#### Transcript A 1

```Midterm Exam 1
Monday 9-9:50a
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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!
Review Sessions:
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).
MO Diagrams for
More Complex Molecules
Chapter 5
Wednesday, October 16, 2013
Boron trifluoride
1. Point group D3h
2.
x y
z
3. Make reducible reps for outer atoms
z x
y
y z
x
4. Get group
orbital
symmetries by
reducing each Γ
Γ2s
3
0
1
3
0
1
Γ2pz
3
0
-1
-3
0
1
Γ2px
3
0
-1
3
0
-1
Γ2py
3
0
1
3
0
1
Γ2s = A1’ + E’
Γ2pz = A2’’ + E’’
Γ2px = A2’ + E’
Γ2py = A1’ + E’
Boron trifluoride
Γ2s = A1’ + E’
Γ2pz = A2’’ + E’’
Γ2px = A2’ + E’
Γ2py = A1’ + E’
What is the shape of the group orbitals?
2s:
A1’
?
?
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).
BF3 - 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.
Fa
Fb
Fc
σh
S3
S32
Fb
Fa
Fb
Fc
Fa
Fc
Fb
1
1
1
1
1
1
1
C32 C2(a) C2(b) C2(c)
AO
E
C3
Fa
Fa
Fb
Fc
Fa
Fc
A1’
1
1
1
1
1
σv(a) σv(b) σv(c)
A 1 ’ = Fa + F b + F c + F a + Fc + Fb + Fa + F b + F c + Fa + F c + F b
A1’ = 4Fa + 4Fb + 4Fc
A1’
The group orbital wavefunctions are determined by multiplying the
projection table values by the characters of each irreducible
representation and summing the results.
BF3 - 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.
Fa
Fb
Fc
σh
S3
S32
Fb
Fa
Fb
Fc
Fa
Fc
Fb
-1
1
1
1
-1
-1
-1
C32 C2(a) C2(b) C2(c)
AO
E
C3
Fa
Fa
Fb
Fc
Fa
Fc
A2’
1
1
1
-1
-1
σv(a) σv(b) σv(c)
A 2 ’ = Fa + F b + F c – Fa – Fc – Fb + Fa + F b + F c – Fa – Fc – Fb
A2 ’ = 0
There is no A2’ 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.
BF3 - 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.
Fa
Fb
Fc
σh
S3
S32
Fb
Fa
Fb
Fc
Fa
Fc
Fb
0
2
-1
-1
0
0
0
C32 C2(a) C2(b) C2(c)
AO
E
C3
Fa
Fa
Fb
Fc
Fa
Fc
E’
2
-1
-1
0
0
σv(a) σv(b) σv(c)
E’ = 2Fa – Fb – Fc + 0 + 0 + 0 + 2Fa – Fb – Fc + 0 + 0 + 0
E’ = 4Fa – 2Fb – 2Fc
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.
BF3 - Projection Operator Method
Γ2s = A1’ + E’
Γ2pz = A2’’ + E’’
Γ2px = A2’ + E’
Γ2py = A1’ + E’
What is the shape of the group orbitals?
2s:
A1’
?
?
E’(y)
E’(x)
We can get the third group orbital, E’(x), by using normalization.
𝜓 2 𝑑𝜏 = 1
Normalization condition
BF3 - Projection Operator Method
Let’s normalize the A1’ group orbital:
𝜓𝐴′1 = 𝑐a [𝜙(2𝑠Fa ) + 𝜙(2𝑠Fb ) + 𝜙(2𝑠Fc )]
𝜓 2 𝑑𝜏 = 1
𝑐a2
𝑐a2
𝑐a2
A1’ wavefunction
Normalization condition for group orbitals
[𝜙(2𝑠Fa ) + 𝜙(2𝑠Fb ) + 𝜙(2𝑠Fc )]2 𝑑𝜏 = 1
𝜙 2 (2𝑠Fa )𝑑𝜏 +
1+1+1 =1
𝜙 2 (2𝑠Fb )𝑑𝜏 +
𝑐a =
So the normalized A1’ GO is: 𝜓𝐴′1 =
nine terms, but the six
overlap (S) terms are zero.
𝜙 2 (2𝑠Fc )𝑑𝜏 = 1
1
3
1
3
[𝜙(2𝑠Fa ) + 𝜙(2𝑠Fb ) + 𝜙(2𝑠Fc )]
BF3 - Projection Operator Method
Now let’s normalize the E’(y) group orbital:
𝜓𝐸′ (𝑦) = 𝑐a [2𝜙(2𝑠Fa ) − 𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )]
𝑐a2
[2𝜙(2𝑠Fa ) − 𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )]2 𝑑𝜏 = 1
𝑐a2 4
𝑐a2
𝜙 2 (2𝑠Fa )𝑑𝜏 +
4+1+1 =1
𝜙 2 (2𝑠Fb )𝑑𝜏 +
𝑐a =
E’(y) wavefunction
nine terms, but the six
overlap (S) terms are zero.
