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GROUP THEORY
Q + A
Dr. Christoph Sontag
Phayao University
1
SYMMETRY ELEMENTS
Please write down all symmetry elements of :
Ammonia
c-axis:
σ :
Acetone
c-axis:
σ :
Dimethyl-cyclopentane
c-axis:
σ :
Ethanediol
c-axis:
σ :
i :
Propadiene
c-axis:
σ :
S-axis: 2
N
H
H
H
O
C
CH3 CH3
C
OH
C
OH
H
H
H
H
POINT GROUPS
Use the flowchart to find the point groups of the 5
molecules before:
Ammonia:
Acetone:
Dimethyl-cyclopentane:
Ethanediol:
Propadiene: 3
Dnd
ANSWERS
Ammonia: C3 (and 2C3), 3 σv => C3v
Acetone: C2, σv , σv => C2v
Dimethyl-cyclopentane: σ => Cs
Ethanediol: C2, i, σh => C2h
Propadiene: C2, C2’ , C2’, σv , σv, S4, 3S4 => D2d
4
SYMMETRY ELEMENTS
Can you draw the S-axis into the propadiene molecule
(remember: S contains C-axis and σ plane !)
Please mark the inversion center in ethanediol:
Mark the S3 axis in
Trihydroxybenzene:
5
ANSWERS
S-axis in propadiene:
Inversion center of ethanediol:
S3 axis in Trihydroxybenzene:
6
CHARACTER TABLES
The water molecule has C2v symmetry:
Which characters does the px orbital of Oxygen have and
what is the name of the representation ?
7
ANSWER
8
=> The px orbital has B1 representation
SYMMETRY OF AN MO OF WATER
The water molecule has C2v symmetry:
One molecular orbital consists
of the px orbital of Oxygen
overlapping with the 2 s-orbitals
of the Hydrogens:
Determine the characters of this MO in the point group
C2v:
This representation Г consists of which irreducible
representation from the character table ?
9
ANSWERS
The MO of the water molecule has the following
symmetry behaviour for the C2v operations:
This representation equals the B1 symmetry from the
character table.
10
BH3 MOLECULE
BH3 is a planar molecule. Can you draw the structure
and all symmetry elements ?
What is the point group of the molecule ?
What are the characters of the pz orbital of Boron and
what is the representation of it ?
11
ANSWERS
The symmetry elements of BH3 are:
C3, 3x C2, 2x S3, 2x σv => belongs to D3h
The pz orbital transforms as follows:
12
Notice that the pz orbital has
the same representation as
the z-axis.
Therefore we can find the
representation directly in the
character table under “z”.
MO DIAGRAM OF BH3
To construct the MO’s of the molecule, we can combine
the atomic orbitals of Boron with combination group
orbitals of 3x H:
What is the characters of these 3 group orbitals ?
and have to be taken together which
leads to the representation e’
Therefore they match the symmetry of which AO of
Boron ?
13
ANSWERS
The combination of H-orbitals with no node has the
symmetry a1’
and the 2 other combinations with one node e’
Therefore the first group orbital can overlap with the s-
orbital of Boron (both have a1’ symmetry):
and the e’ orbitals with px and py of Boron:
The pz orbital of Boron has no partner and stays non-
bonding
14
COMPLETE MO DIAGRAM BH3
15
From the diagram we can conclude that the molecule is a
LEWIS-ACID because the LUMO is low in energy (it’s non-
bonding) and – if we fill electrons into it – will not affect the
B-H bonds.
Π ORBITALS OF CYCLOPROPENYL
The Cyclopropenyl cation is an aromatic ring system
with 3 pπ orbitals:
Determine the representation of these 3 p orbitals in the
D3h point group:
This representation is the sum of which irreducible
representations of the character table ?
16
ANSWERS
The 3 p-orbitals transform under the symmetry
operations of D3h as follows:
This representation is the sum of which representations
of the character table ?
We use the reduction formula to check for example if A2”
is in the representation Г:
With the same method
we find that also e” is
in Г.
17
MOLECULAR ORBITALS
Butadiene has the point group C2h:
Which representation does the HOMO of the molecule
have ?
Draw the C2 axis, inversion center and mirror plane and
check if the MO – not the 4 single AO’s ! – behave.
18
ANSWER
The HOMO has the representation Г = 1 -1 -1 1
and is therefore Bu.
For example: C2 rotation would reverse the phases of all
orbitals => -1 character
19
REDUCING REPRESENTATIONS
When we have a representation of Г = 4 1 -2
for a molecule with C3v symmetry, how many times is
the representation A2 in Г ?
20
ANSWER
21
The representation A2 is 2 times inside Г
HYBRIDIZATION
Trichlorborane is in the point group D3h.
What are the characters of the 3 σ-bonds ?
E 2C3 3C2 σh 2S3 3σv
Г : 3
Г is the sum of which basic representation ?
22
ANSWER
The 3 σ bonds have the characters:
E 2C3 3C2 σh 2S3 3σv
Г 3 0 1 3 0 1
Each of the 3 C2 rotations keep one bond at the same
position, but exchanges the 2 other bonds => 1
The same is true for the 3 σv mirror planes which are each
on one bond.
The horizontal mirror plane leaves all 3 bonds in their
position => 3
Г is the sum of the representations A1’ and E’.
A1 represents also the s-orbital, and E’ represents the px
and py orbital on the central atom.
From this we can conclude that the 3 B-Cl σ-bonds are
formed by the combination of s + px + py = sp2 hybrid.
23

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Grade 9 Quarter 4 Dll Grade 9 Quarter 4 DLL.pdf
 

