Molecular Orbitals in Inorganic Chemistry
Main group MO diagrams
Q: The question asks us to construct a representation table for the z-axis and to determine which irreducible representation it transforms as. I don't understand how an axis can have a representation table. Could you explain this question to me please?A: Draw a unit arrow along the z-axis of your molecule and then consider how it transforms (moves) under each of the operations, thus building a representation table for the axis. For example under C3 it will map onto itself, giving a 1 in the table.
Q: I thought in the H3+ molecule each hydrogen only has 1s orbital so how can it have improper rotation as there won't ever be a phase change?A: You are correct, the 1s orbital does not change phase on rotation and then reflection! Nevertheless the rotation and reflection still occurs, the orbitals are just symmetric for this system. Thus the three H1s AOs cannot transform as the a" irreducible representation, it is just not possible.
Q:Can you explain the diagram depicting the molecule with a p orbital on each hydrogen when its discussing improper rotation axes please?A: It is the molecule that determines the point group, not the fragment orbitals (this will become clear in the tutorial question for lecture 2) where you will see that I am building toward the molecule BH3 and using the H3 as a fragment. B of course does have pAOs. When an electron is excited from an s to a p AO in H, of course the pAO does not instantaneously come into existence on excitation, the orbital is still there even if it is not occupied. (However, taking this statement to its natural conclusion is a bit questionable, i.e. does this mean that the H atom has f, g, h orbitals!! Potentially it does, but they are going to be very high energy and we always ignore them!) I also ask students to put a pAO on the atom because many students have trouble visualising a vector out of the plane, it is easier to imagine a pAO. Of course you don't need to, you can just imagine a vector off both the rotation axis and mirror plane.
Q:I am slightly confused by the purpose of the labels T(x), T(y) and T(z) as well as R(x),R(y) and R(z). I am not sure what they are trying to show?A: Not an uncommon question! The T and R correspond to the motions of the centre of mass of a molecule (or the cartesian components of the origin of axial system) that is translation or rotation of the CoM. I cover this in my spectroscopy course, you can find last year's notes online (you need lecture 1) if you really want to know now. At the moment I just want you to relate the phase change of the axis (which relates to the translation of the CoM in the axial direction) or orbital to the symmetry label. For example, a p(x) orbital has the same phase pattern as the x-axis and the symmetry of the x-axis can be immediately read off the character table by realising it has the same symmetry as "T(x)" (translation in the x direction). I “derive” these components of the character tables in my spectroscopy course.
Q: Why do the 2s AO of oxygen interact with the H2 FOs in H2O? Surely the 2s O orbital is far too low in energy to do this especially given the relative electronegativity of Oxygen vs. Hydrogen? My belief is that the s-p splitting is far too large in O to allow the 2s to interact with the Hs. The diagram from Group Theory for Chemists and agrees with this thinking. A closely related question is also: For the H20 MO diagram why does the a1 from the H2 fragment interact with the a1 on O that is lowest in energy not the a1 that is closest in energy?A:The answer to this is quite complex, here is a detailed answer pdf
Q:Which do we take as the convention to label the symmetry of the fragment's; the fragment point group or the molecular point group?A: In general when using molecular fragments use the molecular point group and when using symmetry fragments use the point group of the final molecule for the fragment orbitals. Don't forget to re-assign the symmetry of the MOs to the irreducible representations (symmetry labels) under the full point group of the molecule (using short-cuts!) when employing molecular fragments.
Q: I've been looking through the MO diagram for CO for the tutorials and I'm confused about which orbitals are bonding. Are 4σ, 5σ and 1π orbitals all bonding or is one non-bonding? Because if they are wouldn't the bond order then be 4?A: A good question! Part of the problem is that CO has a complicated MO diagram due to mixing that does not match exactly what you might expect from a simple analysis. In this case I have computed the MOs (ie solved the Schrodinger equation) and the diagram in your notes is the way the MOs come out. Carry out a calculation yourself and you will see what I mean. Bond order is telling us about the interaction between two atoms C and O. We can assume the 2pi interactions are bonding. The issue arrises with the 4σ and 5σ orbitals. If you look at the 5σ orbital the electron density is polarized out of the internuclear region there is a very weak interaction between the end of the hybrid on the C and the pure pAO on O ... this orbital contributes very little to the formal bonding of C to O. While the 4σ is much more bonding this is a strong s on C and p on O overlap, strongly bonding between C and O. Thus the 5σ is essentially non-bonding and doesn't count. Then we have 3 clearly bonding MOs, the two pi and the the 4σ (the 3σ is non-bonding).
Q: In the CO tutorial (using your terminology) I have placed my non-bonding (C 2pz) sigma orbital (your 6σ) below the antibonding sigma orbital (your 5σ). Will this effect the mixing between these?A: In this case no, when orbitals mix they both contribute equally to a first approximation, and so there is no effect of having the orbitals reversed in this (very specific) case. I'd like to add, that since there is very little interaction between the 2s on the C and the 2pz on the O, this orbital should be thought of lying just above the O2p level, and as the C is significantly less electronegative the non-bonding C2p AO should lie above the antibonding sigma MO!
