10.

1 Artificial Sweeteners: Fooled by Lecture Presentation

Molecular Shape 425
10.2 VSEPR Theory: The Five Basic Chapter 10
Shapes 426
Chemical
10.3 VSEPR Theory: The Effect of Lone
Pairs 430 Bonding II: Molecular
10.4 VSEPR Theory: Predicting Shapes, Valence
Molecular Geometries 435 Bond Theory, and
10.5 Molecular Shape and Polarity 438
Molecular Orbital
Polarity 438
10.6 Valence Bond Theory: Orbital Theory
Orbital Overlap as a Chemical Bond 443
Bond 443 Sherril Soman
10.7 Valence Bond Theory: Grand Valley State University
Hybridization of Atomic Orbitals 445
10.8 Molecular Orbital Theory: Electron
Electron Delocalization 458
© 2014 Pearson Education, Inc.
© 2014 Pearson Education, Inc.
 Example 10.1 VSEPR Theory and the Basic Shapes
Determine the molecular geometry of NO3−.

Solution
The molecular geometry of NO3− is determined by the number of electron groups around the central atom (N).
Begin by drawing a Lewis structure of NO3−.

NO3− has 5 + 3(6) + 1 = 24 valence electrons. The Lewis structure has three resonance structures:

The hybrid structure is intermediate between these three and has three equivalent bonds.

Use any one of the resonance structures to determine the number of electron groups around the central atom.

The nitrogen atom has three electron groups.

Based on the number of electron groups, determine the geometry that minimizes the repulsions between the groups.

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Nivaldo J. Tro
 Example 10.1 VSEPR Theory and the Basic Shapes
Continued
The electron geometry that minimizes the repulsions between three electron groups is trigonal planar.

Since the three bonds are equivalent (because of the resonance structures), they each exert the same repulsion on
the other two and the molecule has three equal bond angles of 120°.

For Practice 10.1

Determine the molecular geometry of CCl4.

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 Example 10.2 Predicting Molecular Geometries
Predict the geometry and bond angles of PCl3.
Procedure For…
Predicting Molecular Geometries

Solution
Step 1 Draw the Lewis structure for the molecule.
PCl3 has 26 valence electrons.

Step 2 Determine the total number of electron groups around the central atom. Lone pairs, single bonds,
double bonds, triple bonds, and single electrons each count as one group.

The central atom (P) has four electron groups.

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 Example 10.2 Predicting Molecular Geometries
Continued
Step 3 Determine the number of bonding groups and the number of lone pairs around the central atom.
These should sum to your result from step 2. Bonding groups include single bonds, double bonds, and
triple bonds.

Three of the four electron groups around P are bonding groups and one is a lone pair.

Step 4 Refer to Table 10.1 to determine the electron geometry and molecular geometry. If no lone pairs
are present around the central atom, the bond angles will be that of the ideal geometry. If lone pairs are
present, the bond angles may be smaller than the ideal geometry.

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 Example 10.2 Predicting Molecular Geometries
Continued

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 Example 10.2 Predicting Molecular Geometries
Continued
The electron geometry is tetrahedral (four electron groups) and the molecular geometry—the shape of
the molecule—is trigonal pyramidal (three bonding groups and one lone pair). Because of the presence
of a lone pair, the bond angles are less than 109.5°.

For Practice 10.2

Predict the molecular geometry and bond angle of ClNO.

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 Example 10.3 Predicting Molecular Geometries
Predict the geometry and bond angles of ICl4−.
Procedure For…
Predicting Molecular Geometries

Solution
Step 1 Draw the Lewis structure for the molecule.
ICl4− has 36 valence electrons.

Step 2 Determine the total number of electron groups around the central atom. Lone pairs, single bonds,
double bonds, triple bonds, and single electrons each count as one group.

The central atom (I) has six electron groups.

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 Example 10.3 Predicting Molecular Geometries
Continued
Step 3 Determine the number of bonding groups and the number of lone pairs around the central atom.
These should sum to your result from step 2. Bonding groups include single bonds, double bonds, and
triple bonds.

Four of the six electron groups around I are bonding groups and two are lone pairs.

Step 4 Refer to Table 10.1 to determine the electron geometry and molecular geometry. If no lone pairs
are present around the central atom, the bond angles will be that of the ideal geometry. If lone pairs are
present, the bond angles may be smaller than the ideal geometry.

