Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 7
7.1-7.5
Molecular Geometry
and Bonding Theories
Homework
7, 9, 11, 13, 19, 21, 35, 37, 39
43, 45, 47, 49, 79 and 83
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Valence Bond (VB) Theory
Deals with the overlap of the atomic orbitals (AO) of the participation atoms to
form a chemical bond. Due to the overlapping, electrons are localized in the
bond region
However, the atomic orbitals for bonding may not be "pure" atomic orbitals
directly from the solution of the Schrodinger Equation. Often, the bonding
atomic orbitals have a character of several possible types of orbitals. The
methods to get an AO with the proper character for the bonding is called
hybridization. The resulting atomic orbitals are called hybridized atomic
orbitals or simply hybrid orbitals.
Valence Shell Electron Pair Repulsion Theory
Valence Shell Electron Pair Repulsion (VSEPR) theory allows the
Chemist to predict the 3-dimensional shape of molecules from
knowledge of their Lewis Dot structure. In VSEPR theory, the
position of bound atoms (ligands) and electron pairs are described
relative to a central atom. Once the ligands and lone pair electrons
are positioned, the resulting geometrical shape presented by the
atoms only (ignoring lone pairs) is used to describe the molecule.
The VSEPR theory is based on the desire of electrons to be as far
Apart from each other as possible in a molecule
Linear Molecules
There are only two places in the valence shell of the
central atom in BeF2 where electrons can be found.
Repulsion between these pairs of electrons can be
minimized by arranging them so that they point in
opposite directions. Thus, the VSEPR theory predicts that
BeF2 should be a linear molecule, with a 180o angle
between the two Be-F bonds.
Electron domains = 2
VSEPER Notation = AX2
Cl
Be
Cl
lone pairs
on to
central
atom
20
atoms
bonded
central
atom
Trigonal Planar Molecules
There are three places on the central atom in boron trifluoride (BF3)
where valence electrons can be found. Repulsion between these
electrons can be minimized by arranging them toward the corners of
an equilateral triangle. The VSEPR theory therefore predicts a
trigonal planar geometry for the BF3 molecule, with a F-B-F bond
angle of 120o.
Electron Domains = 3
VSEPR Notation AX3
Bent (Angular)
There are three regions of electron density two bonding regions and
and lone pair region
S
O
Electron Domains = 3
O
VSEPR Notation AX2E
Tetrahedral Molecules
BeF2 and BF3 are both two-dimensional molecules, in which the
atoms lie in the same plane. If we place the same restriction on
methane (CH4), we would get a square-planar geometry in which
the H-C-H bond angle is 90o. If we let this system expand into three
dimensions, however, we end up with a tetrahedral molecule in
which the H-C-H bond angle is 109o
Electron Domains = 4
VSEPR Notation AX4
Trigonal Pyramidal
Has four regions of electron density. Three bonding regions and one lone pair
Region. In the trigonal pyramidal structure the H-N-H angle is less than
109° because the lone pair likes to spread out as much as possible
N
H
H
H
Electron Domains = 4
VSEPR Notation = AX2E
Bent (Angular)
Has four regions of electron density. Two bonding regions and two lone pair
Region. In the trigonal pyramidal structure the H-O-H angle is even less than
109° than the trigonal pyramidal because the two lone pair like to spread
out as much as possible
O
H
Electron Domains = 4
H
VSEPR Notation = AX2E2
Trigonal Bipyramidal Molecules
Repulsion between the five pairs of valence electrons on the
phosphorus atom in PF5 can be minimized by distributing these
electrons toward the corners of a trigonal bipyramidal. Three of
the positions in a trigonal bipyramidal are labeled equatorial
because they lie along the equator of the molecule. The other
two are axial because they lie along an axis perpendicular to the
equatorial plane. The angle between the three equatorial
positions is 120o, while the angle between an axial and an
equatorial position is 90o.
Electron Domains = 5
VSEPR Notation = AX5
See-Saw
Because if you turn it side-ways it looks like a see saw. It has five regions
of electron density. four bonding regions and one lone pair region.
In the see-saw structure the lone pair is found in the equatorial plane
because the lone pair likes to spread out as much as possible
F
F
S
F
Electron Domains = 5
F
VSEPR Notation = AX4E
T- Shaped
Because it looks like a T. Has five regions of electron density.
three bonding regions and 2 lone pair regions. In the T-Shped structure the
lone pairs are found in the equatorial plane because the lone pairs
likes to spread out as much as possible
F
Cl
F
F
Electron Domains = 5
VSEPR Notation = AX3E2
Linear
Because all the molecules are in a line with the bond angle being 180°.
