7
CHAPTER 1
CHEMICAL BONDING
S
tructure* is the key to everything in chemistry. The properties of a substance
depend on the atoms it contains and the way the atoms are connected. What is less
obvious, but very powerful, is the idea that someone who is trained in chemistry
can look at a structural formula of a substance and tell you a lot about its properties.
This chapter begins your training toward understanding the relationship between struc-
ture and properties in organic compounds. It reviews some fundamental principles of
molecular structure and chemical bonding. By applying these principles you will learn
to recognize the structural patterns that are more stable than others and develop skills in
communicating chemical information by way of structural formulas that will be used
throughout your study of organic chemistry.
1.1 ATOMS, ELECTRONS, AND ORBITALS
Before discussing bonding principles, let’s first review some fundamental relationships
between atoms and electrons. Each element is characterized by a unique atomic number
Z, which is equal to the number of protons in its nucleus. A neutral atom has equal num-
bers of protons, which are positively charged, and electrons, which are negatively charged.
Electrons were believed to be particles from the time of their discovery in 1897
until 1924, when the French physicist Louis de Broglie suggested that they have wave-
like properties as well. Two years later Erwin Schr?dinger took the next step and cal-
culated the energy of an electron in a hydrogen atom by using equations that treated the
electron as if it were a wave. Instead of a single energy, Schr?dinger obtained a series
of energy levels, each of which corresponded to a different mathematical description of
the electron wave. These mathematical descriptions are called wave functions and are
symbolized by the Greek letter H9274 (psi).
*A glossary of important terms may be found immediately before the index at the back of the book.
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According to the Heisenberg uncertainty principle, we can’t tell exactly where an
electron is, but we can tell where it is most likely to be. The probability of finding an
electron at a particular spot relative to an atom’s nucleus is given by the square of the
wave function (H9274
2
) at that point. Figure 1.1 illustrates the probability of finding an elec-
tron at various points in the lowest energy (most stable) state of a hydrogen atom. The
darker the color in a region, the higher the probability. The probability of finding an elec-
tron at a particular point is greatest near the nucleus, and decreases with increasing dis-
tance from the nucleus but never becomes zero. We commonly describe Figure 1.1 as
an “electron cloud” to call attention to the spread-out nature of the electron probability.
Be careful, though. The “electron cloud” of a hydrogen atom, although drawn as a col-
lection of many dots, represents only one electron.
Wave functions are also called orbitals. For convenience, chemists use the term
“orbital” in several different ways. A drawing such as Figure 1.1 is often said to repre-
sent an orbital. We will see other kinds of drawings in this chapter, use the word “orbital”
to describe them too, and accept some imprecision in language as the price to be paid
for simplicity of expression.
Orbitals are described by specifying their size, shape, and directional properties.
Spherically symmetrical ones such as shown in Figure 1.1 are called s orbitals. The let-
ter s is preceded by the principal quantum number n (n H11005 1, 2, 3, etc.) which speci-
fies the shell and is related to the energy of the orbital. An electron in a 1s orbital is
likely to be found closer to the nucleus, is lower in energy, and is more strongly held
than an electron in a 2s orbital.
Regions of a single orbital may be separated by nodal surfaces where the proba-
bility of finding an electron is zero. A 1s orbital has no nodes; a 2s orbital has one. A
1s and a 2s orbital are shown in cross section in Figure 1.2. The 2s wave function changes
sign on passing through the nodal surface as indicated by the plus (H11001) and minus (H11002)
signs in Figure 1.2. Do not confuse these signs with electric charges—they have noth-
ing to do with electron or nuclear charge. Also, be aware that our “orbital” drawings
are really representations of H9274
2
(which must be a positive number), whereas H11001 and H11002
refer to the sign of the wave function (H9274) itself. These customs may seem confusing at
first but turn out not to complicate things in practice. Indeed, most of the time we won’t
8 CHAPTER ONE Chemical Bonding
x
z
y
FIGURE 1.1 Probability dis-
tribution (H9274
2
) for an electron
in a 1s orbital.
Node
(a)(b)
Nucleus
y
x
H11545
H11546
Nucleus
x
y
H11545
FIGURE 1.2 Cross sections of (a) a 1s orbital and (b) a 2s orbital. The wave function has the
same sign over the entire 1s orbital. It is arbitrarily shown as H11001, but could just as well have
been designated as H11002. The 2s orbital has a spherical node where the wave function changes
sign.
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even include H11001 and H11002 signs of wave functions in our drawings but only when they are
necessary for understanding a particular concept.
Instead of probability distributions, it is more common to represent orbitals by their
boundary surfaces, as shown in Figure 1.3 for the 1s and 2s orbitals. The boundary sur-
face encloses the region where the probability of finding an electron is high—on the
order of 90–95%. Like the probability distribution plot from which it is derived, a pic-
ture of a boundary surface is usually described as a drawing of an orbital.
A hydrogen atom (Z H11005 1) has one electron; a helium atom (Z H11005 2) has two. The
single electron of hydrogen occupies a 1s orbital, as do the two electrons of helium. The
respective electron configurations are described as:
Hydrogen: 1s
1
Helium: 1s
2
In addition to being negatively charged, electrons possess the property of spin. The
spin quantum number of an electron can have a value of either H11001
1
2
or H11002
1
2
. According
to the Pauli exclusion principle, two electrons may occupy the same orbital only when
they have opposite, or “paired,” spins. For this reason, no orbital can contain more than
two electrons. Since two electrons fill the 1s orbital, the third electron in lithium
(Z H11005 3) must occupy an orbital of higher energy. After 1s, the next higher energy orbital
is 2s. The third electron in lithium therefore occupies the 2s orbital, and the electron
configuration of lithium is
Lithium: 1s
2
2s
1
The period (or row) of the periodic table in which an element appears corresponds to
the principal quantum number of the highest numbered occupied orbital (n H11005 1 in the
case of hydrogen and helium). Hydrogen and helium are first-row elements; lithium
(n H11005 2) is a second-row element.
With beryllium (Z H11005 4), the 2s level becomes filled, and the next orbitals to be
occupied in it and the remaining second-row elements are the 2p
x
, 2p
y
, and 2p
z
orbitals.
These orbitals, portrayed in Figure 1.4, have a boundary surface that is usually described
as “dumbbell-shaped.” Each orbital consists of two “lobes,” that is, slightly flattened
spheres that touch each other along a nodal plane passing through the nucleus. The 2p
x
,
2p
y
, and 2p
z
orbitals are equal in energy and mutually perpendicular.
The electron configurations of the first 12 elements, hydrogen through magnesium,
are given in Table 1.1. In filling the 2p orbitals, notice that each is singly occupied before
any one is doubly occupied. This is a general principle for orbitals of equal energy known
1.1 Atoms, Electrons, and Orbitals 9
zz
x
y
y
x
1s 2s
FIGURE 1.3 Boundary surfaces of a 1s orbital and a 2s orbital. The boundary surfaces enclose
the volume where there is a 90–95% probability of finding an electron.
A complete periodic table of
the elements is presented on
the inside back cover.
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as Hund’s rule. Of particular importance in Table 1.1 are hydrogen, carbon, nitrogen, and
oxygen. Countless organic compounds contain nitrogen, oxygen, or both in addition to car-
bon, the essential element of organic chemistry. Most of them also contain hydrogen.
It is often convenient to speak of the valence electrons of an atom. These are the
outermost electrons, the ones most likely to be involved in chemical bonding and reac-
tions. For second-row elements these are the 2s and 2p electrons. Because four orbitals
(2s, 2p
x
, 2p
y
, 2p
z
) are involved, the maximum number of electrons in the valence shell
of any second-row element is 8. Neon, with all its 2s and 2p orbitals doubly occupied,
has eight valence electrons and completes the second row of the periodic table.
PROBLEM 1.1 How many valence electrons does carbon have?
Once the 2s and 2p orbitals are filled, the next level is the 3s, followed by the 3p
x
, 3p
y
,
and 3p
z
orbitals. Electrons in these orbitals are farther from the nucleus than those in the
2s and 2p orbitals and are of higher energy.
10 CHAPTER ONE Chemical Bonding
x xx
z
y yy
zz
2p
x
2p
z
2p
y
FIGURE 1.4 Boundary surfaces of the 2p orbitals. The wave function changes sign at the
nucleus. The yz-plane is a nodal surface for the 2p
x
orbital. The probability of finding a 2p
x
electron in the yz-plane is zero. Analogously, the xz-plane is a nodal surface for the 2p
y
orbital,
and the xy-plane is a nodal surface for the 2p
z
orbital.
TABLE 1.1
Electron Configurations of the First Twelve Elements of the
Periodic Table
Number of electrons in indicated orbital
Element
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Sodium
Magnesium
Atomic
number Z
1
2
3
4
5
6
7
8
9
10
11
12
1s
1
2
2
2
2
2
2
2
2
2
2
2
2s
1
2
2
2
2
2
2
2
2
2
2p
x
1
1
1
2
2
2
2
2
2p
y
1
1
1
2
2
2
2
2p
z
1
1
1
2
2
2
3s
1
2
Answers to all problems that
appear within the body of a
chapter are found in Appen-
dix 2. A brief discussion of
the problem and advice on
how to do problems of the
same type are offered in the
Study Guide.
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PROBLEM 1.2 Referring to the periodic table as needed, write electron config-
urations for all the elements in the third period.
SAMPLE SOLUTION The third period begins with sodium and ends with argon.
The atomic number Z of sodium is 11, and so a sodium atom has 11 electrons.
The maximum number of electrons in the 1s, 2s, and 2p orbitals is ten, and so the
eleventh electron of sodium occupies a 3s orbital. The electron configuration of
sodium is 1s
2
2s
2
2p
x
2
2p
y
2
2p
z
2
3s
1
.
Neon, in the second period, and argon, in the third, possess eight electrons in their
valence shell; they are said to have a complete octet of electrons. Helium, neon, and
argon belong to the class of elements known as noble gases or rare gases. The noble
gases are characterized by an extremely stable “closed-shell” electron configuration and
are very unreactive.
1.2 IONIC BONDS
Atoms combine with one another to give compounds having properties different from
the atoms they contain. The attractive force between atoms in a compound is a chemi-
cal bond. One type of chemical bond, called an ionic bond, is the force of attraction
between oppositely charged species (ions) (Figure 1.5). Ions that are positively charged
are referred to as cations; those that are negatively charged are anions.
Whether an element is the source of the cation or anion in an ionic bond depends
on several factors, for which the periodic table can serve as a guide. In forming ionic
compounds, elements at the left of the periodic table typically lose electrons, forming a
cation that has the same electron configuration as the nearest noble gas. Loss of an elec-
tron from sodium, for example, gives the species Na
H11001
, which has the same electron con-
figuration as neon.
A large amount of energy, called the ionization energy, must be added to any atom
in order to dislodge one of its electrons. The ionization energy of sodium, for example,
is 496 kJ/mol (119 kcal/mol). Processes that absorb energy are said to be endothermic.
Compared with other elements, sodium and its relatives in group IA have relatively low
ionization energies. In general, ionization energy increases across a row in the periodic
table.
Elements at the right of the periodic table tend to gain electrons to reach the elec-
tron configuration of the next higher noble gas. Adding an electron to chlorine, for exam-
ple, gives the anion Cl
H11002
, which has the same closed-shell electron configuration as the
noble gas argon.
Energy is released when a chlorine atom captures an electron. Energy-releasing reactions
are described as exothermic, and the energy change for an exothermic process has a
negative sign. The energy change for addition of an electron to an atom is referred to as
its electron affinity and is H11002349 kJ/mol (H1100283.4 kcal/mol) for chlorine.
±￡Cl(g)
Chlorine atom
1s
2
2s
2
2p
6
3s
2
3p
5
Cl
H11002
(g)
Chloride ion
1s
2
2s
2
2p
6
3s
2
3p
6
e
H11002
Electron
H11001
±￡Na(g)
Sodium atom
1s
2
2s
2
2p
6
3s
1
[The (g) indicates that the species is present in the gas phase.]
Na
H11001
(g)
Sodium ion
1s
2
2s
2
2p
6
e
H11002
Electron
H11001
1.2 Ionic Bonds 11
H11546H11545
FIGURE 1.5 An ionic bond
is the force of electrostatic
attraction between oppo-
sitely charged ions, illus-
trated in this case by Na
H11001
(red) and Cl
H11002
(green). In
solid sodium chloride, each
sodium ion is surrounded by
six chloride ions and vice
versa in a crystal lattice.
In-chapter problems that
contain multiple parts are ac-
companied by a sample solu-
tion to part (a). Answers to
the other parts of the prob-
lem are found in Appendix 2,
and detailed solutions are
presented in the Study
Guide.
The SI (Système International
d’Unites) unit of energy is
the joule (J). An older unit is
the calorie (cal). Most or-
ganic chemists still express
energy changes in units of
kilocalories per mole (1
kcal/mol H110054.184 kJ/mol).
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PROBLEM 1.3 Which of the following ions possess a noble gas electron config-
uration?
(a) K
H11001
(d) O
H11002
(b) He
H11001
(e) F
H11002
(c) H
H11002
(f) Ca
2H11001
SAMPLE SOLUTION (a) Potassium has atomic number 19, and so a potassium
atom has 19 electrons. The ion K
H11001
, therefore, has 18 electrons, the same as the
noble gas argon. The electron configurations of K
H11001
and Ar are the same:
1s
2
2s
2
2p
6
3s
2
3p
6
.