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 Example 10.3 Predicting Molecular Geometries
Continued

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 Example 10.3 Predicting Molecular Geometries
Continued
The electron geometry is octahedral (six electron groups) and the molecular geometry—the shape of
the molecule—is square planar (four bonding groups and two lone pairs). Even though lonepairs are
present, the bond angles are 90° because the lone pairs are symmetrically arranged and do not compress
the I− Cl bond angles.

For Practice 10.3

Predict the molecular geometry of I3−.

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 Example 10.4 Predicting the Shape of Larger Molecules
Predict the geometry about each interior atom in methanol (CH 3OH) and make a sketch of the molecule.

Solution
Begin by drawing the Lewis structure of CH3OH. CH3OH contains two interior atoms: one carbon atom and one
oxygen atom. To determine the shape of methanol, determine the geometry about each interior atom as follows:

Using the geometries of each of these, draw a three-dimensional sketch of the molecule as shown here:

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 Example 10.4 Predicting the Shape of Larger Molecules
Continued

For Practice 10.4

Predict the geometry about each interior atom in acetic acid and make a sketch of the molecule.

Chemistry: A Molecular Approach, 3rd Edition © 2014 Pearson Education, Inc.

Nivaldo J. Tro
 Example 10.5 Determining if a Molecule Is Polar
Determine if NH3 is polar.

Solution
Draw the Lewis structure for the molecule and determine its molecular geometry.

The Lewis structure has three bonding groups and one lone pair about the central atom. Therefore the molecular
geometry is trigonal pyramidal.

Determine if the molecule contains polar bonds. Sketch the molecule and superimpose a vector for each polar
bond. The relative length of each vector should be proportional to the electronegativity difference between the
atoms forming each bond. The vector should point in the direction of the more electronegative atom.

The electronegativities of nitrogen and hydrogen are 3.0 and 2.1, respectively. Therefore, the bonds are polar.

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 Example 10.5 Determining if a Molecule Is Polar
Continued

Determine if the polar bonds add together to form a net dipole moment. Examine the symmetry of
the vectors (representing dipole moments) and determine if they cancel each other or sum to a net dipole
moment.

The three dipole moments sum to a net dipole moment. The molecule is polar.

For Practice 10.5

Determine if CF4 is polar.

Chemistry: A Molecular Approach, 3rd Edition © 2014 Pearson Education, Inc.

Nivaldo J. Tro
 Example 10.6 Hybridization and Bonding Scheme
Write a hybridization and bonding scheme for bromine trifluoride, BrF 3.
Procedure For…
Hybridization and Bonding Scheme

Solution
Step 1 Write the Lewis structure for the molecule.
BrF3 has 28 valence electrons and the following Lewis structure:

Step 2 Use VSEPR theory to predict the electron geometry about the central atom (or interior atoms).
The bromine atom has five electron groups and therefore has a trigonal bipyramidal electron geometry.

Step 3 Select the correct hybridization for the central atom (or interior atoms) based on the electron
geometry (see Table 10.3).
A trigonal bipyramidal electron geometry corresponds to sp3d hybridization.

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 Example 10.6 Hybridization and Bonding Scheme
Continued

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 Example 10.6 Hybridization and Bonding Scheme
Continued
Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the
appropriate orbitals on the terminal atoms.

Step 5 Label all bonds using the σ or π notation followed by the type of overlapping orbitals.

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 Example 10.6 Hybridization and Bonding Scheme
Continued

For Practice 10.6

Write a hybridization and bonding scheme for XeF4.

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 Example 10.7 Hybridization and Bonding Scheme
Write a hybridization and bonding scheme for acetaldehyde,

Procedure For…
Hybridization and Bonding Scheme

Solution
Step 1 Write the Lewis structure for the molecule.
Acetaldehyde has 18 valence electrons and the following Lewis structure:

Step 2 Use VSEPR theory to predict the electron geometry about the central atom (or interior atoms).
The leftmost carbon atom has four electron groups and a tetrahedral electron geometry. The rightmost
carbon atom has three electron groups and a trigonal planar geometry.

Step 3 Select the correct hybridization for the central atom (or interior atoms) based on the electron
geometry (see Table 10.3).
The leftmost carbon atom is sp3 hybridized, and the rightmost carbon atom is sp2 hybridized.

