• Chapters 4 and 5
• multiple short answer problems testing basic concepts
• closed book, closed notes
• bring pen, pencil, calculator, ID
• study lectures, book, suggested problems
• look for seating chart on lecture room door
• don‘t cram!
Thurs 6-7:50p, ICS 174 (Sam)
Fri 4:30-6:20p, ICS 174 (Lindsay)

Materials Provided
Character tables also provided
(unless the point of the question is to build the table).

Chapter 5
Wednesday, October 16, 2013

1. Point group D
3h
2. x z y z y y x z x
Boron trifluoride
3. Make reducible reps for outer atoms
Γ
2s
Γ
2pz
Γ
2px
Γ
2py
3 0 1 3 0 1
3 0 -1 -3 0 1
3 0 -1 3 0 -1
3 0 1 3 0 1
4. Get group orbital symmetries by reducing each
Γ
Γ
2s
Γ
2pz
= A
1
’
= A
2
’’
+ E
’
+ E
’’
Γ
Γ
2px
2py
= A
= A
2
’
1
’
+ E
’
+ E
’

Boron trifluoride
Γ
2s
Γ
2pz
= A
1
’
= A
2
’’
+ E
’
+ E
’’
Γ
Γ
2px
2py
What is the shape of the group orbitals?
= A
= A
2
’
1
’
+ E
’
+ E ’
2s:
A
1
’
?
?
E’ ( y ) E’ ( x )
Which combinations of the three AOs are correct?
The projection operator method provides a systematic way to find how the AOs should be combined to give the right group orbitals
(SALCs).

BF
3
- Projection Operator Method
In the projection operator method, we pick one AO in each set of identical AOs and determine how it transforms under each symmetry operation of the point group.
F a
F b
AO E C
3
F a
F a
F b
C
3
2 C
2(a)
C
2(b)
C
2(c)
F c
F a
F c
F b
σ h
F a
S
3
F b
S
3
2
F c
σ v(a)
σ v(b)
σ v(c)
F a
F c
F b
F c A
1
’ 1 1 1 1 1 1 1 1 1 1 1 1
A
1
’
A
1
’
= F a
= 4F a
+ F b
+ F
+ 4F b c
+ F
+ 4F c a
+ F c
+ F b
+ F a
+ F b
+ F c
+ F a
+ F c
+ F b
A
1
’
The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results.

BF
3
- Projection Operator Method
F
In the projection operator method, we pick one AO in each set of identical AOs and determine how it transforms under each symmetry operation of the point group.
F a b
AO E C
3
F a
F a
F b
C
3
2 C
2(a)
C
2(b)
C
2(c)
F c
F a
F c
F b
σ h
F a
S
3
F b
S
3
2
F c
σ v(a)
σ v(b)
σ v(c)
F a
F c
F b
F c A
2
’ 1 1 1 -1 -1 -1 1 1 1 -1 -1 -1
A
2
’
A
2
’
= F a
= 0
+ F b
+ F c
– F a
– F c
– F b
+ F a
+ F b
+ F c
– F a
– F c
– F b
There is no A
2
’ group orbital!
The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results.

BF
3
- Projection Operator Method
In the projection operator method, we pick one AO in each set of identical AOs and determine how it transforms under each symmetry operation of the point group.
F a
F b
F c
AO E C
3
F a
F a
F b
C
3
2 C
2(a)
C
2(b)
C
2(c)
F c
F a
F c
F b
σ h
F a
S
3
F b
S
3
2
F c
σ v(a)
σ v(b)
σ v(c)
F a
F c
F b
E ’ 2 -1 -1 0 0 0 2 -1 -1 0 0 0
E’ = 2F a
E’ = 4F a
– F b
– 2F b
– F c
+ 0 + 0 + 0 + 2F
– 2F c a
– F b
– F c
+ 0 + 0 + 0
E’ ( y )
The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results.

