M. Hargittai, I. Hargittai - Symmetry through the Eyes of a Chemist (793765), страница 41
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Molecules259C4vE2C4A1A2B1B2E1111211−1−10Table 6-1. The C4v Character TableC22v2d1111−21−11−101−1−110x2 +y2 , z2zRz(x, y) (Rx , Ry )x2 –y2xy(xz, yz)for illustrating two-dimensional representations. Taking the three porbitals as basis functions, the symmetry operations of the C4v pointgroup are applied to them. This is shown in Figure 6-14. The matrixrepresentations are given here:⎡⎤1 0 0E = ⎣0 1 0⎦0 0 1⎡0⎣C4 = −10⎤00⎦1100⎡1σv (x z) = ⎣00⎡0⎣σd = 10⎡0 −13⎣C4 = 1 00 00−10⎤00⎦1100⎤00⎦1⎤00⎦1⎡−1 0⎣0 −1C2 =0 0⎡−1σv (yz) = ⎣ 00⎡0⎣σd = −10010⎤00⎦1−100⎤00⎦1⎤00⎦1All these matrices can be simultaneously block-diagonalized intoa 2×2 and a 1×1 matrix. The set of the 1×1 matrices correspondsto pz and the set of the 2×2 matrices corresponds to px and py . Therepresentations are:pz(px , py )E122C4 C2110 −22v102d10A1E2606 Electronic Structure of Atoms and MoleculesFigure 6-14.
The symmetry operations of the C4v point group applied to the 2porbitals.Notice that the operations C4 and d transform px into py and viceversa. They cannot be separated from one another so they togetherbelong to the two-dimensional representation E.If two or more atomic orbitals are interrelated under a symmetryoperation of the point group and, accordingly, they together belong toan irreducible representation, their energies will also be the same.
Inother words, these orbitals are degenerate. Such atomic orbitals areparenthesized in the character tables.The direct connection between symmetry and degeneracy of theatomic orbitals is demonstrated here once again. The higher the6.3. Molecules261symmetry of the molecule, the greater will be the interrelation ofthe orbitals upon symmetry operations.
Consequently, their energies become less and less distinguishable. The following exampleshows how the degeneracy of p orbitals decreases with diminishingsymmetry:Free atomSphericalsymmetry(px , py , pz )Threefold degenerateOh point groupT1u(px , py , pz )Threefold degenerateC4v point groupA1Epz(px , py )NondegenerateTwofold degenerateC2v point groupA1B1B2pzpxpyNondegenerateNondegenerateNondegenerateThe degree of degeneracy of atomic orbitals always correspondsto the dimension of the irreducible representation to which theseatomic orbitals belong.
The same is true for molecular orbitals. Thus,knowing the symmetry of a molecule and looking at the charactertable, one can determine at once the maximum possible degeneracyof its molecular orbitals. The irreducible representation having thehighest dimension will show this.6.3.2.
Electronic StatesThe orbitals and electronic configurations are useful descriptions.However, they are only models, and they employ approximations. Theenergy of an orbital has rigorous physical significance for systems thatcontain only a single electron. In many-electron systems, the energyof the orbitals loses its physical meaning, and only the energies of the(ground and excited) states are real. It is these states that are describedby the total electronic wave functions.
Electronic transitions, in fact,represent changes in the state of an atom or a molecule and not necessarily in the electronic configurations.We shall not be concerned with the atomic states. The systematicway of determining them is given, for example, in References [4]and [7].
Molecular states and the determination of their symmetries,however, will be briefly introduced [23].2626 Electronic Structure of Atoms and MoleculesFirst, let us consider the customary notations. Assume that a hypothetical ground state molecule of the C2v point group has four electrons, two in an A1 symmetry and two in a B1 symmetry orbital. Inshort notation this can be written as a12 b12 . An electron occupying an A1symmetry orbital is represented by a1 , the lower-case letter indicatingthat this is the symmetry of an orbital and not of an electronic state.If two electrons occupy an orbital, the notation is a12 .