Transfer of an electron from a sodium atom to a chlorine atom yields a sodium
cation and a chloride anion, both of which have a noble gas electron configuration:
Were we to simply add the ionization energy of sodium (496 kJ/mol) and the electron
affinity of chlorine (H11002349 kJ/mol), we would conclude that the overall process is
endothermic with H9004H° H11005H11001147 kJ/mol. The energy liberated by adding an electron to
chlorine is insufficient to override the energy required to remove an electron from
sodium. This analysis, however, fails to consider the force of attraction between the
oppositely charged ions Na
H11001
and Cl
–
, which exceeds 500 kJ/mol and is more than suf-
ficient to make the overall process exothermic. Attractive forces between oppositely
charged particles are termed electrostatic, or coulombic, attractions and are what we
mean by an ionic bond between two atoms.
PROBLEM 1.4 What is the electron configuration of C
H11001
? Of C
H11002
? Does either one
of these ions have a noble gas (closed-shell) electron configuration?
Ionic bonds are very common in inorganic compounds, but rare in organic ones.
The ionization energy of carbon is too large and the electron affinity too small for car-
bon to realistically form a C
4H11001
or C
4H11002
ion. What kinds of bonds, then, link carbon to
other elements in millions of organic compounds? Instead of losing or gaining electrons,
carbon shares electrons with other elements (including other carbon atoms) to give what
are called covalent bonds.
1.3 COVALENT BONDS
The covalent, or shared electron pair, model of chemical bonding was first suggested
by G. N. Lewis of the University of California in 1916. Lewis proposed that a sharing
of two electrons by two hydrogen atoms permits each one to have a stable closed-shell
electron configuration analogous to helium.
H
Two hydrogen atoms,
each with a single
electron
H
Hydrogen molecule:
covalent bonding by way of
a shared electron pair
HH
±￡Na(g)
Sodium atom
Na
H11001
Cl
H11002
(g)
Sodium chloride
Cl(g)
Chlorine atom
H11001
12 CHAPTER ONE Chemical Bonding
Ionic bonding was proposed
by the German physicist Wal-
ter Kossel in 1916, in order
to explain the ability of sub-
stances such as sodium chlo-
ride to conduct an electric
current.
Gilbert Newton Lewis (born
Weymouth, Massachusetts,
1875; died Berkeley, Califor-
nia, 1946) has been called
the greatest American
chemist. The January 1984 is-
sue of the Journal of Chemi-
cal Education contains five
articles describing Lewis’ life
and contributions to chem-
istry.
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Structural formulas of this type in which electrons are represented as dots are called
Lewis structures.
The amount of energy required to dissociate a hydrogen molecule H
2
to two sep-
arate hydrogen atoms is called its bond dissociation energy (or bond energy). For H
2
it is quite large, being equal to 435 kJ/mol (104 kcal/mol). The main contributor to the
strength of the covalent bond in H
2
is the increased binding force exerted on its two
electrons. Each electron in H
2
“feels” the attractive force of two nuclei, rather than one
as it would in an isolated hydrogen atom.
Covalent bonding in F
2
gives each fluorine 8 electrons in its valence shell and a
stable electron configuration equivalent to that of the noble gas neon:
PROBLEM 1.5 Hydrogen is bonded to fluorine in hydrogen fluoride by a cova-
lent bond. Write a Lewis formula for hydrogen fluoride.
The Lewis model limits second-row elements (Li, Be, B, C, N, O, F, Ne) to a total
of 8 electrons (shared plus unshared) in their valence shells. Hydrogen is limited to 2.
Most of the elements that we’ll encounter in this text obey the octet rule: in forming
compounds they gain, lose, or share electrons to give a stable electron configuration
characterized by eight valence electrons. When the octet rule is satisfied for carbon,
nitrogen, oxygen, and fluorine, they have an electron configuration analogous to the noble
gas neon.
Now let’s apply the Lewis model to the organic compounds methane and carbon
tetrafluoride.
Carbon has 8 electrons in its valence shell in both methane and carbon tetrafluoride. By
forming covalent bonds to four other atoms, carbon achieves a stable electron configu-
ration analogous to neon. Each covalent bond in methane and carbon tetrafluoride is
quite strong—comparable to the bond between hydrogens in H
2
in bond dissociation
energy.
PROBLEM 1.6 Given the information that it has a carbon–carbon bond, write a
satisfactory Lewis structure for C
2
H
6
(ethane).
Representing a 2-electron covalent bond by a dash (—), the Lewis structures for
hydrogen fluoride, fluorine, methane, and carbon tetrafluoride become:
Combine
to write a
Lewis structure
for methane
and fourC H CH
H
H
H
Combine
to write a
Lewis structure
for carbon
tetrafluoride
and fourC FF
F
F
CF
Fluorine molecule:
covalent bonding by way of
a shared electron pair
FF
Two fluorine atoms, each
with seven electrons in
its valence shell
FF
1.3 Covalent Bonds 13
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1.4 DOUBLE BONDS AND TRIPLE BONDS
Lewis’s concept of shared electron pair bonds allows for 4-electron double bonds and
6-electron triple bonds. Carbon dioxide (CO
2
) has two carbon–oxygen double bonds,
and the octet rule is satisfied for both carbon and oxygen. Similarly, the most stable
Lewis structure for hydrogen cyanide (HCN) has a carbon–nitrogen triple bond.
Multiple bonds are very common in organic chemistry. Ethylene (C
2
H
4
) contains
a carbon–carbon double bond in its most stable Lewis structure, and each carbon has a
completed octet. The most stable Lewis structure for acetylene (C
2
H
2
) contains a car-
bon–carbon triple bond. Here again, the octet rule is satisfied.
PROBLEM 1.7 Write the most stable Lewis structure for each of the following
compounds:
(a) Formaldehyde, CH
2
O. Both hydrogens are bonded to carbon. (A solution of
formaldehyde in water is sometimes used to preserve biological specimens.)
(b) Tetrafluoroethylene, C
2
F
4
. (The starting material for the preparation of Teflon.)
(c) Acrylonitrile, C
3
H
3
N. The atoms are connected in the order CCCN, and all
hydrogens are bonded to carbon. (The starting material for the preparation
of acrylic fibers such as Orlon and Acrilan.)
SAMPLE SOLUTION (a) Each hydrogen contributes 1 valence electron, carbon
contributes 4, and oxygen 6 for a total of 12 valence electrons. We are told that
both hydrogens are bonded to carbon. Since carbon forms four bonds in its sta-
ble compounds, join carbon and oxygen by a double bond. The partial structure
so generated accounts for 8 of the 12 electrons. Add the remaining four electrons
to oxygen as unshared pairs to complete the structure of formaldehyde.
Partial structure showing
covalent bonds
O
X
C
O
X
C
HH
±
±
Complete Lewis structure
of formaldehyde
HH
±
±
orEthylene: C
H
H
H
H
C C?C
H
H
H
H
±
±
±
±
orAcetylene: H±CPC±HC CHH
orCarbon dioxide: O C O O?C?O
orHydrogen cyanide: CH NH±CPN
H±C±H
H
W
W
H
Methane Carbon tetrafluoride
F±C±F
F
W
W
F
Hydrogen
fluoride
H±F
Fluorine
F±F
14 CHAPTER ONE Chemical Bonding
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1.5 POLAR COVALENT BONDS AND ELECTRONEGATIVITY
Electrons in covalent bonds are not necessarily shared equally by the two atoms that they
connect. If one atom has a greater tendency to attract electrons toward itself than the other,
we say the electron distribution is polarized, and the bond is referred to as a polar cova-
lent bond. Hydrogen fluoride, for example, has a polar covalent bond. Because fluorine
attracts electrons more strongly than hydrogen, the electrons in the H±F bond are pulled
toward fluorine, giving it a partial negative charge, and away from hydrogen giving it a
partial positive charge. This polarization of electron density is represented in various ways.
The tendency of an atom to draw the electrons in a covalent bond toward itself is
referred to as its electronegativity. An electronegative element attracts electrons; an
electropositive one donates them. Electronegativity increases across a row in the peri-
odic table. The most electronegative of the second-row elements is fluorine; the most
electropositive is lithium. Electronegativity decreases in going down a column. Fluorine
is more electronegative than chlorine. The most commonly cited electronegativity scale
was devised by Linus Pauling and is presented in Table 1.2.
PROBLEM 1.8 Examples of carbon-containing compounds include methane (CH
4
),
chloromethane (CH
3
Cl), and methyllithium (CH
3
Li). In which one does carbon bear
the greatest partial positive charge? The greatest partial negative charge?
Centers of positive and negative charge that are separated from each other consti-
tute a dipole. The dipole moment H9262 of a molecule is equal to the charge e (either the
positive or the negative charge, since they must be equal) multiplied by the distance
between the centers of charge:
H9262H11005e H11003 d
(The symbols
H9254H11001
and
H9254H11002
indicate partial positive
and partial negative
charge, respectively)
H9254H11001
H±F
H9254H11002
H±F
(The symbol represents
the direction of polarization
of electrons in the H±F bond)
1.5 Polar Covalent Bonds and Electronegativity 15
TABLE 1.2 Selected Values from the Pauling Electronegativity Scale
Group number
Period
1
2
3
4
5
I
H
2.1
Li
1.0
Na
0.9
K
0.8
II
Be
1.5
Mg
1.2
Ca
1.0
III
B
2.0
Al
1.5
IV
C
2.5
Si
1.8
V
N
3.0
P
2.1
VI
O
3.5
S
2.5
VII
F
4.0
Cl
3.0
Br
2.8
I
2.5
Linus Pauling (1901–1994)
was born in Portland, Ore-
gon and was educated at
Oregon State University and
at the California Institute of
Technology, where he
earned a Ph.D. in chemistry
in 1925. In addition to re-
search in bonding theory,
Pauling studied the structure
of proteins and was awarded
the Nobel Prize in chemistry
for that work in 1954. Paul-
ing won a second Nobel
Prize (the Peace Prize) for his
efforts to limit the testing of
nuclear weapons. He was
one of only four scientists to
have won two Nobel Prizes.
The first double winner was
a woman. Can you name
her?
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Because the charge on an electron is 4.80 H11003 10
H1100210
electrostatic units (esu) and the dis-
tances within a molecule typically fall in the 10
H110028
cm range, molecular dipole moments
are on the order of 10
H1100218
esu·cm. In order to simplify the reporting of dipole moments
this value of 10
H1100218
esuH11080cm is defined as a debye, D. Thus the experimentally determined
dipole moment of hydrogen fluoride, 1.7 H11003 10
H1100218
esuH11080cm is stated as 1.7 D.
Table 1.3 lists the dipole moments of various bond types. For H±F, H±Cl,
H±Br, and H±I these “bond dipoles” are really molecular dipole moments. A polar
molecule has a dipole moment, a nonpolar one does not. Thus, all of the hydrogen
halides are polar molecules. In order to be polar, a molecule must have polar bonds, but
can’t have a shape that causes all the individual bond dipoles to cancel. We will have
more to say about this in Section 1.11 after we have developed a feeling for the three-
dimensional shapes of molecules.
The bond dipoles in Table 1.3 depend on the difference in electronegativity of the
bonded atoms and on the bond distance. The polarity of a C±H bond is relatively low;
substantially less than a C±O bond, for example. Don’t lose sight of an even more
important difference between a C±H bond and a C±O bond, and that is the direction
of the dipole moment. In a C±H bond the electrons are drawn away from H, toward
C. In a C±O bond, electrons are drawn from C toward O. As we’ll see in later chap-
ters, the kinds of reactions that a substance undergoes can often be related to the size
and direction of key bond dipoles.
1.6 FORMAL CHARGE
Lewis structures frequently contain atoms that bear a positive or negative charge. If the
molecule as a whole is neutral, the sum of its positive charges must equal the sum of
its negative charges. An example is nitric acid, HNO
3
:
As written, the structural formula for nitric acid depicts different bonding patterns
for its three oxygens. One oxygen is doubly bonded to nitrogen, another is singly bonded
H±O±N
O
O
H11002
H11001
?
±
16 CHAPTER ONE Chemical Bonding
TABLE 1.3 Selected Bond Dipole Moments
Bond*
H±F
H±Cl
H±Br
H±I
H±C
H±N
H±O
Dipole moment, D
1.7
1.1
0.8
0.4
0.3
1.3
1.5
Bond*
C±F
C±O
C±N
C?O
C?N
CPN
Dipole moment, D
1.4
0.7
0.4
2.4
1.4
3.6
*The direction of the dipole moment is toward the more electronegative atom. In the listed examples
hydrogen and carbon are the positive ends of the dipoles. Carbon is the negative end of the dipole
associated with the C±H bond.
The debye unit is named in
honor of Peter Debye, a
Dutch scientist who did im-
portant work in many areas
of chemistry and physics and
was awarded the Nobel Prize
in chemistry in 1936.
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to both nitrogen and hydrogen, and the third has a single bond to nitrogen and a nega-
tive charge. Nitrogen is positively charged. The positive and negative charges are called
formal charges, and the Lewis structure of nitric acid would be incomplete were they
to be omitted.
We calculate formal charges by counting the number of electrons “owned” by each
atom in a Lewis structure and comparing this electron count with that of a neutral atom.
Figure 1.6 illustrates how electrons are counted for each atom in nitric acid. Counting
electrons for the purpose of computing the formal charge differs from counting electrons
to see if the octet rule is satisfied. A second-row element has a filled valence shell if the
sum of all the electrons, shared and unshared, is 8. Electrons that connect two atoms by
a covalent bond count toward filling the valence shell of both atoms. When calculating
the formal charge, however, only half the number of electrons in covalent bonds can be
considered to be “owned” by an atom.