BF
3
- Projection Operator Method
Γ
2s
Γ
2pz
= A
1
’
= A
2
’’
+ E
’
+ E
’’
Γ
2px
Γ
2py
= A
= A
2
’
1
’
+ E
’
+ E ’
What is the shape of the group orbitals?
2s:
A
1
’
?
E’ ( y )
?
E’ ( x )
We can get the third group orbital, E ’(x), by using normalization.
2

BF
3
- Projection Operator Method
Let’s normalize the A
1
’ group orbital: 𝜓
𝐴
′
1
= 𝑐 a
[𝜙(2𝑠
F a
) + 𝜙(2𝑠
F b
) + 𝜙(2𝑠
F c
)]
𝜓 2 𝑑𝜏 = 1
A
1
’ wavefunction
Normalization condition for group orbitals 𝑐 2 a
[𝜙(2𝑠
F a
) + 𝜙(2𝑠
F b
) + 𝜙(2𝑠
F c
)] 2 𝑑𝜏 = 1 nine terms, but the six overlap ( S ) terms are zero.
𝑐 2 a
𝜙 2 (2𝑠
F a
)𝑑𝜏 + 𝜙 2 (2𝑠
F b
)𝑑𝜏 + 𝜙 2 (2𝑠
F c
)𝑑𝜏 = 1 𝑐 2 a
1 + 1 + 1 = 1
So the normalized A
1
’ GO is: 𝑐 a
=
1
3 𝜓
𝐴
′
1
=
1
3
[𝜙(2𝑠
F a
) + 𝜙(2𝑠
F b
) + 𝜙(2𝑠
F c
)]

BF
3
- Projection Operator Method
Now let’s normalize the E’(y) group orbital: 𝜓
𝐸
′
(𝑦)
= 𝑐 a
[2𝜙(2𝑠
F a
) − 𝜙(2𝑠
F b
) − 𝜙(2𝑠
F c
)] E’ ( y ) wavefunction 𝑐 2 a
[2𝜙(2𝑠
F a
) − 𝜙(2𝑠
F b
) − 𝜙(2𝑠
F c
)] 2 𝑑𝜏 = 1 nine terms, but the six overlap ( S ) terms are zero.
𝑐 2 a
4 𝜙 2 (2𝑠
F a
)𝑑𝜏 + 𝜙 2 (2𝑠
F b
)𝑑𝜏 + 𝜙 2 (2𝑠
F c
)𝑑𝜏 = 1 𝑐 2 a
4 + 1 + 1 = 1 𝑐 a
=
1
6
So the normalized E’ ( y ) GO is: 𝜓
𝐸
′
(𝑦)
=
1
6
[2𝜙(2𝑠
F a
) − 𝜙(2𝑠
F b
) − 𝜙(2𝑠
F c
)]

BF
3
- Projection Operator Method 𝜓
𝐴
′
1
= 𝜓
𝐸
′
(𝑦)
1
3
[𝜙(2𝑠
F a
) + 𝜙(2𝑠
F b
) + 𝜙(2𝑠
F c
)]
=
1
6
[2𝜙(2𝑠
F a
) − 𝜙(2𝑠
F b
) − 𝜙(2𝑠
F c
)] 𝒄 𝒊
𝟐 𝐢𝐬 the probability of finding an electron in so 𝒄 𝒊
𝟐
= 𝟏 for a normalized group orbital. 𝝓 𝒊 in a group orbital,
So the normalized E ’( x ) GO is: 𝜓
𝐸
′
(𝑥)
=
1
2
[𝜙(2𝑠
F b
) − 𝜙(2𝑠
F c
)]

BF
3
- Projection Operator Method
Γ
2s
Γ
2pz
= A
1
’
= A
2
’’
+ E
’
+ E
’’
Γ
2px
Γ
2py
= A
= A
2
’
1
’
+ E
’
+ E ’
What is the shape of the group orbitals? notice the GOs are orthogonal ( S = 0).
2s: ?
A
1
’ E’ ( y ) E’ ( x )
Now we have the symmetries and wavefunctions of the 2s GOs.
We could do the same analysis to get the GOs for the p x
, p y
, and p z orbitals (see next slide).