Group Theory Q&A: Symmetry Elements, Point Groups, and Molecular Orbitals

  • 1. GROUP THEORY Q + A Dr. Christoph Sontag Phayao University 1
  • 2. SYMMETRY ELEMENTS Please write down all symmetry elements of : Ammonia c-axis: σ : Acetone c-axis: σ : Dimethyl-cyclopentane c-axis: σ : Ethanediol c-axis: σ : i : Propadiene c-axis: σ : S-axis: 2 N H H H O C CH3 CH3 C OH C OH H H H H
  • 3. POINT GROUPS Use the flowchart to find the point groups of the 5 molecules before: Ammonia: Acetone: Dimethyl-cyclopentane: Ethanediol: Propadiene: 3 Dnd
  • 4. ANSWERS Ammonia: C3 (and 2C3), 3 σv => C3v Acetone: C2, σv , σv => C2v Dimethyl-cyclopentane: σ => Cs Ethanediol: C2, i, σh => C2h Propadiene: C2, C2’ , C2’, σv , σv, S4, 3S4 => D2d 4
  • 5. SYMMETRY ELEMENTS Can you draw the S-axis into the propadiene molecule (remember: S contains C-axis and σ plane !) Please mark the inversion center in ethanediol: Mark the S3 axis in Trihydroxybenzene: 5
  • 6. ANSWERS S-axis in propadiene: Inversion center of ethanediol: S3 axis in Trihydroxybenzene: 6
  • 7. CHARACTER TABLES The water molecule has C2v symmetry: Which characters does the px orbital of Oxygen have and what is the name of the representation ? 7
  • 8. ANSWER 8 => The px orbital has B1 representation
  • 9. SYMMETRY OF AN MO OF WATER The water molecule has C2v symmetry: One molecular orbital consists of the px orbital of Oxygen overlapping with the 2 s-orbitals of the Hydrogens: Determine the characters of this MO in the point group C2v: This representation Г consists of which irreducible representation from the character table ? 9
  • 10. ANSWERS The MO of the water molecule has the following symmetry behaviour for the C2v operations: This representation equals the B1 symmetry from the character table. 10
  • 11. BH3 MOLECULE BH3 is a planar molecule. Can you draw the structure and all symmetry elements ? What is the point group of the molecule ? What are the characters of the pz orbital of Boron and what is the representation of it ? 11
  • 12. ANSWERS The symmetry elements of BH3 are: C3, 3x C2, 2x S3, 2x σv => belongs to D3h The pz orbital transforms as follows: 12 Notice that the pz orbital has the same representation as the z-axis. Therefore we can find the representation directly in the character table under “z”.
  • 13. MO DIAGRAM OF BH3 To construct the MO’s of the molecule, we can combine the atomic orbitals of Boron with combination group orbitals of 3x H: What is the characters of these 3 group orbitals ? and have to be taken together which leads to the representation e’ Therefore they match the symmetry of which AO of Boron ? 13
  • 14. ANSWERS The combination of H-orbitals with no node has the symmetry a1’ and the 2 other combinations with one node e’ Therefore the first group orbital can overlap with the s- orbital of Boron (both have a1’ symmetry): and the e’ orbitals with px and py of Boron: The pz orbital of Boron has no partner and stays non- bonding 14
  • 15. COMPLETE MO DIAGRAM BH3 15 From the diagram we can conclude that the molecule is a LEWIS-ACID because the LUMO is low in energy (it’s non- bonding) and – if we fill electrons into it – will not affect the B-H bonds.
  • 16. Π ORBITALS OF CYCLOPROPENYL The Cyclopropenyl cation is an aromatic ring system with 3 pπ orbitals: Determine the representation of these 3 p orbitals in the D3h point group: This representation is the sum of which irreducible representations of the character table ? 16
  • 17. ANSWERS The 3 p-orbitals transform under the symmetry operations of D3h as follows: This representation is the sum of which representations of the character table ? We use the reduction formula to check for example if A2” is in the representation Г: With the same method we find that also e” is in Г. 17
  • 18. MOLECULAR ORBITALS Butadiene has the point group C2h: Which representation does the HOMO of the molecule have ? Draw the C2 axis, inversion center and mirror plane and check if the MO – not the 4 single AO’s ! – behave. 18
  • 19. ANSWER The HOMO has the representation Г = 1 -1 -1 1 and is therefore Bu. For example: C2 rotation would reverse the phases of all orbitals => -1 character 19
  • 20. REDUCING REPRESENTATIONS When we have a representation of Г = 4 1 -2 for a molecule with C3v symmetry, how many times is the representation A2 in Г ? 20
  • 21. ANSWER 21 The representation A2 is 2 times inside Г
  • 22. HYBRIDIZATION Trichlorborane is in the point group D3h. What are the characters of the 3 σ-bonds ? E 2C3 3C2 σh 2S3 3σv Г : 3 Г is the sum of which basic representation ? 22
  • 23. ANSWER The 3 σ bonds have the characters: E 2C3 3C2 σh 2S3 3σv Г 3 0 1 3 0 1 Each of the 3 C2 rotations keep one bond at the same position, but exchanges the 2 other bonds => 1 The same is true for the 3 σv mirror planes which are each on one bond. The horizontal mirror plane leaves all 3 bonds in their position => 3 Г is the sum of the representations A1’ and E’. A1 represents also the s-orbital, and E’ represents the px and py orbital on the central atom. From this we can conclude that the 3 B-Cl σ-bonds are formed by the combination of s + px + py = sp2 hybrid. 23