Has five regions of electron density, two bonding regions
and three lone pair regions.
F
Xe
F
Electron Domains = 5
VSEPR Notation = AX2E3
Octahedral Molecules
There are six places on the central atom in SF6 where
valence electrons can be found. The repulsion between
these electrons can be minimized by distributing them
toward the corners of an octahedron. The term
octahedron literally means "eight sides," but it is the six
corners, or vertices, that interest us. To imagine the
geometry of an SF6 molecule, locate fluorine atoms on
opposite sides of the sulfur atom along the X, Y, and Z
axes of an XYZ coordinate system.
Electron Domains = 6
Octahedral
Square Pyramidal
Square Planar:
VSEPR notation = AX6
VSEPR notation = AX5E
VSEPR notation = AX4E2
Applying the VSEPR Theory
1. Draw a plausible Lewis Structure
2. Determine the number of electron pairs around the
central atom
3. Identify the electrons and bonding pairs (X) or lone pairs (E)
4. Write the VSEPR Notation for the Molecule
5. Match the shape with the name
Valence shell electron pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic
repulsions between the electron (bonding and nonbonding) pairs.
Steric #
2
Class
AB2
# of bonding
electrons
2
# of lone pair
arrangement
of e- pairs
Molecular
Shape
linear
linear
0
trigonal
planar
3
3
AB3
3
2
trigonal
planar
0
trigonal
planar
AB2E
B
B
1
bent
VSEPR Cont……..
Steric #
Class
# of bonding
electrons
# of lone pair
arrangement
of e- pairs
tetrahedral
4
4
4
AB4
AB3E
AB2E2
4
3
2
Molecular
Shape
tetrahedral
0
tetrahedral
trigonal
pyramidal
tetrahedral
bent
1
2
O
H
H
lone-pair vs. lone pair
lone-pair vs. bonding
bonding-pair vs. bonding
>
>
repulsion
pair repulsion
pair repulsion
VSEPR Cont……..
Steric #
Class
# of bonding
electrons
# of lone pair
arrangement
of e- pairs
trigonal
bipyramidal
5
AB5
5
AB4E
4
trigonal
bipyramidal
0
trigonal
bipyramidal
5
Molecular
Shape
1
See- saw
VSEPR Cont……..
Steric #
Class
# of bonding
electrons
# of lone pair
arrangement
of e- pairs
Molecular
Shape
trigonal
bipyramidal
T-shaped
F
5
AB3E2
3
2
F
Cl
F
trigonal
bipyramidal
linear
I
5
AB2E3
2
3
I
I
VSEPR Cont……..
Steric #
Class
# of bonding
electrons
# of lone pair
arrangement
of e- pairs
octahedral
6
AB6
6
AB5E
5
octahedral
0
octahedral
6
Molecular
Shape
1
square
pyramidal
F
F
F
Br
F
octahedral
6
AB4E2
4
2
F
square
planar
F
F
Xe
F
F
Predicting Molecular Geometry
1. Draw Lewis structure for molecule.
2. Count number of lone pairs on the central atom and
number of atoms bonded to the central atom.
3. Use VSEPR to predict the geometry of the molecule.
What are the molecular geometries of SO2 and SF4?
O
S
AB2E
bent
F
O
F
S
F
AB4E
F
See-Saw
7.1
Molecular shape can be predicted by using the valence-shell electron-pair repulsion
(VSEPR) model.
ABx
A is the central atom surrounded by x B atoms.
x can have integer values of 2 to 6.
The basis of the VSEPR model is that electrons repel each other.
Electrons are found in various domains.
Lone pairs
2 double bonds
Single bonds
Double bonds
Triple bonds
1 single bond
3 single bonds
1 double bond
1 lone pair
1 lone pair
2 electron domains
(on central atom)
3 electron domains
(on central atom)
4 electron domains
(on central atom)
The basis of the VSEPR model is that electrons repel each other.
Electrons will arrange themselves to be as far apart as possible.
Arrangements minimize repulsive interactions.
2 electron domains
Linear
3 electron domains
Trigonal planar
4 electron domains
Tetrahedral
5 electron domains
Trigonal bipyramidal
6 electron domains
Octahedral
The electron domain geometry is the arrangement of electron domains around the central
atom.
