Chemistry part 14, Julia Burdge,2e (2009) pdf

318 CHAPTER 9 Chemical
Bonding
II:
Molecular
Geometry
and
Bonding
Theories
Think
About
It
Compare
these
results with the information
in
Figure 9.2 and Table 9.2. Make sure
that you can draw Lewis structures
cOITectly.
Without a correct
Le
wis
structure, you will
be
unable to
determine the shape
of
a molecule.
molecular geometries where there are one or more lone pairs on the central atom. Note the posi-
tions occupied by the lone pairs in the trigonal bipyramidal electron-domain geometry.
When
there

are lone pairs on the central atom in a trigonal bipyramid, the lone pairs preferentially occupy
equatorial
po
sitions because repulsion is greater
when
the angle between electron domains is 90°
or less. Placing a
lone
pair in an axial
po
sition would put it at 90° to three other electron domains.
Placing it in an equatorial position
put
s it at 90° to only two other domains, thus minimizing the
number
of
strong repulsive interactions.
All
po
sitions are equivalent in the octahedral geometry, so one lone pair on the central atom
can occupy any
of
the
po
sitions.
If
there is a second lone pair in this geometry, though, it
must
occupy the
po

sition opposite the first. This arrangement minimizes the repulsive forces between
the two lone pairs (they are
180° apart instead
of
90° apart).
In summary, the steps to determine the electron-domain and molecular geometries are as
follows:
1.
Draw the Lewis structure
of
the molecule or polyatomic ion.
2.
Count the number
of
electron domains on the central atom.
3.
Determine the electron-domain geometry by applying the
VSEPR
model.
4. Determine the molecular geometry by considering the positions
of
the atoms only.
Sample Problem 9.1 shows how to determine the shape
of
a molecule or poly atomic ion.
Sample Problem 9.1 .
Determine the s
hape
s
of

(a)
50
3
and
(b) ICI
4
.
Strategy Use Lewis structures and the
VSEPR
model
to determine first the electron-domain
geometry and then the molecular geometry (shape).
Setup (a)
The
Lewis structure
of
50
3
is:
••
'0'
• •
• I
•.
:O=S-O:
•
••
There are three electron domains on the central atom: one double
bond
and two single bonds.

(b)
The
Le
wis structure
ofICl
4
is:
••
:CI:
· . .1

:CI-·I
CI:
· . J . . .
:CI:
• •
There are six electron domains on the central atom in
ICl
4
:
four single bonds and two lone pairs.
Solution
(a) According to the
VSEPR
model
, three electron domains will
be
aITanged
in
a trigonal plane.

Since there are no lone pairs on the central atom
in
50
3
,
the molecular geometry is the
same
as the
electron-domain geometry. Therefore, the shape
of
S0
3 is trigonal planar.
Electron-domain geometry: trigonal planar
-_.
Molecular geometry: trigonal planar
(b)
Six electron domains will
be
aITanged in an octahedron. Two
lone
pairs on an octahedron will
be
located
on
opposite s
ide
s
of
the centr'al atom, making the shape
of

ICl
4
square planar.
Electron-domain geometry: octahedral
•
Molecular geometry: square planar
SECTION
9.1
Molecular
Geometry
3
19
Practice Problem A Determine the shapes
of
(a) CO
2
and (b) SCI
2
.
Practice Problem B Determine the shapes
of
(a) SbF
5
and (b) BrF5'
Deviation from Ideal
Bond
Angles
Some
electron domains are better than others at repelling neighboring domains. As a result,
the

bond angles may
be
slightly different from those shown in
Figure
9.2.
For
example, the electron-
domain
geometry
of
ammonia
(NH
3) is tetrahedral, so we
might
predict the
H-N
- H
bond
angles
to
be
109.5°. In fact, the
bond
angles are about 107°, slightly smaller than predicted.
The
lone
pair
on
the nitrogen
atom

repels the N - H
bond
s
more
strongly than the
bond
s repel
one
another.
It
therefore
"squeezes"
them
closer together than the ideal tetrahedral angle
of
109.5°.
In
effect, a lone
pair
takes up
more
space than the bonding pairs.
This
can
be
understood by
considering the attractive forces involved in determining the location
of
the electron pairs. A lone
pair

on
a central
atom
is attracted only to the nucleus
of
that atom. A
bonding
pair
of
electrons,
on
the
other
hand, is simultaneously attracted
by
the nuclei
of
both
of
the
bonding
atoms.
As
a result,
the
lone
pair
ha
s
more

freedom to spread out
and
greater
capa
city to repel
other
electron domains.
Also,
because
they contain
more
electron
den
sity, multiple
bond
s repel
more
strongly than single
bonds.
Consider
the bond angles in
each
of
the following
example
s:
10
1.
5 °
-+ 1

84.8°
BrFS
Geometry
of
Molecules
with
More Than One Central Atom
Thus far
we
have
considered
the geometries
of
molecule
s having only
one
central atom. We can
determine the overall geometry
of
more
complex
molecules by treating them as though they have
multiple central atoms.
Methanol
(CH
3
0 H), for example, has a central C
atom
and a central 0
atom, as shown in the following

Lewis
structure:
H
I

H-C-O-H
I

H
Both the C and the 0
atoms
are sUlTounded by four electron domains. In the case
of
C, they
are three
C-H
bond
s and one
C-O
bond.
In
the
case
of
0 , they are one O- C
bond
, one
O-H
bond, and two lone
pair

s.
In
each case, the electron-domain geometry is tetrahedral. However, the
molecular geometry
of
the C
part
of
the
molecule
is tetrahedral, whereas the
mol
ecular geometry
of the
0
part
of
the
molecule
is bent. Note
that
although the Lewis structure makes it appear as
though there is a
180° angle between
the
O-C
and
O-H
bo
nd

s, the angle is actually approxi-
mately
109.5°, the angle in a tetrahedral arrangement
of
electron domains.
Sample
Problem
9.2
shows
how
to determine when
bond
angles differ f
rom
ideal values.
Acetic acid, the substance that gives vinegar its characteristic smell and sour taste, is sometimes used
in combination with corticosteroids
to
treat certain types
of
ear infections. Its Lewis structure is
• •
H"O'
I II

H-C-C-O-H
I

H
Determine the molecular geometry about each

of
the central atoms, and determine the approximate
value
of
each
of
the bond angles in the molecule. Which if any
of
the bond angles
wo
uld you expect
to be smaller than
th
e ideal values?
(Continued)
When
we
specify
the
geo
metry of a particular
portion
of
a
molecule,
we
refer
to it
as
the

geometry"
about" a particular atom.
In
met
hanol,
for
ex
ample,
we
say
that the
geometry
is
tetrahedral
about the C atom
and
bent about the 0
atom.
320 CHAPTER 9 Chemi
cal
Bo
nding
II
:
Mo
lecular Geomet ry and Bondi
ng
T
heor
i

es
Think
About
It
Compare
these
answers with the information in
Figure 9.2 and Table 9.2.
Strategy
Identify the central atoms and count the number
of
electron domains around each
of
them.
Use
the
VSEPR
model to determine each electron-domain geometry, and the information in Table 9.2
to determine the molecular geometry about each central atom.
Setup
The
leftmost C atom is surrounded
by
four electron domains: one
C-C
bond
and three
C-
H
bonds.