The symmetryof a state is represented by capital letters, just as are the irreduciblerepresentations.The symmetry of the electronic states can be determined from thesymmetry of the occupied orbitals. There are two different cases:l. States with fully occupied orbitals. An electronic configuration inwhich all orbitals are completely filled possesses only one electronic state, and it will be totally symmetric. This can be seen forthe case of nondegenerate orbitals. The wave function describingthe electronic state can be written as the product of the one-electronorbitals.
The symmetry of the product is given by the charactersof the direct product representation. However, the product of anyorbital with itself will always give the totally symmetric representation, no matter what characters it has, both 1 · 1 and (–1)·(–1)equal 1, i.e., in each class of the point group the characters of theproduct will be 1.
The same is true for degenerate orbitals, althoughthe procedure in this case is not as simple.2. States with partially occupied orbitals. First of all, the completelyfilled orbitals are ignored for the reasons described above. Thesymmetry of the state will be given by the direct product of thepartially filled orbitals.Let us consider some examples for the above hypotheticalmolecule. The supposed ground state and the configurations of twodifferent singly excited states are represented in Figure 6-15.The ground state a12 b12 has only fully occupied orbitals, so itssymmetry is A1 . The first excited state, a12 b1 a2 , has one fully occupiedorbital, a12 , so this is not considered.
The symmetry of this state willbe given by the direct product B1 · A2 . Table 6-2 lists the direct products under the C2v character table. The symmetry of the state is B2 .The other excited state in our example has the configuration a12 b1 b2 .The direct product is given in Table 6-2; the state symmetry is A2 .6.3.
Molecules263Figure 6-15. Different states of a molecule with C2v symmetry.C2vTable 6-2. C2v Character Table and Some Direct Product RepresentationsEC2v (xz)σν (yz)A1A2B1B2111111−1−11−11−11−1−11zRzx, Ryy, RxB1 ·A2B1 ·B211−11−1−11−1B2A2x2 , y2 , z2xyxzyzSince we are concerned only with the spatial symmetry properties,the electron spin and its role in determining the electronic states havebeen neglected in the above description.6.3.3. Examples of MO Construction6.3.3.1. Homonuclear Diatomicsa) Hydrogen, H2 . There are two 1s hydrogen atomic orbitals availablefor bonding. The molecular point group is D∞h . This molecule doesnot have a central atom, so the symmetry operations of the point groupare applied to both 1s orbitals, since they together form the basis fora representation of this point group.
The 1s orbital of one hydrogenatom alone does not belong to any irreducible representation of theD∞h point group. Several symmetry operations of this group transform one of the two 1s orbitals into the other rather than into itself (see2646 Electronic Structure of Atoms and Molecules(a)(b)Figure 6-16. Some symmetry operations of the D∞h point group applied to: (a)One 1s orbital in the hydrogen molecule; (b) The two 1s orbitals of the hydrogenmolecule together.Figure 6-16a). Thus, they must be treated together; in this way theyform a basis for a representation.
All symmetry operations are indicated in Figure 6-16b. The D∞h character table is given in Table 5-3.The characters of this representation will beD∞h2 H(1s)E2⌽2C∞2∞v2i0⌽2S∞0∞C20This is a reducible representation of the D∞h point group whichreduces to g + u . Two molecular orbitals must be generated, onewith g and the other with u symmetry. The two possible combinations are the bonding and antibonding orbitals which can be formedfrom the two 1s atomic orbitals.The two electrons in the hydrogen molecule will occupy the lowerenergy bonding orbital, and none will go into the antibonding orbital.Hence, the molecule is stable.6.3.
Molecules265b) Other Homonuclear Diatomic Molecules. The principle utilizedto construct molecular orbitals is the same as that for the hydrogenmolecule. For helium, the MO picture is the same as for hydrogenexcept that here the additional two electrons occupy the antibondingu orbital, and, therefore, the molecule is unstable.In the series from lithium through neon, similar symmetry considerations apply, except that in these examples the second electron shellmust be considered.
The two 2s orbitals, as was found to be the casefor the two 1s orbitals, form MOs that possess g and u symmetry.As regards the 2p orbitals, the two 2pz orbitals lie along the molecular axis and belong to the same irreducible representation as the2s orbitals. They also combine to give MOs that possess g and usymmetry:The 2s and 2pz orbitals of the same atom belong to the same irreducible representation of the D∞h point group. Their energies arealso similar so they cannot be separated completely.
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