1. MO Diagrams for More Complex Molecules
Chapter 5
Wednesday, October 22, 2014

2. Boron trifluoride Γ2s = A1’ + E’ Γ2px = A2’ + E’ Γ2pz = A2’’ + E’’
1. Point group D3h
3. Make reducible reps for outer atoms z 2. x x y y z y z x
Γ2s 3 1
Γ2pz -1 -3
Γ2px
Γ2py
4. Get group orbital symmetries by reducing each Γ
Γ2s = A1’ + E’
Γ2px = A2’ + E’
Γ2pz = A2’’ + E’’
Γ2py = A1’ + E’

3. Which combinations of the three AOs are correct?
Boron trifluoride
Γ2s = A1’ + E’
Γ2px = A2’ + E’
Γ2pz = 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).

4. 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 AO E C3
C32
C2(a)
C2(b)
C2(c) σh S3
S32
σv(a)
σv(b)
σv(c) Fa Fb Fc Fb Fc
A1’ 1
A1’ = Fa + Fb + Fc + Fa + Fc + Fb + Fa + Fb + Fc + Fa + Fc + Fb
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.

5. 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 AO E C3
C32
C2(a)
C2(b)
C2(c) σh S3
S32
σv(a)
σv(b)
σv(c) Fa Fb Fc Fb Fc
A2’ 1 -1
A2’ = Fa + Fb + Fc – Fa – Fc – Fb + Fa + Fb + Fc – 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.

6. 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 AO E C3
C32
C2(a)
C2(b)
C2(c) σh S3
S32
σv(a)
σv(b)
σv(c) Fa Fb Fc Fb Fc E’ 2 -1
E’ = 2Fa – Fb – Fc Fa – Fb – Fc
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.

7. BF3 - Projection Operator Method
Γ2s = A1’ + E’
Γ2px = A2’ + E’
Γ2pz = 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

8. BF3 - Projection Operator Method
Let’s normalize the A1’ group orbital:
𝜓 𝐴 1 ′ = 𝑐 a [𝜙(2𝑠 F a )+ 𝜙(2𝑠 F b )+ 𝜙(2𝑠 F c )]
A1’ wavefunction
𝜓 2 𝑑𝜏=1
Normalization condition for group orbitals
𝑐 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 a 𝜙 2 (2𝑠 F a )𝑑𝜏+ 𝜙 2 (2𝑠 F b )𝑑𝜏 𝜙 2 (2𝑠 F c )𝑑𝜏 =1
𝑐 a = 1 3
𝑐 a =1
𝜓 𝐴 1 ′ = [𝜙(2𝑠 F a )+ 𝜙(2𝑠 F b )+ 𝜙(2𝑠 F c )]
So the normalized A1’ GO is:

9. BF3 - 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 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 a 𝜙 2 (2𝑠 F a )𝑑𝜏+ 𝜙 2 (2𝑠 F b )𝑑𝜏 𝜙 2 (2𝑠 F c )𝑑𝜏 =1
𝑐 a = 1 6
𝑐 a =1
𝜓 𝐸 ′ (𝑦) = [2𝜙(2𝑠 F a )− 𝜙(2𝑠 F b )− 𝜙(2𝑠 F c )]
So the normalized E’(y) GO is:

10. BF3 - Projection Operator Method
𝜓 𝐴 1 ′ = [𝜙(2𝑠 F a )+ 𝜙(2𝑠 F b )+ 𝜙(2𝑠 F c )]
𝜓 𝐸 ′ (𝑦) = [2𝜙(2𝑠 F a )− 𝜙(2𝑠 F b )− 𝜙(2𝑠 F c )]
𝒄 𝒊 𝟐 𝐢𝐬 the probability of finding an electron in 𝝓 𝒊 in a group orbital, so 𝒄 𝒊 𝟐 =𝟏 for a normalized group orbital.
𝜓 𝐸 ′ (𝑥) = [ 𝜙(2𝑠 F b )− 𝜙(2𝑠 F c )]
So the normalized E’(x) GO is:

11. BF3 - Projection Operator Method
Γ2s = A1’ + E’
Γ2px = A2’ + E’
Γ2pz = 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).

12. BF3 - Projection Operator Method
boron orbitals
2s:
A1’
E’(y)
E’(x)
A1’
2py: E’
A1’
E’(y)
E’(x) E’
2px:
A2’
E’(y)
E’(x)
A2’’
2pz:
A2’’
E’’(y)
E’’(x)

13. BF3 - Projection Operator Method
boron orbitals
2s:
A1’
E’(y)
E’(x)
A1’
2py: E’
A1’
E’(y)
E’(x)
little overlap E’
2px:
A2’
E’(y)
E’(x)
A2’’
2pz:
A2’’
E’’(y)
E’’(x)

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

15. Boron Trifluoride Energy σ* σ* π* nb π σ σ nb E′ –8.3 eV A2″ A2″
A1′ + E′
A2″ + E″
A2′ + E′
–18.6 eV
–40.2 eV nb
A2″ π E′ σ
A1′ σ nb

16. 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.

17. 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

18. σ-MOs for Octahedral Complexes2.
3. Make reducible reps for sigma bond vectors
4. This reduces to:
Γσ = A1g + Eg + T1u
six GOs in total

19. 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.

20. 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 + σ A1g (non-degenerate)
σ1 - σ3 , σ2 - σ4 , σ5 - σ T1u (triply degenerate)
σ1 - σ2 + σ3 - σ4 , 2σ6 + 2σ5 - σ1 - σ2 - σ3 - σ4 Eg (doubly degenerate) 5 3 4 2 1 6

21. 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.
T2g are non-bonding

22. σ-MOs for Octahedral Complexes
6. Here is the general MO diagram for σ bonding in Oh complexes:

23. 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