The
middle C atom is surrounded by three electron domains: one C- C bond, one
C-O
bond,
and one
C=O
(double) bond.
The
0 atom is surrounded by four electron domains: one O- C bond,
one
O-H
bond, and two lone pairs.
Solution
The electron-domain geometry
of
the leftmost C
is
tetrahedral. Because all four electron
domains are bonds, the molecular geometry
of
this part
of
the molecule is also tetrahedra
l.
The electron-
domain geometry
of
the middle C is trigonal planar. Again, because all the domains are bonds, the
molecular geometry is also trigonal planar. The electron-domain geometry
of

the 0 atom is tetrahedral.
Because two
of
the domains are lone pairs, the molecular geometry about the 0 atom is bent.
Bond
angles are determined using electron-domain geometry. Therefore, the approximate bond
angles about the leftmost
Care
109.5°, those about the middle
Care
120°, and those about the 0 are
109.5°.
The
angle between the two single bonds on the middle carbon will
be
less than 120° because
the double bond repels the single bonds more strongly than they repel each other. Likewise, the bond
angle between the two bonds on the
0 will
be
less than 109.5° because the lone pairs on 0 repel the
single bonds more strongly than they repel each other and push the two
bond
ing
pa
irs closer together.
The
angles are labeled as follows:
- 109.5°

> 120°
H
'0'
H
< 120° < 109.5°
Practice Problem A Ethanolamine (HOCH
2
CH
2
NH
2
)
has a
sme
ll similar to ammonia and is
commonly found in biological tissue
s.
Its Lewis structure is
H H
. . I I

H-
O-C-C
- N- H
. . I I I
H H H
Determine the molecular geometry about each central atom and label all the bond angles. Cite any
expected deviations from ideal
bond
angles.

,
Practice Problem B Determine the molecular geometry about each central atom in H
2
S0
4
,
Label
any expected deviations from ideal
bond
angles. Its Lewis structure is

'0'
• •
• • I
•.
H-O-S
- O- H

I

:0:

Molecular Geometry
9.1.1
What
are the electron-domain
9.1.2
What
are the electron-domain
geometry and mol

ec
ular geometry
geometry and
mo
lecular geometry
of
CO
~-?
ofC
IO;-?
a)
tetrahedral, trigonal planar a) tetrahedral, trigonal planar
b)
tetrahedral, trigonal pyramidal
b)
tetrahedral, trigonal pyramidal
c)
trigonal pyramidal, trigonal
c)
trigonal pyramidal, trigonal
pyramidal pyramidal
d)
trigonal planar, trigonal planar d) trigonal planar, trigonal planar
e)
tetrahedral, tetrahedral
e)
tetrahedral, tetrahedral
SECTION 9.2 Molecular Geometry and Polarity
321
9.1.3

What
is the approximate value
of
the
bond angle indicated?
H H
I I
H-C=C~H
a)
< 90°
b) < 109.5°
c)
> 109.5°
d)
> 120°
e)
< 120°
9.1.4
What is the approximate value
of
the
bond angle indicated?
H
Ir;:,
H-C-O-H
I
••
H
a)
< 180°

b) > 180°
c)
< 109.5°
d)
>
109S
e)
< 90°
Molecular Geometry and Polarity
Molecular geometry is tremendously important in understanding the physical and chemical behav-
ior
of
a substance.
Molecular
polarity, for example,
is
one
of
the most important consequences
of
molecular geometry, because molecular polarity influences physical, chemical, and biological
properties. Recall from Section 8.4 that a bond between two atoms
of
different electronegativities
is
polar and that a diatomic molecule containing a polar bond is a
polar
molecule. Whether a mol-
ecule made up
of

three or more atoms is polar depends not only on the polarity
of
the individual
bonds, but also on its molecular geometry.
Each
of
the CO
2
and H
2
0 molecules contains two identical atoms bonded to a central atom
and two polar bonds. However, only one
of
these molecules is polar.
To
understand why, think
of
each individual bond dipole as a vector. The overall dipole moment
of
the molecule is determined
by vector addition
of
the individual bond dipoles.
In the case
of
CO
2
,
we have two identical vectors pointing in opposite directions. When the
vectors are placed on a Cartesian coordinate system, they have no

y component and their x compo-
nents are equal in magnitude but opposite in sign. The sum
of
these two vectors is zero in both the
x and y directions. Thus, although the bonds in CO
2
are polar, the molecule is nonpolar.
2:.x
= 0
x t
~-
2:.y
= 0
overall
2:.
= 0
y
. .
The vectors representing the bond dipoles in water, although equal
in
magnitude and opposite
in the x direction, are not opposite in the y direction. Therefore, although their x components sum to
zero, their
y components do not. This means that there is a net resultant dipole and H
2
0 is polm:
x-
'
y
2:.x

= 0
2:.y
* 0
overall
2:.
* 0
Dipole moments can be used to distinguish between molecules that have the same chemical
formula but different arrangements
of
atoms. Such compounds are called structural isomers. For
ex
ample, there are two structural isomers
of
dichloroethylene (C
2
H
2
CI
2
).
Because the individual
bond dipoles sum to zero in trans-dichloroethylene, the
trans isomer is nonpolar:
2:.x
= 0
2:.y
= 0
overall:
2:.
= 0

x
;

-
y
Multimedia
Chemical
bonding-molecu
lar geometry
and polarity (interadi
ve)
.
Recall
that we
can
represent
an
individual
bond
dipole
us
ing
a
crossed
arr
ow that
po
i
nts
toward

the
more
ele
ctronegative atom
[
~~
Section
8.41

Can
Bond
Dipoles
Cancel
One
Another
in
More
Complex
Molecules?
In
ABx
molecules where x
:>
3, it may be less obvious whether
the individual bond dipoles cancel one another. Consider the mol-
ecule
BF
3
,
for example, which has a trigonal planar geometry:

\0
0
4X= 0
x +-7l-;:-+,
~30;:-;
0:-
4Y
= 0
overall 4 = 0
60°
y
We will simplify the math in this analysis by assigning the vec-
tors representing the three identical B - F bonds an arbitrary mag-
nitude
of
1.00. The
x,
y coordinates for the end
of
arrow 1 are (0,
1.00).
Determining the coordinates for the ends
of
arrows 2 and 3
requires the use
of
trigonometric functions.
You
may have learned
the mnemonic

SOH
CAH
TOA, where the letters stand for:
Sin
= Opposite over Hypotenuse
Cos
= Adjacent over Hypotenuse
Tangent
= Opposite over Adjacent
The
x coordinate for the end
of
arrow 2 corresponds to the length
of
the line opposite the 60° angle.
The
hypotenuse
of
the triangle
has a length
of
1.00 (the arbitrarily assigned value). Therefore,
using
SOH,
opposite
sin
60° = 0.866 =
=
~-


hypotenuse
opposite
1
so the
x coordinate for the end
of
arrow 2 is 0.866.
The
magnitude
of
the y coordinate corresponds to the length
of
the line adjacent to the
60
° angle. Using TOA,
tan 60° = 1.73 = opposite = 0.866
adjacent adjacent
. 0.866
adjacent
= = 0.500
1.73
so the
y coordinate for the end
of
arrow 2 is
-0.500.
(The trigo-
nometric formula gives us the length
of
the side. We know from

the diagram that the sign
of
this y component is negative.)
Arrow 3 is similar to arrow
2.
Its x component is equal in
magnitude but opposite in sign, and its
y component is the same
322
magnitude and sign as that for arrow
2.
Therefore, the x and y
coordinates for all three vectors are
x
Y
Arrow 1
0
1
Arrow 2
0.866
-0.500
Arrow 3
-0.866
- 0.500
Sum
= 0 0
Because the individual bond dipoles (represented here as the vec-
tors) sum to zero, the molecule is nonpolar overall.
Although it is somewhat more complicated, a similar anal-
ysis can be done to show that all

x,
y,
and 2 coordinates sum to
zero when there are four identical polar bonds arranged in a tet-
rahedron about a central atom. In fact, any time there are identi-
cal bonds symmetrically distributed around a central atom, with
no lone pairs on the central atom, the molecule will be nonpolar
overall, even
if
the bonds themselves are polar.
In
cases where the bonds are distributed symmetrically
around the central atom, the nature
of
the atoms surrounding the
central atom determines whether the molecule is polar overall.
For example, CCl
4
and CHCl
3
have the same molecular geom-
etry (tetrahedral), but CCl
4
is nonpolar because the bond dipoles
cancel one another.
In
CHCI
3
,
however, the bonds are not all

identical, and therefore the bond dipoles do not sum to zero. The
CHCl
3
molecule is polar.
2
x =
4X
= 0
4Y
= 0
42 = 0
overall 4 = 0
4X
= 0
4Y
* 0
42 = 0
overall: 4 * 0
SECTION 9.2
Molecular
Geometry
and Polarity 323
The bond dipoles
in
the cis isomer do not cancel one another, so cis-dichloroethylene is polar:
2,X
= 0
2,y
* 0
overall:

