Molecular Orbitals in Inorganic Chemistry
Transition metal MO diagrams
Q:I'm confused the rules seemed to have changed between building a main group MO diagram and a TM MO diagram?
A:You are correct the rules have changed! It is important to realise that we treat the TM-MO diagram DIFFERENTLY from main group MO diagrams. We are no-longer building a MO diagram from scratch, TM MO diagrams ALWAYS start from the octahedral case. All changes are now considered as SMALL perturbations to this diagram! (for this course!!)
Q:Why do the metal 4s AO and a1g ligand FOs interact, are they not too far apart in energy?
A:First remember that TM complexes break the rules. So all the nice tidy rules we had for main-group elements do not hold as happily for TMs. So with TM we start with the octahedral diagram as I have given it to you. Why do I know it looks like this? Because when we calculate the MOs this is how they come out (in a generalised way). You should also remember from foundation that the 3d and 4s orbitals are actually very close in energy (go back and look!) so this is a very "expanded" diagram compared to the main group complexes we have studied up to now. Lastly these are sAOs they tend to have strong interactions and this has occurred here. I have plotted the radial extent of some AOs for you, look how far out the 4s are, they are going to overlap well with the ligands while the contracted 3dAOs are not going to overlap so well. So the sAOs have disparate energy but good overlap while the 3dAOs have close energies but poor overlap. It would be nice if the physical world was simple and could be described by simple rules, but it is not and we have to try and make sense of this complexity as best we can. So do not forget that the "rules" are just us trying to organise and make order out of a complex system and they are not "the truth".
Q:Why are the orbitals in L6 Figure 19 and 20 split?
A:These figures refer to the reduction in symmetry from Oh to C4v for the metal and ligand fragment orbitals respectively. As the point group changes, the way the orbitals respond to the symmetry elements stays the same. However, now there are less symmetry elements (look at the top row of C4v compared to Oh). Thus the way we label the symmetry of the orbitals has changed.
Nevertheless, the metal orbitals are still degenerate. Perhaps you missed it in the lecture when I said that I separated the energy lines to help you identify the labeling you should expect? As this has been a common question I've created an updated diagram that better represents the situation, see the first figure below.
The ligand orbitals which do not contain the new apical ligand fragment remain the same energy, however those that contain the new frament will change in energy slightly. The change will depend on the exact ligand and so will the energy ordering of the levels. See the second figure below, here I have used a slightly larger new component which increases the bonding and antibonding interactions slightly.
Q:I'm confused about the two descriptions of the ligand orbitals, L5 Figure 11 and L5 Figure 13?
A:the first thing to realise is that we treat the TM-MO diagram DIFFERENTLY from main group MO diagrams. We only include certain information, it is assumed you are aware of the rest. When constructing a TM-energy diagram to show all the ligand orbitals is too much work, so we only show those that interact with the metal are still there we just assume you know this and then ignore them!) So in this case the only orbital we are interested in is the one with a lobe pointing towards the metal, like that below for EH3.
The second thing to realise is that small perturbations to an orbital do not change the overall symmetry of the orbitals dramatically. Yes the symmetry is formally "broken" but in actual fact there is not a large effect on the MOs. So we can make the simplifications shown above. The "sp-like" FO are simplifications of the real FOs involved, for example in the case of NH3 or PH3 or SiH3 or CH3. The first representation below is using the H 1sAOs and the second is using ligand L "sp-like" FOs.
Q:I'm confused, in L6 we derived first the Oh energy diagram and then the C4v energy diagram by changing the apical sigma ligand, as shown below. Why is the 3d a1 FO not interacting with the lowest energy ligand a1 FO?
A: We always start with the Oh diagram and consider changes as small perturbations Notice how the "same" energy levels interact even though the symmetry labels have changed! For example, the 3d a1 FO is not interacting with the lowest energy ligand a1 FO, it is still interacting with the same energy level as when it was an "eg" label. The "eg" level is split slightly due to the change in symmetry, but the overall the same levels interact. This is the new rule in action for TM-MO diagrams.
Q:I was trying to go over the symmetry elements for the octahedral point group and couldn't find all the 6C2 axes
A: this one is really hard to draw, the best way I have found is to work it out "by symmetry". First, we know there is a C2 axis between each set of bonds. In the equatorial plane this gives us two C2 axes. Then for each of the axial planes (these are defined along the bonds of the equatorial ligands) there are another two C2. As there are two axial planes of this type we count four C2 in the axial planes. Combining these results we have six C2 axes in total.
