only
have two bonding sites. The 2s 22p2
orbitals undergo hybridization, forming four
sp 3 bonding orbitals.
*Balls represent atoms, sticks represent bonds, and lobes represent lone pairs of electrons.
C08_S4_874637.indd
7 AM
S ti
Chemistry Project
3
Sulforaphane The compound
sulforaphane is an ingredient in broccoli
and has been discovered to be a potent
anticancer agent. Have students research
sulforaphane (C 5H 11OS) and draw models
to represent this molecule. AL
84
M l
l Sh
263
Assessment
Performance Have students
write
the
formula of a simple mole2/7/07
11:46:03 AM
cule on a sheet of paper. Have
Molecular Shape Have students predict
another student draw a model of
the molecular shape and bond angle of the
the molecule, predict the geometry,
common molecules CO 2, NO 2, CH 2O,
the hybrid, and the bond angle. OL
CCl 4, H 2S, and HF. OL
Chemistry Journal
Section 8.4 • Molecular Shapes 263
EXAMPLE Problem 8.7
IN-CLASS Example
Question Sulfur dichloride is a
cherry-red liquid at room temperature. What is the shape of sulfur
dichloride? Determine the bond
angle, and identify the type of
hybrid.
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 Problem
You are given the information that a phosphorus trihydride molecule has three, terminal
hydrogen atoms bonded to a central phosphorus atom.
2
Solve for the Unknown
Find the total number of valence electrons and the number of electron pairs.
Answer All atoms have an octet. The
shape is bent, with a bond angle of 104.5°,
and it is an sp 3 hybrid.
5 valence electrons
1 valence electron
1 P atom × __ + 3 H atoms × __ = 8 valence electrons
1P atom
H
PRACTICE Problems
56.
57.
58.
59.
60.
trigonal planar, 120°, sp 2
bent, 104.5°, sp 3
linear, 180°, sp
tetrahedral, 109°, sp 3
tetrahedral, 109°, sp 3
Determine the total number of bonding pairs.
2 electrons/pair
H—P—H
Have 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.
1F atom
8 electrons
__
= 4 pairs
→
Lewis structure
H
P
H
Draw 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.
H
Molecular shape
The molecular shape is trigonal pyramidal with a 107° bond angle and sp 3 hybrid orbitals.
3
Evaluate the Answer
All electron pairs are used and each atom has a stable electron configuration.
PRACTICE Problems
Extra Practice Page 980 and glencoe.com
Determine the molecular shape, bond angle, and hybrid orbitals for each molecule.
56. BF 3
58. BeF 2
57. OCl 2
59. CF 4
60. Challenge For a NH 4 + ion, identify its molecular shape, bond angle, and hybrid orbitals.
Section 8.4
Assessment
Section Summary
61.
◗ VSEPR model theory states that electron pairs repel each other and determine both the shape of and bond
angles in a molecule.
62. Define the term bond angle.
◗ Hybridization explains the observed
shapes of molecules by the presence
of equivalent hybrid orbitals.
MAIN Idea
Summarize the VSEPR bonding theory.
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 PF 3 and PF 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 CS 2, CH 2O, H 20Se, CCl 2F 2, and NCl 3.
Section 8.4
Assessment
C08_S4_874637.indd
4
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. sp 3 and 109°
264 Chapter 8 • Covalent Bonding
2/2/07
66. PF 3 is trigonal pyramidal with sp 3 hybrid orbitals. PF 5 is trigonal
bipyramidal with sp 3d hybrid orbitals. Shape is determined by the type of
hybrid orbital.
67. Refer to the Solutions Manual for Lewis structures.
CS 2: linear, 180°, sp
CH 2O: trigonal planar, 120°, sp 2
H 2Se: bent, 104.5°, sp 3
CCl 2F 2: tetrahedryl, 109°, sp 3
NCl 3: trigonal pyramidal, 107°, sp 3
Self-Check Quiz glencoe.com
10:09:41
Section 8.5
Section 8.5
1 Focus
Electronegativity
and Polarity
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.
Focus Transparency
MAIN 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.
Review Vocabulary
MAIN Idea
Electron Affinity, Electronegativity,
and Bond Character
electronegativity: the relative
ability of an atom to attract electrons
in a chemical bond
Electronegativity and Polarity
The 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 electronegativities—shown in Figure 8.20—allows chemists to evaluate the electron 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.
New Vocabulary
polar covalent bond
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, individual electronegativity values for helium, neon, and argon are not listed. However, larger noble gases, such as xenon, sometimes bond with
highly electronegative atoms, such as fluorine.
Figure 8.20 Electronegativity values
are derived by comparing an atom’s attraction for shared electrons to that of a fluorine’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.