2,
* 0
x
,k
y
Because
of
the difference in polarity, these two isomers can be distinguished experimentally by
measuring the dipole moment.
I
Sample Problem 9.3 shows you how
to
determine whether a molecule is polar.
. . . .
h Samp'le Problem 9.3
.'
Detemune
whether (a)
PCl
s and (b) H
2
CO
(C double bonded to
0)
are polar.
Strategy
For each molecule, draw the Lewis structure, use the
VSEPR
model to detennine its
molecular geometry, and then

deternline whether the individual bond dipoles cancel.
Setup (a)
The
Lewis structure
of
PCIs is

:CI:

tl~c.r
:q-P~
.
It
.cr.
:CI:
. .

With five identical electron domains around the central atom, the electron-domain and molecular
geometries are trigonal bipyramidal.
The
equatorial bond dipoles will cancel one another,
just
as in
the case
of
BF
3
,
and the axial bond dipoles will also cancel each other.
(b)

The
Lewis structure
of
H
1C
O is
'0'
til
C~
H-7
""'
H
The bond dipoles, although symmetrically distributed around the C atom, are not identical and
therefore will not sum to zero.
Solution (a) PCIs is nonpolar.
(b)
H
2
CO is polar.
Practice Problem A Deternline whether (a) CH1Cl
l
and (b) XeF4 are polar.
Practice Problem B Deternline whether (a)
NBr
3 and (b) BCl
3
are polar.
~

Checkpoint

9.2
Molecular Geometry and Polarity
•
9.2.1
Identify the polar molecules in the 9.2.2
Identify the nonpolar molecules in the
following group: HBr,
CH
4
,
CS
2
.
following group:
SO
l,
NH
3
,
XeF
2
.
a)
HBronly
a)
S0
2,
NH
3
,

and
XeF
2
b)
HBr
and
CS
l
b)
S0
2 only
c)
HBr
, CH
4
,
and CS
2
c)
XeF
2
only
d)
CH
4
and CS
2
d)
S0
2

andXeF
2
e)
CH
4
only
e)
S02
and
NH
3
Think
About
It
Make sure that
your Lewis structures are correct
and that you count electron
domains on the central atom
carefully. This will give you the
correct electron-domain and
molecular geometries. Molecular
polarity depends both on individual
bond dipoles and molecular
geometry.
How
r~
EI
ctF
r
• •

•
In Chapter 6 we learned that although an electron is a particle with a
known
ma
ss, it exhibits wavelike properties. The quantum mechan-
ical model
of
the atom, which gives rise to the familiar shapes
of
s
and p atomic orbitals, treats electrons in atoms as waves, rather than
particles. Therefore, rather than use
,
llTOW
S
to
denote the locations
and spins
of
electrons, we will adopt a convention whereby a singly
occupied orbital will appear as a light color and a doubly occupied
orbital will appear as a darker version
of
the same color. In the rep-
resentations
of
orbitals that follow, atomic s orbitals will
be
repre-
sented as yellow, and atomic

p orbitals will be represented as light
blue
if
singly occupied and darker blue
if
doubly occupied. (Empty
p orbitals will appear white.) When two electrons occupy the same
atomic orbital, their spins are paire
d.
Paired electrons occupy the
same orbital and have opposite spins
[I
~~
Section
6.6J.
324
H-H
• • • •
:F-F:
• •
••
••
H-F:

Lewis dot structures
of
H
2
, F
2

, and
HF
-
F2
HF
Ball-and-stick
models
Two
si
ngly occupied s orbitals
each containing one electron
Two singly occupied
p orbita
ls
each containing one electron
T
wo
singly occupi
ed
orbitals (one s, one p)
each contai
nin
g one electron
Overlapped s orbital
s,
sharing
th
e pair
of
electron

s,
both doubly occupied
Overlapped
p orbital
s,
sharing the pair
of
electron
s,
. both doubly occupied
Overlapped orbitals (one s, one
p),
sharing the pair of electrons, both doubly occupied
Valence Bond Theory
The Lewis theory
of
chemical bonding provides a relatively simple way for us to visualize the
arrangement
of
electrons in molecules. It is insufficient, however, to explain the differences
between the covalent bonds in compounds such as
H
z,
Fz,
and HF. Although Lewis theory describes
the bonds in these three molecules in exactly the s
ame
way, they really are quite different from one
another, as evidenced
by

their bond lengths and
bond
enthalpies listed in Table 9.3. Understand-
ing these differences and why covalent bonds form in
th
e first place requires a bonding model that
combines Lewi
s's
notion
of
atoms sharing electron pairs and the quantum mechanical
de
scriptions
of
atomic orbitals.
According to
valence bond theory, atoms share electrons when an atomic orbital on
one
atom overlaps with an atomic orbital on the other. Each
of
the overlapping atomic orbitals
must contain a single, unpaired electron. Furthermore, the two electrons shared by the bonded
atoms
must
have opposite spins
[
~~
Section
6.6] .
The

nuclei
of
both atoms are attracted to the
shared
pair
of
electrons.
It
is this mutual attraction for the shared electrons that holds the atoms
t
oge
th
er
.
0
Bond Length (A)
Bond
Enthalpy
(kJlmol)
H2
0.74
436.4
F2
1.42
150.6
HF
0.92
568.2
SECTION 9.3 Valence Bond Theory 325
The

H - H bond in
H2
forms when the singly occupied
ls
orbitals
of
the two H atoms
.

. . . .


.

.

. . . .

. . . . . . . . . . . . . . .
.,.
.
• • • •
overlap:
H
H
H H
Similarly, the
F-
F bond in F2 forms when the singly occupied 2p orbitals
of

the two F atoms overlap:
Recall that the ground-state electron configuration
of
the F atom is
[He]2i2
p
5
[
~~
Section
6.8] .
(
The
ground-state orbital diagram
of
F is shown in the margin.)
We can also depict the formation
of
an H - F bond using the valence bond model.
In
this
case, the singly occupied
ls
orbital
of
the H atom overlaps with the singly occupied 2p orbital
of
the F atom:
H H
According to the quantum mechanical model, the sizes, shape

s,
and energies
of
the Is orbital
of
H
and the
2p orbital
of
F are different. Therefore, it is not surprising that the bonds in H
2
,
Fb and
HF
"ary in strength and length.
Wh
y do covalent bonds form? According to valence bond theory, a covalent bond will form
between two atoms
if
the potential energy
of
the
re
sulting molecule is lower than that
of
the iso-
lated atoms. Simply
put
, this means that the formation
of

covalent bonds is exothermic. While this
fact may not
seem
intuitively obvious, you know that energy must
be
supplied to a molecule in
order to
break covalent bonds
[
~~
Section
8.9] . Because the formation
of
a bond is the reverse
process,
we
should expect energy to
be
given off.
, . . . . . . . .


. . . . . . , . . .


.

. .
Valence bond theory also introduces the concept
of

directionality to chemical bonds. For
example, we expect the bond formed by the overlap
of
a p orbital to coincide with the axis along
which the
p orbital lies. Consider the molecule H
2
S.
Unlike the other molecules that we have
~
n
countered
,
H
2
S does
not
have the bond angle that Lewis theory and the
VSEPR
model would
lead us to predict. (With four electron domains on the central atom, we would expect the bond
angle to
be
on
the order
of
1 09
.5
°
.)