To illustrate, let’s start with the hydrogen of nitric acid. As shown in Figure 1.6,
hydrogen is associated with only two electrons—those in its covalent bond to oxygen.
It shares those two electrons with oxygen, and so we say that the electron count of each
hydrogen is
1
2
(2) H11005 1. Since this is the same as the number of electrons in a neutral
hydrogen atom, the hydrogen in nitric acid has no formal charge.
Moving now to nitrogen, we see that it has four covalent bonds (two single bonds
H11001 one double bond), and so its electron count is
1
2
(8) H11005 4. A neutral nitrogen has five
electrons in its valence shell. The electron count for nitrogen in nitric acid is 1 less than
that of a neutral nitrogen atom, so its formal charge is H110011.
Electrons in covalent bonds are counted as if they are shared equally by the atoms
they connect, but unshared electrons belong to a single atom. Thus, the oxygen which
is doubly bonded to nitrogen has an electron count of 6 (four electrons as two unshared
pairs H11001 two electrons from the double bond). Since this is the same as a neutral oxy-
gen atom, its formal charge is 0. Similarly, the OH oxygen has two bonds plus two
unshared electron pairs, giving it an electron count of 6 and no formal charge.
The oxygen highlighted in yellow in Figure 1.6 owns three unshared pairs (six
electrons) and shares two electrons with nitrogen to give it an electron count of 7. This
is 1 more than the number of electrons in the valence shell of an oxygen atom, and so
its formal charge is H110021.
The method described for calculating formal charge has been one of reasoning
through a series of logical steps. It can be reduced to the following equation:
Formal charge H11005
group number in
H11002 number of bonds H11002 number of unshared electrons
periodic table
1.6 Formal Charge 17
H±O±N
O
O
H11002
H11001
?
±
Electron count (O) (4) H11001 4 H11005 6H11005
1
2
Electron count (N) (8) H11005 4H11005
1
2
Electron count (O) (2) H11001 6 H11005 7H11005
1
2
Electron count (O) (4) H11001 4 H11005 6H11005
1
2
Electron count (H) (2) H11005 1H11005
1
2
FIGURE 1.6 Counting electrons in nitric acid. The electron count of each atom is equal to half the number of electrons it
shares in covalent bonds plus the number of electrons in its own unshared pairs.
The number of valence elec-
trons in an atom of a main-
group element such as
nitrogen is equal to its group
number. In the case of nitro-
gen this is 5.
It will always be true that a
covalently bonded hydrogen
has no formal charge (formal
charge H11005 0).
It will always be true that a
nitrogen with four covalent
bonds has a formal charge of
H110011. (A nitrogen with four co-
valent bonds cannot have
unshared pairs, because of
the octet rule.)
It will always be true that an
oxygen with two covalent
bonds and two unshared
pairs has no formal charge.
It will always be true that an
oxygen with one covalent
bond and three unshared
pairs has a formal charge of
H110021.
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PROBLEM 1.9 Like nitric acid, each of the following inorganic compounds will
be frequently encountered in this text. Calculate the formal charge on each of
the atoms in the Lewis structures given.
SAMPLE SOLUTION (a) The formal charge is the difference between the num-
ber of valence electrons in the neutral atom and the electron count in the Lewis
structure. (The number of valence electrons is the same as the group number in
the periodic table for the main-group elements.)
The formal charges are shown in the Lewis structure of thionyl chloride
as
Cl±S±Cl
O
W
H11002
H11001
Sulfur:
Oxygen:
Chlorine:
Valence electrons of neutral atom
6
6
7
Formal charge
H110011
H110021
0
Electron count
1
2
(6) H11001 2 H11005 5
1
2
(2) H11001 6 H11005 7
1
2
(2) H11001 6 H11005 7
Cl±S±Cl
O
W
(a) Thionyl chloride
H±O±S±O±H
O
W
W
O
(b) Sulfuric acid
H±O±N?O
(c) Nitrous acid
18 CHAPTER ONE Chemical Bonding
So far we’ve only considered neutral molecules—those in which the sums of the
positive and negative formal charges were equal. With ions, of course, these sums will
not be equal. Ammonium cation and borohydride anion, for example, are ions with net
charges of H110011 and H110021, respectively. Nitrogen has a formal charge of H110011 in ammonium
ion, and boron has a formal charge of H110021 in borohydride. None of the hydrogens in the
Lewis structures shown for these ions bears a formal charge.
PROBLEM 1.10 Verify that the formal charges on nitrogen in ammonium ion
and boron in borohydride ion are as shown.
Formal charges are based on Lewis structures in which electrons are considered to
be shared equally between covalently bonded atoms. Actually, polarization of N±H
bonds in ammonium ion and of B±H bonds in borohydride leads to some transfer of
positive and negative charge, respectively, to the hydrogens.
PROBLEM 1.11 Use H9254H11001 and H9254H11002 notation to show the dispersal of charge to the
hydrogens in NH
4
H11001
and BH
4
H11002
.
Ammonium ion
H±N±H
H
W
W
H
H11001
Borohydride ion
H±B±H
H
W
W
H
H11002
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Determining formal charges on individual atoms of Lewis structures is an impor-
tant element in good “electron bookkeeping.” So much of organic chemistry can be made
more understandable by keeping track of electrons that it is worth taking some time at
the beginning to become proficient at the seemingly simple task of counting electrons.
1.7 STRUCTURAL FORMULAS OF ORGANIC MOLECULES
Table 1.4 outlines a systematic procedure for writing Lewis structures. Notice that the
process depends on knowing not only the molecular formula, but also the order in which
the atoms are attached to one another. This order of attachment is called the constitu-
tion, or connectivity, of the molecule and is determined by experiment. Only rarely is
it possible to deduce the constitution of a molecule from its molecular formula.
Organic chemists have devised a number of shortcuts to speed the writing of struc-
tural formulas. Sometimes we leave out unshared electron pairs, but only when we are
sure enough in our ability to count electrons to know when they are present and when
they’re not. We’ve already mentioned representing covalent bonds by dashes. In con-
densed structural formulas we leave out some, many, or all of the covalent bonds and
use subscripts to indicate the number of identical groups attached to a particular atom.
These successive levels of simplification are illustrated as shown for isopropyl alcohol
(“rubbing alcohol”).
PROBLEM 1.12 Expand the following condensed formulas so as to show all the
bonds and unshared electron pairs.
(a) HOCH
2
CH
2
NH
2
(d) CH
3
CHCl
2
(b) (CH
3
)
3
CH (e) CH
3
NHCH
2
CH
3
(c) ClCH
2
CH
2
Cl (f) (CH
3
)
2
CHCH?O
SAMPLE SOLUTION (a) The molecule contains two carbon atoms, which are
bonded to each other. Both carbons bear two hydrogens. One carbon bears the
group HO±; the other is attached to ±NH
2
.
When writing the constitution of a molecule, it is not necessary to concern your-
self with the spatial orientation of the atoms. There are many other correct ways
to represent the constitution shown. What is important is to show the sequence
OCCN (or its equivalent NCCO) and to have the correct number of hydrogens
present on each atom.
In order to locate unshared electron pairs, first count the total number of
valence electrons brought to the molecule by its component atoms. Each hydro-
gen contributes 1, each carbon 4, nitrogen 5, and oxygen 6, for a total of 26. There
are ten bonds shown, accounting for 20 electrons; therefore 6 electrons must be
contained in unshared pairs. Add pairs of electrons to oxygen and nitrogen so that
their octets are complete, two unshared pairs to oxygen and one to nitrogen.
H±O±C±C±N±H
H
W
W
H
H
W
H
W
W
H
H±C±C±C±H written as or condensed even further to
(CH
3
)
2
CHOH
H
W
W
H
OH
W
CH
3
CHCH
3
H
W
W
H
H
W
W
O
W
H
1.7 Structural Formulas of Organic Molecules 19
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TABLE 1.4 How to Write Lewis Structures
Step
1. The molecular formula and the connectivity are
determined experimentally and are included
among the information given in the statement of
the problem.
2. Count the number of valence electrons available.
For a neutral molecule this is equal to the sum of
the valence electrons of the constituent atoms.
6. If one or more atoms have fewer than 8 electrons,
use unshared pairs on an adjacent atom to form a
double (or triple) bond to complete the octet.
7. Calculate formal charges.
4. Count the number of electrons in shared electron
pair bonds (twice the number of bonds), and sub-
tract this from the total number of electrons to
give the number of electrons to be added to com-
plete the structure.
5. Add electrons in pairs so that as many atoms as
possible have 8 electrons. (Hydrogen is limited to 2
electrons.) When the number of electrons is insuffi-
cient to provide an octet for all atoms, assign elec-
trons to atoms in order of decreasing electronega-
tivity.
Illustration
Methyl nitrite has the molecular formula CH
3
NO
2
. All
hydrogens are bonded to carbon, and the order of
atomic connections is CONO.
Each hydrogen contributes 1 valence electron, car-
bon contributes 4, nitrogen contributes 5, and each
oxygen contributes 6 for a total of 24 in CH
3
NO
2
.
An electron pair on the terminal oxygen is shared
with nitrogen to give a double bond.
For methyl nitrite we write the partial structure
The structure shown is the best (most stable) Lewis
structure for methyl nitrite. All atoms except hydro-
gen have 8 electrons (shared H11001 unshared) in their va-
lence shell.
None of the atoms in the Lewis structure shown in
step 6 possesses a formal charge. An alternative
Lewis structure for methyl nitrite,
although it satisfies the octet rule, is less stable than
the one shown in step 6 because it has a separation
of positive charge from negative charge.
The partial structure in step 3 contains 6 bonds
equivalent to 12 electrons. Since CH
3
NO
2
contains 24
electrons, 12 more electrons need to be added.
With 4 bonds, carbon already has 8 electrons. The re-
maining 12 electrons are added as indicated. Both
oxygens have 8 electrons, but nitrogen (less electro-
negative than oxygen) has only 6.
H±C±O±N±O
H
W
W
H
3. Connect bonded atoms by a shared electron pair
bond ( ) represented by a dash (±).
H±C±O±N±O
H
W
W
H
H±C±O±N?O
H
W
W
H
H±C±O?N±O
H11002
H11001
H
W
W
H
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As you practice, you will begin to remember patterns of electron distribution. A
neutral oxygen with two bonds has two unshared electron pairs. A neutral nitro-
gen with three bonds has one unshared pair.
With practice, writing structural formulas for organic molecules soon becomes rou-
tine and can be simplified even more. For example, a chain of carbon atoms can be rep-
resented by drawing all of the C±C bonds while omitting individual carbons. The result-
ing structural drawings can be simplified still more by stripping away the hydrogens.
In these simplified representations, called bond-line formulas or carbon skeleton dia-
grams, the only atoms specifically written in are those that are neither carbon nor hydro-
gen bound to carbon. Hydrogens bound to these heteroatoms are shown, however.
PROBLEM 1.13 Expand the following bond-line representations to show all the
atoms including carbon and hydrogen.
(a) (c)
(b) (d)
SAMPLE SOLUTION (a) There is a carbon at each bend in the chain and at the
ends of the chain. Each of the ten carbon atoms bears the appropriate number
of hydrogen substituents so that it has four bonds.
Alternatively, the structure could be written as
CH
3
CH
2
CH
2
CH
2
CH
2
CH
2
CH
2
CH
2
CH
2
CH
3
or in condensed form as CH
3
(CH
2
)
8
CH
3
.
H11013 H±C±C±C±C±C±C±C±C±C±C±H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
HO
CH
3
CH
2
CH
2
CH
2
OH becomes
becomes
OH
Cl
W
C
H
2
C
WW
H
2
C
CH
2
CH
2
±
C
H
2
±
H
±
±
Cl
±
±
CH
3
CH
2
CH
2
CH
3
becomes simplified to
H
H
HH
HH
±
±
HH
±
±
±
HH
±
±
±
±
±
H±O±C±C±N±H
H
W
W
H
H
W
H
W
W
H
1.7 Structural Formulas of Organic Molecules 21
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1.8 CONSTITUTIONAL ISOMERS
In the introduction we noted that both Berzelius and W?hler were fascinated by the fact
that two different compounds with different properties, ammonium cyanate and urea, pos-
sessed exactly the same molecular formula, CH
4
N
2
O. Berzelius had studied examples of
similar phenomena earlier and invented the word isomer to describe different compounds
that have the same molecular formula.
We can illustrate isomerism by referring to two different compounds, nitromethane
and methyl nitrite, both of which have the molecular formula CH
3
NO
2
. Nitromethane,
used to power race cars, is a liquid with a boiling point of 101°C. Methyl nitrite is a gas
boiling at H1100212°C, which when inhaled causes dilation of blood vessels. Isomers that dif-
fer in the order in which their atoms are bonded are often referred to as structural iso-
mers. A more modern term is constitutional isomer. As noted in the previous section,
the order of atomic connections that defines a molecule is termed its constitution, and
we say that two compounds are constitutional isomers if they have the same molecular
formula but differ in the order in which their atoms are connected.
PROBLEM 1.14 There are many more isomers of CH
3
NO
2
other than
nitromethane and methyl nitrite. Some, such as carbamic acid, an intermediate in
the commercial preparation of urea for use as a fertilizer, are too unstable to iso-
late. Given the information that the nitrogen and both oxygens of carbamic acid
are bonded to carbon and that one of the carbon–oxygen bonds is a double bond,
write a Lewis structure for carbamic acid.
PROBLEM 1.15 Write structural formulas for all the constitutionally isomeric
compounds having the given molecular formula.
