Symmetry species

Quack M 1985 On the densities and numbers of rovibronic states of a given symmetry species rigid and nonrigid molecules, transition states and scattering channels J. Chem. Phys. 82 3277-83... 

In this assignment, we keep the symmetry species of the vibronic state in D3/, but indicate the vibrational quantum numbers for the Civ normal modes. The energy increases from left to right, and up to down,... 

All molecules possess the identity element of symmetry, for which the symbol is / (some authors use E, but this may cause confusion with the E symmetry species see Section 4.3.2). The symmetry operation / consists of doing nothing to the molecule, so that it may seem too trivial to be of importance but it is a necessary element required by the mles of group theory. Since the C operation is a rotation by 2n radians, Ci = I and the symbol is not used. 

We have seen that any two of the C2, ( Jxz), (r Jyz) elements may be regarded as generating elements. There are four possible combinations of + 1 or — 1 characters with respect to these generating elements, + 1 and + 1, + 1 and -1,-1 and +1,-1 and —1, with respect to C2 and (tJxz). These combinations are entered in columns 3 and 4 of the C2 character table in Table A.l 1 in Appendix A. The character with respect to / must always be + 1 and, just as (r Jyz) is generated from C2 and (tJxz), the character with respect to (r Jyz) is the product of characters with respect to C2 and (tJxz). Each of the four rows of characters is called an irreducible representation of the group and, for convenience, each is represented by a symmetry species Aj, A2, or B2. The A] species is said to be totally symmetric since all the characters are + 1 the other three species are non-totally symmetric. 

The symmetry species labels are conventional A and B indicate symmetry or antisymmetry, respectively, to C2, and the subscripts 1 and 2 indicate symmetry or antisymmetry, respectively, to n (xz). 

In the sixth column of the main body of the character table is indicated the symmetry species of translations (7) of the molecule along and rotations (R) about the cartesian axes. In Figure 4.14 vectors attached to the nuclei of H2O represent these motions which are assigned to symmetry species by their behaviour under the operations C2 and n (xz). Figure 4.14(a) shows that... 

In the final column of the character table are given the assignments to symmetry species and These are the components of the symmetric polarizability tensor... 

The H2O molecule, therefore, has three normal vibrations, which are illustrated in Figure 4.15 in which the vectors attached to the nuclei indicate the directions and relative magnitudes of the motions. Using the C2 character table the wave functions ij/ for each can easily be assigned to symmetry species. The characters of the three vibrations under the operations C2 and (t (xz) are respectively + 1 and +1 for Vj, - - 1 and + 1 for V2, and —1 and —1 for V3. Therefore... 

Using the conventional axis notation given, assign these vibrations to symmetry species of the appropriate point group. 

From these it follows that the symmetry species of the vibrations are given by ... 

There will be many occasions when we shall need to multiply symmetry species or, in the language of group theory, to obtain their direct product. For example, if H2O is vibrationally excited simultaneously with one quantum each of Vj and V3, the symmetry species of the wave function for this vibrational combination state is... 

The second difference is the appearance of a doubly degenerate E symmetry species whose characters are not always either the - - 1 or — 1 that we have encountered in nondegenerate point groups. 

Lower-case letters are recommended for the symmetry species of a vibration (and for an electronic orbital) whereas upper-case letters are recommended for the symmetry species of the corresponding wave function. 

Tables for all degenerate point groups, giving the symmetry species of vibrational combination states resulting from the excitation of one quantum of each of two different degenerate vibrations and of vibrational overtone states resulting from the excitation of two quanta of the same degenerate vibration, are given in the books by Herzberg and by Hollas, referred to in the bibliography.

Multiplication of symmetry species is carried out using the usual mles so that, for example. 

The III character table is given in Table A.46 in Appendix A. The very high symmetry of this point group results in symmetry species with degeneracies of up to five, as in and... 

If we compare the vectors representing a translation of, say, the H2O molecule along the z-axis, as illustrated in Figure 4.14(a), with the dipole moment vector, which is also along the z-axis and shown in Figure 4.18(a), it is clear that they have the same symmetry species [i.e. T(piJ = T T )] and, in general. 

A molecule has a permanent dipole moment if any of the translational symmetry species of the point group to which the molecule belongs is totally symmetric. 