In
fact, the H - S - H bond angle is 92°.
Lo
oking at this in terms
Kee
p in
min
d that alt
ho
ugh
th
er
e
are
still j
ust
two el
ect
ron
s,
eac
h at
om
"thinks" it owns
them
bo
t
h,
so
w

he
n the
si
ng
ly
occ
up
ied
or
bita
ls
o
ve
rl
ap, both o
rbital
s
end
up
doub
ly
o
cc
upied.
Orbital diagram for F
Re
call
tha
t the enthalpy
cha

n
ge
for a forward
process
an
d
th
at for t
he
rev
er
se
p
ro
ces
s
di
ffer
only in
sign:
!!.H
forward
= -
!!.
H
r"""rse
[
~
Section
5.3]

326 CHAPTER 9 Chemical Bonding
II:
Molecular
Geometry
and
Bonding
Theories
In
order
for
you
to
understand
the
material
in
t
his
section
a
nd
Sect
i
on
9.4,
you
must
be
able
to draw orbital

diagrams
for ground-
state
ele
c
tron
configurations
[~
~
Section
6.8].
Think
About
It
Because the 4p
orbitals on the Se atom are all
mutually perpendicular,
we
should
expect the angles between bonds
formed by their overlap to
be
approximately
90
°.
of
valence bond theory, the central atom (S) has two unpaired electrons, each
of
which resides in a

'3p
orbitai.
the
'
orbhai'
dIagram
'fo'[ ·
ti1
e ground-state electron configuration
of
the S atom is
S [Ne]
[TI]
IHI1
11
I
?
3s-
3p4
Remember that p orbitals are mutually perpendicular, lying along the x,
y,
and z axes
[
~~
Section
6.7]. We can rationalize the observed bond angle by envisioning the overlap
of
each
of
the singly

occupied
3p
orbitals with the
Is
orbital of a hydrogen atom:
H
H
H
H
In summary, the important features
of
valence bond theory are
as
follows:
o A bond forms when singly occupied atomic orbitals on two atoms overlap.
o The two electrons shared in the region
of
orbital overlap must be
of
opposite spin.
o Formation
of
a bond results in a lower potential energy for the system.
Sample
Problem 9.4 shows how
to
use valence bond theory to explain the bonding in a
molecule.
Sample Problem 9.4
Hydrogen selenide (H

2
Se) is a foul-smelling
ga
s that can cause eye and respiratory tract
infianunation.
Th
e
H-Se-H
bond
angle in H
2
Se is approximately 92°.
Use
valence
bond
theory to
describe the bonding in this molecule.
Strategy
Consider the central
atom's
ground-state electron configuration, and determine what
orbitals are available for bond formation.
Setup
The
ground-state electron configuration
of
Se
is [Ar]4
i3d
J

0
4l.
Its orbital diagram (showing
only the
4p orbitals) is
IHI1 11 I
4p4
Solution Two
of
the 4p orbitals are singly occupied and therefore available for bonding.
The
bonds in
H
2
Se form
as
the result
of
the overlap
of
a hydrogen Is orbital with each
of
these orbitals on the Se atom.
Practice Problem A
Use
valence bond theory to describe the bonding in phosphine (PH
3
),
which
has H -

P-H
bond
angles
of
approximately 94°.
Practice Problem B
Use
valence
bond
theory to describe the bonding in arsine (AsH
3
),
which has
H-A
s
-H
bond angles
of
approximately 92°.
Checkpoint 9.3
Valence Bond Theory
9.3.1 Which
of
the following atoms, in its
ground state, does not have unpaired
electrons? (Select all that apply.)
a)O
b)Be
c)B
d) F e)

Ne
9.3.2
According to valence bond theory, how
many bonds would you expect a nitrogen
atom (in its ground state) to form?
a)2
b)3
c)4
d)5
e)6
SECTION
9.4
Hybridization
of
Atomic Orbitals
327
Hybridization
of
Atomic
Orbitals
Although valence bond theory is useful and can explain more
of
our experimental observations
than Lewis bond theory, it fails to explain the bonding in many
of
the molecules that we encounter.
According to valence bond theory, for example, an atom must have a singly occupied atomic orbital
in order to form a bond with another atom. How then do we explain the bonding in
BeC1
2

? The cen-
tral atom, Be, has a ground-state electron configuration
of
[He]2s
2,
so it has no unpaired electrons.
With no singly occupied atomic orbitals in its ground state, how does.
Be
form two bonds?
Furthermore, in cases where the ground-state electron configuration
of
the central atom does
have the required number
of
unpaired electrons, how do we explain the observed bond angles?
Carbon, like sulfur, has two unpaired electrons in its ground state.
Using valence bond theory
as
our guide, we might envision the formation
of
two covalent bonds with oxygen, as in
CO
2
,
If
. .

. . . . . .

. . . . . .

. . .

.


.
the two unpaired electrons on C (each residing in a 2p orbital) were to form bonds, however, the
O-C-O
bond angle should be on the order
of
90°, like the bond angle in H
2
S. In fact, the bond
angle in
CO
2
is 180°:
Actual bond angle is 180°.
Bond angle should be 90°.
In order to explain these and other observations, we need to extend our discussion
of
orbital over-
lap to include the concept
of
hybridization or mixing
of
atomic orbitals.
The idea

of
hybridization
of
atomic orbitals begins with the molecular geometry and works
backward to explain the bonds and the observed bond angles in a molecule.
To
extend our dis-
cussion
of
orbital overlap and introduce the concept
of
hybridization
of
atomic orbitals, we first
consider beryllium chloride (BeCI
2
), which has two electron domains on the central atom.
Using
its Lewis structure (shown in the margin) and the VSEPR model,
we
predict that BeC1
2
will have a
. . . . .

. , .


. . . .

. .

.

.
CI-
Be
-Cl
bond angle
of
180°.
If
this is true, though, how does Be form two bonds with no
unpaired electrons, and why is the angle between the two bonds
180°?
In order to answer the first part
of
the question, we envision the promotion
of
one
of
the elec-
trons in the 2s orbital to an empty
2p orbital. Recall that electrons can be promoted from a lower
atomic orbital to a higher one
[~
.
Se
ction
6.3].

The
ground-state electron configuration is the one
in which all the electrons occupy orbitals
of
the lowest possible energy. A configuration in which
one or more electrons occupy a
higher energy orbital is called an excited state. An excited state
generally is denoted with a star (e.g., Be* for an excited-state
Be
atom). Showing only the valence
orbitals, we can represent the promotion
of
one
of
the valence electrons
of
beryllium as
Be
promotion
•
Be
*
2p
With one
of
its valence electrons promoted
to
the 2p subshell, the Be atom now has two unpaired
electrons and therefore can form two bonds. However, the orbitals in which the two unpaired elec-
trons reside are different from each other, so we would expect bonds formed as a result

of
the over-
lap
of
these two orbitals (each with a 3p orbital on a
CI
atom) to be different:
2s
3p
2p
3p
Experimentally, though, the bonds in BeCl
2
are identical
in
length and strength.
•• ••
:CI-Be-Cl:
• • • •
The
ground-state orbital
diagram
for Cis:
[ill [ill 11
11
1
•
Media
Player/
MPEG

Content
Hybrid
orbitals-orbital
hybrid
i
zation
a
nd
valence
bond
theory.

