Covalent Bonding BIG Idea 8.1 The Covalent Bond 8.2 Naming Molecules

Covalent Bonding BIG Idea 8.1  The Covalent Bond 8.2  Naming Molecules
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.
• The chemical and physical properties
of water make it a unique liquid.
238
©BIOS Gilson FranÁois/Peter Arnold, Inc.
H—O
—
• 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
Lewis structure
Start-Up Activities
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).
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?
STEP 3 Staple along
the fold. Label the tabs as
follows: Bond Character,
Nonpolar Covalent,
Polar Covalent, and Ionic.
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.
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Covalent Bonds
Chapter 8 • Covalent Bonding 239
Matt Meadows
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
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.
■
240
Chapter 8 • Covalent Bonding
©Charles Krebs/Getty Images
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.
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.
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?
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.
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?
■
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
F
Fluorine
atom
F F
Bonding pair
of electrons
Lone
pairs
Fluorine
molecule
Section 8.1 • The Covalent Bond 241
7. Turn the temperature knob on the hot plate
Compare Melting Points
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
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.
Analysis
1. State Which solid melted first? Which solid
did not melt?
2. Apply Based on your observations and data,
describe the melting point of each solid as low,
medium, high, or very high.
3. Infer Which compounds are bonded with ionic
bonds? Which are bonded with covalent bonds?
4. Summarize how the type of bonding affects
the melting points of compounds.
Single Covalent Bonds
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.
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
242
Chapter 8 • Covalent Bonding
→
+
+
H
Hydrogen atom
→
HH
Hydrogen molecule
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.
Water
a
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.
O
→ H—O
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.
Two Single Covalent Bonds
Ammonia
b
+
N
→ H—N
—
3H
—
H
H
Three Single Covalent Bonds
Methane
c
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.
+
—
2H
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.
Section 8.1 • The Covalent Bond 243
EXAMPLE Problem 8.1
Lewis Structure of a Molecule The pattern on the glass
shown in Figure 8.6 was made by chemically etching its surface
with hydrogen fluoride (HF). Draw the Lewis structure for a
molecule of hydrogen fluoride.
1
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.
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.
■
2
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
+
Hydrogen
atom
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.
PRACTICE Problems
Extra Practice Page 979 and glencoe.com
Draw the Lewis structure for each molecule.
1.
2.
3.
6.
VOCABULARY
ACADEMIC VOCABULARY
Overlap
to occupy the same area in part
The two driveways overlap at the street
forming a common entrance.
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.
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.
244
Chapter 8 • Covalent Bonding
©Visual Arts Library (London)/Alamy
H
O
N
H
H
H
Water (H2O)
C
H
H
H
H
H
Methane (CH4)
Ammonia (NH3)
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.
■
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
&/,$!",%3
Incorporate information
from this section into
your Foldable.
Figure 8.8b.
The pi bond A multiple covalent bond consists of one sigma bond
and at least one pi bond. A pi bond, represented by the Greek letter
pi (π), forms when parallel orbitals overlap and share electrons. The
shared electron pair of a pi bond occupies the space above and below
the line that represents where the two atoms are joined together.
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.
■
a
O
+
O
→
O—O
b
N
+
N
→
N—
—N
Two shared pairs
of electrons
Three shared pairs
of electrons
Personal Tutor For an online tutorial on
multiple covalent bonds, visit glencoe.com.
Section 8.1 • The Covalent Bond 245
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.
The Strength of Covalent Bonds
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.
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
Bond length
246
Chapter 8 • Covalent Bonding
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
Table 8.2
Bond-Dissociation Energy
Molecule
Bond-Dissociation Energy
F2
159 kJ/mol
O2
498 kJ/mol
N2
945 kJ/mol
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
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.
■
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.
◗ 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.
9. Illustrate the formation of single, double, and triple covalent bonds using
Lewis structures.
10. Compare and contrast ionic bonds and covalent bonds.
11. Contrast sigma bonds and pi bonds.
12. Apply Create a graph using the bond-dissociation energy data in Table 8.2
and the bond-length data in Table 8.1. Describe the relationship between bond
length and bond-dissociation energy.
13. Predict the relative bond-dissociation energies needed to break the bonds in the
structures below.
a. H — C —
— C—H
Self-Check Quiz glencoe.com
b. H
— —
◗ Bond length is measured nucleus-tonucleus. Bond-dissociation energy is
needed to break a covalent bond.
