# D6h symmetry clusters- orbital classification confusion!

I was am simulating high symmetry clusters D6h in several programs such as molcas and gaussian. I understand that the highest symmetry used in these programs is D2h and therefore there would be mixing of the different symmetry representations upon correlating from D6h to D2h. However upon simulating Mn@Si12 cluster, I'm seeing that the silicon atoms in the silicon cage have for all representations: s, px, py and pz orbitals despite there being restrictions according to D2h e.g b3u representation has only px functions and b2u only having py functions. Could anyone please shed some light on this phenomenon?

The symmetry operations are usually defined with respect to the origin of the coordinate system. For single atoms one naturally chooses $(0,0,0)$, which coincides with the symmetry center.

For example the $\sigma_{xy}$ mirror plane in matrix representation (with $(0,0,0)$ being the symmetry center) would be:

$\sigma_{xy}=\pmatrix{1&0&0\\0&1&0\\0&0&-1}$

which means: invert every $z$ component. If applied to some object centered on $(0,0,0)$ (e.g. the AO of an atom on this position), the result would still be centered on $(0,0,0)$.

In case of molecules we can have at most one atom sitting on the symmetry center. Sometimes none, e.g. in $\ce{C2}$ where the symmetry center is in the middle between both atoms. If you apply the above matrix here (assuming the molecule axis is aligned with the $z$ axis), then it would map the AO from one to the other atom. Therefore it is not symmetric (with respect to the above defined symmetry operation) anymore.

Thus you cannot relate the IRREP of an atomic orbital in the single atom, with its IRREP in the molecule. The symmetry center changed! For example take an $s$ atomic orbital of any of your $\ce{Si}$ atoms. The position of the atom does not coincide with the center of the molecule. Therefore any symmetry operation (except for $E$) would map the $s$ orbital to some other $\ce{Si}$. The AO is no longer characterized by an irreducible representation.

What the code does instead, is first generating Symmetry Adapted Linear Combinations (SALC) of the atomic orbitals. Those are then again characterized by IRREPs (of the molecule) and used for the calculation. Now each obtained MO is indeed only constructed from SALC of the same IRREP. In other words: the Hamiltonian, and therefore the MO coefficient matrix as well, become block diagonal. However, in the output the MO coefficients are transformed back from the SALC basis set to the AO basis set, which is not block diagonal and may contain contributions from any AO for each IRREP of the MOs (an exception would be the AOs of an atom located in the center of the molecule).

• Thanks alot for your answer, I believe you have answered my question quite clearly, but I'm still struggling to understand a few things. Would you be kind enough to answer this query please. If it is not possible to relate Irrep of AO to Irrep of MO and each MO may contain contributions from any AO for each IRREP, how is it possible to classify basis functions according to irreps . This done for high symmetry molecules in molcas manual p438. Many thanks for the help. Jan 21 '18 at 11:45
• I only found a HTML version of the molcas manual, so I cannot check what p438 is about. Anyway, you do have a basis which you can classify according to irreps: the SALC, which is build based on the AO basis. Jan 21 '18 at 12:26
• Are you referring to Table 10.5? It has a footnote: "Functions placed on the symmetry center." As there can be at most one atom in the symmetry center, this does not work in general. Jan 21 '18 at 12:33
• Yes i'm referring to table 10.5. Thanks alot for looking at it. So for the example of NiH, Ni is on the symmetry centre? and for C2 and Ni2, one atom of each lies on the centre and by symmetry operations, the functions are applicable to the other atom? Jan 21 '18 at 12:46
• Homonuclear diatomics, like $\ce{C2}$ and $\ce{Ni2}$ have an inversion center inbetween both atoms. That is their symmetry center, hence there is no atom sitting on the symmetry center. For the other example: heteronuclear diatomics don't have inversion symmetry, so there is no symmetry center. Jan 21 '18 at 12:52