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With A and B, it's pretty simple. If you rotate the molecule about its principle axis to where the orbitals line up, and the signs change, it's Mulliken label has a B. otherwise, it's A.

I heard E symmetry means there are two configurations that are equal in energy, and T means there are three configurations that are equal in energy. My professor posted some slides that included examples of E symmetry and T symmetry, but it's still confusing:

enter image description here

Why is this E symmetry? It shows a configuration with one sign along the Z axis and the other sign for the other two axes, and then a configuration with one sign along the Y axis and the other sign for the other two axes. Why can't we have a third configuration, with one sign along the X axis and the other sign for the other two axes, making this have T symmetry? Could somebody explain what E and T symmetry are in their words?

EDIT: I just talked to my professor, and I forgot his exact explanation, but he said the above diagram has a mistake in it. He said two of the orbitals should not be there.

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2 Answers 2

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In point groups with three fold or higher axis of symmetry ($C_n$ where $n>2$) there is the possibility of degeneracy, in other words rotation about a horizontal axis and about a vertical one have the same moments of inertia, and hence angular momentum, and will have identical rotational energy levels. (You can see that the structures you draw differ only in how xyz are labelled). Similarly vibrational levels may have as a result of symmetry identical energy levels. These degenerate motions have to be treated as a set and are labelled E for double degeneracy and T for triple. The symmetry species A, B E etc., are defined by their row of characters in the point group which describes how they behave to symmetry operations such as rotation, reflection, etc. Each row is unique and hence has a unique label, typically 1, -1, 2, 0 etc. this label is a representation of how the symmetry operation affects the molecule.

To be able to generate the characters in a point group and find what orbitals these belong to you need to put vectors on xyz and operate on them with matrices representing rotations, mirror planes etc. so it is not straightforwards but the book by Carter (see below) explains this.

(The Mulliken labels are actually quite complicated, if you want the original see Mulliken, J. Chem. Phys. v23, p1997, (1955), or a summary by R. Carter 'Molecular Symmetry & Group Theory' publ Wiley, or in detail by G. Herzberg "Molecular Spectra and Molecular Structure' vol II van Nostrand Rienhold)

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    $\begingroup$ Just a historical trivia: T is not for triple. Actually, E for Entartet is also dubious. Since Mulliken borrowed someone else's symbols from a German chapter in an obscure book called Handbuch der Radiologie. Someone had shared his unpublished book chapter on group theory with him and Mulliken got the full credit for the symbols he never invented. Actually none of Mulliken symbols are his. $\endgroup$
    – AChem
    Nov 7 at 15:45
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    $\begingroup$ Interesting, I should have used three fold instead of triple. Some author use F instead of T. $\endgroup$
    – porphyrin
    Nov 7 at 17:40
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    $\begingroup$ You are right, F is the older symbol for T. $\endgroup$
    – AChem
    Nov 8 at 19:19
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First of all — it should be noted that this is not a unique representation of the $\mathrm{e_g}$ set. You can take any two linear combinations of these two orbitals (i.e. add and subtract them), and as long as they are linearly independent (i.e. the resulting combinations have no net overlap with each other), they are valid depictions.

But to address the question about why there isn't a third member: the three axes are not entirely equivalent. In fact, there is an example that may be closer to home — there is a similar 'asymmetry' in the set of atomic d-orbitals as well: there is a $\mathrm{d}_{z^2}$ but there isn't a $\mathrm{d}_{x^2}$ or $\mathrm{d}_{y^2}$.

The three axes are equal in the sense that we could have chosen any of the three axes to be 'special', but by convention (and for mathematical reasons) we chose $z$ (see: Why is there a z² label for d orbitals, but no x² and y² labels with corresponding shapes?).

In an octahedral metal complex, the metal d orbitals are split into $\mathrm{e_g}$ and $\mathrm{t_{2g}}$ sets. The former contains the $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ orbitals; the latter the $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$. These can overlap with symmetry-adapted linear combinations (SALCs) of ligand orbitals, as long as the SALCs have the same symmetry as the metal d orbitals.

The SALCs are precisely what are drawn in the diagrams in your question. You'll see that the first of the two SALCs that you drew overlaps with the $\mathrm{d}_{z^2}$ orbital, and the second overlaps with the $\mathrm{d}_{x^2-y^2}$ orbital. In other words, the $\mathrm{e_g}$ SALC overlaps with the $\mathrm{e_g}$ set of d-orbitals, so all is well and good.

I'm aware I'm not directly answering why the SALCs lack a third member. This is mainly because I don't want to dive into that maths. But I hope that the discussion above, and the link, explain why the d-orbitals aren't perfectly 'symmetrical'; and I hope that the symmetry overlap between the SALCs and the d-orbitals at least suggests that the same argument can be extended to the SALCs.

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