The molecular geometry is the arrangement of bonded atoms.
In an ABx molecule, a bond angle is the angle between two adjacent A-B bonds.
180°
109.5°
120°
Linear
90°
Trigonal planar
120°
Tetrahedral
Trigonal bipyramidal
90°
Octahedral
AB5 molecules contain two different bond angles between adjacent bonds.
Axial positions; perpendicular
to the trigonal plane
90°
Equatorial positions; three bonds
arranged in a trigonal plane.
120°
Trigonal bipyramidal
When the central atom in an ABx molecule bears one or more lone pairs, the electrondomain geometry and the molecular geometry are no longer the same.
••
•• ••
••
O =O− O
The steps to determine the electron-domain and molecular geometries are as follows:
Step 1:
Draw the Lewis structure of the molecule or polyatomic ion.
Step 2:
Count the number of electron domains on the central atom.
Step 3:
Determine the electron-domain geometry by applying the VSEPR model.
Step 4:
only.
Determine the molecular geometry by considering the positions of the atoms
Determine the shapes of (a) SO3 and (b) ICl4-.
Strategy Use Lewis structures and the VSEPR model to determine first the
electron-domain geometry and then the molecular geometry (shape).
(a) The Lewis structure of SO3 is:
There are three electron domains on the central atom: one double bond and two
single bonds.
(b) The Lewis structure of ICl4- is:
There are six electron domains on the central atom in ICl4-: four single bonds
and two lone pairs.
Solution
(a) According to the VSEPR model, three electron domains will be arranged in a
trigonal plane. Since there are no lone pairs on the central atom in SO3, the
molecular geometry is the same as the electron-domain geometry. Therefore, the
shape of SO3 is trigonal planar.
Think About It Compare these results with the information in
Figure 7.2 and Table 7.2. Make sure that you can draw Lewis
structures correctly. Without a correct Lewis structure, you will
unable
to determine
the shape
of aoctahedron.
molecule.Two lone pairs on an
(b) Sixbe
electron
domains
will be arranged
in an
octahedron will be located on opposite sides of the central atom, making the
shape of ICl4- square planar.
Some electron domains are better than others at repelling neighboring domains.
Lone pairs take up more space than bonded pairs of electrons.
Multiple bonds repel more strongly than single bonds.
Some electron domains are better than others at repelling neighboring domains.
Lone pairs take up more space than bonded pairs of electrons.
Multiple bonds repel more strongly than single bonds.
The geometry of more complex molecules can be determined by treating them as though
they have multiple central atoms.
Central O atom
No. of electron domains:
4
Electron-domain geometry:
tetrahedral
Molecular geometry:
Central C atom
No. of electron domains:
4
Electron-domain geometry:
tetrahedral
Molecular geometry:
tetrahedral
bent
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
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 would you expect to be smaller than the ideal values?
Strategy 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 bond. The O 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 tetrahedral. 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 O atom is
tetrahedral. Because two of the domains are lone pairs, the molecular geometry
about the O atom is bent.
Solution (cont.) Bond angles are determined using electron-domain geometry.
Therefore, the approximate bond angles about the leftmost C are 109.5°C, those
about the middle C are 120°, and those about the O 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 O will be less than 109.5°
because the lone pairs on O repel the single bonds more strongly than they repel
each other and push the two bonding pairs closer together. The angles are labeled
as follows:
~109.5°
>120°
<120°
<109.5°
Think About It Compare these answers with the information in Figure 7.2 and
Table 7.2
7.2
Molecular polarity is one of the most important consequences of molecular geometry.
A diatomic molecule is polar when the electronegativites of the two atoms are different.
••
H−F
••
δ−
••
••
••
H−F
••
δ+
The polarity of a molecule made up of three or more atoms depends on:
(1)
the polarity of the individual bonds
(2)
the molecular geometry
Carbon dioxide, CO2
The bonds in CO2 are polar but the molecule is nonpolar.
The polarity of a molecule made up of three or more atoms depends on:
(1)
the polarity of the individual bonds
(2)
the molecular geometry
Water, H2O
The bonds in H2O are polar and the molecule is polar.
The polarity of a molecule made up of three or more atoms depends on:
(1)
the polarity of the individual bonds
(2)
the molecular geometry
Boron trifluoride, BF3
The bonds in BF3 are polar but the molecule is nonpolar.
Determine whether PCl5 is polar.