Q:In Lecture 5, looking at the octahedral point group, the labeling of the mirror planes is confusing, one is annotated as sigma-d however I was under the impression this was sigma-h as the mirror plane depicted does not pass between bonds but instead is perpendicular to the principal C4?
A: First of all, this is a tricky one, well spotted! The problem is that we have multiple principle axis perpendicular to each other. So the same mirror plane is "σd" (or σv) for one axis and "σh" for another. So we select each C4 axis, label the mirror plane perpendicular to that axis as "σh", notice how this goes through the center of a face (red mirror plane in (b)). Then what mirror planes are left? Those that go through the edges of the cube (red mirror plane in (a)) these we call σd ("d" because they go between C2 axes).
Q:I have a problem concerning homonuclear diatomic transition metal MO diagrams. In the tutorial question, it asks us to draw a MO diagram for diMolybdenum. From our notes, it shows the s-orbitals to be far in energy from the d-AOs and so the lowest lying bonding orbital from the s-AO overlap is higher in energy than the highest energy antibonding orbital of the d-AO overlap. Therefore, I was wondering what makes the s-AO shift down in energy in diMolybdenum? Would it be due to a similar explanation as the s-p gap increasing across the PT and so the s-d gap increases too?
A: In the notes I wanted to start with a simple example, and so I started with a TM where the nd-(n+1)s gap is large. The real situation is quite complex! The energy of an orbital depends on its occupation, thus if the occupation changes the orbital energy changes. For a given configuration the nd subshell is stablised relative to the (n+1)s subshell across the series. However, the difference in the electron-electron repulsion and exchange energies for a 3dn-14s1 vs a 3dn4s2 configuration means that the energy "difference" between the two configurations is not a simple difference in energy between the frozen 3d and 4s orbital energies. The difference in energy taking into account electron electron exchange and repulsion shows a decrease in the "sd gap" across the series. In addition there is the "half filled shell" stability (maximum stabilisation of the exchange energy) for Cr and Cu where a 3dn4s2 with n=5 and 10 respectively, configuration is favoured. In this question I am expecting you to recognise that the (n+1)s (n=3,4 and 5 respectively) for Cr, Mo and W are doubly occupied and thus to "guess" that (n+1)s orbitals are potentially "stabilised" for these atoms. Of course, in reality it is not just the orbitals that are stabilised but the configurations, which lead to orbital stabilisation. For those looking for a fuller explaination here is an article: Why the 4s orbital is occupied before the 3d by M.P. Melrose and E.R. Scerri, J. Chem. Ed., 1996, 73, p498 http://pubs.acs.org/doi/pdf/10.1021/ed073p498
Q:Could you help explain to me why the antibonding a1g orbitals in the digram from page 2 (shown below) of the L9 hand drawn MO diagram are higher in energy than the b1g orbital?
A: To answer this question you want to think about (a) where else have you seen these orbitals and (b) what is their comparative orbital character?First consider (a), you have seen these orbitals in the ML6 octahedral sigma framework, where they are degenerate. So a good first approximation would be to guess these are very close in energy. We also know that the pi-donor sigma orbitals can be a little higher in energy than the sigma-donor orbitals thus having a slightly reduced energy difference and hence slightly smaller energy gap leading to a slightly stronger interaction and slightly larger splitting energy. Comparatively this means the X2 contribution will be slightly larger leading to a greater antibonding interaction.
Now lets consider (b), there are 2 long range (through 2 bonds) bonding interactions in the b1g and 4 medium range (through space) antibonding interactions. For the a1g there are 3 long range (through 2 bonds) bonding interactions, and 4 medium range (through space) bonding interactions. There are also 8 medium range (through space) antibonding interactions, which are slightly larger due to the larger X2 contribution. If we recognise that the overlap is highly distance dependent the medium range interactions will dominate. Thus we are comparing 4 medium range (through space) antibonding interactions of the b1g with the 8 antibonding and 4 bonding medium range (through space) interactions. Thus, if these were to cancel the a1g also has net 4 antibonding medium range (through space) interactions. If we also recognise the slightly larger X2 contribution the a1g should be slightly higher in energy.