Identify the type of atom that generally forms covalent bonds.
H
— —
◗ Sharing one pair, two pairs, and three
pairs of electrons forms single, double,
and triple covalent bonds, respectively.
-!). )DEA
C —C
H
H
Section 8.1 • The Covalent Bond 247
©Charles O’Rear/CORBIS
Section 8.2
Objectives
◗ Translate molecular formulas into
binary molecular compound names.
◗ Name acidic solutions.
Naming Molecules
MAIN Idea Specific rules are used when naming binary molecular
compounds, binary acids, and oxyacids.
oxyanion: a polyatomic ion in which
an element (usually a nonmetal) is
bonded to one or more oxygen atoms
Real-World Reading Link You probably know that your mother’s mother
is your grandmother, and that your grandmother’s sister is your great-aunt.
But what do you call your grandmother’s brother’s daughter? Naming molecules
requires a set of rules, just as naming family relationships requires rules.
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
Review Vocabulary
first word is dinitrogen and second word is monoxide.
There are exceptions to using the prefixes shown in Table 8.3. The
first element in the compound name never uses the mono- prefix. For
example, CO is carbon monoxide, not monocarbon monoxide. Also, if
using a prefix results in two consecutive vowels, one of the vowels is
usually dropped to avoid an awkward pronunciation. For example,
notice that the oxygen atom in CO is called monoxide, not monooxide.
Table 8.3
248
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-
Chapter 8 • Covalent Bonding
EXAMPLE Problem 8.2
Naming Binary Molecular Compounds Name the compound P 2O 5, which is
used as a drying and dehydrating agent.
1
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.
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
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?
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).
Reading Check Apply What are the scientific names for ammonia,
hydrazine, and nitric oxide?
Section 8.2 • Naming Molecules 249
Naming Acids
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.
Naming binary acids A binary acid contains hydrogen and
one other element. The naming of the common binary acid known as
hydrochloric acid is explained in the following rules.
1. The first word has the prefix hydro- to name the hydrogen part of
the compound. The rest of the first word consists of a form of the
root of the second element plus the suffix -ic. HCl (hydrogen and
chlorine) becomes hydrochloric.
2. The second word is always acid. Thus, HCl in a water solution is
called hydrochloric acid.
Although the term binary indicates exactly two elements, a few acids
that contain more than two elements are named according to the rules
for naming binary acids. If no oxygen is present in the formula for the
acidic compound, the acid is named in the same way as a binary acid,
except that the root of the second part of the name is the root of the
polyatomic ion that the acid contains. For example, HCN, which is
composed of hydrogen and the cyanide ion, is called hydrocyanic acid
in solution.
Naming oxyacids An acid that contains both a hydrogen atom and
an oxyanion is referred to as an oxyacid. Recall from Chapter 7 that an
oxyanion is a polyatomic ion containing one or more oxygen atoms.
The following rules explain the naming of nitric acid (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
250
Chapter 8 • Covalent Bonding
Oxyanion
Acid Suffix
Acid Name
HClO 3
chlorate
-ic
chloric acid
HClO 2
chlorite
-ous
chlorous acid
HNO 3
nitrate
-ic
nitric acid
HNO 2
nitrite
-ous
nitrous acid
Table 8.5
Interactive Table Explore
naming covalent compounds
glencoe.com.
Formulas and Names of
Some Covalent Compounds
Common Name
Formula
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
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
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
Extra Practice Page 979 and glencoe.com
Give the formula for each compound.
25.
26.
27.
28.
29.
30.
silver chloride
dihydrogen oxide
chlorine trifluoride
diphosphorus trioxide
strontium acetate
Challenge What is the formula for carbonic acid?
Section 8.2 • Naming Molecules 251
Look at the
formula of
the molecule.
Examples:
HBr, H2SO3,
and N02
Does the compound
form an acidic
aqueous solution?
No
(NO2)
Yes
(H2SO3 and HBr)
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?
Yes
(H2SO3)
No
(HBr)
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
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
in solution are acids. Binary acids contain
hydrogen and one other element.
Oxyacids contain hydrogen and an
oxyanion.
H+
252 Chapter 8 • Covalent Bonding
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
Self-Check Quiz glencoe.com
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
Molecular Structures
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.