Electronegativity Values for Selected Elements
1
H
2.20
5
6
7
8
9
C
N
O
F
2.04
2.55
3.04
3.44
3.98
13
14
15
16
17
Mg
Al
Si
P
S
Cl
1.31
1.61
1.90
2.19
2.58
3.16
4
Be
0.98
1.57
11
12
Na
0.93
19
Metal
Metalloid
Nonmetal
B
3
Li
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
Before presenting the lesson, project
Section Focus Transparency 31 and
have students answer the accompanying questions. BL EL
2 Teach
35
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
0.82
1.00
1.36
1.54
1.63
1.66
1.55
1.83
1.88
1.91
1.90
1.65
1.81
2.01
2.18
2.55
2.96
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
0.82
0.95
1.22
1.33
1.6
2.16
2.10
2.2
2.28
2.20
1.93
1.69
1.78
1.96
2.05
2.1
2.66
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
0.79
87
0.89
88
1.10
89
1.3
1.5
1.7
1.9
2.2
2.2
2.2
2.4
1.9
1.8
1.8
1.9
2.0
2.2
Fr
Ra
Ac
0.7
0.9
1.1
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 student. Ask the class what this represents when atoms share electrons.
unequal sharing of electrons Have students identify which atoms have a
greater tendency to gain electrons.
the ones with the greatest pull on the electrons Ask students what type of bond
is represented if the electron is completely pulled away from one atom.
an ionic bond OL
Quick Demo
Video Lab
C08_S5_874637.indd
1
DVD Formation of Compounds
2/2/07
10:16:23 AM
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 students 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 8.5 • Electronegativity and Polarity 265
Identify
Misconceptions
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 elements 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 electronegativity difference of zero—meaning that the electrons are equally shared
between the two atoms. This type of bond is considered nonpolar covalent, or a pure covalent bond. On the other hand, because different elements 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 electronegativity difference is greater than 1.70. However, this cutoff is sometimes inconsistent with experimental observations of two nonmetals
bonding together. Figure 8.21 summarizes the range of chemical bonding 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?
■ 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 Check
Determine the percent ionic
charcter of calcium oxide.
In-Text Questions
What percent ionic character is a bond
between two atoms that have an electronegativity difference of 2.00? About
■
Electronegativity and Bond Character
Percent ionic character
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 electronegativities and percent ionic
character in Figure 8.21, have students 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
EN Difference
and Bond Character
Table 8.7
Ionic
CaO
75
NaBr
50
25
0
MgO
HF
N2
0
AlP
Covalent
HCl
1.0
2.0
3.0
Electronegativity difference
60%
Where would LiBr be plotted on the
graph? just to the left of NaBr on the curve,
at an electronegativity difference of 1.98
Reading Check 0%
Graph Check approximately 74%
266
Chapter 8 • Covalent Bonding
Differentiated Instruction
C08_S5_874637.indd
2
Below 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
2/2/07
10:16:27
Electronegativity
Electronegativity
Difference
Cl = 3.16
H = 2.20
= 0.96
δ⁺
δ⁻
Quick Demo
Interactive Figure To see an animation
of bond types, visit glencoe.com.
Polar Molecules Place two
or three stirring bars with
non-stick coating on a flat surface. 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.
H — Cl
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 Bonds
As you just learned, polar covalent bonds form because not all atoms
that share electrons attract them equally. A polar covalent bond is similar 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 + represents 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 covalent 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 molecules, however, are attracted by an electric field. Because polar molecules 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.
Careers In chemistry
Assessment
Flavor Chemist A flavor chemist,
or flavorist, must know how chemicals 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 beverage industries. A certified flavorist
trains for five years in a flavor laboratory, passes an oral examination,
and then works under supervision
for another two years. For more
information on chemistry careers,
visit glencoe.com.
Performance 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 molecules SO 2, CS 2, CH 3Cl, CH 2O, HCl,
and H 2S can be used. OL
Polarity and molecular shape You can learn why some molecules are polar and some are not by comparing water (H 2O) and carbon
tetrachloride (CCl 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+ δH—O
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+ δC—Cl
According to their molecular formulas, both molecules have more than
one polar covalent bond. However, only the water molecule is polar.
7 AMC08_S5_874637.indd
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?
Chemistry Journal
Chemistry Project
3
Bond
Type Using the electronegativity
2/2/07
10:16:30 AM
Chromatography Have student groups
difference, have students determine if the
bonds in the following molecules are polar
covalent, nonpolar covalent, or ionic: H 2S,
CH 4, NH 3, KCl, N 2, CaO, CO 2, HCl. KCl and
CaO are ionic, N 2 is nonpolar covalent, and the rest are
polar covalent. OL
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 perform the lab for the class. OL
Section 8.5 • Electronegativity and Polarity 267
Quick Demo
■
Figure 8.23 A molecule’s shape deter-
a
mines its polarity.