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 Example 10.7 Hybridization and Bonding Scheme
Continued

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 Example 10.7 Hybridization and Bonding Scheme
Continued
Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the
appropriate orbitals on the terminal atoms.

Step 5 Label all bonds using the σ or π notation followed by the type of overlapping orbitals.

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 Example 10.7 Hybridization and Bonding Scheme
Continued
For Practice 10.7
Write a hybridization and bonding scheme for HCN.

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 Example 10.8 Hybridization and Bonding Scheme
Use valence bond theory to write a hybridization and bonding scheme for ethene, H 2C＝CH2
Procedure For…
Hybridization and Bonding Scheme

Solution
Step 1 Write the Lewis structure for the molecule.

Step 2 Use VSEPR theory to predict the electron geometry about the central atom (or interior atoms).

The molecule has two interior atoms. Since each atom has three electron groups (one double bond
and two single bonds), the electron geometry about each atom is trigonal planar.

Step 3 Select the correct hybridization for the central atom (or interior atoms) based on the electron geometry
(see Table 10.3).

A trigonal planar geometry corresponds to sp2 hybridization.

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 Example 10.8 Hybridization and Bonding Scheme
Continued

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 Example 10.8 Hybridization and Bonding Scheme
Continued
Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the
appropriate orbitals on the terminal atoms.

Step 5 Label all bonds using the σ or π notation followed by the type of overlapping orbitals.

Chemistry: A Molecular Approach, 3rd Edition © 2014 Pearson Education, Inc.

Nivaldo J. Tro
 Example 10.8 Hybridization and Bonding Scheme
Continued
For Practice 10.8
Use valence bond theory to write a hybridization and bonding scheme for CO 2.

For More Practice 10.8

What is the hybridization of the central iodine atom in I 3−?

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 Example 10.9 Bond Order
Use molecular orbital theory to predict the bond order in H2−. Is the H2− bond a stronger or weaker bond than
the H2 bond?

Solution
The H2− ion has three electrons. Assign the three electrons to the molecular orbitals, filling lower energy orbitals
first and proceeding to higher energy orbitals.

Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in bonding
orbitals and dividing the result by two.

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 Example 10.9 Bond Order
Continued
Since the bond order is positive, H2− should be stable. However, the bond order of H2− is lower than the bond order
of H2 (which is 1); therefore, the bond in H2− is weaker than in H2.

For Practice 10.9

Use molecular orbital theory to predict the bond order in H2+. Is the H2+ bond a stronger or weaker bond than
the H2 bond?

Chemistry: A Molecular Approach, 3rd Edition © 2014 Pearson Education, Inc.

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 Example 10.10 Molecular Orbital Theory
Draw an MO energy diagram and determine the bond order for the N2− ion. Do you expect the bond to be stronger or
weaker than in the N2 molecule? Is N2− diamagnetic or paramagnetic?

Solution
Write an energy level diagram for the molecular orbitals in N2−. Use the energy ordering for N2.

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 Example 10.10 Molecular Orbital Theory
Continued
The N2− ion has 11 valence electrons (5 for each nitrogen atom plus 1 for the negative charge). Assign the electrons
to the molecular orbitals beginning with the lowest energy orbitals and following Hund’s rule.

Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in
bonding orbitals and dividing the result by two.

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 Example 10.10 Molecular Orbital Theory
Continued
The bond order is 2.5, which is a lower bond order than in the N 2 molecule (bond order = 3); therefore, the bond is
weaker. The MO diagram shows that the N2− ion has one unpaired electron and is therefore paramagnetic.

For Practice 10.10

Draw an MO energy diagram and determine the bond order for the N2+ ion. Do you expect the bond to be stronger
or weaker than in the N2 molecule? Is N2+ diamagnetic or paramagnetic?

For More Practice 10.10

Use molecular orbital theory to determine the bond order of Ne 2.

Chemistry: A Molecular Approach, 3rd Edition © 2014 Pearson Education, Inc.

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 Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic
Molecules and Ions
Use molecular orbital theory to determine the bond order of the CN − ion. Is the ion paramagnetic or diamagnetic?

Solution
Determine the number of valence electrons in the molecule or ion.

Number of valence electrons

= 4 (from C) + 5 (from N) +
1 (from negative charge) = 10

Write an energy level diagram using Figure 10.15 as a guide. Fill

the orbitals beginning with the lowest energy orbital and progressing
upward until all electrons have been assigned to an orbital. Remember
to allow no more an two electrons (with paired spins) per orbital and
to fill degenerate orbitals with single electrons (with parallel spins)
before pairing.