𝜙 2 (2𝑠Fc )𝑑𝜏 = 1
1
6
So the normalized E’(y) GO is: 𝜓𝐸′ (𝑦) =
1
6
[2𝜙(2𝑠Fa ) − 𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )]
BF3 - Projection Operator Method
𝜓𝐴′1 =
1
3
𝜓𝐸′ (𝑦) =
[𝜙(2𝑠Fa ) + 𝜙(2𝑠Fb ) + 𝜙(2𝑠Fc )]
1
6
[2𝜙(2𝑠Fa ) − 𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )]
𝒄𝟐𝒊 𝐢𝐬 the probability of finding an electron in 𝝓𝒊 in a group orbital,
so 𝒄𝟐𝒊 = 𝟏 for a normalized group orbital.
So the normalized E’(x) GO is:
𝜓𝐸′ (𝑥) =
1
2
[𝜙(2𝑠Fb ) − 𝜙(2𝑠Fc )]
BF3 - Projection Operator Method
Γ2s = A1’ + E’
Γ2pz = A2’’ + E’’
Γ2px = A2’ + E’
Γ2py = A1’ + E’
What is the shape of the group orbitals?
notice the GOs are
orthogonal (S = 0).
?
2s:
A1’
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 px, py, and pz
orbitals (see next slide).
BF3 - Projection Operator Method
boron
orbitals
2s:
A1’
E’(y)
E’(x)
2py:
A1’
E’(y)
E’(x)
A1’
E’
E’
2px:
A2’
E’(y)
E’(x)
A2’’
2pz:
A2’’
E’’(y)
E’’(x)
BF3 - Projection Operator Method
boron
orbitals
2s:
A1’
E’(y)
E’(x)
A1’
2py:
A1’
E’(y)
E’
E’(x)
little
overlap
2px:
A2’
E’(y)
E’
E’(x)
A2’’
2pz:
A2’’
E’’(y)
E’’(x)
Boron trifluoride
F 2s is very deep in energy and won’t interact with boron.
B
Li
–8.3 eV
Ca
C
–14.0 eV
N
B
2p
Al
O
F
1s
C
Si
P
Mg
Be
H
Al
–18.6 eV
Ne
2s
He
Si
S
3p
3s
Cl
P
S
N
Cl
Ar
O
–40.2 eV
F
Ne
Ar
Boron Trifluoride
σ*
E′
σ*
π*
–8.3 eV
A2″
Energy
A2″
–14.0 eV
A2′ + E′
A1′
nb
A2″ + E″
–18.6 eV
A1′ + E′
π
A2″
σ
σ
A1′
E′
A1′ + E′
nb
–40.2 eV
d orbitals
• l = 2, so there are 2l + 1 = 5 d-orbitals per shell, enough room
for 10 electrons.
• This is why there are 10 elements in each row of the d-block.
σ-MOs for Octahedral Complexes
1. Point group Oh
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
σ-MOs for Octahedral Complexes
2.
3. Make reducible reps for sigma bond vectors
4. This reduces to:
Γσ = A1g + Eg + T1u
six GOs in total
σ-MOs for Octahedral Complexes
5. Find symmetry matches with central atom.
Γσ = A1g + Eg + T1u
Reading off the character table, we see that the group orbitals match
the metal s orbital (A1g), the metal p orbitals (T1u), and the dz2 and dx2-y2
metal d orbitals (Eg). We expect bonding/antibonding combinations.
The remaining three metal d orbitals are T2g and σ-nonbonding.
σ-MOs 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:
L6 SALC
symmetry label
σ1 + σ2 + σ3 + σ4 + σ5 + σ6
A1g (non-degenerate)
σ1 - σ3 , σ2 - σ4 , σ5 - σ6
T1u (triply degenerate)
σ1 - σ2 + σ3 - σ4 , 2σ6 + 2σ5 - σ1 - σ2 - σ3 - σ4 Eg (doubly degenerate)
σ-MOs for Octahedral Complexes
There is no combination of ligand σ orbitals with the symmetry of
the metal T2g orbitals, so these do not participate in σ bonding.
L
+
T2g orbitals cannot form
sigma bonds with the L6 set.
S = 0.
non-bonding
σ-MOs for Octahedral Complexes
6. Here is the general MO diagram for σ bonding in Oh 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, H3+, FHF-, CO2, H2O, BF3, and σ-ML6
```