(a) C
2
H
6
O (c) C
4
H
10
O
(b) C
3
H
8
O
SAMPLE SOLUTION (a) Begin by considering the ways in which two carbons and
one oxygen may be bonded. There are two possibilities: C±C±O and C±O±C.
Add the six hydrogens so that each carbon has four bonds and each oxygen two.
There are two constitutional isomers: ethyl alcohol and dimethyl ether.
In Chapter 3 another type of isomerism, called stereoisomerism, will be intro-
duced. Stereoisomers have the same constitution but differ in the arrangement of atoms
in space.
Dimethyl ether
H±C±O±C±H
H
W
W
H
H
W
W
H
Ethyl alcohol
H±C±C±O±H
H
W
W
H
H
W
W
H
Nitromethane
H±C±N
H
W
W
H
O
O
H11002
H11001
?
±
Methyl nitrite
H±C±O±N?O
H
W
W
H
22 CHAPTER ONE Chemical Bonding
The suffix -mer in the word
“isomer” is derived from the
Greek word meros, meaning
“part,” “share,” or “por-
tion.” The prefix iso- is also
from Greek (isos, “the
same”). Thus isomers are dif-
ferent molecules that have
the same parts (elemental
composition).
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1.9 RESONANCE
When writing a Lewis structure, we restrict a molecule’s electrons to certain well-defined
locations, either linking two atoms by a covalent bond or as unshared electrons on a sin-
gle atom. Sometimes more than one Lewis structure can be written for a molecule, espe-
cially those that contain multiple bonds. An example often cited in introductory chem-
istry courses is ozone (O
3
). Ozone occurs naturally in large quantities in the upper
atmosphere, where it screens the surface of the earth from much of the sun’s ultraviolet
rays. Were it not for this ozone layer, most forms of surface life on earth would be dam-
aged or even destroyed by the rays of the sun. The following Lewis structure for ozone
satisfies the octet rule; all three oxygens have 8 electrons in their valence shell.
This Lewis structure, however, doesn’t accurately portray the bonding in ozone,
because the two terminal oxygens are bonded differently to the central oxygen. The cen-
tral oxygen is depicted as doubly bonded to one and singly bonded to the other. Since
it is generally true that double bonds are shorter than single bonds, we would expect
ozone to exhibit two different O±O bond lengths, one of them characteristic of the
O±O single bond distance (147 pm in hydrogen peroxide, H±O±O±H) and the other
one characteristic of the O?O double bond distance (121 pm in O
2
). Such is not the
case. Both bond distances in ozone are exactly the same (128 pm)—somewhat shorter
than the single bond distance and somewhat longer than the double bond distance. The
structure of ozone requires that the central oxygen must be identically bonded to both
terminal oxygens.
In order to deal with circumstances such as the bonding in ozone, the notion of
resonance between Lewis structures was developed. According to the resonance con-
cept, when more than one Lewis structure may be written for a molecule, a single struc-
ture is not sufficient to describe it. Rather, the true structure has an electron distribution
that is a “hybrid” of all the possible Lewis structures that can be written for the mole-
cule. In the case of ozone, two equivalent Lewis structures may be written. We use a
double-headed arrow to represent resonance between these two Lewis structures.
It is important to remember that the double-headed resonance arrow does not indi-
cate a process in which the two Lewis structures interconvert. Ozone, for example, has
a single structure; it does not oscillate back and forth between two Lewis structures,
rather its true structure is not adequately represented by any single Lewis structure.
Resonance attempts to correct a fundamental defect in Lewis formulas. Lewis for-
mulas show electrons as being localized; they either are shared between two atoms in a
covalent bond or are unshared electrons belonging to a single atom. In reality, electrons
distribute themselves in the way that leads to their most stable arrangement. This some-
times means that a pair of electrons is delocalized, or shared by several nuclei. What
we try to show by the resonance description of ozone is the delocalization of the lone-
pair electrons of one oxygen and the electrons in the double bond over the three atoms
of the molecule. Organic chemists often use curved arrows to show this electron
￠￡
H11001
?
±
O
OO
H11002
H11001
±
?
O
OO
H11002
H11001
?
±
O
OO
H11002
1.9 Resonance 23
Bond distances in organic
compounds are usually 1 to
2? (1? H11005 10
H1100210
m). Since the
angstrom (?) is not an SI
unit, we will express bond
distances in picometers
(1 pm H11005 10
H1100212
m). Thus,
128 pm H110051.28 ?.
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delocalization. Alternatively, an average of two Lewis structures is sometimes drawn
using a dashed line to represent a “partial” bond. In the dashed-line notation the central
oxygen is linked to the other two by bonds that are halfway between a single bond and
a double bond, and the terminal oxygens each bear one half of a unit negative charge.
The rules to be followed when writing resonance structures are summarized in
Table 1.5.
￠￡
H11001
O
OO
H11002
1
2
H11002
1
2
H11001
?
±
O
OO
H11002
H11001
±
?
O
OO
H11002
Curved arrow notation
Electron delocalization in ozone
Dashed-line notation
24 CHAPTER ONE Chemical Bonding
TABLE 1.5 Introduction to the Rules of Resonance*
Rule
1. Atomic positions (connectivity) must be the same
in all resonance structures; only the electron posi-
tions may vary among the various contributing
structures.
2. Lewis structures in which second-row elements
own or share more than 8 valence electrons are
especially unstable and make no contribution to
the true structure. (The octet rule may be exceed-
ed for elements beyond the second row.)
Illustration
The structural formulas
represent different compounds, not different reso-
nance forms of the same compound. A is a Lewis
structure for nitromethane; B is methyl nitrite.
Structure D has no separation of charge and is more
stable than E, which does. The true structure of
methyl nitrite is more like D than E.
Structural formula C,
has 10 electrons around nitrogen. It is not a permissi-
ble Lewis structure for nitromethane and so cannot
be a valid resonance form.
The two Lewis structures D and E of methyl nitrite
satisfy the octet rule:
(Continued)
3. When two or more structures satisfy the octet rule,
the most stable one is the one with the smallest
separation of oppositely charged atoms.
A
CH
3
±N
O
O
H11002
H11001
?
±
and
B
CH
3
±O±N?O
C
CH
3
±N
O
O
?
?
￠￡
DE
CH
3
±O±N?O CH
3
±O?N±O
H11001
H11002
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1.9 Resonance 25
TABLE 1.5 Introduction to the Rules of Resonance* (Continued)
Rule
5. Each contributing Lewis structure must have the
same number of electrons and the same net
charge, although the formal charges of individual
atoms may vary among the various Lewis struc-
tures.
6. Each contributing Lewis structure must have the
same number of unpaired electrons.
Illustration
The Lewis structures
are not resonance forms of one another. Structure H
has 24 valence electrons and a net charge of 0; I has
26 valence electrons and a net charge of H110022.
The two most stable resonance forms of methyl
nitrite are not equivalent.
Structural formula J is a Lewis structure of nitro-
methane; K is not, even though it has the same
atomic positions and the same number of electrons.
Structure K has 2 unpaired electrons. Structure J has
all its electrons paired and is a more stable structure.
Nitromethane is stabilized by electron delocalization
more than methyl nitrite is. The two most stable res-
onance forms of nitromethane are equivalent to
each other.
7. Electron delocalization stabilizes a molecule. A
molecule in which electrons are delocalized is
more stable than implied by any of the individual
Lewis structures that may be written for it. The
degree of stabilization is greatest when the con-
tributing Lewis structures are of equal stability.
H
CH
3
±N
O
O
H11002
H11001
?
±
and CH
3
±N
O
O
H11002
H11002
±
±
I
J
CH
3
±N
O
O
H11002
H11001
?
±
and CH
3
±N
O
O
±
±
K
CH
3
±N
O
O
H11002
H11001
?
±
CH
3
±N
O
O
H11002
H11001
±
?
￠￡
*These are the most important rules to be concerned with at present. Additional aspects of electron delocalization, as well as additional rules
for its depiction by way of resonance structures, will be developed as needed in subsequent chapters.
￠￡CH
3
±O±N?O CH
3
±O?N±O
H11001
H11002
4. Among structural formulas in which the octet rule
is satisfied for all atoms and one or more of these
atoms bears a formal charge, the most stable reso-
nance form is the one in which negative charge
resides on the most electronegative atom (or posi-
tive charge on the most electropositive one).
The most stable Lewis structure for cyanate ion is F
because the negative charge is on its oxygen.
In G the negative charge is on nitrogen. Oxygen is
more electronegative than nitrogen and can better
support a negative charge.
￠￡
F
NPC±O
H11002
G
N?C?O
H11002
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PROBLEM 1.16 Electron delocalization can be important in ions as well as in
neutral molecules. Using curved arrows, show how an equally stable resonance
structure can be generated for each of the following anions:
(a) (c)
(b) (d)
SAMPLE SOLUTION (a) When using curved arrows to represent the reorganiza-
tion of electrons, begin at a site of high electron density, preferably an atom that
is negatively charged. Move electron pairs until a proper Lewis structure results.
For nitrate ion, this can be accomplished in two ways:
Three equally stable Lewis structures are possible for nitrate ion. The negative
charge in nitrate is shared equally by all three oxygens.
It is good chemical practice to represent molecules by their most stable Lewis
structure. The ability to write alternative resonance forms and to compare their relative
stabilities, however, can provide insight into both molecular structure and chemical
behavior. This will become particularly apparent in the last two thirds of this text, where
the resonance concept will be used regularly.
1.10 THE SHAPES OF SOME SIMPLE MOLECULES
So far our concern has emphasized “electron bookkeeping.” We now turn our attention
to the shapes of molecules.
Methane, for example, is described as a tetrahedral molecule because its four
hydrogens occupy the corners of a tetrahedron with carbon at its center as the various
methane models in Figure 1.7 illustrate. We often show three-dimensionality in struc-
tural formulas by using a solid wedge ( ) to depict a bond projecting from the paper
toward the reader and a dashed wedge ( ) to depict one receding from the paper. A
simple line (±) represents a bond that lies in the plane of the paper (Figure 1.8).
The tetrahedral geometry of methane is often explained in terms of the valence
shell electron-pair repulsion (VSEPR) model. The VSEPR model rests on the idea that
an electron pair, either a bonded pair or an unshared pair, associated with a particular
atom will be as far away from the atom’s other electron pairs as possible. Thus, a tetra-
hedral geometry permits the four bonds of methane to be maximally separated and is
characterized by H±C±H angles of 109.5°, a value referred to as the tetrahedral
angle.
O±N ￠￡
O
O
H11002
H11001
?
±
H11002
O?N
O
O
H11002
H11001
±
±
H11002
O±N ￠￡
O
O
H11002
H11001
?
±
H11002
O±N
O
O
H11001
±
?
H11002
H11002
O±B
O
±
?
H11002
O
H11002
H11002
O±C
O±H
O
?
±
H11002
O±C
O
?
±
H11002
O
H11002
O±N
O
O
H11002
H11001
?
±
H11002
26 CHAPTER ONE Chemical Bonding
Although reservations have
been expressed concerning
VSEPR as an explanation for
molecular geometries, it re-
mains a useful tool for pre-
dicting the shapes of organic
compounds.
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1.10 The Shapes of Some Simple Molecules 27
LEARNING BY MODELING
A
s early as the nineteenth century many
chemists built scale models in order to better
understand molecular structure. We can gain a
clearer idea about the features that affect structure
and reactivity when we examine the three-
dimensional shape of a molecule. Several types of
molecular models are shown for methane in Figure
1.7. Probably the most familiar are ball-and-stick
models (Figure 1.7b), which direct approximately
equal attention to the atoms and the bonds that con-
nect them. Framework models (Figure 1.7a) and
space-filling models (Figure 1.7c) represent opposite
extremes. Framework models emphasize the pattern
of bonds of a molecule while ignoring the sizes of the
atoms. Space-filling models emphasize the volume
occupied by individual atoms at the cost of a clear de-
piction of the bonds; they are most useful in cases in
which one wishes to examine the overall molecular
shape and to assess how closely two nonbonded
atoms approach each other.
The earliest ball-and-stick models were exactly
that: wooden balls in which holes were drilled to ac-
commodate dowels that connected the atoms. Plastic
versions, including relatively inexpensive student
sets, became available in the 1960s and proved to be
a valuable learning aid. Precisely scaled stainless steel
framework and plastic space-filling models, although
relatively expensive, were standard equipment in
most research laboratories.
Computer graphics-based representations are
rapidly replacing classical molecular models. Indeed,
the term “molecular modeling” as now used in or-
ganic chemistry implies computer generation of mod-
els. The methane models shown in Figure 1.7 were all
drawn on a personal computer using software that
possesses the feature of displaying and printing the
same molecule in framework, ball-and-stick, and
space-filling formats. In addition to permitting mod-
els to be constructed rapidly, even the simplest soft-
ware allows the model to be turned and viewed from
a variety of perspectives.
More sophisticated programs not only draw
molecular models, but also incorporate computa-
tional tools that provide useful insights into the elec-
tron distribution. Figure 1.7d illustrates this higher
level approach to molecular modeling by using colors
to display the electric charge distribution within the
boundaries defined by the space-filling model. Fig-
ures such as 1.7d are called electrostatic potential
maps. They show the transition from regions of high-
est to lowest electron density according to the colors
of the rainbow. The most electron-rich regions are
red; the most electron-poor are blue. For methane,
the overall shape of the electrostatic potential map is
similar to the volume occupied by the space-filling
model. The most electron-rich regions are closer to
carbon and the most electron-poor regions closer to
the hydrogen atoms.
