C3 character table

Inspection of this character table, given in Table A. 12 in Appendix A, shows two obvious differences from a character table for any non-degenerate point group. The first is the grouping together of all elements of the same class, namely C3 and C as 2C3, and (t , and 0- as 3o- . 

Two elements, P and Q, are said to belong to the same class if there exists a third element R such that 

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. 

The + 1 and —1 characters of the and A2 species have the same significance as in a non-degenerate point group. The characters of the E species may be understood by using the 

The symmetry properties of a fundamental vibrational wave function are the same as those of the corresponding normal coordinate Q. For example, when the C3 operation is carried out on Qi, the normal coordinate for Vj, it is transformed into Q[, where 

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The difficulty in applying the projection operator for this symmetry group arises from the fact that the C3 character table contains... 

Table 3.14 contains the C3 character table and related data needed for the isotope effect for CHCI3. 

For a symmetric rotor molecule such as methyl fluoride, a prolate symmetric rotor belonging to the C3 point group, in the zero-point level the vibrational selection mle in Equation (6.56) and the character table (Table A. 12 in Appendix A) show that only... 

Follow the procedure used in the text in obtaining the character table for the C2 point group and develop the character table for the C3 point group. 

Table 7.9 The character table for the C3 group. In this table, (o = ...

Let us take the group C3 as an example. Its correct character table is, in part,... 

Consider the group D3. Let the C3 axis coincide with the z axis and one of the C2 axes coincide with the x axis. Write out the complete matrices for all irreducible representations of this group. Derive from these the character table. 

We first note that all types of A orbitals (in D3h) have the same symmetry properties with respect to the rotations constituting the subgroup C3 also, both and " orbitals have the same properties with respect to these rotations. Thus we can use the group C3 to set up some linear combinations that will be correct to this extent. Since these rotations about the C3 axis do not interchange any of the orbitals 0, 02, 03 with those of the set 4, 05, 6, we can, temporarily, treat the two sets separately. We thus first write down linear combinations corresponding to the A and representations of C3. As shown in Section 7.3 for such cyclic systems, the characters are the correct coefficients, and we can thus write, by inspection of the character table for the group C3 ... 

The various symmetry operations will affect our set of C—O stretchings in the same way as they will affect the set of C—O bonds themselves. With this in mind we can determine the desired characters very quickly as follows. For the operation E the character equals 3, since each C—O bond is carried into itself. The same is true for the operation ah. For the operations C3 and S3 the characters are zero because all C—O bonds are shifted by these operations. The operations C2 and av have characters of 1, since each carries one C—O bond into itself but interchanges the other two. The set of characters, listed in the same order as are the symmetry operations at the top of the DVt character table, is thus as follows 3 0 1 3 0 1. The representation reduces to A + , as it should according to our previous discussion. Thus we have shown that normal modes of symmetry types A and E must involve some degree of C—O stretching. Since there is only one A mode, we can state further that this mode must involve entirely C—O stretching. 

For convenience in presentation of, for example, character tables, it is useful to regard many point groups as direct product groups obtained by combining groups of lower order. For example, the group C3 may be considered as arising from the direct product... 

Character tables are what most quantum chemists remember best of their group theory, but it is convenient to complete this review of the fundamentals by examining first the character table for a group and then the full matrix irreps for that group. As an iluustration, we shall examine the group C3 (or the isomorphic D3). Here is the character table in the same format as used in the Tables provided for this course. The... 

Atomic basis functions on B are straightforward to classify. Evidently, an s type function on B will be totally symmetric — an a orbital. A quick inspection of the D3h character table will show that a p set on B, which transforms like the three Cartesian directions, spans the reducible representation a 2 e. Functions centred on the F atoms require more effort. Since the operations in the classes containing C3 and S3 move all three F atoms, their character is necessarily zero for any functions centred on the F atoms. Consider first a set of s functions on each F. These span a reducible representation with character... 

While we have chosen to proceed here by reducing representations for the full group D3h, it would have been simpler to take advantage of the fact that D3h is the direct product of C3u and C where the plane in the latter is perpendicular to the principal axis of the former. The behaviour of any atomic basis functions with respect to the C3 subgroup is trivial to determine, and there are only two classes of non-trivial operations in C3v. In more general cases, it is often worthwhile to look for such simplifications. It is seldom useful, for instance, to employ the full character table for a group that contains the inversion, or a unique horizontal plane, since the symmetry with respect to these operations can be determined by inspection. With these observations and the transformation properties of spherical harmonics given in the Supplementary Notes, it should be possible to determine the symmetries spanned by sets of atomic basis functions for any molecular system. Finally, with access to the appropriate literature the labour can be eliminated entirely for some cases, since... 

Determine correlation relations between the IRs of (a) Td and C3v, and (b) Oh and D3d. [Hints. Use character tables from Appendix A3. For (a), choose the C3 axis along [111] and select the three dihedral planes in Td that are vertical planes in C3v. For (b), choose one of the C3 axes (for example, that along [11 1]) and identify the three C2 axes normal to the C3 axis.]... 

Table 12.8. Character tables for the isomorphous point groups D3 = E 2 C3 3C2 and C3v = 2C3 3uv. ...

Exercise 13.4-7 Determine if time-reversal symmetry introduces any additional degeneracies in systems with symmetry (1) C3 and (2) C4, for (i) Neven and (ii) Nodd. [Hints Do not make use of tabulated PFs but calculate any PFs not already given in the examples in Section 12.4. Characters may be found in the character tables in Appendix A3.]... 

Example A3.1 Find the character table for the cyclic group C3. 

Prior to interpreting the character table, it is necessary to explain the terms reducible and irreducible representations. We can illustrate these concepts using the NH3 molecule as an example. Ammonia belongs to the point group C3V and has six elements of symmetry. These are E (identity), two C3 axes (threefold axes of rotation) and three crv planes (vertical planes of symmetry) as shown in Fig. 1-22. If one performs operations corresponding to these symmetry elements on the three equivalent NH bonds, the results can be expressed mathematically by using 3x3 matrices. ... 

Table 4-4 shows a preliminary character table for the C3v point group. The complete set of symmetry operations is listed in the upper row. Clearly, some of them must belong to the same class since the number of irreducible representations is 3 and the number of symmetry operations is 6. A closer look at this table reveals that the characters of all irreducible representations are equal in C3 and Cf and also in oy, o+ and a", respectively. Thus, according to rule 4 C3 and Cl form one class, and ay, ct and a" together form another class. 

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