I'm fairly confused now. Lets use chromium hexacarbonyl as an example and lets pretend that CO ligands can't back bond. So the chromium atom already has its 4s orbital and 2 of its t_{2g} orbitals filled. The first carbonyl ligand then forms a bond with the remaining empty t_{2g} orbital and the remaining 5 CO ligands bond to the e_{g} orbitals as well as the 4p orbitals. So thinking about molecular orbitals, 4 of the 18 electrons fill the 3s bonding and antibonding orbitals. 12 electrons are required to fill the t_{2g} bonding and antibonding orbitals. There are 2 electrons leftover which fill one of the e_{g} bonding orbitals. In this case, the e_{g} bonding orbital is the LUMO, not the antibonding orbital.

Don't worry CrimpJiggler--I think your response gave me enough information to clear your confusion.

To help out, I think the following MO diagram from wikipedia will be helpful:

http://en.wikipedia.org/wiki/File:LFTi(III).pngMy first comment--for Cr

^{0}, the free metal electron configuration is 4s

^{1}3d

^{5}. This is one of the deviations from the trend, but it is irrelevant for this discussion. Why? Because as soon as you put ligands on that metal, the sigma bonding with the 4s orbital will drop greatly in energy as it combines with ligand orbitals and be filled by electrons. The corresponding antibonding orbital would be much higher in energy and so we don't have to worry about a close energy 4s orbital. To help understand this, take a look at the link above and look where the 4s metal orbital goes. Thus, the d-electron count for Cr

^{0} is d

^{6}. Even if it was d

^{4}, 2 of it's t

_{2g} orbitals would not be filled as you mentioned. That would disobey Hund's first rule. So (assuming low spin because of the carbonyls), two electrons would fill one t

_{2g} orbital while the other two orbitals of the set would be filled with one electron. Since we have a d

^{6} metal in an octahedral low spin environment, all 3 orbitals of t

_{2g} would be filled and e

_{g} would be empty.

Now, what I think you are confused about is about how MO theory works. You mentioned that the first carbonyl would need to bond with the remaining "empty t

_{2g} orbital" which we know now is not there and that the rest of the COs bond to the e

_{g} and 4p orbitals of the metal (empty). This is not how it works. Orbitals don't combine based on complimentary occupancy. They combine based on shared symmetry, overlap, and (I'm not as sure at this last point) similarity in energy. So forget about electrons until the end. Just look at the orbitals. I think at this point, return to the link and look at the metal orbitals and the ligand orbitals on both sides and observe that the orbitals only of similar symmetry combine. The ligand orbitals have already been combined into 3 sets of orbitals from making symmetry adapted linear combinations (SALCs) of the 6 ligand s-orbitals. This generated 3 sets of orbitals from the 6 ligand s-orbitals, one of a

_{1g} symmetry, two for a degenerate e

_{g} set, and 3 for a t

_{1u} set. With octahedral symmetry, the metal 4s is a

_{1g}, 4p orbitals is t

_{1u}, and the 5 d orbitals are split into e

_{g} and t

_{2g} orbitals. Thus all metals orbitals except for the t

_{2g} set combine with the ligand orbitals (NOTE: I'm respecting your assumption that the CO ligands can't back bond--which means that we just are looking at the sigma-donating ligands only, of course there would be more to say if we include the back-bonding). Okay--so now look at the MO diagram again in the link--we have six orbitals much lower in energy, the 5 from the d-set and 4 more antibonding orbitals above that. Note that the number of orbitals hasn't changed--a good sign. The t

_{2g} set is nonbonding since it had no interaction with anything. All above is antibonding and all below is bonding. Let's fill in electrons. From the ligands, we have 12 electrons. They already fill all the bonding orbitals. What's left--the d electrons from the metal. This is why we only care about the (mostly antibonding) d-set. Because this is the area where electrons are limited and show us the frontier orbitals. Then fill accordingly like any other MO diagram. For our d

_{6} Cr

^{0}, and assuming a large splitting energy because CO ligands are strong field ligands, all 6 electrons fill the t

_{2g} set. With that, we described the MO diagram for a sigma-donating only version of CrCO

_{6}.

I hope that explanation clears things up for you. MO theory is pretty abstract and challenging to understand. I'd be happy to continue the discussion, to explain the parts you are confused about or to continue in other directions. I don't want you walking away without understanding the theory!

I have a question regarding the CFSE too. I have a UV-Vis absorption spectrum of a nickel (II) octahedral complex and I was asked to calculate the CFSE.

Is this the right method to calculate the CFSE?

E = (h)*(c/wavelength) = E in J

convert J to kJ/mol

E(kJ/mol) = [E(J) * 6.02*10^23] / 1000

is that correct?

However, there are two obvious peaks in my absorption spectrum, does that mean there were two electronic transition and they each have different CFSE?

What about the Racah parameter? I was asked to calculate the Racah parameter too.

Thanks!

zs3889, I haven't ignored your question. I haven't done such calculations in a long time so I'm hesitant to answer. I'm not sure what level of theory you have had. In general chemistry, professors tend to teach that the CFSE is the energy of the peak you see in a spectrum, assuming that it is a d-d transition from t

_{2g} to e

_{g}. This is a huge oversimplification and in more advanced inorganic courses you learn more details. In your spectrum, the two peaks could arise from different d-d transitions that are both allowed, there could be Jahn-Teller distortion effects potentially, and also, depending on your ligands, there could be ligand to metal charge transfer or other absorption bands from the ligands themselves if they are colorful stand alone. I'm foggy on my notion of the Racah parameter, but I believe you use Tanabe-Sugano diagrams for that more advanced analysis of electronic transitions.

In the future, it'd be best to start a new thread I believe since I think your questions are different in nature and it would be confusing to have two conversations in one thread.