- _ _
Multimedia
Hybrid
or
bi
tal
s-o
rbi
tal
hy
br
id
i
zation
(interactive).
Remember
that
Be

is
one
of the
atoms
that
does
not
obey
the
octet
rule
[H~
Section 8.
11
.
The
Lewis
structure of
BeCl,
is
• • • •
:CI-Be-CI:
• • • •
[TIJ
IHIHI1
I
3s
2
3
p

s
Orbital diagram for CI
,
328
CHAPTER
9 Chemical
Bonding
II:
Molecular
Geometry
and
Bonding
Theories
Hybr
id orbitals
are
another
type
of
electron
domain.
In picturing
the
shapes
of
the
resulting
hybrid
orbitals
it

may
help
to
remember
that
orbitals,
l
ike
waves,
can
combine
constructively
or
de
structively
.
We
can
think of the
large
lobe
of
each
sp
hybr
id orbital
as
the
result
of a

constructive
combination
and
the
small
lobe
of
ea
ch
as
the r
esult
of a
destructive
combination.
T
he
ener
gy
requir
ed to promote
an
electron
in
an
atom is
more
than
compensated
for

by
the
e
ner
gy
giv
en
off
when
a
bond
forms
.
Figure 9.3 (
a)
An
atomic s orbital
(yellow) and one atomic
p orbital
(blue) combine
to
form two
sp
hybrid
orbitals (green). The realistic hybrid
orbital shapes are shown
fir
s
t.
The

thinner representations are used
to
keep diagrams clear.
(b)
The 2s
orbital and one of the 2p orbitals
on
Be combine to form two sp hybrid
orbitals. Unoccupied orbitals are shown
in white.
(c
) Like
an
y two electron
domains, the hybrid orbitals
on
Be
are
180
0
apart.
(d)
The hybrid orbitals on
Be each overlap with a singly occupied
3p
orbital
on
a
CI
atom,

Hybridization
of
sand
p Orbitals
In order to explain
ho
w
be
ry
llium
form
s two identical
bond
s,
we
mu
st mix the orbitals
in
which
the unpaired electrons
re
side, thus yielding two
equi
valent orbitals.
The
mixing
of
beryllium's 2s
orbital
with

one
of
its 2p orbitals, a process
known
as hybridization, yields two
hybrid
orbitals
· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . , . . .
that are neither s
nor
p,
but
ha
ve s
ome
character
of
each.
The
hybrid orbitals are designated 2sp
or
simply sp.
Be
*
r
i
IT]
r
1
-+'

, ,
: 2s
2p
I 2p
2p
r-'
2p
2p
,
I I
: 1 1 :
I
I
sp sp :
l T

hybridization
•
I
I

Mixing
of
one
s orbital
and
one
p orbital to yield two sp orbitals
The
mathematical combination

of
the
quantum
mechanical wave functions for an s orbital
and a
p orbital gives rise to two new, equivalent wave functions. As s
hown
in
Figure 9.3(a),
each
· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. -
· . . . .
sp hybrid orbital
ha
s
one
small
lobe
and
one
large lobe and, like any two electron domains
on
an
atom
,
the
y are oriented in opposite directions with a 180
0

angle between them.
The
figure shows
the atomic and hybrid orbitals separately for clarity.
Note
that the hybrid orbitals are s
hown
in two
wa
ys: the first is a mo
re
realistic s
hape
, w
herea
s the s
econd
is a simplified shape that
we
use to
ke
ep
the
figures
clear
and
make the visualization
of
orbitals
ea

sier. Note also that the representa-
tions
of
hybrid orbitals are green. Figure 9.3(b)
and
(c) show the locations
of
the atomic orbitals
and
the
hybrid orbitals, respectivel
y,
relative to the beryllium nucleus.
With two
sp hybrid orbitals,
each
containing a single unpaired electron,
we
can see
how
the
Be
atom
is able to
form
two identical bonds with two CI
atom
s [Figure 9.3(d)].
Each
of

the
singly
occupied
sp hybrid orbitals
on
the
Be
atom
overlaps with the singly occupied 3p atomic orbital
on
a CI atom.
We
can
do a s
imilar
analysis
of
the
bond
s and the trigonal-planar geometry
of
boron
trifluo-
ride (
BF
3)
'
The
ground-state electron configuration
of

the
B atom is [He]2s22pl, containing
just
+
I>
I>
(,>0
c><J
2s
2p
sp sp
sp sp
(a)
y y
I>
x
x
z
(b)
(c)
Cl_
(d)
SECTION
9.4
Hybridization
of
Atomic
Orbitals 329
one unpaired electron. Promotion
of

one
of
the 2s electrons to an empty 2p orbital gives the three
unpaired electrons needed to explain the formation
of
three bonds. The ground-state and excited-
state electron configurations can be represented by
promotion
•
11
1
B B*
Because the three bonds in
BF
3 are identical, we must hybridize the three singly occupied atomic
.


.
orbitals (the one s and two p orbitals) to give three singly occupied hybrid orbitals:
1
1
1 ,
hybridization
•
B*
1
[]
, ,1" 11
f J

111 1
111110
: 2s 2p 2p
12p
1
:
sp2 sp2
sp2 : 2p
1
r

1
______

______
1
Mixing
of
one s orbital and two
p orbitals to yield three sp2 orbitals
Figure 9.4 illustrates the hybridization and bond formation in
BF
3
.
In both cases (i.e., for BeCl? and
BF
3),
some but not all
of
the p orbitals are hybridized.

When the remaining unhybridized atomic
p orbitals do not contain electrons, as in the case
of
BF
3
,
+
+
2s 2p 2p
(a)
y
y
x
x -
-;;
z
z
(b)
i
(c)

'F'
• •
I
B
. /
'
.
F'
.

F.

• •
••
A
hy
brid
orbit
al
anal
ys
is
s
ta
rts
with a known
molecul
ar
geom
et
ry
and
known bo
nd
an
gles.
It
is
not
used

to predict geometri
es.
Figure 9.4 (
a)
An s atomic orbital
and two
p atomic orbitals combine to
form three
Sp 2 hybrid orbitals. (b) The
three
Sp
2 hybrid orbitals on B are
arranged in a trigonal plane. (Empty
atomic orbitals are shown in white.)
(c) Hybrid orbitals on B overlap with
2p orbitals on
F.
330
CHAPTER
9 Chemical Bonding
II:
Molecular
Geometry
and
Bonding
Theories
Em
pty
un
h

yb
rid
ized p or
bi
t
als
are
important in
certa
in
typ
es
of b
on
d formati
on,
as
we will s
ee
in
Ch
ap
t
er
s 10 and 16.
H
I
H-C-H
I
H

CH
4
,.
:Cl:


/\;1:
:CI-P

I
~Cl:

:Cl:

Note th
at
t
he
s
up
e
rscr
ip
ted
n
umbers
in h
ybr
id
o

rbita
l notati
on
ar
e
used
to
designa
te the
nu
mb
er
of ato
mi
c
orbitals
t
ha
t
ha
ve
undergone .
hyb
rid
iz
atio
n.
When
the
su

per
scr
ipt
is
1, it
is
not
shown
(
anal
ogous to
the
su
bs
cr
ip
ts
in
chemic
al
for
mul
a
s).

they'
wiii'
n6't
be
'part 6f the 'dls'

cu
sslon '
cit
bondln'g' in this chapter. As
we
will see in Section 9.5,
though, unhybridlzed atomic orbitals that
do contain electrons are important in our description
of
the bonding in a molecule.
We can now apply the same kind
of
analysis to the methane molecule (
CH
4
).
The
Lewis
structure
of
CH
4
ha
s four electron domains around the central carbon atom. This means that we
need four hybrid orbitals, which in turn means that four atomic orbitals must be hybridized.
The
ground-state electron configuration
of
the C atom contains two unpaired electrons. Promotion
of

one electron from the 2s orbital to the empty 2p orbital yields the four unpaired electrons needed
for the formation
of
four bonds:
C
promotion
•
1
111111
C*
2
p
3
Hybridization
of
the s orbital and the three p orbitals yields four hybrid orbitals designated
Sp
3. We
can then place the electrons that were originally in the
sand
p atomic orbitals into the
Sp
3 hybrid
orbitals.
C*
1 1
hybridiz
ation.
i
11

