M. Hargittai, I. Hargittai - Symmetry through the Eyes of a Chemist (793765), страница 36

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In some cases, as in H2 O,the symmetry coordinates are good representations of the normalvibrations. In other cases they are not. An example for the latteris Au2 Cl6 where the pure symmetry coordinate vibrations would beclose in energy, so the real normal vibrations are mixtures of thedifferent vibrations of the same symmetry type [7]. The relationshipbetween the symmetry coordinates and the normal vibrations can be2265 Molecular Vibrationsestablished only by calculations called normal coordinate analysis[8, 9].

These calculations necessitate further data in addition to theknowledge of molecular symmetry and are not pursued here.Return now to the symmetry coordinates of the water molecule.They can be generated using the projection operator. As has beenmentioned before, here we are interested only in the symmetryaspects of the symmetry coordinates. Thus, the numerical factors areomitted, and normalization is not considered.

First, let us work out thesymmetry coordinate involving the stretching vibrations:P̂ A1 ⌬r1 ≈ 1 · E · ⌬r1 + 1 · C2 · ⌬r1 + 1 · σ · ⌬r1 + 1 · σ · ⌬r1= ⌬r1 + ⌬r2 + ⌬r2 + ⌬r1 ≈ ⌬r1 + ⌬r2P̂ B2 ⌬r1 ≈ 1 · E · ⌬r1 + (−1) · C2 · ⌬r1 + (−1) · σ · ⌬r1+ 1 · σ · ⌬r1 == ⌬r1 − ⌬r2 − ⌬r2 + ⌬r1 ≈ ⌬r1 − ⌬r2 .Figure 5-6. Generation of the symmetry coordinates representing bond stretchingfor H2 O.5.3. Selection Rules227The same procedure is presented pictorially in Figure 5-6. Thebending mode of the water molecule stands alone (see the v2 modein Figure 5-4), so it will be a symmetry coordinate by itself.Since the symmetry coordinates of water are good approximationsof the normal vibrations, the pictorial representations are applicableto them as well.

Indeed, the three normal modes of Figure 5-4 are thesame as the symmetry coordinates we just derived. The A1 symmetrystretching mode is called the symmetric stretch while the B2 mode isthe antisymmetric stretch.5.3. Selection RulesThe vibrational wave function, as any wave function, must forma basis for an irreducible representation of the molecular pointgroup [3].The total vibrational wave function, ⌿v , can be written as theproduct of the wave functions ⌿i (ni ), where ⌿i is the wave functionof the ith normal vibration (i = 1 through m) in the nth state.⌿v = ⌿1 (n 1 ) · ⌿2 (n 2 ) · ⌿3 (n 3 )......⌿m (n m )In general, at any time, each of the normal modes may be in anystate.

There is, however, a situation when all the normal modes arein their ground states and only one of them gets excited into thefirst excited state. Such a transition is called a fundamental transition. The intensity of the fundamental transitions is much higher thanthe intensity of the other kinds of transitions.∗ Therefore, these are ofparticular interest.The vibrational wave function of the ground state belongs to thetotally symmetric irreducible representation of the point group of themolecule.

The wave function of the first excited state will belong tothe irreducible representation to which the normal mode undergoingthe particular transition belongs.∗Were the vibrations strictly harmonic, only fundamental transitions would beobservable.2285 Molecular VibrationsA fundamental transition willintegrals has nonzero value: ◦⌿ v◦⌿ v◦⌿voccur only if one of the following|x| ⌿iv|y| ⌿iv|z| ⌿ivHere ⌿◦v is the total vibrational wave function for the ground state,⌿iv is the total vibrational wave function for the first excited state referring to the ith normal mode and x, y and z are Cartesian coordinates.The condition for an integral of product functions to have a nonzerovalue was given in Chapter 4.

For the vibrational transitions thiscondition can be expressed in the following way:⌫⌿◦v · ⌫⌿iv ⊂ ⌫xor⌫⌿◦v · ⌫⌿iv ⊂ ⌫ yor⌫⌿◦v · ⌫⌿iv ⊂ ⌫zThe considerations on the symmetries of the ground and excitedstates and the above conditions lead to the selection rule for infraredspectroscopy: A fundamental vibration will be infrared active if thecorresponding normal mode belongs to the same irreducible representation as one or more of the Cartesian coordinates.The selection rule for Raman spectroscopy can also be derived bysimilar reasoning. It says: A fundamental vibration will be Ramanactive if the normal mode undergoing the vibration belongs to thesame irreducible representation as one or more of the components ofthe polarizability tensor of the molecule. These components are thequadratic functions of the Cartesian coordinates given in the fourtharea of the character tables. The Cartesian coordinates themselves aregiven in the third area.

Thus, the symmetry of the normal modes ofa molecule is sufficient information to tell what transitions will beinfrared and what transitions will be Raman active. The normal modesof the water molecule belong to the A1 and B2 irreducible representation of the C2v point group. By using merely the C2v character table, itcan be deduced that all three vibrational modes will be active in boththe infrared and Raman spectra.Since a particular normal mode may belong to different symmetryspecies in different point groups, its behavior depends strongly on themolecular symmetry. Just to mention one example, the vl symmetric5.4. Examples229stretching mode of an AX3 molecule is not infrared active if themolecule is planar (D3h ).