(a) (b) (c) (d)
FIGURE 1.7 (a) A framework (tube) molecular model of methane (CH
4
). A framework model shows the bonds
connecting the atoms of a molecule, but not the atoms themselves. (b) A ball-and-stick (ball-and-spoke) model of methane.
(c) A space-filling model of methane. (d ) An electrostatic potential map superimposed on a ball-and-stick model of methane.
The electrostatic potential map corresponds to the space-filling model, but with an added feature. The colors identify regions
according to their electric charge, with red being the most negative and blue the most positive.
—Cont.
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Water, ammonia, and methane share the common feature of an approximately tetra-
hedral arrangement of four electron pairs. Because we describe the shape of a molecule
according to the positions of its atoms rather than the disposition of its electron pairs,
however, water is said to be bent, and ammonia is trigonal pyramidal (Figure 1.9). The
H±O±H angle in water (105°) and the H±N±H angle in ammonia (107°) are slightly
less than the tetrahedral angle.
Boron trifluoride (BF
3
; Figure 1.10) is a trigonal planar molecule. There are 6 elec-
trons, 2 for each B±F bond, associated with the valence shell of boron. These three bonded
pairs are farthest apart when they are coplanar, with F±B±F bond angles of 120°.
PROBLEM 1.17 The salt sodium borohydride, NaBH
4
, has an ionic bond between
Na
H11001
and the anion BH
4
–
. What are the H±B±H angles in the borohydride anion?
Multiple bonds are treated as a single unit in the VSEPR model. Formaldehyde
(Figure 1.11) is a trigonal planar molecule in which the electrons of the double bond and
those of the two single bonds are maximally separated. A linear arrangement of atoms
in carbon dioxide (Figure 1.12) allows the electrons in one double bond to be as far
away as possible from the electrons in the other double bond.
PROBLEM 1.18 Specify the shape of the following:
(a) (c)
(b) H
4
N
H11001
(Ammonium ion) (d) CO
3
2H11002
(Carbonate ion)
SAMPLE SOLUTION (a) The structure shown accounts for all the electrons in
hydrogen cyanide. There are no unshared electron pairs associated with carbon, and
so the structure is determined by maximizing the separation between its single bond
to hydrogen and the triple bond to nitrogen. Hydrogen cyanide is a linear molecule.
N?N?N
H11002H11002H11001
(Azide ion)H±CPN (Hydrogen cyanide)
28 CHAPTER ONE Chemical Bonding
Organic chemistry is a very visual science and
computer modeling is making it even more so. Ac-
companying this text is a CD-ROM entitled Learning
By Modeling. As its name implies, it is a learning tool,
designed to help you better understand molecular
structure and properties, and contains two major
components:
? SpartanBuild software that you can use to build
molecular models of various types include tube,
ball-and-spoke, and space-filling. This text in-
cludes a number of modeling exercises for you
to do, but don’t limit yourself to them. You can
learn a lot by simply experimenting with
SpartanBuild to see what you can make.
? SpartanView software with which you can
browse through an archive of already-prepared
models on the Learning By Modeling CD. These
models include many of the same substances
that appear in this text. SpartanView is the tool
you will use to view electrostatic potential
maps as well as animations of many organic
chemical transformations.
All of the models, those you make yourself and those
already provided on Learning By Modeling, can be
viewed in different formats and rotated in three di-
mensions.
Immediately preceding the Glossary at the back
of this text is a tutorial showing you how to use Spar-
tanBuild and SpartanView, and describing some addi-
tional features.
As you go through this text, you will see two dif-
ferent modeling icons. The SpartanBuild icon alerts
you to a model-building opportunity, the Spartan-
View icon indicates that the Learning By Modeling CD
includes a related model or animation.
SpartanBuild icon SpartanView icon
109.5H11034
C H
H
H
H
109.5H11034
109.5H11034
109.5H11034
FIGURE 1.8 A wedge-and-
dash drawing of the struc-
ture of methane. A solid
wedge projects from the
plane of the paper toward
you; a dashed wedge pro-
jects away from you. A bond
represented by a line drawn
in the customary way lies in
the plane of the paper.
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1.10 The Shapes of Some Simple Molecules 29
O
H
H
(a) Water (H
2
O) has a bent structure.
105H11034
:
:
N
H
H
(b) Ammonia (NH
3
) has a trigonal pyramidal structure.
107°
:
H
FIGURE 1.9 Ball-and-spoke and space-filling models and wedge-and-dash draw-
ings of (a) water and (b) ammonia. The shape of a molecule is described in terms of its atoms. An
approximately tetrahedral arrangement of electron pairs translates into a bent geometry for wa-
ter and a trigonal pyramidal geometry for ammonia.
FIGURE 1.10 Representations of the trigonal planar geometry of boron trifluoride (BF
3
).
There are 6 electrons in the valence shell of boron, a pair for each covalent bond to fluorine. The
three pairs of electrons are farthest apart when the F±B±F angle is 120°.
FIGURE 1.11 Models of formaldehyde (H
2
C?O) showing the trigonal planar
geometry of the bonds to carbon. Many molecular models, including those shown here, show
only the connections between the atoms without differentiating among single bonds, double
bonds, and triple bonds.
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1.11 MOLECULAR DIPOLE MOMENTS
We can combine our knowledge of molecular geometry with a feel for the polarity of
chemical bonds to predict whether a molecule has a dipole moment or not. The molec-
ular dipole moment is the resultant of all of the individual bond dipole moments of a
substance. Some molecules, such as carbon dioxide, have polar bonds, but lack a dipole
moment because their shape (see Figure 1.12) causes the individual C?O bond dipoles
to cancel.
Carbon tetrachloride, with four polar C±Cl bonds and a tetrahedral shape, has no net
dipole moment, because the resultant of the four bond dipoles, as shown in Figure 1.13,
is zero. Dichloromethane, on the other hand, has a dipole moment of 1.62 D. The C±H
bond dipoles reinforce the C±Cl bond dipoles.
Carbon dioxide
Dipole moment H11005 0 DO?C?O
30 CHAPTER ONE Chemical Bonding
Resultant of these two
Cl±C bond dipoles is
in plane of paper
Resultant of these two
C±Cl bond dipoles is
in plane of paper
(a) There is a mutual cancellation of individual bond dipoles in carbon tetrachloride.
It has no dipole moment.
H11002
H11002
H11002
H11002
H11001
Cl
Cl
Cl
Cl
C
δ
δ
δ
δ
δ
Cl
Cl
C
H
H
Resultant of these two
H±C bond dipoles is
in plane of paper
(b) The H±C bond dipoles reinforce the C±Cl bond moment in dichloromethane.
The molecule has a dipole moment of 1.62 D.
H11002
H11002
H11001
H11001
H11001
Resultant of these two
C±Cl bond dipoles is
in plane of paper
δ
δ
δ
δ
δ
FIGURE 1.13 Contri-
bution of individual bond
dipole moments to the mo-
lecular dipole moments of (a)
carbon tetrachloride (CCl
4
)
and (b) dichloromethane
(CH
2
Cl
2
).
FIGURE 1.12
Ball-and-spoke and space-
filling models showing the
linear geometry of carbon
dioxide (O?C?O).
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PROBLEM 1.19 Which of the following compounds would you expect to have
a dipole moment? If the molecule has a dipole moment, specify its direction.
(a) BF
3
(d) CH
3
Cl
(b) H
2
O (e) CH
2
O
(c) CH
4
(f) HCN
SAMPLE SOLUTION (a) Boron trifluoride is planar with 120° bond angles.
Although each boron–fluorine bond is polar, their combined effects cancel and
the molecule has no dipole moment.
1.12 ELECTRON WAVES AND CHEMICAL BONDS
Lewis proposed his shared electron-pair model of bonding in 1916, almost a decade
before de Broglie’s theory of wave–particle duality. De Broglie’s radically different view
of an electron, and Schr?dinger’s success in using wave equations to calculate the energy
of an electron in a hydrogen atom, encouraged the belief that bonding in molecules could
be explained on the basis of interactions between electron waves. This thinking produced
two widely used theories of chemical bonding: one is called the valence bond model,
the other the molecular orbital model.
Before we describe these theories, let’s first think about bonding between two
hydrogen atoms in the most fundamental terms. We’ll begin with two hydrogen atoms
that are far apart and see what happens as the distance between them decreases. The
forces involved are electron–electron (H11002H11002) repulsions, nucleus–nucleus (H11001H11001) repul-
sions, and electron–nucleus (H11002H11001) attractions. All of these forces increase as the dis-
tance between the two hydrogens decreases. Because the electrons are so mobile, how-
ever, they can choreograph their motions so as to minimize their mutual repulsion while
maximizing their attractive forces with the protons. Thus, as shown in Figure 1.14, there
is a net, albeit weak, attractive force between the two hydrogens even when the atoms
are far apart. This interaction becomes stronger as the two atoms approach each other—
the electron of each hydrogen increasingly feels the attractive force of two protons rather
than one, the total energy decreases, and the system becomes more stable. A potential
energy minimum is reached when the separation between the nuclei reaches 74 pm,
which corresponds to the H±H bond length in H
2
. At distances shorter than this, the
nucleus–nucleus and electron–electron repulsions dominate, and the system becomes less
stable.
The valence bond and molecular orbital theories differ in how they use the orbitals
of two hydrogen atoms to describe the orbital that contains the electron pair in H
2
. Both
theories assume that electron waves behave much like more familiar waves, such as
sound and light waves. One property of waves that is important here is called “interfer-
ence” in physics. Constructive interference occurs when two waves combine so as to
reinforce each other (“in phase”); destructive interference occurs when they oppose each
other (“out of phase”) (Figure 1.15). In the valence bond model constructive interference
between two electron waves is seen as the basis for the shared electron-pair bond. In the
molecular orbital model, the wave functions of molecules are derived by combining wave
functions of atoms.
H9262 H11005 0 D
F
W
B
±
±
F F
1.12 Electron Waves and Chemical Bonds 31
All of the forces in chemistry,
except for nuclear chemistry,
are electrical. Opposite
charges attract; like charges
repel. This simple fact can
take you a long way.
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1.13 BONDING IN H
2
: THE VALENCE BOND MODEL
The characteristic feature of valence bond theory is that it describes a covalent bond
between two atoms in terms of an in-phase overlap of a half-filled orbital of one atom
with a half-filled orbital of the other, illustrated for the case of H
2
in Figure 1.16. Two
hydrogen atoms, each containing an electron in a 1s orbital, combine so that their orbitals
overlap to give a new orbital associated with both of them. In-phase orbital overlap (con-
structive interference) increases the probability of finding an electron in the region of
overlap.
Figure 1.17 uses electrostatic potential maps to show the buildup of electron den-
sity in the region between the atoms as two hydrogen atoms approach each other closely
enough for their orbitals to overlap.
Were we to slice through the H
2
molecule perpendicular to the internuclear axis,
its cross section would appear as a circle. We describe the electron distribution in such
a bond as having rotational symmetry and refer to it as a sigma (H9268) bond.
32 CHAPTER ONE Chemical Bonding
(a) Amplitudes of wave functions added
H11545H11001
0
H11002
H11545
Waves reinforce
Nuclei
Distance
(b) Amplitudes of wave functions subtracted
H11545H11001
0
H11002
H11546
Waves cancel
Node
Distance
FIGURE 1.15 Interference between waves. (a) Constructive interference occurs when two
waves combine in phase with each other. The amplitude of the resulting wave at each point
is the sum of the amplitudes of the original waves. (b) Destructive interference in the case of
two phases out of phase with each other causes a mutual cancellation.
Potential energy
Internuclear distance74 pm
H11002436 kJ/mol
(H11002104 kcal/mol)
0 H? H11001 H?
H
-----------
H
H
---------
H
H
-----
H
H±H
FIGURE 1.14 Plot of
potential energy versus dis-
tance for two hydrogen
atoms. At long distances,
there is a weak attractive
force. As the distance de-
creases, the potential energy
decreases, and the system be-
comes more stable because
each electron now feels the
attractive force of two pro-
tons rather than one. The op-
timum distance of separation
(74 pm) corresponds to the
normal bond distance of an
H
2
molecule. At shorter dis-
tances, nucleus–nucleus and
electron–electron repulsions
are greater than electron–
nucleus attractions, and the
system becomes less stable.
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1.13 Bonding in H
2
: The Valence Bond Model 33
H11545H11545
1s orbitals of two
hydrogen atoms
in-phase overlap
of two 1s orbitals
gives new orbital
encompassing both
hydrogen atoms
FIGURE 1.16 Valence bond picture of bonding in H
2
. Overlap of half-filled 1s orbitals of two
hydrogen atoms gives a new orbital encompassing both atoms. This new orbital contains the
two original electrons. The electron density (electron probability) is highest in the region
between the two atoms. The black dots correspond to the nuclei, and the H11001 signs to the signs
of the wave functions. When the wave functions are of the same sign, constructive interfer-
ence leads to an increase in the probability of finding an electron in the region where the two
orbitals overlap.
(a) The 1s orbitals of two separated hydrogen
atoms, sufficiently far apart so that
essentially no interaction takes place
between them. Each electron is associated
with only a single proton.
(b) As the hydrogen atoms approach each
other, their 1s orbitals begin to overlap and
each electron begins to feel the attractive
force of both protons.
(c) The hydrogen atoms are close enough so
that appreciable overlap of the two 1s
orbitals occurs. The concentration of
electron density in the region between the
two protons is more readily apparent.
(d) A molecule of H
2
. The center-to-center
distance between the hydrogen atoms is 74
pm. The two individual 1s orbitals have
been replaced by a new orbital that
encompasses both hydrogens and contains
both electrons. The electron density is
greatest in the region between the two
hydrogens.
