Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr.,
and Bruce E. Bursten
Chapter 9
Molecular Geometries
and Bonding Theories
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Overview of Chapter 9 – Molecular Geometry
Review/On your own:
•
•
•
•
the basics of VSEPR theory – the
reasoning (valence electrons repel one
another)
shapes based on VSEPR - linear (two
electron domains), trigonal planar (three
domains), tetrahedral (four domains),
trigonal bipyramidal (five domains) and
octahedral (six domains)
use Lewis Structures to predict molecular
geometry
predict polarity of simple molecules,
knowing their shapes, presence of lone
pairs and differences in electronegativity
New Material:
•
the use of electron domains, and
especially nonbonding pairs of electrons
(lone pairs) to determine a molecule’s
shape, including shapes based on
trigonal bipyramidal and octahedral
•
the effect of lone pairs on bond angles
due to their greater repulsion than
bonded pairs of electrons
•
application of VSEPR to larger
molecules
•
express molecular polarity in terms of
dipole moments
•
valence bond theory
•
hybrid orbitals
•
sigma and pi bonds in multiple bonds Molecular
Geometries
and resonance structures
and Bonding
© 2009, Prentice-Hall, Inc.
Molecular Shapes
• The shape of a
molecule plays an
important role in its
reactivity.
• By noting the number
of bonding and
nonbonding electron
pairs we can easily
predict the shape of
Molecular
the molecule.
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
What Determines the Shape of a
Molecule?
• Simply put, electron
pairs, whether they be
bonding or nonbonding,
repel each other.
• By assuming the electron
pairs are placed as far as
possible from each other,
we can predict the shape
of the molecule.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Electron Domains
• The central atom in
this molecule, A,
has four electron
domains.
• We can refer to the
electron pairs as electron
domains.
• In a double or triple bond,
all electrons shared
between those two atoms
are on the same side of
the central atom;
therefore, they count as
one electron domain.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Valence Shell Electron Pair
Repulsion Theory (VSEPR)
“The best
arrangement of a
given number of
electron domains is
the one that
minimizes the
repulsions among
them.”
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Electron-Domain
Geometries
These are the
electron-domain
geometries for two
through six electron
domains around a
central atom.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Electron-Domain Geometries
• All one must do is
count the number of
electron domains in
the Lewis structure.
• The geometry will
be that which
corresponds to the
number of electron
domains.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Molecular Geometries
• The electron-domain geometry is often not
the shape of the molecule, however.
• The molecular geometry is that defined by the
positions of only the atoms in the molecules,
Molecular
not the nonbonding pairs.
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Exercise 9.1 (p. 350)
Use the VSEPR model to predict the
molecular geometries of
a) O3
b) SnCl3-
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Exercise 9.1 (p. 350)
Use the VSEPR model to predict the
molecular geometries of
a) O3 bent (single lone pair)
b) SnCl3trigonal pyramidal
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.1)
Consider the AB3 molecules and ions: PCl3,
SO3, AlCl3, SO32-, and CH3+. How many of
these molecules and ions do you predict to
have a trigonal-planar geometry?
a)1
b)2
c)3
d)4
e)5
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.1)
Consider the AB3 molecules and ions: PCl3,
SO3, AlCl3, SO32-, and CH3+. How many of
these molecules and ions do you predict to
have a trigonal-planar geometry?
a)1
b)2
c)3
d)4
e)5
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 2 (9.1)
Predict the electron-domain geometry
and the molecular geometry for
a) SeCl2
bent (2 lone pairs)
b) CO32trigonal planar
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Molecular Geometries
Within each electron
domain, then, there
might be more than
one molecular
geometry.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Linear Electron Domain
• In the linear domain, there is only one
molecular geometry: linear.
• NOTE: If there are only two atoms in the
molecule, the molecule will be linear no
matter what the electron domain is.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Trigonal Planar Electron Domain
• There are two molecular geometries:
– Trigonal planar, if all the electron domains are
bonding,
– Bent, if one of the domains is a nonbonding pair. Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Nonbonding Pairs and Bond Angle
• Nonbonding pairs are physically
larger than bonding pairs.