δ⁺
H
Polarity Made Apparent
H2O
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.
■
Cl δ⁻
b
O
δ⁻
Cl
δ⁻
H δ⁺
The bent shape of
a water molecule
makes it polar.
CCl4
C δ⁺
c
Cl
δ⁻
Cl δ⁻
The symmetry of a CCl 4
molecule results in an equal
distribution of charge, and
the molecule is nonpolar.
Nδ
⁻
δ⁺
H
⁺
Hδ
H δ⁺
NH3
The asymmetric shape of an
ammonia molecule results in an
unequal charge distribution and
the molecule is polar.
The shape of a H 2O 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 negative end. Thus, it is polar.
A CCl 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, CCl 4 is a nonpolar molecule. Note that symmetric molecules are usually nonpolar, and molecules that are asymmetric are polar as long as the bond type is polar.
Is the molecule of ammonia (NH 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.
Caption Question Fig. 8.24
No, detergents must be used because the
polar water molecules cannot dissolve the
nonpolar oil molecules.
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 solubility. 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 substances usually do not mix.
Infer Will water alone
clean oil from a fabric?
■
Chemistry Journal
4
Like Dissolves
Like Add 20 mL of carbon
C08_S5_874637.indd
tetrachloride (CCl 4)and 20 mL of water to a
beaker. The liquids form layers and do not
mix. CCl 4 is more dense than water and
forms the lower layer. In a second beaker,
add 20 ml of CCl 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 CCl 4. The
potassium permanganate dissolved in the
268 Chapter 8 • Covalent Bonding
polar water. Water is polar and attracts the
ions present in the ionic compound potassium 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 permanganate ion. The central oxygen atom has a partial negative charge and attracts the potassium ion. The two
hydrogen atoms create a partial positive region that
attracts the permanganate ion. OL
2/2/07
10:16:32
Properties of Covalent Compounds
The ChemLab
located at the end of the chapter can
be used at this point in the lesson.
Table 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 discussed 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 molecules 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.
Data-Analysis Lab
About the Lab
• Phenol and benzoic acid have similar structures as shown below.
HO
HO
Phenol
2 AMC08_S5_874637.indd
Based on Real Data*
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 moving 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 solvent are retained longer by the moving phase
and tend to appear early on a chromatograph.
Several scientists performed HPLC using a methanol-water mixture as the extracting solvent to
separate a phenol-benzoic acid mixture. Their
results are shown in the graph to the right.
Think Critically
1. 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
Chromatograms of Phenol and Benzoic Acid in
Different Compositions of Mobile Phase Solvent
60
Intensity
How does the polarity of the mobile phase
affect chromatograms? Chromatography is a
75% methanol/
25% water mobile phase
Phenol
Phenol
40
Benzoic
acid
20
Benzoic
acid
Phenol
0
–20
240
480
50% methanol/
50% water mobile
phase
25% methanol/
75% water
mobile phase
720
960
1200
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 mobilephase 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.
Benzoic
acid
1440
Time (s)
*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 separating phenol from benzoic acid. Explain.
Cultural Diversity
5
Medical
Research The use of herbal remedies has been well documented in China
Benzoic acid
• For additional details about original 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 benzoic acid) become more equal as
the concentration of water in the
mobile base decreases.
Data Analysis lab
Interpret Data
O
2/2/07
10:16:39 AM
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.
Section 8.5 • Electronegativity and Polarity 269
3 Assess
Check for Understanding
Ask students to compare the characteristics of ionic solids to those of
covalent and network solids. OL
Reteach
Have students make a chart and discuss the properties of ionic, metallic,
covalent, and network solids. Point
out that a high melting point is a
result of a three-dimensional attraction of an atom or ion for electrons.
This occurs in network solids, metallic solids, and ionic solids. Covalent
solids are held together by intermolecular attractions and have lower
melting points. OL
■ 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.
Extension
Have students investigate intermolecular bonds. Students should use
drawings to represent each type of
van der Waals force and give an
example. Display students’ work. OL
Assessment
Knowledge Write several
chemical formulas on the board.
Have students predict the general
physical properties of each compound based on its bonding. OL
Section 8.5
Forces and properties The properties of covalent molecular compounds 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 (CO 2), and hydrogen sulfide
(H 2S) are examples of covalent gases. Because the hardness of a substance depends on the intermolecular forces between individual molecules, 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 attraction between particles. The structure of the lattice is affected by molecular shape and the type of intermolecular force. Most molecular
information has been determined by studying molecular solids.