FIGURE 1.17 Valence
bond picture of bonding in
H
2
. The drawings illustrate
how the 1s orbitals of two hy-
drogen atoms overlap to give
the orbital that contains both
electrons of a hydrogen mole-
cule. The colors of the rain-
bow, red through violet, are
used to depict highest to low-
est electrostatic potential,
respectively.
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We will use the valence bond approach extensively in our discussion of organic
molecules and expand on it later in this chapter. First though, let’s introduce the molec-
ular orbital method to see how it uses the 1s orbitals of two hydrogen atoms to gener-
ate the orbitals of an H
2
molecule.
1.14 BONDING IN H
2
: THE MOLECULAR ORBITAL MODEL
The molecular orbital approach to chemical bonding is based on the notion that, as elec-
trons in atoms occupy atomic orbitals, electrons in molecules occupy molecular orbitals.
Just as the first task in writing the electron configuration of an atom is to identify the
atomic orbitals that are available to it, so too must we first describe the orbitals avail-
able to a molecule. In the molecular orbital method this is accomplished by represent-
ing molecular orbitals as combinations of atomic orbitals, the linear combination of
atomic orbitals-molecular orbital (LCAO-MO) method.
Take H
2
for example. Two molecular orbitals (MOs) are generated by combining
the 1s atomic orbitals (AOs) of two hydrogen atoms. In one combination, the two wave
functions are added; in the other they are subtracted. The two new orbitals that are pro-
duced are portrayed in Figure 1.18. The additive combination generates a bonding
orbital; the subtractive combination generates an antibonding orbital. Both the bond-
ing and antibonding orbitals have rotational symmetry around the line connecting the
two atoms; they have H9268 symmetry. The two are differentiated by calling the bonding
orbital H9268 and the antibonding orbital H9268* (“sigma star”). The bonding orbital is charac-
terized by a region of high electron probability between the two atoms, and the anti-
bonding orbital has a nodal surface between them.
A molecular orbital diagram for H
2
is shown in Figure 1.19. The customary for-
mat shows the starting AOs at the left and right sides and the MOs in the middle. It must
always be true that the number of MOs is the same as the number of AOs that combine
to produce them. Thus, when the 1s AOs of two hydrogen atoms combine, two MOs
result. The bonding MO (H9268) is lower in energy and the antibonding MO (H9268*) higher in
energy than either of the original 1s orbitals.
34 CHAPTER ONE Chemical Bonding
H11545H11545
(a) Add the 1s wave functions of two hydrogen atoms to
generate a bonding molecular orbital (H9268) of H
2
. There is a
high probability of finding both electrons in the region
between the two nuclei.
H11545H11546
(b) Subtract the 1s wave function of one hydrogen atom from the
other to generate an antibonding molecular orbital (H9268*) of H
2
.
There is a nodal surface where there is a zero probability of
finding the electrons in the region between the two nuclei.
node
FIGURE 1.18 Genera-
tion of H9268 and H9268* molecular or-
bitals of H
2
by combining 1s
orbitals of two hydrogen atoms.
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When assigning electrons to MOs, the same rules apply as for writing electron con-
figurations of atoms. Electrons fill the MOs in order of increasing orbital energy, and the
maximum number of electrons in any orbital is 2. The 2 electrons of H
2
occupy the
bonding orbital, have opposite spins, and both are held more strongly than they would
be in separated hydrogen atoms. There are no electrons in the antibonding orbital.
For a molecule as simple as H
2
, it is hard to see much difference between the
valence bond and molecular orbital methods. The most important differences appear in
molecules with more than two atoms—a very common situation indeed. In those cases,
the valence bond method continues to view a molecule as a collection of bonds between
connected atoms. The molecular orbital method, however, leads to a picture in which the
same electron can be associated with many, or even all, of the atoms in a molecule.
In the remaining sections of this chapter we will use a modification of valence
bond theory to describe CH and CC bonds in some fundamental types of organic com-
pounds.
1.15 BONDING IN METHANE AND ORBITAL HYBRIDIZATION
A vexing puzzle in the early days of valence bond theory concerned the bonding in
methane (CH
4
). Since covalent bonding requires the overlap of half-filled orbitals of the
connected atoms, carbon with an electron configuration of 1s
2
2s
2
2p
x
1
2p
y
1
has only two
half-filled orbitals (Figure 1.20a), so how can it have bonds to four hydrogens?
1.15 Bonding in Methane and Orbital Hybridization 35
Increasing energy
Antibonding
Bonding
1s 1s
Hydrogen 1s
atomic orbital
Molecular orbitals
of H
2
Hydrogen 1s
atomic orbital
FIGURE 1.19 Two molecu-
lar orbitals are generated by
combining two hydrogen 1s
orbitals. One molecular
orbital is a bonding molecu-
lar orbital and is lower in
energy than either of the
atomic orbitals that combine
to produce it. The other
molecular orbital is anti-
bonding and is of higher
energy than either atomic
orbital. Each arrow indicates
one electron; the electron
spins are opposite in sign.
The bonding orbital contains
both electrons of H
2
.
2p
2s
2p
2s
2sp
3
Energy
Ground electronic
state of carbon
Higher energy electronic
state of carbon
sp
3
hybrid
state of carbon
(c)(b)(a)
FIGURE 1.20 (a) Electron
configuration of carbon in
its most stable state. (b) An
electron is “promoted” from
the 2s orbital to the vacant
2p orbital. (c) The 2s orbital
and the three 2p orbitals are
combined to give a set of
four equal-energy sp
3
-
hybridized orbitals, each of
which contains one electron.
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In the 1930s Linus Pauling offered an ingenious solution to the puzzle. He began
with a simple idea: “promoting” one of the 2s electrons to the empty 2p
z
orbital gives
four half-filled orbitals and allows for four C±H bonds (Figure 1.20b). The electron
configuration that results (1s
2
2s
1
2p
x
1
2p
y
1
2p
z
1
), however, is inconsistent with the fact that
all of these bonds are equivalent and directed toward the corners of a tetrahedron. The
second part of Pauling’s idea was novel: mix together (hybridize) the four valence
orbitals of carbon (2s, 2p
x
, 2p
y
, and 2p
z
) to give four half-filled orbitals of equal energy
(Figure 1.20c). The four new orbitals in Pauling’s scheme are called sp
3
hybrid orbitals
because they come from one s orbital and three p orbitals.
Figure 1.21 depicts some of the spatial aspects of orbital hybridization. Each sp
3
hybrid orbital has two lobes of unequal size, making the electron density greater on one
side of the nucleus than the other. In a bond to hydrogen, it is the larger lobe of a car-
bon sp
3
orbital that overlaps with a hydrogen 1s orbital. The orbital overlaps corre-
sponding to the four C±H bonds of methane are portrayed in Figure 1.22. Orbital over-
lap along the internuclear axis generates a bond with rotational symmetry—in this case
a C(2sp
3
)±H(1s) H9268 bond. A tetrahedral arrangement of four H9268 bonds is characteristic
of sp
3
-hybridized carbon.
The peculiar shape of sp
3
hybrid orbitals turn out to have an important consequence.
Since most of the electron density in an sp
3
hybrid orbital lies to one side of a carbon
atom, overlap with a half-filled 1s orbital of hydrogen, for example, on that side produces
a stronger bond than would result otherwise. If the electron probabilities were equal on
both sides of the nucleus, as it would be in a p orbital, half of the time the electron would
be remote from the region between the bonded atoms, and the bond would be weaker.
Thus, not only does Pauling’s orbital hybridization proposal account for carbon forming
four bonds rather than two, these bonds are also stronger than they would be otherwise.
36 CHAPTER ONE Chemical Bonding
2s
sp
3
Combine one 2s and three 2p orbitals to give four equivalent sp
3
hybrid orbitals:
xxxx
y
zzzz
yyy
H11546
H11546
H11546
H11545
H11545
2p
x
2p
y
2p
z
The two lobes of each sp
3
hybrid orbital are of
different size. More of
the electron density is
concentrated on one side of
the nucleus than on the other.
H11545
H11545
H11545
H11546
4
FIGURE 1.21 Representa-
tion of orbital mixing in sp
3
hybridization. Mixing of one
s orbital with three p
orbitals generates four sp
3
hybrid orbitals. Each sp
3
hybrid orbital has 25% s
character and 75% p charac-
ter. The four sp
3
hybrid
orbitals have their major
lobes directed toward the
corners of a tetrahedron,
which has the carbon atom
at its center.
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PROBLEM 1.20 Construct an orbital diagram like that of Figure 1.20 for nitro-
gen in ammonia, assuming sp
3
hybridization. In what kind of orbital is the
unshared pair? What orbital overlaps are involved in the N±H bonds?
1.16 sp
3
HYBRIDIZATION AND BONDING IN ETHANE
The orbital hybridization model of covalent bonding is readily extended to carbon–
carbon bonds. As Figure 1.23 illustrates, ethane is described in terms of a carbon–
carbon H9268 bond joining two CH
3
(methyl) groups. Each methyl group consists of an
sp
3
-hybridized carbon attached to three hydrogens by sp
3
–1s H9268 bonds. Overlap of the
remaining half-filled orbital of one carbon with that of the other generates a H9268 bond
between them. Here is a third kind of H9268 bond, one that has as its basis the overlap of
two sp
3
-hybridized orbitals. In general, you can expect that carbon will be sp
3
-hybridized
when it is directly bonded to four atoms.
PROBLEM 1.21 Describe the bonding in methylsilane (H
3
CSiH
3
), assuming that
it is analogous to that of ethane. What is the principal quantum number of the
orbitals of silicon that are hybridized?
The orbital hybridization model of bonding is not limited to compounds in which
all the bonds are single, but can be adapted to compounds with double and triple bonds,
as described in the following two sections.
1.16 sp
3
Hybridization and Bonding in Ethane 37
H
H
H
H
C
Coming
toward you
Going away
from you
In the plane
of the paper
In the plane
of the paper
109.5H11034
H(1s)±C(2sp
3
)
H9268 bond
FIGURE 1.22 The sp
3
hy-
brid orbitals are arranged
in a tetrahedral fashion
around carbon. Each orbital
contains one electron and
can form a bond with a
hydrogen atom to give a
tetrahedral methane mole-
cule. (Note: Only the major
lobe of each sp
3
orbital is
shown. As indicated in Fig-
ure 1.21, each orbital con-
tains a smaller back lobe,
which has been omitted for
the sake of clarity.)
FIGURE 1.23 Orbital over-
lap description of the
sp
3
–sp
3
H9268 bond between the
two carbon atoms of
ethane.
The C±H and C±C bond
distances in ethane are 111
and 153 pm, respectively,
and the bond angles are
close to tetrahedral.
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1.17 sp
2
HYBRIDIZATION AND BONDING IN ETHYLENE
Ethylene is a planar molecule, as the structural representations of Figure 1.24 indi-
cate. Because sp
3
hybridization is associated with a tetrahedral geometry at carbon,
it is not appropriate for ethylene, which has a trigonal planar geometry at both of its
carbons. The hybridization scheme is determined by the number of atoms to which
the carbon is directly attached. In ethane, four atoms are attached to carbon by H9268
bonds, and so four equivalent sp
3
hybrid orbitals are required. In ethylene, three atoms
are attached to each carbon, so three equivalent hybrid orbitals are required. As shown
in Figure 1.25, these three orbitals are generated by mixing the carbon 2s orbital with
two of the 2p orbitals and are called sp
2
hybrid orbitals. One of the 2p orbitals is
left unhybridized.
38 CHAPTER ONE Chemical Bonding
2s 2s
2sp
2
Energy
Ground electronic
state of carbon
Higher energy electronic
state of carbon
sp
2
hybrid
state of carbon
(c)(b)(a)
2p 2p 2p
FIGURE 1.25 (a) Electron configuration of carbon in its most stable state. (b) An electron is
“promoted” from the 2s orbital to the vacant 2p orbital. (c) The 2s orbital and two of the three
2p orbitals are combined to give a set of three equal-energy sp
2
-hybridized orbitals. One of
the 2p orbitals remains unchanged.
Another name for ethylene
is ethene.
134 pm
121.4H11034
110 pm
C?C
H
H
H
H
(a)
117.2H11034
(b)
FIGURE 1.24 (a) All
the atoms of ethylene lie in
the same plane. All the bond
angles are close to 120°, and
the carbon–carbon bond dis-
tance is significantly shorter
than that of ethane. (b) A
space-filling model of ethyl-
ene.
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Figure 1.26 illustrates the mixing of orbitals in sp
2
hybridization. The three sp
2
orbitals are of equal energy; each has one-third s character and two-thirds p character.
Their axes are coplanar, and each has a shape much like that of an sp
3
orbital.
Each carbon of ethylene uses two of its sp
2
hybrid orbitals to form H9268 bonds to two
hydrogen atoms, as illustrated in the first part of Figure 1.27. The remaining sp
2
orbitals,
one on each carbon, overlap along the internuclear axis to give a H9268 bond connecting the
two carbons.
As Figure 1.27 shows, each carbon atom still has, at this point, an unhybridized
2p orbital available for bonding. These two half-filled 2p orbitals have their axes per-
pendicular to the framework of H9268 bonds of the molecule and overlap in a side-by-side
manner to give what is called a pi (H9266) bond. According to this analysis, the carbon–car-
bon double bond of ethylene is viewed as a combination of a H9268 bond plus a H9266 bond.
The additional increment of bonding makes a carbon–carbon double bond both stronger
and shorter than a carbon–carbon single bond.
