Totally symmetric symmetry species

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... 

Torque constant, 245 Totally symmetric representation, 402 Totally symmetric symmetry species, 60-61 Totally symmetric vibrations, 247 Townes, C., 137 Trace, 82-83 Transition, 114 Transition intensity, 122 Transition moment, 120, 297-298, 299-301... 

We shall not give the numbers c,y,R that specify these symmetry species, except to note that Aig is the totally symmetric symmetry species, with all c s equal to 1. 

The ground-state electronic wave function of QHg belongs to the totally symmetric symmetry species and is an eigenfunction of with eigenvalue -i-l.The functions I and II combine with equal coefficients to form a single symmetry function, and the functions III, IV, and V combine to form a second symmetry function. 

A g is the totally symmetric symmetry species, with all c s equal to 1. 

As stated in Sections 15.3 and 15.5, the canonical MOs of a molecule can be chosen so that each one transforms according to one of the symmetry species of the molecule when the molecular symmetry operators are applied. The same is true of the normal-mode vibrations. For example, when each of the four symmetry operators of H2O is applied to the W) vibration in Fig. 15.16, the vibration displacement vectors end up looking unchanged. This normal mode therefore belongs to the totally symmetric symmetry species a,. Give the symmetry species of the other two normal modes in Fig. 15.16. Take the axes as in Fig. 15.1. 

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. 

If the product of two symmetry species is totally symmetric those species must be the same. Therefore we can rewrite Equation (7.120) as... 

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. 

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. 

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. 

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... 

Consider the CFI stretching vibrations of benzene, for example. Since there are six identical C—FI bonds there are six CFI stretching vibrations. These belong to various symmetry species but only one, V2, illustrated in Figure 6.39, is totally symmetric 0. ... 

It was explained in Section 4.3.2 that the direct product of two identical degenerate symmetry species contains a symmetric part and an antisymmetric part. The antisymmetric part is an A (or 2") species and, where possible, not the totally symmetric species. Therefore, in the product in Equation (7.79), Ig is the antisymmetric and Ig + Ag the symmetric part. 

The A A2 X Ai, n -n system of formaldehyde (see Section 7.3.1.2) is also electronically forbidden since A2 is not a symmetry species of a translation (see Table A.l 1 in Appendix A). The main non-totally symmetric vibration which is active is Vq, the hj out-of-plane bending vibration (see Worked example 4.1, page 90) in 4q and d transitions. 

Vibrations of the symmetry class Ai are totally symmetrical, that means all symmetry elements are conserved during the vibrational motion of the atoms. Vibrations of type B are anti-symmetrical with respect to the principal axis. The species of symmetry E are symmetrical with respect to the two in-plane molecular C2 axes and, therefore, two-fold degenerate. In consequence, the free molecule should have 11 observable vibrations. From the character table of the point group 04a the activity of the vibrations is as follows modes of Ai, E2, and 3 symmetry are Raman active, modes of B2 and El are infrared active, and Bi modes are inactive in the free molecule therefore, the number of observable vibrations is reduced to 10. 

When deriving selection rules from character tables it is noted that vibrations are usually excited from the ground state which is totally symmetric. The excited state has the symmetry of the vibration being excited. Hence A vibration will be spectroscopically active if the vibration has the same symmetry species as the relevant operator. 

For the homonuclear (HON) species, the permutation-symmetry operator had the following form Y = 83) <8) Ye S2), where 83) is a Young operator for the third-order symmetric group which permutes the nuclear coordinates and 82) is a Young operator for the second-order symmetric group which permutes the electronic coordinates. For the fermionic nuclei (H and T, spin = 1/2) the Young operators corresponded to doublet-type representations, while for the bosonic D nuclei we use operators that correspond to the totally symmetric representation. In all cases the electronic operator corresponded to a singlet representation. 

For C2v, the representation Ai is symmetric with respect to every operation. Such a one-dimensional representation, in which every element is assigned the number 1, occurs for every group, and is called the totally symmetric representation. Two representations are symmetric with respect to C2 each is denoted A. Two are symmetric with respect to symmetry species a unique symbol, and " need not be used in this case. 

Simplification of secular equations. Because the Hamiltonian is totally symmetric - that is, for a molecule of C2v symmetry such as H2O, of symmetry species Ai - the matrix elements Hij = ipi, Ti. ipj) as well as the overlap integrals Sij = (tpi, ipj) will be equal to zero unless the direct product representation r. contains Ai. This is the basis for the assertion that states of different symmetry do not mix. ... 

The polarizability tensor of a molecule related the components of the induced dipole moment of the molecule to the components of the electric field doing the inducing. It therefore has 9 components, axx, ctxy, etc., only 6 of which are independent. The theory of the Raman effect shows that a vibrational transition, from the totally symmetric ground state to an excited state of symmetry species F, will he Raman active if at least one of the following direct products contains the totally symmetric representation ... 

All doubly occupied orbitals have aY symmetry and if all the orbitals have paired electrons for any given molecular configuration, the state is totally symmetric and belongs to the species Species A is symmetric with respect to rotation about the z-axis and species B antisymmetric. If there are more than one A and B species they are further given numerical subscripts. 

Once the state symmetries have been established it only remains to be shown that the direct product of the species of the ground state symmetry, the coordinate translational symmetries and the excited state symmetry belong to the totally symmetric species A. Let us take the n-wr transition in formaldehyde. The ground state total wave function has the symmetry Ax. The coordinate vectors x, y and z transform as By, Bf and Ax respectively (refer Character Table for C2k Section 2.9, Table 2.2), The excited state transforms as symmetry species A2. The direct products are ... 

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