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
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.
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.
To do this, divide the number of electrons available for bonding by
two.
4. Place the bonding pairs.
Place one bonding pair (single bond) between the central atom and
each of the terminal atoms.
5. Determine the number of bonding pairs remaining.
To do this, subtract the number of pairs used in Step 4 from the total
number of bonding pairs determined in Step 3. These remaining
pairs include lone pairs as well as pairs used in double and triple
bonds. Place lone pairs around each terminal atom (except H atoms)
bonded to the central atom to satisfy the octet rule. Any remaining
pairs will be assigned to the central atom.
6. Determine whether the central atom satisfies the octet rule.
Is the central atom surrounded by four electron pairs? If not, it does
not satisfy the octet rule. To satisfy the octet rule, convert one or two
of the lone pairs on the terminal atoms into a double bond or a triple
bond between the terminal atom and the central atom. These pairs
are still associated with the terminal atom as well as with the central
atom. Remember that carbon, nitrogen, oxygen, and sulfur often
form double and triple bonds.
Apply the Strategy
Study Example Problems 8.3 through 8.5 to see how the steps in
the Problem-Solving Strategy are applied.
254
Chapter 8 • Covalent Bonding
EXAMPLE Problem 8.3
Lewis Structure for a Covalent Compound with
Single Bonds Ammonia is a raw material used in the
manufacture of many materials, including fertilizers,
cleaning products, and explosives. Draw the Lewis
structure for ammonia (NH 3).
1
2
Analyze the Problem
Math Handbook
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
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
Determine the total number of
bonding pairs. To do this, divide the
number of available electrons by two.
Four pairs of electrons are available for bonding.
—
H—N—H
H
Place a bonding pair (a single bond)
between the central nitrogen atom
and each terminal hydrogen atom.
Determine the number of bonding pairs remaining.
4 pairs total - 3 pairs used
= 1 pair available
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
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.
Section 8.3 • Molecular Structures 255
EXAMPLE Problem 8.4
Lewis Structure for a Covalent Compound with Multiple
Bonds Carbon dioxide is a product of all cellular respiration.
Draw the Lewis structure for carbon dioxide (CO 2).
1
Analyze the Problem
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
Solve for the Unknown
Find the total number of valence electrons available for bonding.
4 valence electrons
6 valence electrons
1 C atom × __ + 2 O atoms × __
1C atom
1O atom
= 16 valence electrons
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.
Determine the total number of
bonding pairs by dividing the number
of available electrons by two.
Eight pairs of electrons are available for bonding.
O—C—O
Place a bonding pair (a single bond)
between the central carbon atom and
each terminal oxygen atom.
Determine the number of bonding pairs remaining. Subtract the
number of pairs used in these bonds from the total number of
pairs of electrons available.
8 pairs total - 2 pairs used
= 6 pairs available
O—C—O
Subtract the number of pairs used in
these bonds from the total number of
pairs of electrons available.
Add three lone pairs to each terminal
oxygen atom.
Determine the number of bonding pairs remaining.
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.
—C —
—O
O—
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.
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.
256 Chapter 8 • Covalent Bonding
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.
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
Find the total number of valence electrons available for bonding.
5 valence electrons
6 valence electrons
1 P atom × __ + 4 O atoms × __
P atom
Real-World Chemistry
Phosphorus and Nitrogen
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.
Draw single bonds from each
terminal oxygen atom to the
central phosphorus atom.
—
O—P—O
O
16 pairs total - 4 pairs used
= 12 pairs available
Subtract the number of pairs
used from the total number of
pairs of electrons available.
Add three lone pairs to each terminal oxygen atom.
12 pairs available - 12 lone pairs used = 0
—
O
3-
—
O—P—O
O
3
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
All of the atoms have an octet, and the group has a net charge of 3-.
PRACTICE Problems
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.
Algal blooms Phosphorus and
nitrogen are nutrients required for
algae growth. Both can enter lakes
and streams from discharges of
sewage and industrial waste, and in
fertilizer runoff. If these substances
build up in a body of water, a rapid
growth of algae, known as an algal
bloom, can occur, forming a thick
layer of green slime over the water’s
surface. When the algae use up the
supply of nutrients, they die and
decompose. This process reduces the
amount of dissolved oxygen in the
water that is available to other
aquatic organisms.