• Therefore, their repulsions are
greater; this tends to decrease
bond angles in a molecule.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Multiple Bonds and Bond Angles
• Double and triple
bonds place greater
electron density on
one side of the
central atom than do
single bonds.
• Therefore, they also
affect bond angles.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Tetrahedral Electron Domain
• There are three molecular geometries:
– Tetrahedral, if all are bonding pairs,
– Trigonal pyramidal if one is a nonbonding pair,
– Bent if there are two nonbonding pairs.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Trigonal Bipyramidal Electron
Domain
• There are two
distinct positions in
this geometry:
– Axial
– Equatorial
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Trigonal Bipyramidal Electron
Domain
Lower-energy conformations result from
having nonbonding electron pairs in
equatorial, rather than axial, positions in this
geometry.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Trigonal Bipyramidal Electron
Domain
• There are four
distinct molecular
geometries in this
domain:
–
–
–
–
Trigonal bipyramidal
Seesaw
T-shaped
Linear
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Octahedral Electron Domain
• All positions are
equivalent in the
octahedral domain.
• There are three
molecular
geometries:
– Octahedral
– Square pyramidal
– Square planar
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Exercise 9.2 (p. 354)
Use the VSEPR model to predict the
molecular geometry of
a) SF4
b) IF5
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.2)
A certain AB4 molecule has a square-planar molecular geometry.
Which of the following statements about the molecule is or are true?
i.The molecule has four electron domains about the central atom
ii.The B-A-B angles between neighboring B atoms is 90o.
iii. The molecule has two nonbonding pairs of electrons on atom A.
a) Only one of the statements is true.
b) Statements (i) and (ii) are true.
c) Statements (i) and (iii) are true.
d) Statements (ii) and (iii) are true.
e) All three statements are true.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.2)
A certain AB4 molecule has a square-planar molecular geometry.
Which of the following statements about the molecule is or are true?
i.The molecule has four electron domains about the central atom
ii.The B-A-B angles between neighboring B atoms is 90o.
iii. The molecule has two nonbonding pairs of electrons on atom A.
a) Only one of the statements is true.
b) Statements (i) and (ii) are true.
c) Statements (i) and (iii) are true.
d) Statements (ii) and (iii) are true.
e) All three statements are true.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 2 (9.2)
Predict the electron-domain geometry
and molecular geometry of
a) BrF5+
b) SF5+
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Larger Molecules
In larger molecules,
it makes more
sense to talk about
the geometry about
a particular atom
rather than the
geometry of the
molecule as a
whole.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Larger Molecules
This approach
makes sense,
especially because
larger molecules
tend to react at a
particular site in the
molecule.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Exercise 9.3 (p. 355)
Eyedrops for dry eyes usually contain a
water-soluble polymer called polyvinylalcohol,
which is based on the unstable organic
molecule called vinyl alcohol. Predict the
approximate values for the H-O-C and O-C-C
bond angles in vinyl alcohol.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.3)
The atoms of the compound methylhydrazine, CH6N2,
which is used as a rocket propellant, are connected as
follows (note that lone pairs are not shown): See p. 356
in your textbook.
a)What do you predict for the ideal values of the C-N-N
and H-N-H angles, respectively?
b)109.5o and 109.5o
c)109.5o and 120o
d)120o and 109.5o
e)120o and 120o
Molecular
Geometries
f)None of the above
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.3)
The atoms of the compound methylhydrazine, CH6N2,
which is used as a rocket propellant, are connected as
follows (note that lone pairs are not shown): See p. 356
in your textbook.
a)What do you predict for the ideal values of the C-N-N
and H-N-H angles, respectively?
b)109.5o and 109.5o
c)109.5o and 120o
d)120o and 109.5o
e)120o and 120o
Molecular
Geometries
f)None of the above
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 2 (9.3)
Predict the H—C—H and C—C—C
bond angles in the molecule shown,
called propyne.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Polarity
• In Chapter 8 we
discussed bond dipoles.