Covalent Network Solids
There 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, nonconductors 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 tetrahedral arrangement, which is shown in Figure 8.25, forms a strongly
bonded crystal system that is extremely hard and has a very high melting point.
Assessment
Section Summary
68.
◗ The electronegativity difference determines the character of a bond
between atoms.
69. Describe a polar covalent bond.
◗ Polar bonds occur when electrons are
not shared equally forming a dipole.
71. List three properties of a covalent compound in the solid phase.
◗ 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.
MAIN Idea Summarize how electronegativity difference is related to
bond character.
70. Describe a polar molecule.
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: SCl 2, CS 2, and CF 4.
76. Determine whether a compound made of hydrogen and sulfur atoms is polar
or nonpolar.
77. Draw the Lewis structures for the molecules SF 4 and SF 6. Analyze each structure
to determine whether the molecule is polar or nonpolar.
Section 8.5
Assessment
C08_S5_874637.indd
6
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
270
Chapter 8 • Covalent Bonding
2/2/07
74. a. mostly covalent
b. mostly covalent
75. SCl 2, polar; CS 2, nonpiolar; CF 4, nonpolar,
76. polar; because it is asymmetric
77.
F
F
F
F
S
F S
F
F
F
F
F
polar
c. polar covalent
nonpolar
Self-Check Quiz glencoe.com
10:16:41
Sticky Feet: How Geckos Grip
For 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.
Purpose
Students will learn how van der
Waals forces provide the sticking
power for geckos’ toes.
Background
2
Spatulae Setae are
complex structures. The end
of each seta has microscopic
branches called spatulae
Gecko toes look like suction cups,
but they do not function that way.
Suction was ruled out as an explanation for how they stick to surfaces by
testing gecko feet in a vacuum. Other
ideas, such as static cling or waterbased attraction, have also been
ruled out by experiment.
1
Gecko toe The bottom
of a gecko’s toe is covered
with millions of tiny hairs,
called setae, arranged in
rows.
Teaching Strategies
The 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 containing 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 surface is important factor in determining the holding power.
• Number 4 explains how large surface 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 surface at the optimal angle. This
results in decreased van der Waals
forces and reduced gripping
strength.
3
Surface area Each seta
has a relatively enormous surface
area because of its vast number
of spatulae.
Surface
δ
-
4
Spatula
δ
δ
+
-
Attraction
Temporary dipole
δ
+
Temporary dipole
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.
5
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.
Chemistry
Invent 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.
Chemistry
C08_HIW_874637.indd
available at glencoe.com
RUBRIC
1
2/2/07
1:18:14 PM
Invent Some other applications that have been dis-
cussed include less painful adhesive bandages, super-grip
athletic equipment, and sticky tools for nanosurgery.
How It Works
271
MODEL MOLECULAR SHAPES
See the ChemLab worksheet in your
FAST FILE.
Background: Covalent bonding occurs when atoms
RUBRIC available at glencoe.com
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.
Preparation
Time Allotment one class period
Process Skills applying concepts,
comparing and contrasting, formulating models
Question: How do the Lewis structure and the positions
of valence electrons affect the shape of the covalent
compound?
Safety Precautions Approve lab
safety forms before work begins.
Materials
molecular model kit
Procedure
Safety Precautions
• 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.
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 2
molecules.
Analyze and Conclude
1. increasing flexibility: triple, double, single; increasing strength: single, double,
triple
2. The H 2O 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 electrons, causing the bent shape. The CO 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 SO 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 2O, PH 3, HCN, and CO. All others are
nonpolar.
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 molecules 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 2O), and
construct its molecule.
Students should assemble two models, each
with a central oxygen atom jointed to two terminal oxygen atoms. One terminal atom is
joined by a single bond, the other by a double
bond. The location of these two bonds determines the two resonance structures. The Lewis
structures should show that you can convert
272 ChemLab
Analyze and Conclude
1. Think Critically Based on the molecular models
you built and observed in this lab, rank single, double, and triple bonds in order of increasing flexibility
and increasing strength.
2. Observe and Infer Explain why H 2O and CO 2
molecules 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 structures, and explain why resonance occurs.
4. Recognize Cause and Effect Use the electronegativity 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.
INQUIRY EXTENSION
Model 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.
between the two resonance structures by swapping the
position of a lone pair and a covalent bond.
C08_CL_874637.indd
1
2/2/07
O
Inquiry Extension
10. Classify the shape of the H 2O molecule using information in Table 8.6.
11. Repeat Steps 9 and 10 for the PH 3, CF 4, CO 2, SO 3,
HCN, and CO molecules.
O
O
O
Customize this lab with
the LabManager™
CD-ROM.
O
O
10:20:10
Download quizzes, key
terms, and flash cards
from glencoe.com.