Section 8.3 • Molecular Structures 257
©Suzanne Long/Alamy
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.
■
a
-
O
b
N
O
O
-
O
N
-
O
N
O
-
O
O
O
N
O
O
O
Resonance Structures
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.
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.
PRACTICE Problems
Extra Practice Page 980 and glencoe.com
Draw the Lewis resonance structures for the following molecules.
43. NO 2 44. SO 2
45. O 3
46. Challenge Draw the Lewis resonance structure for the ion SO 3 2-.
Exceptions to the Octet Rule
■
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
O
N
O
258 Chapter 8 • Covalent Bonding
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.
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.
—
—
—
—
—
H
→
N—H
H
H
H—B—N—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.
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?
■
H
—
—
H
H—B +
H
H
—
H
—
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.
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.
Reading Check Summarize three reasons why some molecules do
not conform to the octet rule.
■ 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.
Cl
+
P
Cl
Cl
Cl
Cl
Cl
Cl
Cl
P
Cl
Cl
Expanded octet
Section 8.3 • Molecular Structures 259
EXAMPLE Problem 8.6
Lewis Structure: Exception to the Octet Rule Xenon is a noble gas that will
form a few compounds with nonmetals that strongly attract electrons. Draw the
correct Lewis structure for xenon tetrafluoride (XeF 4).
1
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.
2
Solve for the Unknown
First, find the total number of valence electrons.
8 valence electrons
7 valence electrons
1 Xe atom × __ + 4 F atoms × __ = 36 valence electrons
1Xe atom
1F atom
36 electrons
__
= 18 pairs
2 electrons/pair
Determine the total number of bonding pairs.
F
F
Use four bonding pairs to bond the four F atoms to
the central Xe atom.
Xe
F
F
18 pairs available - 4 pairs used = 14 pairs available
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
Xe
F
3
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.
PRACTICE Problems
Extra Practice Page 980 and glencoe.com
Draw the expanded octet Lewis structure for each molecule.
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.
Section 8.3
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 -.
260 Chapter 8 • Covalent Bonding
Self-Check Quiz glencoe.com
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
New Vocabulary
VSEPR model
hybridization
Molecular Shapes
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.
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.
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
Section 8.4 • Molecular Shapes 261
Matt Meadows
VOCABULARY
WORD ORIGIN
Trigonal planar
comes from the Latin words
trigonum, which means triangular,
and plan-, which means flat
Connection
Biology The shape of food molecules is important to
our sense of taste. The surface of your tongue is covered with taste buds,
each of which contains from 50 to 100 taste receptor cells. Taste receptor
cells can detect five distinct tastes—sweet, bitter, salty, sour, and umami
(the taste of MSG, monosodium glutamate)—but each receptor cell
responds best to only one taste.
The shapes of food molecules are determined by their chemical
structures. When a molecule enters a taste bud, it must have the correct
shape for the nerve in each receptor cell to respond and send a message
to the brain. The brain then interprets the message as a certain taste.
When such molecules bind to sweet receptors, they are sensed as sweet.
The greater the number of food molecules that fit a sweet receptor cell,
the sweeter the food tastes. Sugars and artificial sweeteners are not the
only sweet molecules. Some proteins found in fruits are also sweet molecules. Some common molecular shapes are illustrated in Table 8.5.
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
C
sp3
H
sp3
H
→
→
→
→
p2
→
→
→
CH4
sp3
→
Energy
H
sp3
s2
Carbon
Interactive Figure To see an animation
of molecular shapes, visit glencoe.com.
262
Chapter 8 • Covalent Bonding
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.
Table 8.6 Molecular Shapes
Molecule
Total
Pairs
Shared
Pairs
Lone
Pairs
Hybrid
Orbitals
Interactive Table Explore molecular shapes at glencoe.com.
Molecular Shape*
180°
BeCI 2
2
2
0
sp
Linear
AICI 3
3
3
0
120°
sp 2
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.
The three bonding electron pairs in
AlCl 3 have maximum separation in a
trigonal planar shape with 120° bond
angles.
Trigonal planar
109.5°
CH 4
4
4
0
sp 3
Tetrahedral
PH 3
4
3
1
sp 3
107.3°
Trigonal pyramidal
H 2O
4
2
2
90°
5
5
0
sp 3d
120°
Trigonal bipyramidal
SF 6
6
6
0
90°
sp 3d 2
90°
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.
