Transcript Chapter 9 Orbitals and Covalent Bond Atomic Orbitals Don’t Work to explain molecular geometry.  In methane, CH4 , the shape s tetrahedral.  The.

Chapter 9
Orbitals and Covalent Bond
1
Atomic Orbitals Don’t Work
to explain 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

2
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.


3
sp3 hybridization has tetrahedral
geometry.
In terms of energy
2p
Energy
Hybridization
2s
4
sp3
How we get to hybridization
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
 This includes bent and trigonal pyramidal
molecules because one of the sp3 lobes
holds the lone pair.

5
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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7
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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
H
C
H
12
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

21
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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