• But just because a
molecule possesses
polar bonds does not
mean the molecule as a
whole will be polar.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Polarity
By adding the
individual bond
dipoles, one can
determine the
overall dipole
moment for the
molecule.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Polarity
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Exercise 9.4 (p. 357)
Predict whether the following molecules
are polar or nonpolar:
a) BrCl
b) SO2
c) SF6
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.4)
Consider an AB3 molecule in which A and B differ in
electronegativity. You are told that the molecule has an
overall dipole moment of zero. Which of the following
could be the molecular geometry of the molecule?
a)Trigonal pyramidal
b)Trigonal planar
c)T-shaped
d)Tetrahedral
e)More than one of the above
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.4)
Consider an AB3 molecule in which A and B differ in
electronegativity. You are told that the molecule has an
overall dipole moment of zero. Which of the following
could be the molecular geometry of the molecule?
a)Trigonal pyramidal
b)Trigonal planar
c)T-shaped
d)Tetrahedral
e)More than one of the above
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 2 (9.4)
Determine whether the following
molecules are polar or nonpolar:
a) SF4
b) SiCl4
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Overlap and Bonding
• We think of covalent bonds forming through
the sharing of electrons by adjacent atoms.
• In such an approach this can only occur when
orbitals on the two atoms overlap.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Overlap and Bonding
• Increased overlap brings
the electrons and nuclei
closer together while
simultaneously
decreasing electronelectron repulsion.
• However, if atoms get too
close, the internuclear
repulsion greatly raises
the energy.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
But it’s hard to imagine tetrahedral, trigonal
bipyramidal, and other geometries arising
from the atomic orbitals we recognize.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
• Consider beryllium:
– In its ground electronic
state, it would not be
able to form bonds
because it has no
singly-occupied orbitals.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
But if it absorbs the
small amount of
energy needed to
promote an electron
from the 2s to the 2p
orbital, it can form two
bonds.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
• Mixing the s and p orbitals yields two degenerate
orbitals that are hybrids of the two orbitals.
– These sp hybrid orbitals have two lobes like a p orbital.
– One of the lobes is larger and more rounded as is the s
orbital.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
• These two degenerate orbitals would align
themselves 180 from each other.
• This is consistent with the observed geometry of
beryllium compounds: linear.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
• With hybrid orbitals the orbital diagram for
beryllium would look like this.
• The sp orbitals are higher in energy than the
1s orbital but lower than the 2p.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
Using a similar model for boron leads to…
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
…three degenerate sp2 orbitals.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
With carbon we get…
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
…four degenerate
sp3 orbitals.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
Current evidence suggests hybridization
involving d orbitals does not exist.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Exercise 9.5 (p. 365)
Indicate the hybridization of orbitals
employed by the central atom in each of
the following:
a) NH2-
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.5)
For which of the following molecules or ions does the
following description apply? “The bonding can be
explained using a set of sp2 hybrid orbitals on the
central atom, with one of the hybrid orbitals holding a
nonbonding pair of electrons.”
a)CO2
b)H2S
c)O3
d)CO32e)more than one of the above
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.5)
For which of the following molecules or ions does the
following description apply? “The bonding can be
explained using a set of sp2 hybrid orbitals on the
central atom, with one of the hybrid orbitals holding a
nonbonding pair of electrons.”
a) CO2
b) H2S
c) O3
d) CO32e) more than one of the above
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 2 (9.5)
Predict the electron-domain geometry
and the hybridization of the central atom
in
a) SO32-
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Hybrid Orbitals
Once you know the
electron-domain
geometry, you know
the hybridization
state of the atom.
Note: ignore
hybridizations involving
d orbitals – probably do
not exist
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Valence Bond Theory
• Hybridization is a major player in this
approach to bonding.
• There are two ways orbitals can overlap
to form bonds between atoms.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sigma () Bonds
• Sigma bonds are characterized by
– Head-to-head overlap.
– Cylindrical symmetry of electron density about the
internuclear axis.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Pi () Bonds
• Pi bonds are
characterized by
– Side-to-side overlap.