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 Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic
Molecules and Ions
Continued

FIGURE 10.15 Molecular Orbital Energy Diagrams for Second-Row p-block Homonuclear Diatomic Molecules

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 Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic
Molecules and Ions
Continued

Solution
Calculate the bond order using the appropriate formula:

If the MO diagram has unpaired electrons, the molecule or ion is paramagnetic. If the electrons are all paired, the
molecule or ion is diamagnetic.

Since the MO diagram has no unpaired electrons, the ion is diamagnetic.

For Practice 10.11

Use molecular orbital theory to determine the bond order of NO. (Use the energy ordering of O 2.) Is the molecule
paramagnetic or diamagnetic?

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10.1 Artificial Sweeteners: Fooled by
Molecular Shape
Taste

Using Lewis Theory to Predict Molecular Shapes

© 2014 Pearson Education, Inc.

 38

Molecular Geometry

• Molecular geometry is simply

the shape of a molecule. A carbon
dioxide
• Molecular geometry is molecule is
described by the geometric linear.
figure formed when the atomic
nuclei are joined by (imaginary)
straight lines.
• Molecular geometry is found
using the Lewis structure, but
the Lewis structure itself does A water
NOT necessarily represent the molecule is
molecule’s shape. angular or
bent.
Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Ten
 10.2 VSEPR Theory: The Five Basic Shapes
VSEPR Theory

• Valence-Shell Electron-Pair Repulsion (VSEPR) is

a simple method for determining geometry.
• Basis: pairs of valence electrons in bonded atoms
repel one another.
• These mutual repulsions push electron pairs as far
from one another as possible. When the electron
B B pairs (bonds) are as
far apart as they can
B A B get, what will be the
A
B-A-B angle?

B
B

© 2014 Pearson Education, Inc.

 40

Electron-Group Geometries

• An electron group is a collection of Electron Electron-group

valence electrons, localized in a groups geometry
region around a central atom.
2 Linear
• One electron group:
– an unshared pair of valence electrons or 3 Trigonal planar
– a bond (single, double, or triple)
4 Tetrahedral
• The repulsions among electron
groups lead to an orientation of the 5 Trigonal
groups that is called the electron- bipyramidal
group geometry. 6 Octahedral
• These geometries are based on the
number of electron groups:
Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Ten
 41

A Balloon Analogy

• Electron groups repel

one another in the
same way that
balloons push one
another apart.
• When four balloons,
tied at the middle,
push themselves
apart as much as
possible, they make a
tetrahedral shape.

Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Ten
 42

Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Ten
 Electron Groups
•• •• •• There are three electron groups on N:
•O N O• • • Three lone pair
• • One single bond
••
• One double bond
• In the VSEPR notation
• A, central atom
• X, terminal atoms, bond
pairs.
• E, electrons ,lone pairs
• The H2O molecule would
therefore carry the designation
AX2E2.

© 2014 Pearson Education, Inc.

 Two Electron Groups: Linear Electron
Valence shell electron pair repulsion (VSEPR) model:
Geometry
Predict the geometry of the molecule from the electrostatic
repulsions between the electron (bonding and nonbonding) pairs.

# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

AB2 2 0 linear linear

B B

10.1
Linear Geometry

Cl Be Cl

20 lone bonded
atoms pairs on to
central atom
central atom 10.1
Three Electron Groups:
Trigonal Planar Electron Geometry VSEPR
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

AB2 2 0 linear linear

trigonal trigonal
AB3 3 0
planar planar

10.1
 Three Electron Groups:
Trigonal Planar Electron Geometry

Boron trifluride

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Four Electron Groups: VSEPR
Tetrahedral Electron Geometry
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

AB2 2 0 linear linear

trigonal trigonal
AB3 3 0
planar planar
AB4 4 0 tetrahedral tetrahedral

10.1
 Tetrahedral Geometry

CH 4
NH4+

Methane

© 2014 Pearson Education, Inc.

