M. Hargittai, I. Hargittai - Symmetry through the Eyes of a Chemist (793765), страница 37
Текст из файла (страница 37)
The forms of thesymmetry coordinates of HNNH are shown in Figure 5-8. They mightapproximate well the normal modes of the molecule, and again, theymight not.Finally, let us decide which normal modes will be infrared activeand which ones will be Raman active. The Cartesian coordinatesbelong to the Au and Bu irreducible representation of the C2h pointgroup, while their binary products belong to Ag and Bg . Consequently,the selection rules are:Infrared active: Au , BuRaman active: Ag .This means that the Ag symmetry stretching modes and the Agsymmetry bending mode will be Raman active, while the Bu symmetrystretching and bending modes will be infrared active. Similarly, the Ausymmetry out-of-plane deformation mode will be infrared active.Carbon dioxide, CO2 . The molecule is linear and belongs to theD∞h point group.
The number of atoms is 3, so the number of normalvibrations is: 3 × 3 – 5 = 4.2325 Molecular VibrationsFigure 5-7. Generation of some symmetry coordinates of HNNH; (a) Symmetrycoordinates corresponding to N–H bond stretches; (b) Symmetry coordinates representing in-plane deformation.5.4. Examples233Figure 5-8. Symmetry coordinates for the HNNH molecule.The set of Cartesian displacement vectors as basis for a representation is shown in Figure 5-9. The symmetry operations of the pointgroup are also shown.
The D∞h character table is given in Table 5-3.Recall (Chapter 4) that the matrix of rotation by an angle ⌽ is⌽C =cos ⌽ − sin ⌽sin ⌽cos ⌽The rotation by an arbitrary angle ⌽ will leave the three z coordinates unchanged and the x and y coordinates mixed according to theabove expression. The following matrix represents the C ⌽ rotation:†Figure 5-9. Cartesian displacement vectors of CO2 .†cos is abbreviated by c and sin by s in the matrix.2345 Molecular VibrationsTable 5-3.
The D∞h Character TableaD∞hE⌽2C∞...∞v⌺+g⌺–g⌸g⌬g...⌺+u⌺–u⌸u⌬u1122..1122112c⌽2c2⌽...112c⌽2c2⌽...........................1–100..1–1001122..–1–1–2–2...............y1−s⌽c⌽z100x200y200z200x300y300000000010000000c⌽s⌽00000−s⌽c⌽000000010000000c⌽s⌽00000−s⌽c⌽0ai⌽2S∞...∞C211–2c⌽2c2⌽...–1–12c⌽–2c2⌽...........................1–100..–1100........x2 , y2 , z2Rz(Rx , Ry )(xz, yz)(x2 -y2 , xy)z(x, y)c stands for cos.x1y1z 1x2y2z 2x3y3z 3x1c⌽⎢ s⌽⎢⎢⎢ 0⎢⎢ 0⎢⎢ 0⎢⎢⎢ 0⎢⎢ 0⎢⎢⎣ 00⎡z3⎤00⎥⎥⎥0⎥⎥0⎥⎥0⎥⎥⎥0⎥⎥0⎥⎥⎥0⎦1The character will be: 3 + 6 cos⌽. The other relatively complicated operation is the mirror-rotation by an arbitrary angle, S ⌽ .
Thisoperation means a rotation around the z axis by angle ⌽, followed byreflection through the xy plane. This reflection interchanges the positions of the two oxygen atoms so they need not be considered. Theblock matrix of the S ⌽ operation will be:5.4. Examples235y2z2 ⎤⎡ x20x2 cos ⌽ − sin ⌽⎥ ⎢y2 ⎣ sin ⌽cos ⌽0⎦z 200−1The character is: –1 + 2 cos ⌽.Omitting the details of the determination of the remaining characters, the representation of the Cartesian displacement vectors is:⌫tot3 + 6 cos ⌽9−33− 1 + 2 cos ⌽−1Subtract the characters of the translational and rotational representations. Remember that CO2 is linear and the rotation around themolecular axis need not be taken into account.⌫tot−(⌫tran−(⌫rot===9 3 + 6 cos ⌽3 1 + 2 cos ⌽22 cos ⌽310−3 −1 + 2 cos ⌽−3 −1 + 2 cos ⌽2−2 cos ⌽⌫vib=4 2 + 2 cos ⌽2−2−1−1)0)2 cos ⌽0The reduction formula cannot be applied to the infinite point groups(Chapter 4).
Here inspection of the character table may help. Since⌽2 cos⌽ at S∞appears with the ⌸u irreducible representation, it isworth a try to subtract this one from ⌫vib :⌫vib−(⌫⌸u==2 + 2 cos ⌽2 cos ⌽422220−2−2202 cos ⌽2 cos ⌽00)00This representation can be resolved as the sum of ⌺g and ⌺u :⌺g⌺u==1111111 −1⌺g + ⌺u=222011−1 −1002365 Molecular VibrationsThus, the normal modes of the CO2 molecule will be⌫vib = ⌺g + ⌺u + ⌸uSince ⌸u is a degenerate vibration, it counts as two, and so weindeed have the four necessary normal vibrations.The obvious choice for the three internal coordinate changes is thestretching of the two C=O bonds and the bending of the O=C=O angle.Using these as bases for representations we can build up the symmetrycoordinates.⌫str222000We have already seen before that this representation reducesas ⌺g + ⌺u .