– Electron density
above and below the
internuclear axis.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Single Bonds
Single bonds are always  bonds, because 
overlap is greater, resulting in a stronger bond
and more energy lowering.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Multiple Bonds
In a multiple bond one of the bonds is a  bond
and the rest are  bonds.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Multiple Bonds
• In a molecule like
formaldehyde (shown
at left) an sp2 orbital
on carbon overlaps in
 fashion with the
corresponding orbital
on the oxygen.
• The unhybridized p
orbitals overlap in 
fashion.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Multiple Bonds
In triple bonds, as in
acetylene, two sp
orbitals form a 
bond between the
carbons, and two
pairs of p orbitals
overlap in  fashion
to form the two 
bonds.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Exercise 9.6 (p. 368)
Describe how the bonds in
formaldehyde are formed in terms of
overlaps of appropriate hybridized and
unhybridized orbitals.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.6)
We have just arrived at a bonding description for the formaldehyde
molecule. Which of the following statements about the molecule is or are
true?
i.Two of the electrons in the molecule are used to make the p bond in the
molecule.
ii.Six of the electrons in the molecule are used to make the s bonds in the
molecule.
iii.The C-O bond length in formaldehyde should be shorter than that in
methanol, H3COH.
a)Only one of the statements is true.
b)Statements (i) and (ii) are true.
c)Statements (i) and (iii) are true.
d)Statements (ii) and (iii) are true.
e)All three statements are true.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.6)
We have just arrived at a bonding description for the formaldehyde
molecule. Which of the following statements about the molecule is or are
true?
i.Two of the electrons in the molecule are used to make the p bond in the
molecule.
ii.Six of the electrons in the molecule are used to make the s bonds in the
molecule.
iii.The C-O bond length in formaldehyde should be shorter than that in
methanol, H3COH.
a)Only one of the statements is true.
b)Statements (i) and (ii) are true.
c)Statements (i) and (iii) are true.
d)Statements (ii) and (iii) are true.
e)All three statements are true.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 2 (9.6)
Consider the acetonitrile molecule:
a) Predict the bond angles around each carbon atom
b) Describe the hybridization at each of the carbon
atoms
c) Determine the total number s and p bonds in the
molecule.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Delocalized Electrons: Resonance
When writing Lewis structures for species like
the nitrate ion, we draw resonance structures to
more accurately reflect the structure of the
molecule or ion.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Delocalized Electrons: Resonance
• In reality, each of the four
atoms in the nitrate ion has a
p orbital.
• The p orbitals on all three
oxygens overlap with the p
orbital on the central nitrogen.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Delocalized Electrons: Resonance
This means the  electrons are
not localized between the
nitrogen and one of the
oxygens, but rather are
delocalized throughout the ion.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Resonance
The organic molecule
benzene has six 
bonds and a p orbital
on each carbon atom.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Resonance
• In reality the  electrons in benzene are not
localized, but delocalized.
• The even distribution of the electrons in benzene
makes the molecule unusually stable.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Exercise 9.7 (p. 371)
Describe the bonding in the nitrate ion,
NO3-. Does this ion have delocalized 
bonds?
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.7)
How many electrons are in the p system of the
ozone molecule, O3?
a)2
b)4
c)6
d)14
e)18
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 1 (9.7)
How many electrons are in the  system of the
ozone molecule, O3?
a)2
b)4
c)6
d)14
e)18
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Practice Exercise 2 (9.7)
Which of the following molecules or ions
will exhibit delocalized bonding?
SO2, SO3, SO32-, H2CO, NH4+.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Integrative Exercise (p. 386)
Elemental sulfur is a yellow solid that consists of
S8 molecules. The structure of the S8 molecule
is a puckered, eight-membered ring.
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.
Sample Integrative Exercise (p. 386)
Heating elemental sulfur to high temperatures
produces gaseous S2 molecules.:
S8(s)  4 S2(g)
a) With respect to electronic structure, which
element in the second row of the periodic table is
most similar to sulfur?
b) Use the VSEPR model to predict the S-S-S bond
angles in S8 and the hybridization at S in S8.
d) Use bond enthalpies (Table 8.4) to estimate the
enthalpy change for the reaction just described.
Is the reaction exothermic or endothermic?
Molecular
Geometries
and Bonding
© 2009, Prentice-Hall, Inc.