Five Electron Groups: Trigonal
Bipyramidal Electron Geometry
VSEPR
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

AB2 2 0 linear linear

trigonal trigonal
AB3 3 0
planar planar
AB4 4 0 tetrahedral tetrahedral
trigonal trigonal
AB5 5 0
bipyramidal bipyramidal

10.1
 Five Electron Groups: Trigonal Bipyramidal
Electron Geometry

Phosphorus Pentachoride

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Octahedral Electron VSEPR
Geometry
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

AB2 2 0 linear linear

trigonal trigonal
AB3 3 0
planar planar
AB4 4 0 tetrahedral tetrahedral
trigonal trigonal
AB5 5 0
bipyramidal bipyramidal
AB6 6 0 octahedral octahedral

10.1
 Octahedral Geometry

Sulfur Hexafluoride

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10.1
 10.3 VSEPR Theory: The Effect of Lone Pairs
Molecular Geometry

lone-pair vs. lone pair lone-pair vs. bonding bonding-pair vs. bonding
repulsion
> pair repulsion
> pair repulsion
Bent Molecular Geometry: Derivative of
Trigonal Planar Electron Geometry
VSEPR
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

trigonal trigonal
AB3 3 0
planar planar
trigonal
AB2E 2 1 bent
planar

10.1
Pyramidal and Bent Molecular
Geometries: VSEPR
Derivatives of Tetrahedral Electron
Geometry # of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

AB4 4 0 tetrahedral tetrahedral

trigonal
AB3E 3 1 tetrahedral
pyramidal

10.1
Pyramidal and Bent Molecular
Geometries: VSEPR
Derivatives of Tetrahedral Electron
Geometry # of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

AB4 4 0 tetrahedral tetrahedral

trigonal
AB3E 3 1 tetrahedral
pyramidal

AB2E2 2 2 tetrahedral bent

O
H H

10.1
Derivatives of the Trigonal VSEPR
Bipyramidal Electron Geometry
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry
trigonal trigonal
AB5 5 0
bipyramidal bipyramidal
trigonal distorted
AB4E 4 1
bipyramidal tetrahedron

seesaw shape

10.1
Derivatives of the Trigonal VSEPR
Bipyramidal Electron Geometry
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry
trigonal trigonal
AB5 5 0
bipyramidal bipyramidal
trigonal distorted
AB4E 4 1
bipyramidal tetrahedron
trigonal
AB3E2 3 2 T-shaped
bipyramidal
F

F Cl

10.1
Derivatives of the Trigonal VSEPR
Bipyramidal Electron Geometry
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry
trigonal trigonal
AB5 5 0
bipyramidal bipyramidal
trigonal distorted
AB4E 4 1
bipyramidal tetrahedron
trigonal
AB3E2 3 2 T-shaped
bipyramidal
trigonal
AB2E3 2 3 linear
bipyramidal
I

I 10.1
Derivatives of the Octahedral VSEPR
Geometry
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

AB6 6 0 octahedral octahedral

square
AB5E 5 1 octahedral
pyramidal
F
F F
Br
F F

10.1
Derivatives of the Octahedral VSEPR
Geometry
# of atoms # lone
bonded to pairs on Arrangement of Molecular
Class central atom central atom electron pairs Geometry

AB6 6 0 octahedral octahedral

square
AB5E 5 1 octahedral
pyramidal
square
AB4E2 4 2 octahedral
planar
F F
Xe
F F

10.1
© 2014 Pearson Education, Inc.
© 2014 Pearson Education, Inc.
10.4 Predicting the Shapes around Central Atoms
1. Draw Lewis structure for molecule.
2. Determine the number of electron groups around the
central atom.
3. Classify each electron group as a bonding or lone
pair, and count each type.
– Remember, multiple bonds count as one group.
4. Use VSEPR to predict the geometry of the molecule.
What are the molecular geometries of SO2 and SF4?

O S O F
AB4E
AB2E F S F
distorted
bent tetrahedron
F
10.4
 Multiple Central Atoms

The shape around left N is tetrahedral–trigonal pyramidal.

The shape around left C is tetrahedral.

The shape around center C is trigonal planar.

The shape around right O is tetrahedral–bent.