It is infrared active, however, if the moleculeis pyramidal (C3v ). Vibrational spectroscopy is obviously one of thebest experimental tools to determine the symmetry of molecules.5.4. ExamplesThe utilization of symmetry rules in the description of molecularvibrations will be further illustrated by a few examples.Diimide, HNNH. This molecule belongs to the C2h point group (seeFigure 4-7). The number of atoms is 4, so the number of normal vibrations is 3 × 4 − 6 = 6.Our first task is to generate the representation of the Cartesian displacement vectors of the four atoms of the molecule (seeFigure 4-8a–c). As was shown in Chapter 4 (Section 4.7), the representation is⌫tot 12 0 0 4The reduction of this representation is also given in Chapter 4 (seep.

208). The result is⌫tot = 4A g + 2Bg + 2Au + 4BuThese 12 irreducible representations account for the 12 degrees ofmotional freedom of HNNH. Subtracting the irreducible representations corresponding to the translation and rotation of the molecule(see C2h character table, Table 5-2) leaves us the symmetry species ofthe normal modes of vibration:⌫tot− (⌫tran− (⌫rot⌫vib= 4A g + 2Bg + 2Au + 4Bu=+ Au + 2Bu )= A g + 2Bg)= 3A g+ Au + 2BuNext we will see what kind of internal coordinate changes canaccount for each of these normal modes.

There will have to be twoN–H stretching modes and one N–N stretching mode. For deformation2305 Molecular VibrationsTable 5-2. The C2h Character Table and the Representations of the Internal Coordinates of DiimideC2hEC2i␴hAgBgAuBu11111–11–111–1–11–1–11RzRx , Ryzx, y⌫NH200⌫NN111⌫NNH200⌫HNNH a11–1aOut-of plane deformation mode.212–1=Ag + Bu=Ag=Ag + Bu=Aux2 , y2 , z2 , xyxz, yzmodes the two N–N–H angle bending modes are obvious choices, andthey will be in-plane deformation modes. These constitute five normalvibrations so one is left to be accounted for.

In deciding the nature ofthis normal mode, inspection of the character table may help. Of theabove three different kinds of irreducible representations, Ag and Buare symmetric with respect to ␴h so they must be vibrations withinthe molecular plane. The five vibrational modes suggested above thenaccount for 3Ag + 2Bu . The remaining Au normal mode, however, isantisymmetric with respect to ␴h , so it must involve out-of-planemotion. Consequently, this normal mode will be an out-of-planedeformation mode.We will work out next the representations of the internal coordinates. The representation of the two N–H distance changes has beengiven in Chapter 4 (Section 4.3). This and the other representationsare all shown in Table 5-2, together with the C2h character table.

The⌫NH representation has been reduced to Ag + Bu in Chapter 4 (Section4.8). The reduction of the ⌫NNH representation is the same. Both theN–N stretching and the out-of-plane deformation are already irreducible representations by themselves. Since Ag occurs three times,we cannot tell without calculation whether there will be three pure Agmodes, one N–H stretch, one N–N stretch, and one N–N–H bend, oreach of the three Ag modes will be a mixture of these three vibrations.Similarly, there are two Bu symmetry normal vibrations, and they willbe either pure N–H antisymmetric stretching and N–N–H bendingmodes or their mixtures. The only unambiguous assignment is that5.4. Examples231the Au symmetry normal mode will be the out-of-plane deformationmode.Generate now the symmetry coordinates of HNNH by means of theprojection operator (α is the N–N–H angle):P̂ Ag ⌬r1 ≈ 1 · E · ⌬r1 + 1 · C2 · ⌬r1 + 1 · i · ⌬r1 + 1 · σh · ⌬r1= ⌬r1 + ⌬r2 + ⌬r2 + ⌬r1 ≈ ⌬r1 + ⌬r2P̂ Bu ⌬r1 ≈ 1 · E · ⌬r1 + (−1) · C2 · ⌬r1 + (−1) · i · ⌬r1+ 1 · σh · ⌬r1 == ⌬r1 − ⌬r2 − ⌬r2 + ⌬r1 ≈ ⌬r1 − ⌬r2P̂ Ag ⌬α1 ≈ 1 · E · ⌬α1 + 1 · C2 · ⌬α1 + 1 · i · ⌬α1 + 1 · σh · ⌬α1= ⌬α1 + ⌬α2 + ⌬α2 + ⌬α1 ≈ ⌬α1 + ⌬α2P̂ Bu ⌬α1 ≈ 1 · E · ⌬α1 + (−1) · C2 · ⌬α1 + (−1) · i · ⌬α1+ 1 · σh · ⌬α1 == ⌬α1 − ⌬α2 − ⌬α2 + ⌬α1 ≈ ⌬α1 − ⌬α2 .The same procedure is depicted in Figure 5-7.

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