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
NbBr 5
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.
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
*Balls represent atoms, sticks represent bonds, and lobes represent lone pairs of electrons.
Section 8.4 • Molecular Shapes 263
EXAMPLE Problem 8.7
Find the Shape of a Molecule Phosphorus trihydride, a colorless gas, is produced
when organic materials, such as fish flesh, rot. What is the shape of a phosphorus
trihydride molecule? Identify the bond angle size and hybrid orbitals.
1
Analyze the 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.
5 valence electrons
1 valence electron
1 P atom × __ + 3 H atoms × __ = 8 valence electrons
1P atom
1F atom
8 electrons
__
= 4 pairs
Determine the total number of bonding pairs.
2 electrons/pair
H—P—H
H
→
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.
264
Chapter 8 • Covalent Bonding
Self-Check Quiz glencoe.com
Section 8.5
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.
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
Electron Affinity, Electronegativity,
and Bond Character
electronegativity: the relative
ability of an atom to attract electrons
in a chemical bond
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
Metal
Metalloid
Nonmetal
5
6
7
8
9
B
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
3
4
Li
Be
0.98
1.57
11
12
Na
0.93
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
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
Section 8.5 • Electronegativity and Polarity 265
EN Difference
and Bond Character
Table 8.7
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.
■
charcter of calcium oxide.
Percent ionic character
Graph Check
Determine the percent ionic
Electronegativity and Bond Character
Ionic
CaO
75
NaBr
50
HF
25
0
MgO
N2
0
AlP
Covalent
HCl
1.0
2.0
Electronegativity difference
266
Chapter 8 • Covalent Bonding
3.0
Electronegativity
Electronegativity
Difference
Cl = 3.16
H = 2.20
= 0.96
δ⁺
δ⁻
Interactive Figure To see an animation
of bond types, visit glencoe.com.
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.
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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
δ+ δC—Cl
According to their molecular formulas, both molecules have more than
one polar covalent bond. However, only the water molecule is polar.
Reading Check Apply Why does a statically charged balloon cause a
slow stream of water from a faucet to bend when placed next to it?
Section 8.5 • Electronegativity and Polarity 267
■
Figure 8.23 A molecule’s shape deter-
mines its polarity.
a
Cl δ⁻
b
δ⁺
H
H2O
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.
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?
268
Chapter 8 • Covalent Bonding
©Tony Craddock/Photo Researchers, Inc.
Properties of Covalent Compounds
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
Based on Real Data*
Interpret 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.
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
25% methanol/
75% water
mobile phase
0
–20
240
480
50% methanol/
50% water mobile
phase
720
960
1200
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.
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.
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.
Section 8.5 • Electronegativity and Polarity 269
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.
■
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.
270
Chapter 8 • Covalent Bonding
©Scientifica/Visuals Unlimited
Self-Check Quiz glencoe.com
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.
2
Spatulae Setae are
complex structures. The end
of each seta has microscopic
branches called spatulae
1
Gecko toe The bottom
of a gecko’s toe is covered
with millions of tiny hairs,
called setae, arranged in
rows.
3
Surface area Each seta
has a relatively enormous surface
area because of its vast number
of spatulae.
Surface
δ-
Spatula
δ+
Attraction
Temporary dipole
δ-
4
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.
How It Works 271
(t)©Peter Weber/Getty Images, (tcl)©Perennou Nuridsany/Photo Researchers, Inc, (cr)©Susumu Nishinaga/Photo Researchers, Inc, (b)(bcl)©Prof. Kellar Autumn, Lewis & Clark College
MODEL MOLECULAR SHAPES
Background: Covalent bonding occurs when atoms
share valence electrons. In the Valence Shell Electron
Pair Repulsion (VSEPR) theory, the way in which
valence electrons of bonding atoms are positioned
is the basis for predicting a molecule’s shape. This
method of visualizing shape is also based on the
molecule’s Lewis structure.
Question: How do the Lewis structure and the positions
of valence electrons affect the shape of the covalent
compound?
Materials
molecular model kit
Safety Precautions
Procedure
1. Read and complete the lab safety form.
2. Create a table to record your data.
3. Note and record the color used to represent each of
the following atoms in the molecular model kit:
hydrogen (H), oxygen (O), phosphorus (P), carbon
(C), fluorine (F), sulfur (S), and nitrogen (N).
