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Chapter 10
Hybridization and Molecular Orbitals
1
Atomic Orbitals Don’t Work
to explain some molecular geometry.
 In methane, CH4 , the shape s
tetrahedral.
 The valence electrons of carbon should
be two in s, and two in p.
 the p orbitals would have to be at right
angles.
 The atomic orbitals change when
making a molecule

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Hybridization
We blend the s and p orbitals of the
valence electrons and end up with the
tetrahedral geometry.
 We combine one s orbital and 3 p
orbitals.
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3
sp3 hybridization has tetrahedral
geometry.
In terms of energy
2p
Energy
Hybridization
2s
4
sp3
How we get to hybridization
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We know the geometry from experiment.
We know the orbitals of the atom.
hybridizing atomic orbitals can explain the
geometry.
So if the geometry requires a tetrahedral
shape, it is sp3 hybridized.
If the molecule has four single bonds, it is sp3
hybridized.
This includes bent and trigonal pyramidal
molecules because one of the sp3 lobes holds
the lone pair.
sp2 hybridization
C2H4
 Double bond acts as one pair.
 trigonal planar
 Have to end up with three blended
orbitals.
 Use one s and two p orbitals to make
sp2 orbitals.
 Leaves one p orbital perpendicular.

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In terms of energy
2p
Energy
Hybridization
2s
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2p
sp2
Where is the P orbital?
Perpendicular
 The overlap of
orbitals makes a
sigma bond (s
bond)
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Two types of Bonds
Sigma bonds from overlap of orbitals.
 Between the atoms.
 Pi bond (p bond) above and below atoms
 Between adjacent p orbitals.
 The two bonds of a
double bond.

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H
C
H
10
H
∏
C
H
sp2 hybridization
When three things come off atom.
 trigonal planar
 120º
 on s one p bond
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What about two
When two things come off.
 One s and one p hybridize.
 linear
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sp hybridization
End up with two lobes 180º
apart.
 p orbitals are at right
angles
 Makes room for two p
bonds and two sigma
bonds.
 A triple bond or two double
bonds.

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In terms of energy
2p
Energy
Hybridization
2s
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2p
sp
CO2
C can make two s and two p
 O can make one s and one p

O
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C O
Breaking the octet
PCl5
 The model predicts that we must use
the d orbitals.
 dsp3 hybridization
 There is some controversy about how
involved the d orbitals are.
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dsp3
Trigonal
bipyrimidal
 can only s bond.
 can’t p bond.
 basic shape for
five things.
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PCl5
Can’t tell the
hybridization of Cl
Assume sp3 to
minimize repulsion of
electron pairs.
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d2sp3
gets us to six things
around
 octahedral
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Molecular Orbital Model
Localized Model we have learned explains
much about bonding.
 It doesn’t deal well with the ideal of
resonance, unpaired electrons, and bond
energy.
 The MO model is a parallel of the atomic
orbital, using quantum mechanics.
 Each MO can hold two electrons with
opposite spins
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 Square of wave function tells probability

What do you get?

Solve the equations for H2
HA HB
 get two orbitals
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
MO1 = 1sA - 1sB

MO2 = 1sA + 1sB
The Molecular Orbital Model
The molecular orbitals are centered on
a line through the nuclei
– MO1 the greatest probability is
between the nuclei
– MO2 it is on either side of the nuclei
– this shape is called a sigma molecular
orbital
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The Molecular Orbital Model
In the molecule only the molecular
orbitals exist, the atomic orbitals are gone
MO1 is lower in energy than the 1s
orbitals they came from.
– This favors molecule formation
– Called an bonding orbital
MO2 is higher in energy
– This goes against bonding
– antibonding orbital
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The Molecular Orbital Model
Energy
MO2
1s
1s
MO1
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The Molecular Orbital Model
We use labels to indicate shapes, and
whether the MO’s are bonding or
antibonding.
– MO1 = s1s
– MO2 = s1s* (* indicates antibonding)
Can write them the same way as
atomic orbitals
– H2 = s1s2
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The Molecular Orbital Model
Each MO can hold two electrons, but
they must have opposite spins
Orbitals are conserved. The number of
molecular orbitals must equal the
number atomic orbitals that are used to
make them.
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-
H2
Energy
s1s*
1s
1s
s1s
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Bond Order

The difference between the number of
bonding electrons and the number of
antibonding electrons divided by two
# bonding-#antibonding
Bond Order =
2
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