BIG Idea Covalent bonds form when atoms share electrons.
Section 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 8.2 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 8.3 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 8.4 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 8.5 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)
C08_CA_874637.indd
1
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.
What’s CHEMISTRY
Got To Do With It?
DVD The Plastics
Vocabulary Puzzlemaker
Study Guide
Using the Vocabulary
To 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 molecular 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
Use the ExamView®
Assessment Suite
CD-ROM to:
• create multiple versions of tests
• create modified tests with one
2/2/07
10:24:16 AM
mouse click
• edit existing questions and add
your own questions
• build tests aligned with state
standards using built-in state
curriculum tags
• change English tests to Spanish
with one mouse click
For additional practice with vocabulary, have
• track students’ progress using the
students access the Vocabulary Puzzlemaker
Teacher Management System
online at glencoe.com.
Vocabulary Puzzlemaker glencoe.com
Chapter 8 • Study Guide
273
78. What is the octet rule, and how is it used in covalent
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 SiO 2.
Mastering Problems
92. HClO 2: chlorous acid; H 3PO 4: phosphoric
acid; H 2Se: hydroselenic acid; HClO 3:
chloric acid
93. a. nitrogen trifluoride
b. nitrogen monoxide
c. sulfur trioxide
274
Chapter 8 • Assessment
Formula
79. Describe the formation of a covalent bond.
80. Describe the bonding in molecules.
HCIO 2
81. Describe the forces, both attractive and repulsive, that
H 3PO 4
occur as two atoms move closer together.
HCIO 3
bond in a molecule?
93. Name each molecule.
Mastering Problems
a. NF 3
b. NO
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.
94. Name each molecule.
84. Locate the sigma and pi bonds in each of the molecules
shown below.
a.
Name
H 2Se
82. How could you predict the presence of a sigma or pi
a. SeO 2
b. SeO 3
c. SO 3
d. SiF 4
c. N 2F 4
d. S 4N 4
95. Write the formula for each molecule.
a. sulfur difluoride
b. silicon tetrachloride
O
H—C—H
c. carbon tetrafluoride
d. sulfurous acid
96. Write the formula for each molecule.
a. silicon dioxide
b. bromous acid
b. H — C —
— C—H
c. chlorine trifluoride
d. hydrobromic acid
85. In the molecules CO, CO 2, and CH 2O, which C—O
bond is the shortest? Which C—O bond is the strongest?
86. Consider the carbon-nitrogen bonds shown below:
H
Section 8.3
Mastering Concepts
H
—
C ≡N- and
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
Table 8.8 Acid Names
bonding?
97. What must you know in order to draw the Lewis struc-
—
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 attractive force is maximized. If the atoms
move closer than the optimal distance,
repulsive force exceeds attractive force.
82. single covalent bond: sigma bond; double bond: a sigma bond and a pi bond;
triple bond: one sigma and two pi bonds
92. Complete Table 8.8.
Mastering Concepts
—
Mastering Concepts
Mastering Problems
—
Section 8.1
Section 8.1
—
Assessment
98. Doping Agent Material scientists are studying the prop-
H —C — N
H
ture for a molecule?
H
erties of polymer plastics doped with AsF 5. Explain why
the compound AsF 5 is an exception to the octet rule.
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. SO 2
b. SO 3 2—
c. SO 4 2—
99. Reducing Agent Boron trihydride (BH 3) is used as
reducing agent in organic chemistry. Explain why BH 3
often forms coordinate covalent bonds with other
molecules.
100. Antimony and chlorine can form antimony trichloride
or antimony pentachloride. Explain how these two elements can form two different compounds.
Section 8.2
Mastering Concepts
Mastering Problems
88. Explain how molecular compounds are named.
101. Draw three resonance structures for the polyatomic
89. When is a molecular compound named as an acid?
ion CO 3 2-.
90. Explain the difference between sulfur hexafluoride and
102. Draw the Lewis structures for these molecules, each
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.
d. silicon tetrafluoride
94. a. selenium dioxide
C08_CA_874637.indd
2
b. selenium trioxide
95. a. SF 2
b. SiCl 4
96. a. SiO 2
b. HBrO 2
c.
d.
c.
c.
dinitrogen tetrafluoride
tetrasulfur tetranitride
CF 4
d. H 2SO 3
ClF 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.
of which has a central atom that does not obey the
octet rule.
a. PCl 5
c. ClF 5
b. BF 3
d. BeH 2
99. BH 3 only has six electrons and does not have an
electron arrangement with a low amount of potential 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 SbCl 3.
Antimony can also expand its octet and bond with
all five valence electrons, forming SbCl 5.
