Irreducible representations gerade

In the MuUiken notation, the subscripts u (ungerade = odd) and g (gerade = even) indicate whether an irreducible representation is symmetric (g) or anti-symmetric(M), in respect to the inversion operation (/). 

The freedom of a possible transition is restricted by the selection rules, a kind of traffic regulations for spectroscopic transitions, which are based on Eq. (5). The transition moment being zero or nonzero predicts whether a transition is forbidden or allowed, i.e., that the intensity of a forbidden band is much lower in magnitude than that of an allowed band. In the case of an allowed transition, the integral must not vanish, i.e., the integrand must be an even or gerade function, or, in terms of symmetry, must be or contain the totally symmetric irreducible representation. 

If the object has even symmetry under i, then the product of each member of the subgroup with i will lead to the same character as the subgroup member itself. These are the even irreducible representations given the additional subscript g . For the gerade representations in the first three characters for the rotational subgroup are simply repeated under the operations generated by their product with i, so that the E, 2C3, 3C2 characters are repeated under i, 2Se and 3<7d-... 

The Cg label indicates a doubly degenerate level, it represents two d-orbitals, and so the right-hand side of this equation does contain five orbitals as required. The four irreducible representations all have gerade symmefiy, in line with the d-orbitals that underlie them. 

For the d-orbitals we will apply the reduction formula. To make the job easier note that these functions are not changed by the inversion centre, since they all contain only even products of the x, y and z basis. This means that the d-orbitals have gerade symmetry, and so we only include irreducible representations with the g subscript in the reduction. The application of the reduction formula is laid out in Table 5.16, which shows that... 

Table 5.16 Application of the reduction formula to r(d) in Oh. In this case only the gerade irreducible representations need be considered, as each of the d-orbitals is gerade. 

The correct combination must be two XI representations. At this point, any of these would be allowed because they each have character 1 under 2Coo in Table 5.21. However, under the ooo-v class the 11 (both gerade and ungerade cases) have -1. This means that the inclusion of these in our linear combination will lead to a total character of less than the required 2. So the set of irreducible representations we seek can only contain X1+ types. 

Next, looking at the inversion centre column, in F we find 0, and since gerade representations have character 1 under i and ungerade —1, the only possible combination of irreducible representations remaining is... 

Now we will consider each class of operations to narrow down the possible standard irreducible representations that can be present until we arrive at only one option. The 3 under the E class reminds us that there are three orbitals being represented, and so our combination must consist of either three E-type representations or one E and one doubly degenerate representation. Under the inversion centre i the total character is —3, and so all three orbitals must be reversed by the inversion. This means that any irreducible representation present must have ungerade symmetry if we assigned a gerade representation, then it would contribute positively under i. This eliminates all gerade representations from further consideration. 

For more complex molecules in point groups with the inversion, centre vibrations that are IR active will also have irreducible representations that are ungerade while Raman-active modes will be gerade. Since different vibrations will usually occur at different frequencies, it is unlikely that bands in the two spectra will appear to be coincident. For example, rra 5-l,2-dichloroethene has an inversion centre as it belongs to the point group C2h, the IR and Raman spectra for this molecule are compared in Figure 6.13a. This molecule has six atoms, and so 3x6 — 6 = 12 vibrational modes. The total number of bands in the Raman and IR spectra is fewer than 12 because some vibrations are too low frequency to be detected in the range shown. However, it can be seen that the Raman and IR frequencies are indeed different to one another. 

Table 6.6 Reduction of the reducible representation for C—H stretch modes of 1,4 difluorobenzene (a) gerade and (b) ungerade irreducible representations. Note that only classes with nonzero character for F from Table 6.5 are considered here.

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