11 11 11
I i
: sp3
sp3
sp3 sp3 :
L T l
1 1
L
___________________________
_
r
1 1
:
[]
11
11 11
I:
1 1
1 2s 2p 2p 2p 1
I
_______________
J
Mixing
of
one s orbital and
three
p orbitals to yield four sp3 orbitals
The
s
et

of
four sp3 hybrid orbitals on carbon, like any four electron domains on a central atom,
as
sume
s a tetrahedral arrangement. Figure 9.5 illustrates how the hybridization
of
the C atom
results in the formation
of
the four bonds and the 109.5° bond angles observed in
CH
4
.
Hybridization
of
5,
p, and d Orbitals
Recall that
element
s in the third
period
of
the periodic table and beyond do
not
necessarily obey
the octet rule because they have
d orbitals that
can
hold additional electrons
[I

••
Section
8.5].
In
order
to explain the bonding and geometry
of
molecules in which there are
more
than four
electron
domain
s on the central atom, we
must
include d orbitals in
our
hybridization scheme.
PCl
s,
for
example
,
ha
s five electron domains around the P atom. In
order
to explain the five
bonds in this molecule, we will need five singly occupied hybrid orbitals.
The
ground-state
electron configuration

of
the P atom is [Ne]3s
2
3p
3, which contains three unpaired electrons. In
this
ca
se, though, because all three
of
the p orbitals are occupied, promotion
of
an electron from
the
3s
orbital to a 3p orbital would not res
ult
in additional unpaired electrons. However, we can
promote
an electron from the 3s orbital to an
empty
3d
orbital, thus forming the five unpaired
electrons needed:
promotion
•
11 11
11
I
, I
, I

, , I
~
3
p
3
3d
Hybridization
of
the s orbital, the three p orbitals, and one
of
the d orbitals yields hybrid orbitals
. that are designated ·sp
3
d.
After placing the five electrons in the five hybrid orbitals, we can rational-
ize the formation
of
five bonds in the molecule:
1
P * i
[]
11 11
11 I
' 1
-T r r ,
,
i
" 11
-' : 11
' :-I1-'-I

1
: TI ' '1
Ii
"'-1
' 1"-1"-
hybridization
•
: 3s 3p 3p 3p 3d 1
3d 3d 3d
3d

______
1
: sp
3
d :
3d
I
J
1

Mixing
of
one s orbital, three p orbitals,
and one d orbital to yield five
sp
3
d orbitals
1
The

sp
3
d
orbital
s have
shape
s
similar
to those
we
have s
een
for
the s
p,
Sp
2,
and
Sp
3
hybrid
orbitals; that is,
one
large
lobe
and
one
s
mall
lobe.

In
addition,
the
five
hybrid
orbitals
SECTION
9.4
Hybridization
of
Atomic
Orbitals
331
adopt a trigonal bipyramidal arrangement, enabling us to explain the geometry and bond
angles in
PCls:
2s 2p
y
x
z
H
H
H
H
2p
(a)
(b)
(c)
+
x

Cl
2p
y
z
H-
•
•
Figure 9.5 (a) An s atomic orbital
and three
p atomic orbitals combine
to
form four sp3 hybrid orbitals. (b) The
four
Sp3 hybrid orbitals on C are
arranged in a tetrahedron. (c) Hybrid
orbitals on C overlap with
Is
orbitals
on H. For clarity, the small lobes
of
the
hybrid orbitals are not shown.
•
332
CHAPTER
9 Chemical Bonding II:
Molecular
Geometry
and
Bonding

Theories
A similar analysis can be done with the SF
6
molecule. The ground-state electron configura-
tion
of
the S atom is
[Ne]3i3p4,
giving it only two unpaired electrons. In order to obtain the six
unpaired electrons needed
to
form six S
-F
bonds, we must promote two electrons to empty d
orbital
s:
one from the 3s orbital and one from the doubly occupied 3p orbital. The resulting hybrid
orbitals are designated
sp
3
d
2
.
3d
promotion.
S*
[]
3s
1
11

11 11
1
11 11
1 1
3
p
3
3d
2


I
S* i
[]
I 3s
I
I
11 11 11
1
h
ybr
idization
•
1
1 1 1 1 1
3p
3p
3p
3d
3d

13d
3d 3d
I
sp
3
d
2
I
3d
I I I

- -
1-
- - -

L -
r
- -

The
elect
r
ons
that
you
pl
ace
in
hybrid orbitals
ar

e th
ose
that or
ig
ina
lly
resided
in
the atomic
orbitals
th
at
have
undergone
hybr
idizat
ion.
I I

Mixing
of
one s orbital, three p orbital
s,
and two d orbitals
to
yield six sp
3
d
2
orbitals

The six bonds in
SF
6
form, therefore, when each
sp
3
d
2
hybrid orbital on the S atom overlaps with
a singly occupied
2p orbital on an F atom:
F
Table 9.4 shows how the number of electron domains on a central atom corresponds to a
set
of
hybrid orbitals. In general, the hybridized bonding in a molecule can be described using the
following steps:
.


1.
Draw the Lewis structure.
2.
Count the number
of
electron domains on the central atom. This
is
the number
of
hybrid

orbitals necessary
to
account for the molecule's geometry. (This
is
also the number
of
atomic
orbitals that must undergo hybridization.)
3.
Draw the ground-state orbital diagram for the central atom.
4.
Maximize the number
of
unpaired valence electrons by promotion.
S.
Combine the necessary number
of
atomic orbitals to generate the required number
of
hybrid
orbitals.
6.
Place electrons in the hybrid orbitals, putting one electron in each orbital before pairing any
electrons.
It
is
important
to
recognize that we do not use hybrid orbitals to
pr

e
dict
molecular geometries,
but rather to
explain geometries that are already known. As we saw in Section 9.3, the bonding
in many molecules can be explained without the use
of
hybrid orbital
s.
Hydrogen sulfide (H
2
S),
for example, has a bond angle
of
92°. This bond angle
is
best explained without the use
of
hybrid
orbitals.
Number
of
Electron Domains
on
Central
Atom
2
3
4
5

6
Hybrid
Orbitals
sp
SECTION
9.4
Hybridization
of
Atomic
Orbitals 333
Geometry
Linear
Trigonal planar
Tetrahedral
Trigonal blpyramidal
Octahedral
Sample
Problem 9.5 shows how to use hybridization to explain the bonding and geometry
in
a molecule.
I
Sample Problem 9.5
Ammonia (NH3) is a trigonal pyramidal molecule with
H-N
- H bond angles
of
about 107°.
Describe the formation
of
three equivalent N - H bond

s,
and explain the angles between them.
Strategy Starting with a Lewis structure, determine the number and type
of
hybrid orbitals
necessary to rationalize the bonding in NH
3
.
Setup
The
Lewis structure
of
NH3
is

H-N-H
~07°
H
The ground-state electron configuration
of
the N atom is [He]2
i2
p
3 Its valence orbital diagram is
11 11
11
1
2
p
3

Solution Although the N atom has the three unpaired electrons needed to form three
N-H
bond
s,
I we would expect bond angles
of
- 90° (not 107
°)
to form from the overlap
of
the three mutually
perpendicular
2p orbitals. Hybridization, therefore, is necessaty to explain the bonding in NH
3
. (Recall
from Section 9.3 that valence bond theory alone could be used to explain the bonding in H
2
S, where the
bond angles are
- 90°
-hybridization
was unnecessaty.) Although we often need to promote an electron
to maximize the number
of
unpaired electron
s,
no promotion is necessaty for the nitrogen
in
NH
3

. We
already have the three unpaired electrons necessary, and the promotion
of
an electron from the
25
orbital
(Continued)
334
CHAPTER
9 Chemical
Bonding
II:
Mo
lecular
Geometry
and
Bonding
Theories
Think About
It This analysis
agrees with
the
experimentally
observed geometry
and
bond
angles
of
107° in
NH

3
.
to
one
of
the 2p orbitals would not result
in
any additional unpaired electrons. Furthermore, there are no
empty
d orbitals in the second shell. According to the Lewis structure, there are four electron domains
on the central atom (three bonds
and
a lone pair
of
electrons).
Four
electron domains
on
the central
atom require four hybrid orbitals,
and
four hybrid orbitals require the hybridization
of
four atomic
orbitals: one
s and three p. This corresponds to
Sp
3 hybridization. Because the atomic orbitals involved
in the hybridization contain a total
of

five electrons,
we
place five electrons in the resulting hybrid
orbitals. This means that one
of
the hybrid orbitals will contain a
lone
pair
of
electrons:
r- - -


I
i
I""-H
1"""-'1
1"""-'1
1"""""'-'1
I i
: sp3 sp3 sp3 sp3 :
L-

T


~
-
-
-

-,
I I
N:[TIJ
1111111:
I I
I 2s
2p
2p 2p I
I
__
__
__
-
,
~
hybridization
•
I I


- -


Each N - H bond is formed by the overlap between an sp3 hybrid orbital on the N atom and the
Is
atomic orbital on an H atom. Because there are four electron domains on the central atom,
we
expect
them to be arranged in a tetrahedron.