Strategy Draw the Lewis structure, use the VSEPR model to determine its
molecular geometry, and then determine whether the individual bond dipoles
cancel.
(a) The Lewis structure of PCl5 is
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 BF3, and the axial bond dipoles will
also cancel each other.
Solution PCl5 is nonpolar.
Determine whether (b) H2CO (C double bonded to O) is polar.
Strategy Draw the Lewis structure, use the VSEPR model to determine its
molecular geometry, and then determine whether the individual bond dipoles
cancel.
(b) The Lewis structure of H2CO is
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 on both the individual bond
dipoles
and although
the molecular
geometries.
The bond
dipoles,
symmetrically
distributed around the C atom, are not
identical and therefore will not sum to zero.
Solution H2CO is polar.
Dipole moments can be used to distinguish between structural isomers.
trans-dichloroethylene
nonpolar
cis-dichloroethylene
polar
7.3
According to valence bond theory, atoms share electrons when atomic orbitals overlap.
(1) A bond forms when single occupied atomic orbitals on two atoms overlap.
(2) The two electrons shared in the region of orbital overlap must be of opposite spin.
(3) Formation of a bond results in a lower potential energy for the system.
The H−H bond in H2 forms when the singly occupied 1s orbitals of the two H atoms
overlap:
The F−F bond in F2 forms when the singly occupied 2p orbitals of the two F atoms
overlap:
The H−F bond in HF forms when the singly occupied 1s orbital on the H atom overlaps with
the single occupied 2p orbital of the F atom:
Hybridization – mixing of two or more atomic
orbitals to form a new set of hybrid orbitals.
1. Mix at least 2 nonequivalent atomic orbitals (e.g. s
and p). Hybrid orbitals have very different shape
from original atomic orbitals.
2. Number of hybrid orbitals is equal to number of
pure atomic orbitals used in the hybridization
process.
3. Covalent bonds are formed by:
a. Overlap of hybrid orbitals with atomic orbitals
b. Overlap of hybrid orbitals with other hybrid
orbitals
Hybrid Orbitals
Leaving out the math, we can say that an imaginary mixing process
converts a set of atomic orbitals to a new set of hybrid atomic orbitals or
hybrid orbitals.
At this level, we consider the following hybrid orbitals:
sp
sp2
sp3
sp3d
sp3d2
Formation of sp Hybrid Orbitals
The sp Hybrid Atomic Orbitals
The sp hybrid atomic orbitals are possible states of
electron in an atom, especially when it is bonded to
others. These electron states have half 2s and half 2p
characters. From a mathematical view point, there are
two ways to combine the 2s and 2p atomic orbitals:
sp1 = 2s + 2p (yellow)
sp2 = 2s - 2p (green)
These energy states (sp1 and sp2) have a region
of high electron probability each, and the two
atomic orbitals are located opposite to each other,
centered on the atom. The sp hybrid orbitals are
represented by this photograph. A molecule that
is sp hybridized has a steric number of 2
meaning that it has 2 regions of electron density
either as bonds or lone pairs.
H-Be-H
s-sp bond
Be is sp hybridized so the bond between
H and Be is a s-sp bond
Formation of sp2 Hybrid Orbitals
The sp2 Hybrid Orbitals
The energy states of the valence electrons in atoms
of the second period are in the 2s and 2p orbitals. If
we mix two of the 2p orbitals with a 2s orbital, we
end up with three sp2 hybridized orbitals. These
three orbitals lie on a plane, and they point to the
vertices of a equilateral triangle as shown here.
When the central atom has a steric number of 3
it makes use of sp2 hybridized orbitals, and has
a trigonal shape. BF3 is such a molecule:
:
: F:
sp2-sp3 bond
:
:
:
:F
B
F:
:
The sp3 Hybrid Orbitals
Mixing one s and all three p atomic orbitals
produces a set of four equivalent sp3 hybrid atomic
orbitals. The four sp3 hybrid orbitals points towards
the vertices of a tetrahedron, as shown here in this
photograph.
When sp3 hybrid orbitals are used for the central atom in
the formation of molecule, the molecule is said to have
the shape of a tetrahedron and a steric number of 4 and
has the molecular geometry of a tetrahedron.
The typical molecule is CH4, in which the 1s orbital of a H
atom overlap with one of the sp3 hybrid orbitals to form a
C-H bond. Four H atoms form four such bonds, and they
are all equivalent. The CH4 molecule is the most cited
molecule to have a tetrahedral shape. Other molecules
and ions having tetrahedral shapes are SiO44-, SO42-,
As are the cases with sp2, hybrid orbitals, one or two of
the sp3 hybrid orbitals may be occupied by non-bonding
electrons. Water and ammonia are such molecules.