Mastering Problems
101. Refer to the Solutions Manual for structures.
102. Refer to the Solutions Manual for structures.
Chapter Test glencoe.com
2/2/07
10:24:27
Mastering Problems
103. Draw two resonance structures for the polyatomic
ion HCO 2
111. Predict the molecular shape of each molecule.
a. COS
—.
104. Draw the Lewis structure for a molecule of each of these
compounds and ions.
a. H 2S
b. BF 4 —
shape and bond angle, and identify the hybrid orbitals.
Drawing the Lewis structure might help you.
a. SCl 2
c. HOF
b. NH 2Cl
d. BF 3
c. SO 2
d. SeCl 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
b. CF 2Cl 2
112. For each molecule listed below, predict its molecular
Section 8.4
Section 8.5
Mastering Concepts
Mastering Concepts
113. Describe electronegativity trends in the periodic table.
114. Explain the difference between nonpolar molecules and
Section 8.4
polar molecules.
115. Compare the location of bonding electrons in a polar
Mastering Concepts
covalent bond with those in a nonpolar covalent bond.
Explain your answer.
106. What is the basis of the VSEPR model?
107. What is the maximum number of hybrid orbitals a
carbon atom can form?
116. What is the difference between a covalent molecular
solid and a covalent network solid? Do their physical
properties differ? Explain your answer.
108. What is the molecular shape of each molecule? Estimate
the bond angle for each molecule, assuming that there
is not a lone pair.
Mastering Problems
a. A— B
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
b. A— B—A
—
c. A— B—A
A
A
negatively charged.
a. C—H
b. C—N
A—B—A
A
109. Parent Compound PCl 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 PCl 5.
on the central atom. You might find drawing the molecule’s Lewis structure helpful.
a. C—H
b. N—H
c. Si—H
Hybrid Orbital
Lewis Structure
c. C—Cl
d. C—Br
d. O—H
e. Cl—H
121. Refrigerant The refrigerant known as freon-14 is an
ozone-damaging compound with the formula CF 4. Why
is the CF 4 molecule nonpolar even though it contains
polar bonds?
Table 8.9 Structures
Formula
c. C—S
d. C—O
120. Rank the bonds according to increasing polarity.
Mastering Problems
110. Complete Table 8.9 by identifying the expected hybrid
110. XeF 4: sp 3d 2; TeF 4: sp 3d; KrF 2: sp 3d;
OF 2: sp 3
111. a. linear
b. tetrahedral
112. a. bent, 104.5°, sp 3
b. trigonal pyramidal, 107°, sp 3
c. bent, 104.5°, sp 3
d. trigonal planar, 120°, sp 2
119. Predict which bond is the most polar.
a. C—O
b. Si—O
122. Determine if these molecules and ion are polar. Explain
your answers.
a. H 3O +
b. PCl 5
XeF 4
TeF 4
c. H 2S
d. CF 4
123. Use Lewis structures to predict the molecular polarities
KrF 2
for sulfur difluoride, sulfur tetrafluoride, and sulfur
hexafluoride.
OF 2
Mastering Problems
117. The following are circled:
3
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. SF 2 and SF 4 are polar. SF 6 is nonpolar.
7 AMC08_CA_874637.indd
Chapter Test glencoe.com
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 molecule; five identical sp 3d orbitals
Mastering Problems
118. For each of the bonds listed, tell which atom is more
— —
d.
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
2/2/07
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 electrons 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
10:24:43 AM
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.
Chapter 8 • Assessment
275
Mixed Review
ClO
c. PCl 5
H 3AsO 4
d. H 2S
phosphorus trichloride
dichlorine heptoxide
tetraphosphorus hexoxide
nitrogen monoxide
F Se
F
b. O
Cl
–
P
O 3–
O
Cl
d.
Cl
P
O
127.
128.
129.
130.
Ge
F
F
The polar molecules are CH 3Cl, ClF, and
NCl 3 because each molecule is asymmetric and the charge is not distributed
evenly.
in order of least to greatest bond
character: e, d, a, c, b
F
Cl F
F
sp 3d
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 formula is As 2O 3. The name diarsenic trioxide
states that there are two atoms of
arsenic and three atoms of oxygen.
The correct formula is As 2O 3. Even
276
a.
b.
c.
d.
131. Organize Design a concept map that explains how
chlorine monoxide
arsenic acid
phosphorus pentachloride
hydrosulfuric acid
VSEPR model theory, hybridization theory, and molecular shape are related.
132. Compare and contrast the two covalent compounds
identified by the names arsenic(III) oxide and diarsenic
trioxide.
125. Name each molecule.
a.
b.
c.
d.
PCl 3
C l 2O 7
P 4O 6
NO
133. Make and Use Tables Complete Table 8.11, using what
you learned in Chapters 7 and 8.