4. Draw the Lewis structures of the H 2, O 2, and N 2
molecules.
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.
272 Chapter 8 • Covalent Bonding
Matt Meadows
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.
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.
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)
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.
Vocabulary PuzzleMaker glencoe.com
Chapter 8 • Study Guide 273
Mastering Problems
Section 8.1
92. Complete Table 8.8.
Mastering Concepts
78. What is the octet rule, and how is it used in covalent
bonding?
79. Describe the formation of a covalent bond.
80. Describe the bonding in molecules.
81. Describe the forces, both attractive and repulsive, that
occur as two atoms move closer together.
82. How could you predict the presence of a sigma or pi
bond in a molecule?
Formula
Name
HCIO 2
H 3PO 4
H 2Se
HCIO 3
93. Name each molecule.
Mastering Problems
83. Give the number of valence electrons in N, As, Br, and
Se. Predict the number of covalent bonds needed for
each of these elements to satisfy the octet rule.
84. Locate the sigma and pi bonds in each of the molecules
shown below.
a. NF 3
b. NO
c. SO 3
d. SiF 4
94. Name each molecule.
a. SeO 2
b. SeO 3
c. N 2F 4
d. S 4N 4
95. Write the formula for each molecule.
O
a. sulfur difluoride
b. silicon tetrachloride
—
a.
Table 8.8 Acid Names
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
H
—
C ≡N- and
H
—
—
H —C — N
H
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—
Section 8.3
Mastering Concepts
97. What must you know in order to draw the Lewis struc-
ture for a molecule?
98. Doping Agent Material scientists are studying the prop-
erties of polymer plastics doped with AsF 5. Explain why
the compound AsF 5 is an exception to the octet rule.
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
Section 8.2
Mastering Concepts
88. Explain how molecular compounds are named.
89. When is a molecular compound named as an acid?
90. Explain the difference between sulfur hexafluoride and
disulfur tetrafluoride.
91. Watches The quartz crystals used in watches are made
of silicon dioxide. Explain how you use the name to
determine the formula for silicon dioxide.
274
Chapter 8 • Covalent Bonding
or antimony pentachloride. Explain how these two elements can form two different compounds.
Mastering Problems
101. Draw three resonance structures for the polyatomic
ion CO 3 2-.
102. Draw the Lewis structures for these molecules, each
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
Chapter Test glencoe.com
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 —
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
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
Section 8.5
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
106. What is the basis of the VSEPR model?
107. What is the maximum number of hybrid orbitals a
carbon atom can form?
108. What is the molecular shape of each molecule? Estimate
covalent bond with those in a nonpolar covalent bond.
Explain your answer.
116. What is the difference between a covalent molecular
solid and a covalent network solid? Do their physical
properties differ? Explain your answer.
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
118. For each of the bonds listed, tell which atom is more
negatively charged.
a. C—H
b. C—N
— —
d.
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.
Mastering Problems
110. Complete Table 8.9 by identifying the expected hybrid
on the central atom. You might find drawing the molecule’s Lewis structure helpful.
Table 8.9 Structures
Formula
Hybrid Orbital
Lewis Structure
119. Predict which bond is the most polar.
a. C—O
b. Si—O
TeF 4
c. C—Cl
d. C—Br
120. Rank the bonds according to increasing polarity.
a. C—H
b. N—H
c. Si—H
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?
122. Determine if these molecules and ion are polar. Explain
your answers.
a. H 3O +
b. PCl 5
XeF 4
c. C—S
d. C—O
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
Chapter Test glencoe.com
Chapter 8 • Assessment 275
Mixed Review
Think Critically
124. Write the formula for each molecule.
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
Cl 2O 7
P 4O 6
NO
133. Make and Use Tables Complete Table 8.11, using what
you learned in Chapters 7 and 8.
126. Draw the Lewis structure for each molecule or ion.
a.
b.
c.
d.
e.
