Chapter 8
Molecular Structure,
Valence Bond Theory, and
Hybridization
Valence
Shell
Electron
Pair
Repulsion
Theory
Trigonal
planar
Tetrahedral
Trigonal
bipyramidal
Octahedral
VSEPR Theory
• Explains how molecules obtain their
shapes.
• Regions of High Electron Density
(bonds or lone pairs) arrange
themselves around a central atom as
far away from each other as possible
so as to minimize repulsive forces.
From Chapter 8 Note Supplement
•
Molecular Shape: When explaining molecular shape
include the following 3 steps in your answer:
–
What is the geometry of the molecule: based on the number of
regions of high electron density. (See Table 8.1 pp. 215).
Discuss VSEPR theory: The regions of high electron density will get
as far apart as possible from one another in order to minimize
repulsive forces. (The underlined statement is an essential part of
any explanation of molecular shape and is the essence of VSEPR
theory).
What is the shape of the molecule: based on the number of bonds
vs. unshared pairs. (See Table 8.2 pp. 216).
–
–
•
Explain placement of unshared pairs and/or bonds. (Easiest to use bond
angles here).
Table 8.1
page 215
Trigonal Planar
Table 8.1 Page 215
The Number of Regions of High Electron Density
around the Central Atom Determine Molecular
Geometry....
2 regions  linear (Bond angle = 180o)
3 regions  trigonal planar (Bond angle = 120o)
4 regions  tetrahedral (Bond angle = 109.5o)
5 regionso trigonal bipyramidal (Bond angles = 90o
and 120 )
6 regions  octahedral (Bond angle = 90o)
Trigonal Bipyramidal
Axial atoms are above
and below the plane of
the triangle on opposite
sides of the molecule.
Equatorial atoms are
120° apart and are
within the plane of the
triangle.
Molecular Shape and VSEPR
Theory Helps to Determine
1. Bond Angles
2. Molecular Polarity
Table 8.1
page 215
Trigonal Planar
Bond Angles
• Bond angles are based on the geometry of the
molecule
• Slight adjustments may be necessary because
of various space requirements for different types
of regions of high electron density.
• Lone pairs have the greatest space requirement
followed by triple bonds then double bonds and
finally single bonds which require the least
amount of space around the central atom.
Linear
• Two atom molecule.
• 2/0 central atom count.
– Two atoms attached to the central atom and
no lone pairs on the central atom.
• Bond angle of 180°
Linear
Lewis Structure Molecular Geometry
O=C=O
Linear
Lewis Structure Molecular Geometry
O=C=O
CO2 has two regions of high electron density resulting in
a linear geometry. In order to minimize the repulsive
forces the two oxygen atoms are bonded 180˚apart
resulting in a linear shape.
3/0 or 2/1 count
on central atom.
•The geometry of the molecule is trigonal planar.
•The base angle for this geometry is 120°.
•The shape can be either trigonal planar or angular.
Trigonal Planar
Trigonal Planar
BF3 has three regions of high electron density resulting
in a trigonal planar geometry. In order to minimize the
repulsive forces the three fluorine atoms are bonded
120° apart resulting in a trigonal planar shape.
Explain why NO2- has a bond angle of 115.4°?
NO2- has three regions of high electron density which
results in a trigonal planar geometry and a base angle of
120°. However the greater space requirement of the
unshared pair repels the bonds and results in a bond
angle of 115.4°.
4/0 or 3/1 or 2/2
count on central
atom.
•The geometry of the molecule is tetrahedral.
•The base angle for this geometry is 109.5°.
•The shapes can be either tetrahedral, trigonal pyramidal, or
angular.
Why does water have a
bond angle of 105°?
Why does water have a
bond angle of 105°?
• The four regions of high electron density
surrounding the oxygen tend to arrange
themselves as far from each other as possible in
order to minimize repulsive forces. This results
in a tetrahedral geometry in which the H-O-H
bond angle would be 109.5°. However, the two
lone pairs around the oxygen atom, have a
greater space requirement, effectively pushing
the two hydrogen atoms closer together. The
result is a H—O—H angle of 105°.
5/0 or 4/1 or 3/2 or
2/3 count on
central atom.
•The geometry of the molecule is trigonal bipyramidal.
•The base angles for this geometry are 120° and 90°.
However be careful with the angles on this geometry as they
can vary depending upon the shape.
•The shapes can be either trigonal bipyramidal, seesaw,
T- shaped, or linear.
It will be helpful when explaining shapes based on a Trigonal
Bipyramidal geometry to remember the difference between
axial and equatorial.
Axial atoms are above
and below the plane of
the triangle on opposite
sides of the molecule.
Equatorial atoms are
120° apart and are
within the plane of the
triangle.
a
a
a
I3 -
I3 -
I3 -
a
6/0 or 5/1 or 4/2
count on central
atom.