The ⌸u normal mode will correspond to the bendingvibration.Since each of the three symmetry species, ⌺g , ⌺u , and ⌸uappears only once, the symmetry coordinates will be good representations of the normal modes. There is no possibility for mixing.Figure 5-10 shows the forms of the normal vibrations of the CO2molecule. The two bending modes are degenerate; they are of equalenergy.Figure 5-10. Normal modes of vibration of the CO2 molecule.Finally, apply the vibrational selection rules to CO2 :Infrared active: ⌺u , ⌸uRaman active: ⌺gReferences237Accordingly, the symmetric stretching C=O normal mode shouldappear in the Raman spectrum, while the antisymmetric stretching andthe degenerate bending modes are expected to appear in the infraredspectrum.References1.
A. L. Mackay, A Dictionary of Scientific Quotations, Adam Hilger, Bristol, 1991.p. 84/55.2. D. C. Harris, M. D. Bertolucci, Symmetry and Spectroscopy: An Introduction toVibrational and Electronic Spectroscopy, Oxford University Press, New York,1978.3. F. A. Cotton, Chemical Application of Group Theory, Third Edition, WileyInterscience, New York, 1990.4. M. Orchin, H. H.
Jaffe, Symmetry, Orbitals, and Spectra (S.O.S), WileyInterscience, New York, 1971.5. G. Herzberg, Infrared and Raman Spectra, Van Nostrand Company, Princeton,NJ, 1959.6. E. B. Wilson, Jr., J. C. Decius, P. C. Cross, Molecular Vibrations, McGraw-HillBook Co., New York, 1955.7. Harris, Bertolucci, Symmetry and Spectroscopy, pp. 171–173.8. Wilson, Decius, Cross, Molecular Vibrations.9. K. Nakamoto, Infrared Spectra of Inorganic and Coordination Compounds,Second Edition, John Wiley and Sons, New York, 1970.Chapter 6Electronic Structure of Atomsand MoleculesAn atom must be at least as complex as a grand piano.William K.
Clifford [1]Everything that counts in chemistry is related to the electronic structure of atoms and molecules. The formation of molecules from atoms,their behavior and reactivity all depend on the electronic structure.What is the role of symmetry in all this? In various aspects of the electronic structure, symmetry can tell us a good deal; why certain bondscan form and others cannot, why certain electronic transitions areallowed and others are not, and why certain chemical reactions occurand others do not.
Our discussion of these points is based primarilyon some monographs listed in References [2–8].To describe the electronic structure, the electronic wave function⌿(x, y, z, t) is used, which depends, in general, on both space andtime. Here, however, only its spatial dependence will be considered,⌿(x, y, z). For detailed discussions of the nature of the electronic wavefunction, we refer to texts on the principles of quantum mechanics[9–12]. For a one-electron system the physical meaning of the electronic wave function is expressed by the product of ⌿ with its complexconjugate ⌿∗ . The product ⌿∗ · ⌿ d gives the probability of findingan electron in the volume d = dx dy dz about the point (x, y, z).A many-electron system is described by a similar but multivariablewave function⌿(x1 , y1 , z 1 , .
. . , xi , yi , z i , . . . , xn , yn , z n ).M. Hargittai, I. Hargittai, Symmetry through the Eyes of a Chemist, 3rd ed.,C Springer Science+Business Media B.V. 2009DOI: 10.1007/978-1-4020-5628-4 6, 2392406 Electronic Structure of Atoms and MoleculesThe product ⌿∗ ·⌿ d gives the probability of finding the first electronin dτ1 , about the point (x1 , y1 , z 1 ), and the ith electron in dτi about thepoint (xi , yi , zi ), all at the same time.The symmetry properties of the electronic wave function andthe energy of the system are two determining factors in chemicalbehavior. The relationship between the wave function characterizingthe behavior of the electrons and the energy of the system—atoms andmolecules—is expressed by the Schrödinger equation.
In its generaland time-independent form, it is usually written as follows,Ĥ ⌿ = E⌿(6-1)where Ĥ is the Hamiltonian operator and E is the energy of thesystem.The Hamiltonian operator is an energy operator, which includesboth kinetic and potential energy terms for all particles of the system.In our discussion, only its symmetry behavior will be considered.With respect to the interchange of like particles (either nuclei or electrons) the Hamiltonian must be unchanged under a symmetry operation. A symmetry operation carries the system into an equivalentconfiguration, which is indistinguishable from the original.
However,if nothing changes with the system, its energy must be the samebefore and after the symmetry operation. Thus, the Hamiltonian ofa molecule is invariant to any symmetry operation of the point groupof the molecule. This means that it belongs to the totally symmetricrepresentation of the molecular point group.A fundamental property of the wave function is that it can beused as basis for irreducible representations of the point group of amolecule [13].
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