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 10.5 Molecular Shape and Polarity
Polarity of Molecules

m=Qxr
Q is the charge
r is the distance between charges
1 D = 3.36 x 10-30 C m
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Dipole Moments and Polar Molecules

electron rich
electron poor
region
region

H F

d+ d-

m=Qxr
Q is the charge
r is the distance between charges
1 D = 3.36 x 10-30 C m 10.2
10.2
10.2
 Vector Addition

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© 2014 Pearson Education, Inc.
 Which of the following molecules have a dipole moment?
H2O, CO2, SO2, and CH4

O S

dipole moment dipole moment

polar molecule polar molecule

H C H
O C O

no dipole moment H
nonpolar molecule no dipole moment
nonpolar molecule
10.2
Does CH2Cl2 have
a dipole moment?

10.2
 Molecular Polarity Affects Solubility
in Water

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 10.6 Valence Bond Theory: Orbital Overlap
as a Chemical Bond
How does Lewis theory explain the bonds in H2 and F2?

Sharing of two electrons between the two atoms.

Bond Dissociation Energy Bond Length Overlap Of

H2 436.4 kJ/mole 74 pm 2 1s

F2 150.6 kJ/mole 142 pm 2 2p

Valence bond theory – bonds are formed by sharing of e-

from overlapping atomic orbitals.
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 Orbital
OrbitalInteraction
Interaction

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 10.7 Valence Bond Theory: Hybridization
of Atomic Orbitals─混成只有在中心原子
Valence Bond Theory and NH3
N – 1s22s22p3

3 H – 1s1
If the bonds form from overlap of 3 2p orbitals on nitrogen
with the 1s orbital on each hydrogen atom, what would
the molecular geometry of NH3 be? If use the
3 2p orbitals
predict 900

Actual H-N-H
bond angle is
107.30
© 2014 Pearson Education, Inc. 7
 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
10.4
 sp3 Hybridization

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 sp3 Hybridization

Predict correc
bond angle

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 sp2 Hybridization

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 sp2 Hybridization

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Sigma bond (s) – electron density between the 2 atoms

Pi bond (p) – electron density above and below plane of nuclei

of the bonding atoms 10.5
10.5
 Orbital Diagrams of Bonding cont.

Hybrid orbitals overlap to form a s bond.

Unhybridized p orbitals overlap to form a
p bond.
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 sp Hybridization

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 sp Hybridization

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sp Hybridization

10.5
 sp Hybridization

p
s
p

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 sp3d Hybridization

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 sp3d2 Hybridization

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© 2014 Pearson Education, Inc.
 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

# 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
10.4
 Sigma (s) and Pi Bonds (p)

Single bond 1 sigma 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?
O

H
s bonds = 6 + 1 = 7
H C C O H
p bonds = 1
H
10.5
 Problems with Valence Bond (VB) Theory
O Experiments show O2 is paramagnetic
O
No unpaired e-
Should be diamagnetic

Molecular orbital theory – bonds are formed from

interaction of atomic orbitals to form molecular orbitals.

© 2014 Pearson Education, Inc. 10.8

 10.8 Molecular Orbital (MO) Theory
Energy levels of bonding and antibonding molecular
orbitals in hydrogen (H2).

A bonding molecular orbital has lower energy and greater

stability than the atomic orbitals from which it was formed.

An antibonding molecular orbital has higher energy and

lower stability than the atomic orbitals from which it was
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10.6
 Interaction of 1s Orbitals

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bond
½ 1 ½ 0
order
10.7
10.6
Interaction of p Orbitals
10.6
Molecular Orbital (MO) Configurations

1. The number of molecular orbitals (MOs) formed is always

equal to the number of atomic orbitals combined.
2. The more stable the bonding MO, the less stable the
corresponding antibonding MO.
3. The filling of MOs proceeds from low to high energies.
4. Each MO can accommodate up to two electrons.
5. Use Hund’s rule when adding electrons to MOs of the
same energy.
6. The number of electrons in the MOs is equal to the sum of
all the electrons on the bonding atoms.

10.7
 Period Two Homonuclear Diatomic
Molecules

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© 2014 Pearson Education, Inc.
10.7
 Heteronuclear Diatomic Molecules and Ions

s2s bonding MO
shows more electron
density near O because it is
mostly O’s 2s atomic orbital.

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 HF

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 Polyatomic Molecules
Ozone, O3

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Delocalized molecular orbitals are not confined between
two adjacent bonding atoms, but actually extend over three
or more atoms.

Electron density above and below the plane of the

benzene molecule.

10.8
 Representing Three-Dimensional Shapes
on Paper

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10.4
 Bond Rotation

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10.7
© 2014 Pearson Education, Inc.
10.7