131. Organize Design a concept map that explains how
SeF 2
ClO 2 PO 3 3—
POCl 3
GeF 4
Table 8.11 Properties and Bonding
Solid
Bond
Description
Characteristic
of Solid
Example
Ionic
127. Determine which of the molecules are polar. Explain
your answers.
a. CH 3Cl
b. ClF
c. NCl 3
d. BF 3
e. CS 2
Covalent
molecular
Metallic
Covalent
network
134. Apply Urea, whose structure is shown below, is a com-
128. Arrange the bonds in order of least to greatest polar
character.
a. C—O
b. Si—O
c. Ge—O
d. C—Cl
e. C—Br
pound used in manufacturing plastics and fertilizers.
Identify the sigma bond, pi bonds, and lone pairs present in a molecule of urea.
O
H
N
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.
H
—
a.
b.
c.
d.
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
Single covalent
Double
covalent
Triple
covalent
276 Chapter 8 • Covalent Bonding
Atoms that Form
the Bond
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.
Chapter Test glencoe.com
Challenge Problem
137. Examine the bond-dissociation energies for the various
bonds listed in Table 8.12.
Additional Assessment
Chemistry
Table 8.12 Bond-Dissociation Energies
141. Antifreeze Research ethylene glycol, an antifreeze-
Bond
Bond-Dissociation
Energy (kJ/mol)
Bond
Bond-Dissociation
Energy (kJ/mol)
C—C
348
O—H
467
C=C
614
C—N
305
C≡C
839
O=O
498
N—N
163
C—H
416
N=N
418
C—O
358
N≡N
945
C=O
745
a. Draw the correct Lewis structures for C 2H 2 and
HCOOH.
b. Determine the amount of energy needed to break
apart each of these molecules.
coolant, to learn its chemical formula. Draw its Lewis
structure and identify the sigma and pi bonds.
142. Detergents Choose a laundry detergent to research
and write an essay about its chemical composition.
Explain how it removes oil and grease from fabrics.
Document-Based 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.
Data obtained from: Fleming, Declan., 2002. The Chemiluminescence of Luminol,
Exemplarchem, Royal Society of Chemistry.
a
Cumulative Review
b
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)
c
Table 8.13 Mass v. Volume
Volume
Mass
4.1 mL
9.36 g
6.0 mL
14.04 g
8.0 mL
18.72 g
10.0 mL
23.40 g
139. Write the correct chemical formula for each compound.
(Chapter 7)
a. calcium carbonate
b. potassium chlorate
c. silver acetate
d. copper(II) sulfate
e. ammonium phosphate
140. Write the correct chemical name for each compound.
(Chapter 7)
a. NaI
b. Fe(NO 3) 3
c. Sr(OH) 2
d. CoCl 2
e. Mg(BrO 3) 2
■ Figure
8.26
143. Determine the molecular formula for luminol and
draw its Lewis structure.
144. Indicate the hybrid present on the atoms labeled A, B,
and C in Figure 8.26.
H
H
C
C
NH2
O
C
C
C
H
C
-
O
-
C
C
O
O
APA ion
■ Figure
8.27
145. When luminol comes in contact with the iron ion in
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.
Chapter Test glencoe.com
Chapter 8 • Assessment 277
Cumulative
Standardized Test Practice
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.
4
Bond
kJ/mol
Bond
kJ/mol
Cl–Cl
242
N≡N
945
C–C
345
O–H
467
C–H
416
C–O
358
C–N
305
C=O
745
H–I
299
O=O
498
H–N
391
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.
S Si S
278 Chapter 8 • Assessment
H
H
—
N
—
—
H
C
O
—
H C H C
H
H
O
A.
B.
C.
D.
—
Electronegativity
5
Bond Dissociation Energies at 298 K
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
D. SeH 2
B. H 2S
10. Which compound is nonpolar?
C. SiH 3Cl
A. H 2S
D. AsH 3
B. CCl 4
Standardized Test Practice glencoe.com
Short Answer
SAT Subject Test: Chemistry
11. Oxyacids contain hydrogen and an oxyanion. There
are two different oxyacids that contain hydrogen,
nitrogen, and oxygen. Identify these two oxyacids.
How can they be distinguished on the basis of their
names and formulas?
Use the 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
500
600
Nanometers
D. chromatography
E. sublimation
15. Which technique separates components of a
mixture with different boiling points?
16. Which technique separates components of a
mixture based on the size of its particles?
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
Extended Response
1
2
13
14
15
16
17
18
Li
Be
B
C
N
O
F
Ne
Group
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.
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 . . .
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
Standardized Test Practice glencoe.com
Chapter 8 • Assessment 279
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