•The geometry of the molecule is octahedral.
•The base angles for this geometry 90°.
•The shapes can be either octahedral, square pyramidal, or
square planar.
a
a
Xenon Tetrafluoride
Octahedral Geometry
Square Planar Shape
Explain the shape of IBr3
•
•
•
•
•
Draw Lewis Structure
State VSEPR Theory (MRF) and Geometry
Place the Lone Pairs
Place the Bonds
State the Shape
Explain the shape of Xenon
Tetrafluoride
Molecular Polarity
• A dipole is anything with a positive end and a
negative end. Another word for dipole is polar.
• A bond is a dipole (polar) if it connects different atoms.
• A polar molecule (dipole) is a molecule where the
polar bonds are asymmetrically (not symmetrically)
arranged (the dipoles do not cancel).
• A nonpolar molecule is a molecule with no polar
bonds or a molecule where the polar bonds are
symmetrically arranged (the dipoles cancel).
Dipole Moment
• The dipole moment is the measurement of a
molecules polarity.
• Arrow points toward the more
electronegative atom.
+

H
Cl

Determining Molecular Polarity
• Polar Molecule
– Dipoles (Polar Bonds) are asymmetrically
arranged and don’t cancel in this bent
(angular) molecule.
O
H2O
H
H
net
dipole
moment
Determining Molecular Polarity
• Therefore, polar molecules have...
– asymmetrical charge distribution resulting in a
positive and negative end to the molecule and
a net dipole moment.
– the dipoles don’t cancel
H
CHCl3
Cl
Cl
Cl
net
dipole
moment
Determining Molecular Polarity
• Nonpolar Molecules
– Dipoles (Polar Bonds) are symmetrically
arranged and cancel out in this trigonal
planar shaped molecule.
F
BF3
B
F
F
Hybridization – The Methane Dilemma
Carbon 1s22s22p2
Fig. 10.7
Fig. 10.8
Hybridization takes a certain number of different
atomic orbitals and mixes (hybridizes) them to create
an equal number of equivalent hybrid orbitals.
1. ONLY the CENTRAL ATOM(s) hybridize.
2. The number of regions of high electron density
must equal the number of orbitals that hybridize.
Atomic orbitals
hybridization
# hybrid of orbitals
# of regions of HED
Atomic orbitals
hybridization
# of regions of HED
# hybrid of orbitals
What type of hybridization does each
central atom exhibit?
What type of hybridization does B exhibit?
What type of hybridization does Be exhibit?
What type of hybridization does Xe exhibit?
Covalent Bonding (Review)
• Covalent bonds result when:
– atomic orbitals of adjacent
atoms overlap.
Sigma () Bonds
• Result from the overlap of:
– Two s orbitals
– An s orbital and a p orbital
– End to end overlap of two p orbitals
• Single bonds are sigma bonds.
Pi () Bonds
• Result from the side to side overlap of two
p orbitals.
 overlap
Pi () Bonds
• Result from the side to side overlap of two
p orbitals.
• Double bonds are a sigma bond and a pi
bond.
• Triple bonds are a sigma bond and two pi
bonds.
C-C
1  bond
C=C
1  bond
1  bond
C
1  bond
2  bonds
C
How many sigma bonds are in this molecule?
How many pi bonds?
H
H
C=C
H
H
Hybridization and
Type of Bonding (sigma and pi)
in Ethylene, CH2CH2 or (C2H4)
Determine the hybridization for each central atom.
H
H
C=C
H
H
sp2 hybrids and unhybridized p-orbital
 bond involve the hybrid orbitals on the central atoms
1 electron from the sp2 hybrid on
C, the other from the hydrogen
1s orbital
 bond results from the side-by-side overlap
of the unhybridized p-orbitals
Electron from the unhybridized
p-orbital on the C atom
•
•
Sigma () Bonding in Ethylene
Pi () Bonding in Ethylene
Acetylene C2H2
Acetylene C2H2
Molecular Orbitals
• A molecular orbital is an orbital within a
molecule whereas an atomic orbital is an
orbital within an individual atom.
• Both types of orbitals can hold no more
than a pair of electrons.
Paramagnetic and Diamagnetic
• Paramagnetic materials have unpaired
electrons in their structures and are
attracted to a magnetic field.
• Diamagnetic materials have no unpaired
electrons in their structures and are
repelled by a magnetic field.
Write the molecular
orbital diagram for O2.
Is O2 paramagnetic or
diamagnetic? Explain.
Write the molecular
orbital diagram for N2.
Is N2 paramagnetic or
diamagnetic? Explain.
Review Section of Chapter 8 Test
• Percentage Yield
• Effective Nuclear Charge
• Quantum Numbers