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Topic: Sigma Bonding Reducible Representations?  (Read 11222 times)

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Offline CopperSmurf

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Sigma Bonding Reducible Representations?
« on: November 17, 2009, 11:49:06 PM »
If I had a metal M and 3 ligands L with D3h symmetry (planar triangle), how would I get the sigma bonding reducible representation?

In the end, I tried it and eventually got 2A1' + E' + A2" as my irreducible representation but I don't think it's right. I also looked at a previously (really old) post here and that person got this same irreducible representation for ML5 with D3h symmetry.

Also, if my (big) negative-charged ligands were really good pi donors, how can I find out which d-orbitals will be doing the pi-bondings?

And one more thing about the geometry, if I added 2 more ligands above and below this triangular complex, then I know I'd get a triangle-bipyramid, I don't understand how this can change the d-orbital diagram nor the SALC (symmetry adapted linear combination).

hints or anything would be nice. :)

Offline cth

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Re: Sigma Bonding Reducible Representations?
« Reply #1 on: November 18, 2009, 10:06:33 AM »
In the end, I tried it and eventually got 2A1' + E' + A2" as my irreducible representation but I don't think it's right. I also looked at a previously (really old) post here and that person got this same irreducible representation for ML5 with D3h symmetry.

I don't think 2A1' + E' + A2" is right, as it would mean you have 5 bonds (one from each A and two from E). But for 3 ligands, we expect only 3 bonds.

In case you don't have any point group table at hand, you can have a look at http://www.webqc.org/symmetrypointgroup-d3h.html. There, you see that 3dz2 belongs to group A1'. Orbitals 3dx2-y2 and 3dxy belong to E'. And so on. It is all tabulated.

Now, let's imagine hydrogen as a ligand. The simplest way to construct the molecular diagram is first to combine the three 1s orbitals from hydrogens together to form a H3 subunit. Afterwards, you combine this H3 with the metal to form MH3. It may look artificial to proceed like that in two steps, but it is easier to do.
So, the three 1s orbitals will combine to give three hybrid H3 orbitals.


Look at the right side of this diagram for D3h symmetry.
The lowest orbital symmetry belongs to group A1'. The two degenerate orbitals above belong to group E'. OK?

Now, we bring H3 and the transition metal M together. Only orbitals within same group will interact. Orbitals from different groups are orthogonal and do not interaction with one another. So, the lowest orbital from H3 will interact with 3dz2 (same group A1'). The two degenerate orbitals from H3 will interact with 3dx2-y2 and 3dxy (same group E'). Orbitals 3dxz and 3dyz remain unchanged, as no H3 orbital belong to group E''.
Once this is done, you add the electrons and find the representation A1'+E'.

Also, if my (big) negative-charged ligands were really good pi donors, how can I find out which d-orbitals will be doing the pi-bondings?
It depends on the pi orbital symmetry. For example, let's consider on each ligand a 2pz orbital perpendicular to the plan of the molecule. When you take the three ligands together to form L3, the lowest resulting orbital is in the group A2" and the two other degenerate orbitals are in E". In ML3 diagram, the orbital in A2" will not interact, but the ones in E" will interact with 3dxz and 3dyz.

And one more thing about the geometry, if I added 2 more ligands above and below this triangular complex, then I know I'd get a triangle-bipyramid, I don't understand how this can change the d-orbital diagram nor the SALC (symmetry adapted linear combination).
With two more ligands above and below, the orbital 3dz2 which is pointing directly at them will interact with their orbitals. It will change its energy. And as well, you bring two new orbitals from the new ligands into the diagram.

Offline CopperSmurf

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Re: Sigma Bonding Reducible Representations?
« Reply #2 on: November 21, 2009, 11:03:32 PM »
Wouldn't the whole story be different if the ligands weren't simply hydrogen atoms?
i.e. ligand = N(R2) ? Because ligands would have symmetries of their own, or are those simply ignored when looking at their SALC (symmetry adapted linear combo) with the metal's d-orbitals?

And are A2" and E" coming from the metal?
I'm having trouble seeing how the d-orbitals are splitting like this.

