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- 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