However, as these orbitals are very close in energy subtle effects may come into play, such as the Hij term which we cannot “guess” so easily. Thus, these energy levels should be close together, but I would accept the a1g lying slightly below the b1g as well, we would need to carry out a calculation for a specific ligand and metal system to determine the exact ordering. In-fact this is essentially stated on the next page of the notes in the large MO diagram (regarding MOs C and D).
Q: In the handout, you say that pi donor ligands have no pi* antibonding orbitals so the metal is forced to interact with the pi fragment orbital of the pi donor ligand, which lies just underneath the d AO. But in the problems class you made an MO diagram of ML4X2 where X are pi donor ligands. In the final MO diagram you've included pi and pi* fragment orbitals!!! And you've interacted the pi* FO! This is just confusing!
A: Here there is some confusion about two different types of orbital, both have pi symmetry but they are different orbitals. Remember pi is a symmetry label it is not a particular type of orbital. So when there is one pi donor there is only a single orbital, the pAO on the pi donor thus there is only one orbital to interact (no pi* FO). BUT when we have two pi donors, the two pi-dornor pAOs can interact to produce two sets of pi orbitals (now we do have a pi* set). The pi* anti bonding set are not the same as the pi* orbital on a single ligand but are the antibonding combination from two different ligands. I have attached a diagram which I hope makes this much clearer.
Q: In lecture 8 the diagram shows that the π* (pi star) orbital is above the 3d orbitals of the metal however, this is the opposite for the example you did in the problem class - why are they not of the same energy?
A: The difference is the type and number of ligands. In the problems class we looked at a pi-donor ligand, in L8 we treated a pi-acceptor ligand. Also be very careful, there is a difference between treating a single ligand with two pi orbitals, and two ligands each with one pi orbital!
Q: In TM MO diagrams are a1 and b1 meant to be degenerate?
A: When the symmetry of a TM complex is Oh these orbitals form a degenerate e symmetry set, however once the symmetry drops the a1 and b1 (C4v) or a1g and b1g (D4h) are no longer formally degenerate. Typically the symmetry had dropped because we have different and/or new types of ligand, this means the orbital energies are also slightly different. For example, axial X(pi) ligand sigma orbitals are slightly different from the L(sigma) ligand orbitals in ML4X2.
Q:In your MO diagram of the ML4 square planar molecule (tutorial question L7) the 3a1g and 2b1g anti-bonding orbitals are raised above the 4s level of the TM. I am wondering why this is the case, as in octahedral molecule, the splitting energy is not so large, and the anti-bonding orbitals still lie below the 4s?
A: In this case you need to bring in some knowledge from your first year foundation and coordination chemistry courses. The TMs that form square planar complexes are Ni, Pd and Pt. In the electronic filling of the orbitals the (n+1)s is typically considered to be very close in energy to the nd AOs. For example, look at the electronic structure of Ni (and Pd and Pt) from web-elements. Thus, we start with a fragment orbital pattern assuming that the (n+1)s level is very low for these TMs (adding an annotation explaining this in your MO diagram would be a great idea). When I draw the "standard" ML6 diagram we assume the (n+1)s lies sufficiently above the nd that interactions are minor.The above description is substantially "simplified", and the actual situation is far more complex, see the discussion in Q2 above. We will be addressing configurations in my Spring term course.
Q: I have a question about one of the exam questions. As illustrated below, the first question (i) was to draw the 4e and 4a1 orbitals. The 4e1 orbitals seem logical, and so does the 4a1 orbital, until I read in the answer key, which stated that the metal contribution to the 4a1 orbital was bigger than the ligand contribution. In normal circumstances I would follow the same reasoning, but due to the presence of the pi acceptor ligand (which is slightly higher in energy than the metal d orbitals), I was unsure as to why the metal contribution was higher to the MO than the ligand contribution.
A: Careful, you need to look very carefully at the diagram, it tells you what to do and which ligand orbitals are involved. The 4a1 is an antibonding MO formed with the a1 ligand **sigma framework** and the dAOs, as this orbital is antibonding the higher lying fragment is the largest contributior and in this case it is the dAO. The sigma orbitals are almost always deeper than the dAOs, it is only the pi orbitals that come close to the dAOs in energy. The 4e MO is overall antibonding and so the highest energy FO contributes most, in this case it is the p(pi)antibonding FO of the ligand which is higher in energy than the dAOs.