Electrons in a H9266 bond are called H9266 electrons. The probability of finding a H9266 elec-
tron is highest in the region above and below the plane of the molecule. The plane of
the molecule corresponds to a nodal plane, where the probability of finding a H9266 electron
is zero.
In general, you can expect that carbon will be sp
2
-hybridized when it is directly
bonded to three atoms.
1.17 sp
2
Hybridization and Bonding in Ethylene 39
x
z
y
x
z
y
x
z
y
Combine one 2s and two 2p orbitals
x
z
y
Leave this
orbital alone
3
H11001
Three sp
2
hybrid orbitals
x
z
y
2p
z
FIGURE 1.26 Representation of orbital mixing in sp
2
hybridization. Mixing of one s orbital
with two p orbitals generates three sp
2
hybrid orbitals. Each sp
2
hybrid orbital has one-third s
character and two-thirds p character. The axes of the three sp
2
hybrid orbitals are coplanar.
One 2p orbital remains unhybridized, and its axis is perpendicular to the plane defined by the
axes of the sp
2
orbitals.
One measure of the strength
of a bond is its bond dissoci-
ation energy. This topic will
be introduced in Section 4.17
and applied to ethylene in
Section 5.2.
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1.18 sp HYBRIDIZATION AND BONDING IN ACETYLENE
One more hybridization scheme is important in organic chemistry. It is called sp
hybridization and applies when carbon is directly bonded to two atoms, as it is in acety-
lene. The structure of acetylene is shown in Figure 1.28 along with its bond distances
and bond angles.
Since each carbon in acetylene is bonded to two other atoms, the orbital hybridiza-
tion model requires each carbon to have two equivalent orbitals available for the for-
mation of H9268 bonds as outlined in Figures 1.29 and 1.30. According to this model the car-
bon 2s orbital and one of the 2p orbitals combine to generate a pair of two equivalent
sp hybrid orbitals. Each sp hybrid orbital has 50% s character and 50% p character. These
two sp orbitals share a common axis, but their major lobes are oriented at an angle of
180° to each other. Two of the original 2p orbitals remain unhybridized. Their axes are
perpendicular to each other and to the common axis of the pair of sp hybrid orbitals.
40 CHAPTER ONE Chemical Bonding
Another name for acetylene
is ethyne.
Begin with two sp
2
hybridized carbon atoms and four hydrogen atoms:
sp
2
sp
2
H
sp
2
sp
2
sp
2
sp
2
sp
2
hybrid orbitals of carbon
overlap to form H9268 bonds to
hydrogens and to each other
C(2sp
2
) –H(1s)
H9268 bond
C(2sp
2
) –C(2sp
2
) H9268 bond
C(2p) –C(2p) H9266 bond
p orbitals that remain on carbons
overlap to form H9266 bond
Half-filled 2p
orbital
In plane of
paper
HH
H
FIGURE 1.27 The carbon–
carbon double bond in ethyl-
ene has a H9268 component and a
H9266 component. The H9268 compo-
nent arises from overlap of
sp
2
-hybridized orbitals along
the internuclear axis. The H9266
component results from a
side-by-side overlap of 2p
orbitals.
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As portrayed in Figure 1.31, the two carbons of acetylene are connected to each
other by a 2sp–2sp H9268 bond, and each is attached to a hydrogen substituent by a 2sp–1s
H9268 bond. The unhybridized 2p orbitals on one carbon overlap with their counterparts on
the other to form two H9266 bonds. The carbon–carbon triple bond in acetylene is viewed as
a multiple bond of the H9268H11001H9266H11001H9266type.
In general, you can expect that carbon will be sp-hybridized when it is directly
bonded to two atoms.
1.18 sp Hybridization and Bonding in Acetylene 41
H ± C P C ± H
(a)(b)
180H11034
106
pm
106
pm
120
pm
180H11034
2s 2s
2sp
Energy
2p 2p 2p
(c)(b)(a)
Ground electronic
state of carbon
Higher energy electronic
state of carbon
sp hybrid
state of carbon
2s
x
y
z
2p
x
2p
y
2p
z
2p
y
2p
z
Combine one 2s and one 2p orbital
Unhybridized p orbitals
Leave these two orbitals alone
2
Two 2sp hybrid orbitals
FIGURE 1.29 (a) Electron
configuration of carbon in
its most stable state. (b) An
electron is “promoted” from
the 2s orbital to the vacant
2p orbital. (c) The 2s orbital
and one of the three 2p
orbitals are combined to
give a set of two equal-energy
sp-hybridized orbitals. Two
of the 2p orbitals remain
unchanged.
FIGURE 1.30 Representa-
tion of orbital mixing in sp
hybridization. Mixing of the
2s orbital with one of the
p orbitals generates two sp
hybrid orbitals. Each sp
hybrid orbital has 50% s
character and 50% p charac-
ter. The axes of the two sp
hybrid orbitals are colinear.
Two 2p orbitals remain
unhybridized, and their axes
are perpendicular to each
other and to the long axis of
the molecule.
FIGURE 1.28 Acety-
lene is a linear molecule as
indicated in the (a) structural
formula and a (b) space-fill-
ing model.
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PROBLEM 1.22 Give the hybridization state of each carbon in the following
compounds:
(a) Carbon dioxide (O?C?O) (d) Propene (CH
3
CH?CH
2
)
(b) Formaldehyde (H
2
C?O) (e) Acetone [(CH
3
)
2
C?O]
(c) Ketene (H
2
C?C?O) (f) Acrylonitrile (CH
2
?CHCPN)
SAMPLE SOLUTION (a) Carbon in CO
2
is directly bonded to two other atoms. It
is sp-hybridized.
1.19 WHICH THEORY OF CHEMICAL BONDING IS BEST?
We have introduced three approaches to chemical bonding in this chapter:
1. The Lewis model
2. The orbital hybridization model (which is a type of valence bond model)
3. The molecular orbital model
Which one should you learn?
Generally speaking, the three models offer complementary information. Organic
chemists use all three, emphasizing whichever one best suits a particular feature of struc-
ture or reactivity. Until recently, the Lewis and orbital hybridization models were used
far more than the molecular orbital model. But that is changing.
42 CHAPTER ONE Chemical Bonding
H
C
C(2sp) –––– H(1s)
H9268 bond
Carbons are connected by a
C(2sp) –––– C(2sp) H9268 bond
C(2p
z
) –––– C(2p
z
) H9266 bond
C(2p
y
) –––– C(2p
y
) H9266 bond
1s
H
C
1s
HHCC
CHHC
2sp
2p
y
2sp
2p
z
2sp
2p
y
2sp
2p
z
FIGURE 1.31 A description
of bonding in acetylene
based on sp hybridization of
carbon. The carbon–carbon
triple bond is viewed as con-
sisting of one H9268 bond and
two H9266 bonds.
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The Lewis rules are relatively straightforward, easiest to master, and the most
familiar. You will find that your ability to write Lewis formulas increases rapidly with
experience. Get as much practice as you can early in the course. Success in organic
chemistry depends on writing correct Lewis structures.
Orbital hybridization descriptions, since they too are based on the shared electron-
pair bond, enhance the information content of Lewis formulas by distinguishing among
various types of atoms, electrons, and bonds. As you become more familiar with a vari-
ety of structural types, you will find that the term “sp
3
-hybridized carbon” triggers a
group of associations in your mind that are different from those of some other term, such
as “sp
2
-hybridized carbon,” for example.
Molecular orbital theory can provide insights into structure and reactivity that the
Lewis and orbital hybridization models can’t. It is the least intuitive of the three meth-
ods, however, and requires the most training, background, and chemical knowledge to
apply. We have discussed molecular orbital theory so far only in the context of the bond-
ing in H
2
. We have used the results of molecular orbital theory, however, several times
without acknowledging it until now. The electrostatic potential map of methane that
opened this chapter and was repeated as Figure 1.7d was obtained by a molecular orbital
calculation. Four molecular orbital calculations provided the drawings that illustrated
how electron density builds up between the atoms in the valence bond (!) treatment of
H
2
(see Figure 1.17). Molecular orbital theory is well suited to quantitative applications
and is becoming increasingly available for routine use via software that runs on personal
computers. You will see the results of molecular orbital theory often in this text, but the
theory itself will be developed only at an introductory level.
1.20 SUMMARY
The first half of this chapter reviews the Lewis model of chemical bonding and the pro-
cedures for writing structural formulas of chemical compounds, especially organic ones.
The second half discusses bonding in terms of the wave nature of electrons and con-
cludes with its application to compounds that contain carbon–carbon single bonds, dou-
ble bonds, and triple bonds.
Section 1.1 A review of some fundamental knowledge about atoms and electrons
leads to a discussion of wave functions, orbitals, and the electron con-
figurations of atoms. Neutral atoms have as many electrons as the num-
ber of protons in the nucleus. These electrons occupy orbitals in order of
increasing energy, with no more than two electrons in any one orbital.
The most frequently encountered atomic orbitals in this text are s orbitals
(spherically symmetrical) and p orbitals (“dumbbell”-shaped).
1.20 Summary 43
Boundary surface of an s orbital
with carbon at its center
Boundary surface of a p orbital
with carbon at its center
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Section 1.2 An ionic bond is the force of electrostatic attraction between two oppo-
sitely charged ions. Atoms at the upper right of the periodic table, espe-
cially fluorine and oxygen, tend to gain electrons to form anions. Ele-
ments toward the left of the periodic table, especially metals such as
sodium, tend to lose electrons to form cations. Ionic bonds in which car-
bon is the cation or anion are rare.
Section 1.3 The most common kind of bonding involving carbon is covalent bond-
ing. A covalent bond is the sharing of a pair of electrons between two
atoms. Lewis structures are written on the basis of the octet rule, which
limits second-row elements to no more than 8 electrons in their valence
shells. In most of its compounds, carbon has four bonds.
Section 1.4 Many organic compounds have double or triple bonds to carbon. Four
electrons are involved in a double bond; six in a triple bond.
Section 1.5 When two atoms that differ in electronegativity are covalently bonded,
the electrons in the bond are drawn toward the more electronegative ele-
ment.
Section 1.6 Counting electrons and assessing charge distribution in molecules is
essential to understanding how structure affects properties. A particular
atom in a Lewis structure may be neutral, positively charged, or nega-
tively charged. The formal charge of an atom in the Lewis structure of
a molecule can be calculated by comparing its electron count with that
of the neutral atom itself.
Formal charge H11005 (number of electrons in neutral atom)
H11002 (number of electrons in unshared pairs)
H11002
1
2
(number of electrons in covalent bonds)
Section 1.7 Table 1.4 in this section sets forth the procedure to be followed in writ-
ing Lewis structures for organic molecules. It begins with experimentally
The electrons in a carbon fluorine bond are
drawn away from carbon, toward fluorine.
H9254H11001
C±F
H9254H11002
±
±
±
Ethylene has a carbon carbon double bond;
acetylene has a carbon carbon triple bond.
C?C
H
H
H
H
±
±
±
±
H±CPC±H
Each carbon has four bonds in ethyl alcohol;
oxygen and each carbon are surrounded by eight electrons.
H±C±C±O±H
H
W
W
H
H
W
W
H
44 CHAPTER ONE Chemical Bonding
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determined information: the molecular formula and the constitution
(order in which the atoms are connected).
Section 1.8 Different compounds that have the same molecular formula are called
isomers. If they are different because their atoms are connected in a dif-
ferent order, they are called constitutional isomers.
Formamide (left) and formaldoxime (right) are constitutional isomers;
both have the same molecular formula (CH
3
NO), but the atoms are con-
nected in a different order.
Section 1.9 Many molecules can be represented by two or more Lewis structures that
differ only in the placement of electrons. In such cases the electrons are
delocalized, and the real electron distribution is a composite of the con-
tributing Lewis structures, each of which is called a resonance form. The
rules for resonance are summarized in Table 1.5.
Section 1.10 The shapes of molecules can often be predicted on the basis of valence
shell electron-pair repulsions. A tetrahedral arrangement gives the max-
imum separation of four electron pairs (left); a trigonal planar arrange-
ment is best for three electron pairs (center), and a linear arrangement
for two electron pairs (right).
￠￡C±N
O
H
H
H
?
±
±
±
C?N
H11001
H11002
O
H
H
H
±
±
±
±
Two Lewis structures (resonance forms) of formamide; the atoms are
connected in the same order, but the arrangment of the electrons is different.
C±N
O
H
H
H
?
±
±
±
C?N
H
H
O±H
±
±
±
The Lewis structure of acetic acid
H±C±C±O±H
H
W
W
H
O
X
1.20 Summary 45
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Section 1.11 Knowing the shape of a molecule and the polarity of its various bonds
allows the presence or absence of a molecular dipole moment and its
direction to be predicted.
Section 1.12 Both modern theories of bonding, valence bond and molecular orbital
theory, are based on the wave nature of an electron. Constructive inter-
ference between the electron wave of one atom and that of another gives
a region between the two atoms in which the probability of sharing an
electron is high—a bond.
Section 1.13 In valence bond theory a covalent bond is described in terms of in-phase
overlap of a half-filled orbital of one atom with a half-filled orbital of
another.
Both water and carbon dioxide have polar bonds,
but water is a polar molecule and carbon dioxide is not.
O?C?O
O
HH
±
±
46 CHAPTER ONE Chemical Bonding
H11001H11002H11001H11001
Overlap of two p orbitals along internuclear axis
gives a H9268 bond.
H11002 H11002 H11002H11001
Section 1.14 In molecular orbital theory, molecular wave functions (MOs) are approx-
imated by combining the wave functions of the molecule’s atoms (AOs).