Q: For the intermediate B2H4 diagram how did you get the FOs for BH2 and how did you get the symmetry labels for b1 and b2 as when I tried to use the shortcut (looking at Tx and Ty) I got these the other way roundA: The orbitals for BH2 are made from a "known" MO diagram, that of water. Thus these are molecular fragments. For the second part of this question, my guess is that you have the x and y axis interchanged compared to my definition. Well done, you have found through experimentation that if the axes are rotated then the labels change. The principle axis orientation is well defined, but the x and y axis can be interchanged with no effect on the overall MO diagram. This is the reason I always require an axis definition, as long as you are consistent with the axis definition then your labels are correct. Note, that in the exam I will often give you the axis orientation to avoid precisely this problem. (I’m assuming this is what has happened if not do scan your answer and e-mail me and I will see where you have gone wrong.)
Q:In the MO diagram for B2H6, the 3a1 FO, which has the pAO on the oxygen and sAOs on the hydrogens confuses me slightly. As far as I am aware, mixing should occur and this orbital should be stabilised as it is occupied. But, I was wondering why the orbital on the oxygen looks like a pAO rather than a sp hybrid type orbital?A: Good you have noticed that the a1 FO is drawn slightly differently from the water MO diagram, well done! I have drawn it this way to be consistent with some of the text-books which simplify the mixing on this MO down to the pz AO on the O (ie no "sp" like hybridization). Thus, this is a mixed MO however, the sAO contribution from the O atom has been ignored. It makes writing the subsequent MO diagram easier. I would accept either type of FO in an exam situation.
There is a second and more complex issue going on here, which is beyond the scope of this course. The mixing is due to a reduction in symmetry, this means that the relevant FOs under D4h must have the same symmetry and be allowed to mix.
Q:In the MO diagram for PH5 (this MO diagram is generalised to PL5 in the workshop) why is the splitting energy between the a1' orbitals so large that the resulting anti-bonding orbital is above the e' symmetry orbitals in the final MO diagram. This is quite confusing because the original orbitals (FOs) for a1' are far below the starting orbitals of e' before forming the MOs.A: Each MO diagram has a different vertical scale which is not well defined (until you do a calculation!) So compared to for example H2O this MO diagram is very expanded. That is, the spread of the FOs has been exaggerated in this diagram.
We know P is second row and so has a smaller sp gap than the first row. We also know that the electronegativity of P is almost identical to that of H . Now consider the ligand FOs, in reality because the H orbitals are two bonds apart there is not a large amount of splitting between them. We also know that the totally bonding set of ligand (H 1AOs) is quite stabilising. Thus, we can rationalise that the spread of energies for the ligand FOs is moderate and that the sp gap in P is moderate. We also predict that the two FOs (totally bonding ligand FO and 3s FO of P) are almost identical in energy giving rise to a large interaction, which is larger than the spread of the ligand FOs or the sp splitting of P.
This is all confirmed when you look at the MOs for PH5
Additional comment: the spread of the sp splitting for the first row is larger than that of the second row, this can be related back to the shielding of the 3s and 3p by the inner 2s and 2p for the third row elements, while there are no inner shielding electrons for the 2p electrons. This can be illustrated by considering the orbital energies of the 2s and 2p for N which are -25.56 and -13.18 eV respectively, giving an sp gap of 12.38 eV. Compare this to the orbital energies of the 3s and 3p for P which are -18.84 and -9.6 5eV respectively, giving an sp gap of 9.19 eV. (Data from Inorganic Chemistry by G. Miessler and D. Tarr, 4th edition, Prentice Hall, Boston, 2011, Table 5.2, p145) One could also compare the orbital ionisation energies for these elements (for example in Essential Trends in Inorganic Chemistry by D. Mingos, Oxford University Press, Oxford, 1998, Table 2.4, p44). The sp gap of P is approximately 70-75 percent that of N (depending on the parameter you evaluate).
Q: For some MO diagrams you've got a gap between the lower lying orbital interactions from the ones above. How do we know when these spacings arise?A:In general for the electronegative elements of the first row the energy of the sAOs are well separated from the pAOs, ie the sp gap is larger for elements on the right hand side of the periodic table. However, because we need to condense a lot of information onto a single digram this gap between the s and p based MOs often gets contracted. Thus ideally it should be there and in the exam, this would be the type of thing you annotate or mention on your diagram, particularly if you have had to squash all the MOs up a bit.
Q: How do I know where to put the fragment energy levels?A: If this cannot be determined from knoweledge of the relative electronegativity of the fragments involved you will be given either (a) a basic energy level diagram or (b) the necessary information to make a good guess (such as information on where energy levels should lie, ionisation potentials or PES data).
Q:How far apart should the splitting be?A:The most important point is that you explain the reasoning behind your estimated splitting in the exam. As long as your argument is reasonable it will be accepted. Much of research in science is about making an informed prediction, and then checking it against reality (in this case a calculation). Thus outside of the exam you can always carry out a calculation to check!