In
addition, because one
of
the electron domains is a lone pair,
we
expect the H - N - H
bond
angles to
be
slightly smaller than the ideal tetrahedral
bond
angle
of
109.5°:
H
H
H
Practice Problem A
Use
hybrid orbital theory
to
describe the
bonding
and
explain the
bond
angles
in
bromine
pentafluoride (BrFs).

Practice Problem B
Use
hybrid orbital theory to describe the
bonding
and explain the
bond
angles
in
BeF
2
:
8°
F
,
:
·F
Checkpoint
9.4
Hybridization
of
Atomic Orbitals
9.4.1
How
many orbitals does a
set
of
Sp
2
hybrid
orbitals contain?

a)
2
b)
3
c) 4
d)
5
e)
6
9.4.2
How
many p atomic orbitals are
required to generate a set
of
Sp
3 hybrid
orbitals?
a) 0
b) 1
c) 2
d) 3
e)
4
SECTION
9.S
Hybridization
in Molecules
Containing
Multiple
Bonds 335

Hybridization in Molecules Containing
Multiple Bonds
The concept
of
valence bond theory and hybridization can also
be
used to describe the bonding in
molecules containing double and triple bonds, such as ethylene (C
2
H
4
)
and acetylene (C
2
H
2
).
The
Lewis structure
of
ethylene is
H H
\ /
l C\
H H
Each carbon atom is surrounded by three electron domains (two single bonds and one double
bond).
T~us,
we expect the hybridization about each C atom
to

be s/,
just
like the B atom in
BF
3
.
Applying the procedure described in Section 9.4, we first maximize the number
of
unpaired elec-
trons by promoting an electron from the 2s orbital to the empty
2p
orbital:
promotion
IT]
' +.
C*
11
11
11 1
•
2s1
2
p
3
We
then hybridize the required number
of
atomic orbitals, which in this case is three (one for each
electron domain on the C atom):
,

I


I I
:11 11 1
11:
: sp2 sp2 sp2 :
- - - -
-1-
- - - I
hybridization
•
IT]
2p
C*:
IT]
1 1 1
: 2s 2p
2p
'2p
'T J
~
- - - - - - - - - - - - - - - - - - - - - - - -
_I
The three equivalent
Sp2
hybrid orbitals, arranged in a trigonal plane, enable us to explain the
three bonds about each C atom. In this case, however, each C atom is left with a singly occupied,
unhybridized atomic orbital. As we will see, it is the singly occupied p orbitals
not

involved in
hybridization that give rise to multiple bonds in molecules.
In the bonding schemes that we have described thus far, the overlap
of
atomic orbitals or
hybrid orbitals occurs directly between the two nuclei involved in bonding. Such bonds, in which
the shared electron density is concentrated directly along the internuclear axis, are called
sigma
(a) bonds. The ethylene molecule (also known
as
ethene) contains five sigma bonds: one between
the two C atoms (the result
of
the overlap of one
of
the
Sp2
hybrid orbitals on each C atom) and
four between the C and H atoms (each the result
of
the overlap
of
an sp? hybrid orbital on a C atom
and the
Is
orbital on an H atom). The leftover unhybridized p orbital
is
perpendicular to the plane
in which the atoms
of

the molecule lie. Figure 9.6(a) illustrates the formation and the overlap of
s
p2
hybrid orbitals and shows the positions
of
the remaining p orbital on each C atom.
Hybridization
Remaining
p orbitals
shown
as
simplified shape
Formation
of
sigma bond
(a)
Actual shape
of
remaining p orbitals
(b)
7t
Two lobes
of
a pi bond
Figure 9.6 (a) A sigma bond
forms when
sl
hybrid orbitals on the
C atoms overlap. Each C atom has
one remaining unhybridized

p orbital.
(b) The remaining
p orbitals overlap to
form a
pi
bond.
336 CHAPTER 9 Chemical Bonding
II
:
Molecular
Geometry
and
Bonding
Theories
Figure 9.7 Ethylene.
Multimedia
C
hemical
bonding-sigma-pi bondi
ng
.

-~
:
Multimed
ia
Chem
ical bo
ndi
ng-

s
igma
and
pi
bonding
in
be
nzen
e.
For
more
on
the history of thalidomide,
see
t
he
Chapter
Opener
in
Chapter
10.
Think
About
It
The Lewis
structure given for thalidomide
is one
of
two possible resonance
structures. Draw the other

resonance structure, and count
sigma and pi bonds again. Make
sure you get the same answer.
H
Remember that the shapes used
to
represent the atomic and hybrid orbitals are simplified to
make visualization
of
the molecules easier. The actual shapes
of
both atomic and hybrid orbitals
are such that when the
sp2 hybrid orbitals overlap
to
form a sigma bond between the two C atoms,
the remaining unhybridized
p orbitals also overlap, although to a smaller extent. Figure 9.6(b)
shows the overlap
of
the unhybridized p orbitals on the two C atoms in ethylene. The resulting
regions
of
electron density are concentrated above and below the plane
of
the molecule, in contrast
to a sigma bond in which the electron density is concentrated directly along the internuclear axis.
Bonds that form from the sideways overlap
of
p orbitals are called

pi
(7T)
bonds. The two regions
of
overlap shown in Figure 9.6(b) together make up one pi bond. It
is
the formation
of
the pi bond
that makes the ethylene molecule planar. Figure 9.7 shows all the bonds in the ethylene molecule.
Note how the regions
of
electron density in a pi bond are shown when the simplified shapes are
used
to
represent orbitals in a molecule.
A sigma bond and a pi bond together constitute a
double bond. Because the sideways over-
lap
of
p orbitals is not
as
effective
as
the overlap
of
hybrid orbitals that point directly toward each
other, the contribution
of
the pi bond to the overall strength

of
the bond is less than that of the
sigma bond. The bond enthalpy
of
a carbon-carbon double bond (620 kJ/mol) is greater than that
of
a carbon-carbon single bond (347 kJ/mol), but it is not twice
as
large
[
~~
Section
8.9] .
Sample
Problem 9.6 shows how to use a Lewis structure to determine the number
of
sigma
and pi bonds in a molecule.
. . .
Sample
Pro151em
9.6
Thalidomide
(C
13
H
IO
N
2
0

4
)
is
a sedative and antiemetic that was widely prescribed during the 1950s,
although not in the
United States, for pregnant women suffering from morning sickness. Its use was
largely discontinued when it was determined to
be
responsible for thousands
of
devastating birth
defect
s.
Determine the number
of
carbon-carbon sigma bonds and the total number
of
pi bonds in
thalidomide.
a
N <
~N
a a
Thalidomide
Fa
Strategy
Use the Lewis structure to determine the number
of
single and double bonds. Then, to
convert the number

of
single and double bonds
to
the number
of
sigma and pi bonds, remember that a
single bond is composed
of
a sigma bond, whereas a double bond is usually composed
of
one sigma
bond and one pi bond.
Setup
There are nine carbon-carbon single bonds and three carbon-carbon double bonds. Overall
there are seven double bonds in the molecule (three
C=C
and four
C=O).
Solution Thalidomide
cOI!.tains
12 carbon-carbon sigma bonds and a total
of
seven pi bonds (three
in carbon-carbon double bonds and four in carbon-oxygen double bonds).
•
SECTION 9.5
Hybridization
in Molecules
Containing
Multiple

Bonds 337
Practice Problem A The active ingredient
in
Tylenoll!nd a host
of
other over-the-counter pain
relievers is acetaminophen
(C
8
H
9
N0
2
).
Determine the total number
of
sigma and pi bonds in the
acetaminophen molecule.
o
HC)lN
3 H
Acetaminophen
I Practice Problem B Determine the total number
of
sigma and pi bonds in a molecule
of
aspirin
(C
9
H

s
0
4
) ·
o
OH
Aspirin
Because the
p orbitals that form pi bonds must be parallel
to
each other, pi bonds restrict
the rotation
of
a molecule in a way that sigma bonds do not. For example, the molecule 1,2-
. . . . . . . . . .

