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Tetrahedral (Td) is a molecule which has a bond angle of approximately 109.5°. Its molecular orbital can be determined using reducible representations, SALCs and projection operator techniques.
Td Character Table[edit]
Tetrahedral has a symmetry operation of E, 8C3, 3C2, 6S4, and 6σd. Examples of molecule that have Td point group are methane, phosphorus pentoxide, adamantane, etc. In order to understand the molecular orbital of Td, character tables are used to find the reducible representations which enables to determine the bond type of Td. The character table for Td is the following:
| Td | E | 8C3 | 3C2 | 6S4 | 6σd | ||
|---|---|---|---|---|---|---|---|
| A1 | 1 | 1 | 1 | 1 | 1 | x2+y2+z2 | |
| A2 | 1 | 1 | 1 | -1 | -1 | ||
| E | 2 | -1 | 2 | 0 | 0 | (2z2-x2-y2, x2-y2) | |
| T1 | 3 | 0 | -1 | 1 | -1 | (Rx, Ry, Rz) | |
| T2 | 3 | 0 | -1 | -1 | 1 | (x, y, z) | (xy, xz, yz) |
Other character tables could be found here: List of character tables for chemically important 3D point groups
Reducible Representation[edit]
Reducible representation is a mix of the irreducible representations. In other words, it is the reduced form of a combination of the irreducible representation. This will explain how many σ molecular orbitals and π molecular orbitals are going to be found in the molecule. The type of orbital could be illustrated by using the character table, such as (x,y,z) which represents px, py pz orbitals and so on by reading the right two column of the table.
Representation for σ and π orbitals could be found in Td. There are two different methods to find the reducible representations.
Example with methane:
Method one
1. Look for each C-H bonds representation.
- E - there are four C-H bond in methane
- C3 - three C-H bond rotates but one C-H bond will not move
- C2 - all the C-H bonds move (it is shown as zero when all the bonds are moving)
- S4 - all the C-H bonds move
- σd - two C-H bonds will remain at its position
| Td | E | 8C3 | 3C2 | 6S4 | 6σd |
|---|---|---|---|---|---|
| ΓCH | 4 | 1 | 0 | 0 | 2 |
If it is reducible representation for C-H bond, at this point the following equation can be used in order to determine which irreducible representation is present to find the σ orbitals: ni=1/h∑NχRχI
ni represents the total number of the irreducible representation found in the reducible representation. h is the order of the group. This could be found by counting all the coefficient of the symmetry operation. N is the coefficient. χRχI is the reducible and irreducible representations character.[1][2]
ΓCH is reduced to A1+T2. This reduced form shows that total four σ molecular orbitals are found by looking the identity (E) of A1 and T2. A1 has x2+y2+z2 which indicates that it has a s orbital. p(x,y,z), p(xy,xz,yz) is seen in T2.
2.Follow the same step as above, but by looking only at x, y, z coordinates
- E - there are total three different coordinates
- C3 - x moves to y, y moves to z, and z moves to x
- C2 - goes negative itself
- S4 - x to -y, y to x, z to -z
- σd - if it was xz plane, x and z will remain still but y will go negative itself. Overall it will be one.
| Td | E | 8C3 | 3C2 | 6S4 | 6σd |
|---|---|---|---|---|---|
| Γx,y,z | 3 | 0 | -1 | -1 | 1 |
3. Multiply the two representation which could be expressed as Γσ+π = 12 0 0 0 2. Then subtract the Γσ which is ΓCH = 4 1 0 0 2.
The final representation then will be the Γπ only.
Γπ = 8 -1 0 0 0 is reduced to E+T1+T2.
Method two
Follow the same step as above, but focus on p orbital with π symmetry only. π orbitals are the two arrows pointing out from the hydrogen atom which is shown on the right figure. The results will be the same which is E+T1+T2. As a result, total eight π molecular orbital is found due to the total counts of identity of E+T1+T2 from the character table again.
If it was said to reduce methane it would be Γ = 15 0 -1 -1 3. Example for identity: methane has total five atoms and it has three coordination for each atoms. Therefore the E is 15 by multiplication. The representation will equal A1+E+T1+3T2.[3]
SALCs and MO diagram[edit]
SALCs is Symmetry Adapted Linear Combination and it is a useful technique to draw molecular diagram. As an example, Γσ was found to be A1+T2 so if the hydrogen atoms are numbered from one through four and the symmetry operation is occurred, A1 will yield (Φ1+Φ2+Φ3+Φ4) by multiplying the operation for hydrogen and its character of irreducible representation which is 1 1 1 1 1. Figure is shown in the right. T2 will be (Φ1-Φ2-Φ3+Φ4), (Φ1-Φ2+Φ3-Φ4), and (Φ1+Φ2-Φ3-Φ4). In order to draw the diagram, find the number of the nodes to determine the energy level. Least number of nodes would be the least energy. The energies for Td would be found a1(σ), t2(σ) < a1(σ) < t2(σ) < e(π) < t2(π) < t1(π). Note that the first a1(σ), t2(σ) is from the σ SALCs and the second is from the pσ SALCs. HOMO and LUMO could be found by drawing the diagram. The highest energy with orbital occupied is HOMO and the highest energy with orbital unoccupied is LUMO.
References[edit]
- ^ Miessler, Gary. Inorganic Chemistry. Prentice Hall, Inc. ISBN 9780321811059.
- ^ Pfennig, Brian (2015). Principles of Inorganic Chemistry. Wiley. ISBN 9781118859100.
- ^ Pfennig, Brian (2015). Principles of Inorganic Chemistry. Wiley. ISBN 9781118859100.