Table 8.11 Properties and Bonding
a.
b.
c.
d.
e.
SeF 2
ClO 2 PO 3 3—
POCl 3
GeF 4
Solid
Chapter 8 • Assessment
Bond
Description
Characteristic
of Solid
Example
Ionic
Covalent
molecular
127. Determine which of the molecules are polar. Explain
Metallic
your answers.
a. CH 3Cl
b. ClF
c. NCl 3
d. BF 3
e. CS 2
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 present in a molecule of urea.
character.
a. C—O
b. Si—O
c. Ge—O
d. C—Cl
e. C—Br
F
F
124. Write the formula for each molecule.
128. Arrange the bonds in order of least to greatest polar
Cl
e.
Think Critically
126. Draw the Lewis structure for each molecule or ion.
O
c. O
Mixed Review
O
H
N
H
129. Rocket Fuel In the 1950s, the reaction of hydrazine
with chlorine trifluoride (ClF 3) was used as a rocket
fuel. Draw the Lewis structure for ClF 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.
—
124. a.
b.
125. a.
b.
c.
d.
126. a.
C
H
N
H
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, CH 3CN,
is shown below.
Table 8.10 Shared Pairs
Bond Type
Number of Shared
Electrons
Atoms that Form
the Bond
Single covalent
Double
covalent
Triple
covalent
though they are named differently, they both represent the same formula.
C08_CA_874637.indd
4 electrostatic attraction of a positive
133.
Column 2: the
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,
CO 2, Ag, diamond.
H
H C
H
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.
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 sp 3 hybrid because it has 4 bonding positions. The second carbon atom (bonded to
one C atom and one N atom) is an sp hybrid because
it has two bonding positions.
Chapter Test glencoe.com
2/2/07
10:24:57
Challenge Problem
137. a. H
Challenge Problem
137. Examine the bond-dissociation energies for the various
bonds listed in Table 8.12.
Table 8.12 Bond-Dissociation Energies
Bond
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 2H 2 and
HCOOH.
b. Determine the amount of energy needed to break
apart each of these molecules.
H
coolant, to learn its chemical formula. Draw its Lewis
structure and identify the sigma and pi bonds.
and write an essay about its chemical composition.
Explain how it removes oil and grease from fabrics.
Document-Based Questions
Luminol Crime-scene investigators often use the covalent
compound luminol to find blood evidence. The reaction
between luminol, certain chemicals, and hemoglobin, a protein in blood, produces light. Figure 8.26 shows a ball-andstick model of luminol.
b
c
Table 8.13 Mass v. Volume
■ Figure
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
144. Indicate the hybrid present on the atoms labeled A, B,
and C in Figure 8.26.
H
H
O
C
C
C
C
C
C
RUBRIC available at glencoe.com
141.
-
O
O
H
O
8.27
hemoglobin, it reacts to produce Na 2APA, water,
nitrogen, and light energy. Given the structural formula of the APA ion in Figure 8.27, write the chemical formula for the polyatomic APA ion.
143. C 8H 7O 2N 3
H
H
N
H
C C
C
C
H
C
C
O
H
H H
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.
145. When luminol comes in contact with the iron ion in
Data obtained from: Fleming, Declan., 2002. The Chemiluminescence
5
Luminol,
Exemplarchem, Royal Society of Chemistry.
O
H
-
APA ion
7 AMC08_CA_874637.indd
of
H
C
■ Figure
Document-Based Questions
H
C
NH2
H
140. Write the correct chemical name for each compound.
C
Chemistry
draw its Lewis structure.
(Chapter 7)
a. calcium carbonate
b. potassium chlorate
c. silver acetate
d. copper(II) sulfate
e. ammonium phosphate
C
Additional Assessment
8.26
143. Determine the molecular formula for luminol and
139. Write the correct chemical formula for each compound.
(Chapter 7)
a. NaI
b. Fe(NO 3) 3
c. Sr(OH) 2
d. CoCl 2
e. Mg(BrO 3) 2
H
138. 2.34 g/mL; density
139. a. CaCO 3
b. KClO 3
c. AgC 2H 3O 2
d. CuSO 4
e. (NH 4) 3PO 4
140. a. sodium iodide
b. iron(III) nitrat
c. strontium hydroxide
d. cobalt(II) chloride
e. magnesium bromate
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)
O
Cumulative Review
Data obtained from: Fleming, Declan., 2002. The Chemiluminescence of Luminol,
Exemplarchem, Royal Society of Chemistry.