In the Cl, C, C and C point groups the totally symmetric symmetry species is A, A, A and Ai (or 2"+), respectively. For example, CHFClBr (Figure 4.7) belongs to the Ci point group therefore /r 7 0 and, since all three translations are totally symmetric, the dipole... 

The BF3 molecule, shown in Figure 4.18(i), is now seen to have /r = 0 because it belongs to the point group for which none of the translational symmetry species is totally symmetric. Alternatively, we can show that /r = 0 by using the concept of bond moments. If the B-F bond moment is /Tgp and we resolve the three bond moments along, say, the direction of one of the B-F bonds we get... 

The molecule tran5 -l,2-difluoroethylene, in Figure 4.18(h), belongs to the C2 , point group in which none of the translational symmetry species is totally symmetric therefore the molecule has no dipole moment. Arguments using bond moments would reach the same conclusion. 

A molecule has a permanent dipole moment if any of the symmetry species of the translations and/or T( and/or 1/ is totally symmetric. Using the appropriate character table apply this principle to each of these molecules and indicate the direction of any non-zero dipole moment. 

Assign the allene molecule to a point group and use the character table to form the direct products A-2 x5i,5i X 82,82 xE and E X E. Show how the symmetry species of the point group to which 1,1-dilluoroallene belongs correlate with those of allene. 

Each of these can be assigned to one of the symmetry species of the point group to which the molecule belongs. These assignments are indicated in the right-hand column of each character table given in Appendix A and will be required when we consider vibrational Raman spectra in Section 6.2.3.2. 

Number of normal vibrations of each symmetry species... 

In Section 4.3.f it was shown that there are 3N — 5 normal vibrations in a linear molecule and 3N — 6 in a non-linear molecule, where N is the number of atoms in the molecule. There is a set of fairly simple rules for determining the number of vibrations belonging to each of the symmetry species of the point group to which the molecule belongs. These rules involve the concept of sets of equivalent nuclei. Nuclei form a set if they can be transformed into one another by any of the symmetry operations of the point group. For example, in the C2 point group there can be, as illustrated in Figure 6.18, four kinds of set ... 

For example, in the C2 point group each of the four types of sets of nuclei illustrated in Figure 6.18 will contribute degrees of freedom to the symmetry species as follows ... 

Each set of this type will contribute three degrees of freedom to each symmetry species. If there are m sets they will contribute 3m degrees of freedom to each symmetry species, as indicated in Table 6.5. 

Table 6.5 Number of normal vibrations of each symmetry species (Spec.) in the C2 point group...

In Table B. 1 in Appendix B are given formulae, analogous to those derived for the C2 point group, for determining the number of normal vibrations belonging to the various symmetry species in all non-degenerate point groups. 

In Table 6.6 the results for the point group are summarized and the translational and rotational degrees of freedom are subtracted to give, in the final column, the number of vibrations of each symmetry species. 

There are simple symmetry requirements for the integral of Equation (6.44) to be non-zero and therefore for the transition to be allowed. If both vibrational states are non-degenerate, the requirement is that the symmetry species of the quantity to be integrated is totally symmetric this can be written as... 

Since the dipole moment is a vector in a particular direction it has the same symmetry species as a translation of the molecule in the same direction. Figure 6.21 shows this for FI2O in which the dipole moment and the translation in the same direction have the same symmetry species, the totally symmetric dj species. In general. 

Equation (4.14) tells us that, if H2O is vibrating with two quanta of V3, then is Aj and, in general, if it is vibrating with n quanta of a vibration with symmetry species S then... 

Figure 6.22 shows, for example, that the symmetry species of vibrational fundamental and overtone levels for V3 alternate, being Aj for u even and B2 for v odd. It follows that the 3q, 3q, 3q,. .. transitions are allowed and polarized along the y,z,y,... axes (see Figure 4.14 for axis labelling). 

Figure 6.22 Symmetry species of some overtone and combination levels of H2O together with directions of polarization of transition moments. The vibration wavenumbers are cO] = 3657.1 cm a>2 = 1594.8 cm m3 = 3755.8 cm ...

The H2O molecule has no 2 or bi vibrations but selection mles for, say, CH2F2, which has vibrations of all symmetry species, could be applied in an analogous way. 

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