. . . . .

.
dichloroethane exists
as
a single isomer. Although we can draw the molecule in several different
ways, including the two shown in Figure 9.8(a), all
of
them are equivalent because the molecule
can rotate freely about the sigma bond between the two carbon atoms.
On the other hand, 1,2-dichloroethylene exists
as
two distinct isomers cis and
trans-as

shown in Figure 9.8(b). The double bond between the carbon atoms consists
of
one sigma bond
Cl H
I I
H-C-
C
-Cl
I I
H H
H
H
Cl
H H
I I .
H-C-C-Cl
I I
Cl H
(a)
Cl
(b)
H
H
Cl
Cl
H
•
Remember
that
isomers

are
molecules
with the
same
chemical
formula but different structural
arrangements
of
atoms
(Section
9.2).
H H
I I
Cl-C-C-Cl
I I
H H
Cl
H
Figure 9.8 (a) There is free rotation about the
C-C
sing
le
bond. All three Lewis structures represent the same molecule. (b) There is no rotation
about the
C-C
double bond. There are two isomers
of
CHCICHCl.
338
CHAPTER

9 Chemical
Bonding
II:
Molecular
Geometry
and
Bonding
Theories
§o
c~
_''\ r~

''_
Media
Pl
ayer/MPEG
Animation:
Figure
9.10,
Formation
of
pi
Bonds
in
Ethylene
and
Acet
y
lene
,

pp
.
340-341
.
•
(a)
H
H
(b)
Figure 9.9 (a) Formation
of
the sigma bond in acetylene. (b) Formation of the pi bonds in acetylene.
and one pi bond.
The
pi
bond
restricts rotation about the
sigma
bond, making the molecules rigid,
planar, and not interchangeable. To change one
isomer
into the other, the pi
bond
would have to
be broken and rotation would have to occur about the sigma
bond
and the pi bond. This process
would require a significant input
of
energy.

The
acetylene molecule (C
1
H
2
)
is linear. Because each carbon atom has two electron domains
around it in the Lewis structure, the carbon atoms are
sp hybridized. As before, promotion
of
an
electron first maximizes the number
of
unpaired electrons:
11 11 1
2p2
promotion. C*
[1J
2s1
11 11
11
1
2
p
3
The
2s orbital and one
of
the 2p orbitals then mix to form two
sp

hybrid orbital
s:
1
r ,
C
l
[]
hybridization
1 1
1
1
1
1
1 1
1
1 1
"'I
•
1 1
1
: 2s 2p 1
2p
2p
1 1
2p
1
sp sp
1
_
_______

J
L
__
-r-_ J
1
L
_________________________
1
This leaves two unhybtidized p orbitals (each containing an electron) on each C atom. Figure
9.9 shows the
sigma
and pi bonds in the acetylene molecule (also known as ethyne). Just as one
sigma
bond
and one pi bond make up a
double
bond, one
sigma
bond
and two pi bonds make up
a
triple bond. Figure 9.10 (pp. 340-341) summarizes the formation
of
bonds in ethane, ethylene,
and acetylene.
Sample
Problem 9.7 shows how hybrid orbitals and pi bonds can be used to explain the
bonding in formaldehyde, a molecule with a carbon-oxygen double bond.
Samp'le
In addition to its use in aqueous solution as a preservative for laboratory specimens, formaldehyde

gas is used as an antibacterial fumigant.
Use hybridization to explain the bonding in formaldehyde
(CH
2
0).
-
SECTION
9.S
Hybridization
in Molecules Co
ntaining
Mu
l
tip
le Bonds 339
Strategy
Draw
the
Lewis
structure
of
formaldehyde, determine the hybridization
of
the C
and
0
atoms,
and
describe
the

formation
of
the
sigma
and pi bonds
in
the
molecule.
Setup
The
Lewis structure
of
fo
rmaldehyde
is
'0'
II
H-
C-H
The
C
and
0 atoms
each
have three electron
domain
s around them. [Carbon has two single
bond
s
(C-H)

and a
double
bond
(C=O);
oxygen
has a
double
bond
(O=C)
and
two
lone
pairs.]
Solution
Three
electron domains correspond to
Sp2
hybridization.
For
carbon,
promotion
of
an
electron
from
the 2s orbital to the
empty
2p orbital is necessary to
maximize
the

number
of
unpaired
electrons.
For
oxygen, no
promotion
is necessary.
Each
undergoes hybridization to
produce
Sp2
hybrid orbitals;
and
each
is left with a singly occupied, unhybridized p orbital:
1-
- - - - - - - - - - I


hybridization
•
I I
11111
1
11
1
OJ
:
sp2 sp2

sp2:
2p
- - - - -, - _

I I
L
____
_
___
______
______
_ _ _ _ _
r-
-

-

I I
0:
[ill
IHI1
11
1
I
I
2s
2p
2p ' 2p
L _
_________

.J
,

-,
I
liHIHI1
I:
OJ
I
:
sp2
sp2
sp2
I 2p
hybridization
•
_____
1_ - - - J
I I
L
____________
_____
__
_____
_
A
sigma
bond
is
formed

between
the C
and
0
atoms
by the overlap
of
one
of
the
Sp2
hybrid orbitals
from
each
of
them. Two
more
sigma
bonds
form
between
the C
atom
and
the
H atoms by the overlap
of
carbon's
remaining
Sp2

hybrid orbitals with the
ls
orbital
on
each H atom. Finally, the remaining p
. . . . . . .

. . . . . . . . . . .


. . . . . . . .



.
orbitals
on
C and 0 overlap to form a pi bond:
H
H
H
Practice
Problem
A
Use
valence
bond
theory and hybrid orbitals to explain the bonding
in
hydrogen

cyanide
(HCN)
.
Practice
Problem
B
Use
valence
bond
theory
and
hybrid orbitals to explain the
bonding
in
diatomic
nitrogen (N2)'
~!
_

9.5.1
Hybridization in Molecules Containing
Multiple Bonds
Which
of
the fo
ll
owing molecules
9.5.2
From
left to right, give the

contain one or
more
pi
bonds?
(Select
hybridization
of
each
carbon atom in
all
that
apply.)
the allene
molecule
(H
2
C=C=CH
2
).
a)
N2
a)
2 2 2
sp , sp , sp
b)
Cl
2
b)
Sp3,
Sp

2,
Sp
3
c)
CO
2
c)
? ?
sp-, sp, sp-
•
d)
CH
3
0H
d)
Sp
3, sp, sp3
e)
CCl
4
e)
3 3 3
sp , sp , sp
The
two
lone
pairs
on
the
0 atom

are
the
electrons
in
the doubly
occupied
Sp2
hybrid
orbitals.
Think
About
It
Our
analysis
describes the formation
of
both
a
sigma
bond
and a pi
bond
between
the C
and
0 atoms. This
corresponds correctly to the
double
bond
predicted by the

Lewis
structure.