.. ..
O
H
H
s-sp3 bond
Predict correct
bond angle
The sp3d Hybrid Orbitals
The five dsp3 hybrid orbitals resulted when one 3d, one
3s, and three 3p atomic orbitals are mixed. When an atom
makes use of five dsp3 hybrid orbitals the steric number of
the molecule is 5 and it has the molecular geometry of a
trigonal bipyramidal. For example, PClF4 displayed here
forms such a structure. Some of the sp3d hybrid orbitals
may be occupied by electron pairs. The shapes of these
molecules are interesting. In TeCl4, only one of the hybrid
dsp3 orbitals is occupied by a lone pair. This structure may
be represented by TeCl4E, where E represents a lone pair
of electrons. Two lone pairs occupy two such orbitals in
the molecule BrF3, or BrF3E2.
The sp3d2 hybrid orbitals
The six sp3d2 hybrid orbitals resulted when two 3d,
one 3s, and three 3p atomic orbitals are mixed. When
an atom makes use of six sp3d2 hybrid orbitals for
bonds or lone pairs it has a steric number of 6 and
has the molecular geometry of an octahedron.
Number of
Electron Domains
2
Hybridization
3
sp2
4
sp3
5
sp3d
6
sp3d2
sp
How do I predict the hybridization of the central atom?
Count the number of lone pairs AND the number
of atoms bonded to the central atom
(This is also called the steric number)
# of Lone Pairs
+
# of Bonded Atoms
Hybridization
Examples
2
sp
BeCl2
3
sp2
BF3
4
sp3
CH4, NH3, H2O
5
sp3d
PCl5
6
sp3d2
SF6
7.5
Valence bond theory and hybridization can be used to describe the bonding in molecules
containing double and triple bonds.
Each carbon has three electron domains:
2 single bonds
1 double bond
Expect sp2 hybridization
ethylene (C2H4)
A sigma (σ) bond forms when sp2 hybrid orbitals on the C atoms overlap.
In a sigma bond, the shared electron density lies directly along the internuclear axis.
The ethylene molecule contains five sigma bonds:
1 between the two carbon atoms (sp2 and sp2 overlap)
4 between the C and H atoms (sp2 and 1s overlap)
Multiple Bonds
Any single bond between atoms is termed a sigma (s) bonds
Multiple bonds are termed pi (p) bonds
s bond
p bond
Sigma bonds form between hybridized orbitals
p or d orbitals must be free to form multiple bonds
Fig 9-1
Pg 383
Schematic drawing of the
bonding framework of
ethylene. The orbitals of
each carbon atom can be
represented as sp2 hybrids.
p-bonds form from the side by side overlap of adjacent
p orbitals above and below the sigma plane
Fig 9-2
Pg 383
Formation of an sp2 hybrid set leaves one unused valence p orbital.
In ethylene, these orbitals overlap to form a second bond between the
carbon atoms.
Pi bond (p) – electron density above and below plane of nuclei
Sigma bond (s) – electron density between the 2 atoms
of the bonding atoms
A localized p bond can form through
side-by-side overlap between a
d and a p orbital. As with other p
bonds, electron density is
concentrated between the nuclei,
above and below the intermolecular
axis.
All four oxygen atoms are equivalent, so six resonance structures are
needed to represent the p bonding of the sulfate anion. These structures
indicate that delocalized orbitals span the entire anion.
Resonance and Multiple Double Bonds
Delocalized molecular orbitals are not confined between
two adjacent bonding atoms, but actually extend over three
or more atoms.
Resonance structures like ozone or
benzene (right) have overlapping
p orbitals that all lie in the same
plane
Triple Bonds
Acetylene has a triple bond between the two carbons. The sigma
Bond is sp hybridized leaving two p orbitals to overlap
Pz
Pz
Px
Px
Px
Pz
Sigma (s) and Pi Bonds (p)
1 sigma bond
Single bond
Double bond
1 sigma bond and 1 pi bond
Triple bond
1 sigma bond and 2 pi bonds
How many s and p bonds are in the acetic acid
(vinegar) molecule CH3COOH?
H
C
H
O
H
C
O
H
s bonds = 6 + 1 = 7
p bonds = 1