2s:
BF
3
- Projection Operator Method boron orbitals
A
1
’ E’ ( y ) E’ ( x )
A
1
’
2p y
:
2p x
:
A
1
’ E’ ( y ) E’ ( x )
E’
E’
A
2
’ E’ ( y ) E’ ( x )
2p z
: A
2
’’
A
2
’’ E’’ ( y ) E’’ ( x )

2s:
BF
3
- Projection Operator Method boron orbitals
A
1
’ E’ ( y ) E’ ( x )
A
1
’
2p y
:
2p x
:
A
1
’
A
2
’
E’ ( y )
E’ ( y )
E’ ( x ) little overlap
E’ ( x )
E’
E’
2p z
: A
2
’’
A
2
’’ E’’ ( y ) E’’ ( x )

Boron trifluoride
F 2s is very deep in energy and won’t interact with boron.
H
1s
Li
Be
–14.0 eV
B
B
C
He
–8.3 eV
C
N
2p
2s
N
O
F
Ca
Mg
–18.6 eV
Ne
Al
Al
Si
Si
P
3s
3p
P
S
S
Cl
Cl
Ar
Ar
O
–40.2 eV F
Ne

A
1
′
–8.3 eV
E ′
A
2
″
Boron Trifluoride
σ*
σ*
π*
–14.0 eV
A
1
′
A
2
″ nb
σ
π
σ nb
A
2
″
A
2
′ + E ′
A
2
″ + E ″
A
1
′ + E ′
–18.6 eV
E ′
A
1
′ + E ′ –40.2 eV

l
l
d
d

σMOs for Octahedral Complexes
1. Point group O h
The six ligands can interact with the metal in a sigma or pi fashion.
Let’s consider only sigma interactions for now.
2. pi sigma

2.
σMOs for Octahedral Complexes
3. Make reducible reps for sigma bond vectors
4. This reduces to: Γ
σ
= A
1g
+ E g
+ T
1u six GOs in total

σMOs for Octahedral Complexes
5. Find symmetry matches with central atom.
σ
= A
1g
+ E g
+ T
1u
Reading off the character table, we see that the group orbitals match the metal s orbital (A
1g
), the metal p orbitals (T
1u
), and the d z2 and d x2-y2 metal d orbitals (E g
). We expect bonding/antibonding combinations.
The remaining three metal d orbitals are T
2g and σ-nonbonding.

σMOs for Octahedral Complexes
We can use the projection operator method to deduce the shape of the ligand group orbitals, but let’s skip to the results:
σ
1
σ
1
σ
1
-
L
6
+ σ
2
SALC
+ σ
3
+ σ
4
σ
3
, σ
2
σ
4
, σ
+
5
σ
5
+ σ
6
σ
6
σ
2
+ σ
3
σ
4
, 2 σ
6
+ 2 σ
5
σ
1
σ
2
σ
3
σ
4 symmetry label
A
1g
(non-degenerate)
T
1u
(triply degenerate)
E g
(doubly degenerate)

σMOs for Octahedral Complexes
There is no combination of ligand σ orbitals with the symmetry of the metal T
2g orbitals, so these do not participate in σ bonding.
L
+
T
2g orbitals cannot form sigma bonds with the L
6
S = 0 .
set.
non-bonding

σMOs for Octahedral Complexes
6. Here is the general MO diagram for σ bonding in O h complexes:

Summary
MO Theory
• MO diagrams can be built from group orbitals and central atom orbitals by considering orbital symmetries and energies.
• The symmetry of group orbitals is determined by reducing a reducible representation of the orbitals in question. This approach is used only when the group orbitals are not obvious by inspection.
• The wavefunctions of properly-formed group orbitals can be deduced using the projection operator method.
• We showed the following examples: homonuclear diatomics, HF,
CO, H
3
+ , FHF , CO
2
, H
2
O, BF
3
, and σ-ML
6