The number of MOs must equal the number of AOs in the molecule’s
atoms.
Section 1.15 Bonding in methane is most often described by an orbital hybridization
model, which is a modified form of valence bond theory. Four equiva-
lent sp
3
hybrid orbitals of carbon are generated by mixing the 2s, 2p
x
,
2p
y
, and 2p
z
orbitals. The C±H H9268 bonds are formed by overlap of each
half-filled sp
3
hybrid orbital with a half-filled hydrogen 1s orbital.
H11001H11001
H11001
Overlap of an sp
3
-hybridized orbital of carbon with
the 2s orbital of hydrogen to give a C±H H9268 bond.
H11002 H11002
H11001
Section 1.16 The carbon–carbon bond in ethane (CH
3
CH
3
) is a H9268 bond generated by
overlap of an sp
3
orbital of one carbon with an sp
3
orbital of the other.
H11001H11001H11001 H11002 H11002
Overlap of an sp
3
-hybridized orbital of each
of two carbon atoms to give a C±C H9268 bond.
H11002 H11002 H11001
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Section 1.17 Carbon is sp
2
-hybridized in ethylene, and the double bond is considered
to have a H9268 component and a H9266 component. The sp
2
hybridization state
of carbon is derived by mixing the 2s and two of the three 2p orbitals.
Three equivalent sp
2
orbitals result, and the axes of these orbitals are
coplanar. Overlap of an sp
2
orbital of one carbon with an sp
2
orbital of
another produces a H9268 bond between them. Each carbon still has one unhy-
bridized p orbital available for bonding, and “side-by-side” overlap of the
p orbitals of adjacent carbons gives a H9266 bond between them.
1.20 Summary 47
The H9266 bond in ethylene generated by
overlap of p orbitals of adjacent carbons
Section 1.18 Carbon is sp-hybridized in acetylene, and the triple bond is of the H9268H11001
H9266H11001H9266type. The 2s orbital and one of the 2p orbitals combine to give
two equivalent sp orbitals that have their axes in a straight line. A H9268 bond
between the two carbons is supplemented by two H9266 bonds formed by
overlap of the remaining half-filled p orbitals.
The triple bond of acetylene has a H9268 bond component and two H9266 bonds;
the two H9266 bonds are shown here and are perpendicular to each other.
Section 1.19 Lewis structures, orbital hybridization, and molecular orbital descriptions
of bonding are all used in organic chemistry. Lewis structures are used
the most, MO descriptions the least. All will be used in this text.
PROBLEMS
1.23 Each of the following species will be encountered at some point in this text. They all have
the same number of electrons binding the same number of atoms and the same arrangement of
bonds; they are isoelectronic. Specify which atoms, if any, bear a formal charge in the Lewis struc-
ture given and the net charge for each species.
(a) (d)
(b) (e)
(c) C
PC
CPOCPN
NPONPN
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1.24 You will meet all the following isoelectronic species in this text. Repeat the previous prob-
lem for these three structures.
(a) (b) (c)
1.25 All the following compounds are characterized by ionic bonding between a group I metal
cation and a tetrahedral anion. Write an appropriate Lewis structure for each anion, remembering
to specify formal charges where they exist.
(a) NaBF
4
(c) K
2
SO
4
(b) LiAIH
4
(d) Na
3
PO
4
1.26 Determine the formal charge at all the atoms in each of the following species and the net
charge on the species as a whole.
(a) (d)
(b) (e)
(c)
1.27 What is the formal charge of oxygen in each of the following Lewis structures?
(a) (b) (c)
1.28 Write a Lewis structure for each of the following organic molecules:
(a) C
2
H
5
Cl (ethyl chloride: sprayed from aerosol cans onto skin to relieve pain)
(b) C
2
H
3
Cl [vinyl chloride: starting material for the preparation of poly(vinyl chloride),
or PVC, plastics]
(c) C
2
HBrClF
3
(halothane: a nonflammable inhalation anesthetic; all three fluorines are
bonded to the same carbon)
(d) C
2
Cl
2
F
4
(Freon 114: formerly used as a refrigerant and as an aerosol propellant; each
carbon bears one chlorine)
1.29 Write a structural formula for the CH
3
NO isomer characterized by the structural unit indi-
cated. None of the atoms in the final structure should have a formal charge.
(a) C±N?O (c) O±C?N
(b) C?N±O (d) O?C±N
1.30 Consider structural formulas A, B, and C:
(a) Are A, B, and C constitutional isomers, or are they resonance forms?
(b) Which structures have a negatively charged carbon?
(c) Which structures have a positively charged carbon?
(d) Which structures have a positively charged nitrogen?
(e) Which structures have a negatively charged nitrogen?
(f) What is the net charge on each structure?
(g) Which is a more stable structure, A or B? Why?
A
H
2
C±NPN
B
H
2
C?N?N
C
H
2
C±N?N
(CH
3
)
3
O(CH
3
)
2
OCH
3
O
H±C±H
W
H
H±C±HH±C±H
W
H
H±C±H
W
H
H±O±H
W
H
O?N?ON?N?NO?C?O
48 CHAPTER ONE Chemical Bonding
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(h) Which is a more stable structure, B or C? Why?
(i) What is the CNN geometry in each structure according to VSEPR?
1.31 Consider structural formulas A, B, C, and D:
(a) Which structures contain a positively charged carbon?
(b) Which structures contain a positively charged nitrogen?
(c) Which structures contain a positively charged oxygen?
(d) Which structures contain a negatively charged carbon?
(e) Which structures contain a negatively charged nitrogen?
(f) Which structures contain a negatively charged oxygen?
(g) Which structures are electrically neutral (contain equal numbers of positive and nega-
tive charges)? Are any of them cations? Anions?
(h) Which structure is the most stable?
(i) Which structure is the least stable?
1.32 In each of the following pairs, determine whether the two represent resonance forms of a
single species or depict different substances. If two structures are not resonance forms, explain
why.
(a)
(b)
(c)
1.33 Among the following four structures, one is not a permissible resonance form. Identify the
wrong structure. Why is it incorrect?
1.34 Keeping the same atomic connections and moving only electrons, write a more stable Lewis
structure for each of the following. Be sure to specify formal charges, if any, in the new structure.
(a) (d) (g)
(b) (e) (h)
(c) (f ) (i)
H11002H11001
C±N?NH
2
H
H
±
±
H11002
C±C
H
H
O
H
±
±
±
?
H11001H11002
C±C
H
H
H
H
±
±
±
±
H11001
C±OH
H
H
±
±
H11001
C±C?C±O
W
H
W
H
H
H
±
±
H11002
H±C
O±H
?
±
O
H11002
H11001
H11001
H±C?O
H11001
C±C?C±C
W
H
W
H
H
H
±
±
H
H
±
±
H11002
H±C±N?N
H11001
H
W
W
H
A
CH
2
±N±O
H11001 H11002
W
CH
3
B
CH
2
?N±O
H11001 H11002
W
CH
3
D
CH
2
±N?O
H11001H11002
W
CH
3
C
CH
2
?N?O
W
CH
3
N±NPN and N±N±N
N±NPN and N±N?N
N±NPN and N?N?N
A
H±C?N?O
B
H±CPN±O
C
H±CPN?O
D
H±C?N±O
Problems 49
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1.35 (a) Write a Lewis structure for sulfur dioxide in which the octet rule is satisfied for all three
atoms. Show all electron pairs and include any formal charges. The atoms are connected
in the order OSO.
(b) The octet rule may be violated for elements beyond the second period of the periodic
table. Write a Lewis structure for sulfur dioxide in which each oxygen is connected to
sulfur by a double bond. Show all electron pairs and formal charges.
1.36 Write structural formulas for all the constitutionally isomeric compounds having the given
molecular formula.
(a) C
4
H
10
(d) C
4
H
9
Br
(b) C
5
H
12
(e) C
3
H
9
N
(c) C
2
H
4
Cl
2
1.37 Write structural formulas for all the constitutional isomers of
(a) C
3
H
8
(b) C
3
H
6
(c) C
3
H
4
1.38 Write structural formulas for all the constitutional isomers of molecular formula C
3
H
6
O that
contain
(a) Only single bonds (b) One double bond
1.39 For each of the following molecules that contain polar covalent bonds, indicate the positive
and negative ends of the dipole, using the symbol v. Refer to Table 1.2 as needed.
(a) HCl (c) HI (e) HOCl
(b) ICl (d) H
2
O
1.40 The compounds FCl and ICl have dipole moments H9262 that are similar in magnitude (0.9 and
0.7 D, respectively) but opposite in direction. In one compound, chlorine is the positive end of the
dipole; in the other it is the negative end. Specify the direction of the dipole moment in each com-
pound, and explain your reasoning.
1.41 Which compound in each of the following pairs would you expect to have the greater dipole
moment H9262? Why?
(a) NaCl or HCl (e) CHCl
3
or CCl
3
F
(b) HF or HCl (f) CH
3
NH
2
or CH
3
OH
(c) HF or BF
3
(g) CH
3
NH
2
or CH
3
NO
2
(d) (CH
3
)
3
CH or (CH
3
)
3
CCl
1.42 Apply the VSEPR method to deduce the geometry around carbon in each of the following
species:
(a) (b) (c)
1.43 Expand the following structural representations so as to more clearly show all the atoms
and any unshared electron pairs.
(a)
(b) Occurs in bay and verbena oil
A component of high-octane gasoline
CH
2
H11001
CH
3
H11002
CH
3
50 CHAPTER ONE Chemical Bonding
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(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
1.44 Molecular formulas of organic compounds are customarily presented in the fashion
C
2
H
5
BrO
2
. The number of carbon and hydrogen atoms are presented first, followed by the other
atoms in alphabetical order. Give the molecular formulas corresponding to each of the compounds
in the preceding problem. Are any of them isomers?
1.45 Select the compounds in Problem 1.43 in which all the carbons are
(a) sp
3
-hybridized (b) sp
2
-hybridized
Do any of the compounds in Problem 1.43 contain an sp-hybridized carbon?
Hexachlorophene:
an antiseptic
Cl Cl
ClCl
W
Cl
W
Cl
W
OH
W
OH
±
±
Tyrian purple: a purple
dye extracted from a
species of Mediterranean
sea snail
O
O
Br
Br
±
±
?
?
?
H
N
N
H
Nicotine: a toxic substance
present in tobacco
N
N
CH
3
W
Aspirin
OCCH
3
COH
O
X
X
O
Naphthalene: sometimes used as a
moth repellent
Benzene: parent compound of a large
family of organic substances
Found in Roquefort cheese
O
Present in oil of cloves
OH
Pleasant-smelling substance
found in marjoram oil
Problems 51
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1.46 Account for all the electrons in each of the following species, assuming sp
3
hybridization
of the second-row element in each case. Which electrons are found in sp
3
-hybridized orbitals?
Which are found in H9268 bonds?
(a) Ammonia (NH
3
) (e) Borohydride anion (BH
4
H11002
)
(b) Water (H
2
O) (f) Amide anion ( )
(c) Hydrogen fluoride (HF) (g) Methyl anion ( )
(d) Ammonium ion (NH
4
H11001
)
1.47 Imagine describing the bonding in ammonia as arising by overlap of the half-filled unhy-
bridized 2p
x
, 2p
y
, and 2p
z
orbitals of nitrogen with the half-filled 1s orbitals of three hydrogen
atoms.
(a) What kind of orbital would the unshared pair occupy?
(b) What would you expect the bond angles to be?
1.48 Of the orbital overlaps shown in the illustration, one is bonding, one is antibonding, and
the third is nonbonding (neither bonding nor antibonding). Which orbital overlap corresponds to
which interaction? Why?
1.49 Practice working with your Learning By Modeling software. Construct molecular models of
ethane, ethylene, and acetylene, and compare them with respect to their geometry, bond angles,
and C±H and C±C bond distances.
1.50 How many different structures (isomers) can you make that have the formula (a) CH
2
Cl
2
;
(b) Cl
2
C?CH
2
; and (c) ClCH?CHCl?
1.51 Examine the molecular models of H
2
, HF, CH
4
, CH
3
F, and CF
4
. Find the calculated dipole
moment of each compound, and examine their electrostatic potential maps.
1.52 Examine the electrostatic potential map of ethylene. Where is the most negative region?
What kinds of electrons are most responsible for the high electron density in this region? Are they
electrons in H9268 bonds or in the H9266 bond?
1.53 (a) Find the models of I±Br and Cl±F, and compare their calculated dipole moments.
Which is more important, the difference in electronegativity between the bonded halo-
gens or the length of the bond between them? [Remember that the dipole moment
depends on both charge and distance (H9262H11005e H11003 d ).]
(b) Compare the electrostatic potential maps of IBr and ClF. How do they correspond to
the information provided by the dipole moment calculations?
1.54 Compare the dipole moments of cyanogen bromide (BrCPN) and cyanogen chloride
(ClCPN). Which is larger? Why? What does this tell you about the electronegativity of the CN
group?
1.55 Problem 1.8 concerned the charge distribution in methane (CH
4
), chloromethane (CH
3
Cl),
and methyllithium (CH
3
Li). Inspect molecular models of each of these compounds, and compare
them with respect to how charge is distributed among the various atoms (carbon, hydrogen, chlo-
rine, and lithium). Compare their electrostatic potential maps.
H11002
H11001
H11001 H11002 H11002 H11001
(a)(b)(c)
H11001 H11001 H11001
H11002
CH
3
H11002
NH
2
52 CHAPTER ONE Chemical Bonding
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