Cumulative Review
C
b. C 2H 2 = 1671 kJ/mol;
HCOOH = 1986 kJ/mol
142. Detergents Choose a laundry detergent to research
a
Volume
H
O
Chemistry
141. Antifreeze Research ethylene glycol, an antifreeze-
Bond-Dissociation
Energy (kJ/mol)
C C
Additional Assessment
144. A, sp 2; B, sp 3; C, sp 2
145. C 8H 5NO 4 22/2/07
10:25:06 AM
O
C
C
N H
N
H
O
Chapter Test glencoe.com
Chapter 8 • Assessment 277
Cumulative
Standardized Test Practice
Multiple Choice
D
A
C
D
B
C
D
D
A
B
Multiple Choice
1. The common name of SiI 4 is tetraiodosilane. What is
its molecular compound name?
A. silane tetraiodide
B. silane tetraiodine
C. silicon iodide
D. silicon tetraiodide
6. The central selenium atom in selenium hexafluoride forms an expanded octet. How many electron
pairs surround the central Se atom?
A. 4
C. 6
B. 5
D. 7
Use the table below to answer Questions 7 and 8.
2. Which compound contains at least one pi bond?
A. CO 2
B. CHCl 3
C. AsI 3
D. BeF 2
Use the graph below to answer Questions 3 and 4.
Electronegativity
5
4
Bond Dissociation Energies at 298 K
Bond
kJ/mol
Bond
kJ/mol
Cl–Cl
242
N≡N
945
467
C–C
345
O–H
C–H
416
C–O
358
C–N
305
C=O
745
H–I
299
O=O
498
H–N
391
3
7. Which diatomic gas has the shortest bond between
its two atoms?
A. HI
C. Cl 2
D. N 2
B. O 2
2
1
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Atomic number
8. Approximately how much energy will it take to
break all the bonds present in the molecule below?
3. What is the electronegativity of the element with
atomic number 14?
A. 1.5
B. 1.9
C. 2.0
D. 2.2
4. An ionic bond would form between which pairs of
elements?
A. atomic number 3 and atomic number 4
B. atomic number 7 and atomic number 8
C. atomic number 4 and atomic number 18
D. atomic number 8 and atomic number 12
5. Which is the Lewis structure for silicon disulfide?
A. S Si S
B. S Si S
C. S Si S
D.
C08_STP_874637.indd
278
Chapter 8 • Assessment
S Si S
H
H
—
N
—
—
H
C
O
—
H C H C
H
O
H
A.
B.
C.
D.
—
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Standardized Test Practice
3024 kJ/mol
4318 kJ/mol
4621 kJ/mol
5011 kJ/mol
9. Which compound does NOT have a bent molecular
shape?
C. H 2O
A. BeH 2
B. H 2S
D. SeH 2
10. Which compound is nonpolar?
A. H 2S
C. SiH 3Cl
D. AsH 3
B. CCl 4
1
2/2/07
Standardized Test Practice glencoe.com
10:27:22
Short Answer
Short Answer
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 list of separation techniques below to answer
Questions 15 to 17.
A. filtration
B. distillation
C. crystallization
Use the atomic emission spectrum below to answer
Questions 12 and 13.
400
11. Nitric acid (HNO 3) and nitrous acid
(NHO 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
SAT Subject Test: Chemistry
500
600
Nanometers
D. chromatography
E. sublimation
Extended Response
15. Which technique separates components of a
mixture with different boiling points?
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%).
16. Which technique separates components of a
mixture based on the size of its particles?
700
17. Which technique is based on the stronger attraction
some components have for the stationary phase
compared to the mobile phase?
12. Estimate the wavelength of the photons being
emitted by this element.
Use the table below to answer Questions 18 to 19.
13. Find the frequency of the photons being emitted by
this element.
Electron-Dot Structures
1
2
13
14
15
16
17
18
SAT Subject Test: Chemistry
Li
Be
B
C
N
O
F
Ne
15.
16.
17.
18.
19.
Group
Extended Response
Diagram
Use the table below to answer Question 14.
Percent Abundance of Silicon Isotopes
Isotope
Mass
Percent Abundance
28Si
27.98 amu
92.21 %
29Si
28.98 amu
4.70 %
30Si
29.97 amu
3.09 %
18. Based on the Lewis structures shown, which
elements will combine in a 2:3 ratio?
A. lithium and carbon
B. beryllium and fluorine
C. beryllium and nitrogen
D. boron and oxygen
E. boron and carbon
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.
B
A
D
D
A
19. How
many electrons will beryllium have in its outer
14.X
energy level after it forms an ion to become chemically stable?
A. 0
D. 6
B. 2
E. 8
C. 4
NEED EXTRA HELP?
If You Missed
Question . . .
Review Section . . .
2 AMC08_STP_874637.indd
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
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
2
2/2/07
Standardized Test Practice glencoe.com
10:27:28 AM
Chapter 8 • Assessment 279