Offline cth

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Re: Sigma Bonding Reducible Representations?
« Reply #3 on: November 22, 2009, 08:26:16 AM »
Wouldn't the whole story be different if the ligands weren't simply hydrogen atoms?
i.e. ligand = N(R2) ? Because ligands would have symmetries of their own, or are those simply ignored when looking at their SALC (symmetry adapted linear combo) with the metal's d-orbitals?
Yes, possibly. It all depends on the ligand orbitals symmetry.

The overall ligand symmetry has a direct effect on its own molecular orbitals. Depending on the ligand configuration, atomic orbitals (coming from atoms constituting the ligand) will interact differently  :rarrow: molecular orbitals shape and energy will be affected.
For example, if you have two sp2 hybridized carbon atoms next to each other, their 2pz orbitals will interact to form a π-bond. Now, if you force one of the carbon atom to rotate by 90º angle (by some steric hindrance), then the 2pz atomic orbitals become perpendicular to each other  :rarrow: the π-bond is broken.
Of course, the example I took is extreme. But, it illustrates that ligand spatial configuration has an influence on its molecular orbitals.

In turn, molecular orbitals symmetry has an influence on their interactions with 3d metal orbitals. If a molecular orbital belongs to one symmetry representation or to another one, it interacts differently with 3d orbitals.

So, to summary, ligand symmetry has an indirect influence on the interactions ligand/metal. But, what really matters is the symmetry of ligand molecular orbitals.

For NR2 case, the nitrogen atom is hybridised sp3 with two lone pairs and two R groups. One of the two lone pairs will coordinate to the metal centre while the other one remains unchanged (it is pointing away from the metal). The 3 lone pairs coordinating to the metal (coming from the three ligands) will interact in the D3h symmetry to give A1' and E' representations. So, it turns out the example NR2 is not very different.
It can get more complicated with unsymmetrical ligands.

And are A2" and E" coming from the metal?
I am not sure I understand the question.
A2" and E" come from symmetry considerations. It is more general than metal atoms. Ligands can also belong to A2" and E" if they satisfy to the symmetrical requirements.

I'm having trouble seeing how the d-orbitals are splitting like this.
Yes, D3h is not easy to work with. D4h would be easier to see because most 3d orbitals have 4 lobes.

In D3h, ligands are located in the mirror plane. 3dx2-y2 and 3dxy are also within the mirror plane. Their interactions will be different than with 3dxz and 3dyz which are perpendicular to the mirror plane.

Offline CopperSmurf

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Re: Sigma Bonding Reducible Representations?
« Reply #4 on: November 22, 2009, 06:49:35 PM »
For NR2 case, the nitrogen atom is hybridised sp3 with two lone pairs and two R groups. One of the two lone pairs will coordinate to the metal centre while the other one remains unchanged (it is pointing away from the metal). The 3 lone pairs coordinating to the metal (coming from the three ligands) will interact in the D3h symmetry to give A1' and E' representations. So, it turns out the example NR2 is not very different.
It can get more complicated with unsymmetrical ligands.

Wait a sec, I thought A1' and E' representations were used for the sigma bonding? It can't be in both sigma and pi bonding, can it?

Offline cth

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Re: Sigma Bonding Reducible Representations?
« Reply #5 on: November 23, 2009, 04:52:30 AM »
It can't be in both sigma and pi bonding, can it?
A lone pair pointing towards the metal centre would form a sigma bond. The electronic density is located in between the metal and the nitrogen, as for any sigma bond. The two R groups and the remaining lone pair on the nitrogen are pointing away from the metal, so they don't interact with it. In the case of the anionic ligand NR2-, the nitrogen is hybridised sp3:
   _
R-N-R

so it will form only sigma bonds with R and the metal centre, and have a lone pair in its orbitals.

Another simple example:
If NR2- comes in contact with H+, it will form the amine compound NHR2. The N-H bond is a sigma bond. And the nitrogen still has one remaining lone pair pointing away from the hydrogen in a tetrahedral geometry. (I like taking examples with hydrogen, it is so much easier to understand  :D)

Wait a sec, I thought A1' and E' representations were used for the sigma bonding?
In D3h, yes. But it is not a general rule. It is a consequence of the orbital symmetry that fulfill the symmetry operations of A1' and E'. In different symmetry point group than D3h, it could be different. You have to work out the symmetry of the orbitals (is it symmetrical or antisymmetric with the inversion centre, with the mirrors, the rotation axises,...? Then, refer to the corresponding point group table.).

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