M. Hargittai, I. Hargittai - Symmetry through the Eyes of a Chemist (793765), страница 35
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This is the vibration ofthe molecule:5.1. Normal Modes219Figure 5-2. The three translational degrees of freedom of a diatomic molecule.Figure 5-3. Rotation of a diatomic molecule; (a) Two rotational degrees of freedomdescribe the rotation of the molecule around the center of mass; (b) Rotation aroundthe molecular axis does not change the position of the molecule.2205 Molecular VibrationsThe complete nuclear motion of an N-atomic molecule can bedescribed with 3N parameters; that is an N-atomic molecule has 3Ndegrees of freedom.
The translation of a molecule can always bedescribed by three parameters. The rotation of a diatomic or any linearmolecule will be described by two parameters and the rotation of anonlinear molecule by three parameters. This means that there arealways 3 translational and 3 (for linear molecules 2) rotational degreesof freedom. The remaining 3N – 6 (for the linear case 3N – 5) degreesof freedom account for the vibrational motion of the molecule. Theygive the number of normal vibrations.The translational and rotational degrees of freedom, which do notchange the relative positions of the atoms in the molecule, are oftencalled nongenuine modes.
The remaining 3N – 6 (or 3N – 5) degreesof freedom are called genuine vibrations or genuine modes.5.1.2. Their SymmetryThe close relationship between symmetry and vibration is expressedby the following rule: Each normal mode of vibration forms a basisfor an irreducible representation of the point group of the molecule.Let us use the water molecule to illustrate the above statement. Thenormal modes of this molecule are shown in Figure 5-4. The pointgroup is C2v , and the character table is given in Table 5-1.
It is seenthat all operations bring vl and v2 into themselves so their characterswill be:⌫v1⌫v2σv11111111The behavior of the third normal mode, v3 , is different. While E andleave it unchanged, both C2 and v bring it into its negative self:Figure 5-4. Normal modes of vibration for the water molecule. The lengths of thearrows indicate the relative displacements of the atoms.5.1. Normal ModesC2vEA1A2B1B21111221Table 5-1. The C2v Character TableC2v (xz)v (yz)11–1–11–11–11–1–11zRzx, Ryy, Rxx2 , y2 , z2xyxzyzeach atom moves in the opposite direction after the operation. Thismeans that v3 is antisymmetric to these operations.
The characters are:⌫v3 1 −1 −11Looking at the C2v character table, we can say that vl and v2 belongto the totally symmetric irreducible representation A1 , and v3 belongsto B2 .It was easy to determine the symmetry of the normal modes ofthe water molecule because we already knew their forms. Can thesymmetry of the normal modes of a molecule be determined withoutany previous knowledge of the actual forms of the normal modes? Theanswer is fortunately yes.
From the symmetry group of the moleculethe symmetry species of the normal modes can be determined withoutany additional information.First, an appropriate basis set has to be found. Considering that amolecule has 3N degrees of motional freedom, a system of 3N socalled Cartesian displacement vectors is a convenient choice. A set ofsuch vectors is shown in Figure 5-5 for the water molecule. A separateCartesian coordinate system is attached to each atom of the molecule,with the atoms at the origin. The orientation of the axes is the samein each system.
Any displacement of the atoms can be expressed by avector, and in turn this vector can be expressed as the vector sum ofthe Cartesian displacement vectors.Next, the set of Cartesian displacement vectors is used as a basisfor the representation of the point group. As discussed in Chapter 4,the vectors connected with atoms that change their position duringan operation will not contribute to the character and thus they can beignored.Continuing with the water molecule as an example, the basisof the Cartesian displacement vectors will consist of nine vectors2225 Molecular VibrationsFigure 5-5. Cartesian displacement vectors as basis for representation of the watermolecule.(see Figure 5-5). Operation E brings all of them into themselves, andthe character is 9.
Operation C2 changes the position of the twohydrogen atoms, so only the three coordinates of the oxygen atomhave to be considered. The corresponding block of the matrix representation is:x2C2 = y2z 2x2 y2−10⎢⎣ 0 −100⎡z2 ⎤0⎥0 ⎦1The character is (– l) + (– 1) + 1 = – 1.The next operation is v . Again, only the oxygen coordinates haveto be considered. Reflection through the xz plane leaves x2 and z2unchanged and brings y2 into – y2 .
The character is 1 + 1 + (– 1) = 1.Finally, operation v leaves all three atoms in their place, so all thenine coordinates have to be taken into account. Reflection throughthe yz plane leaves all y and z coordinates unchanged and takes all xcoordinates into their negative selves. The character will be: (– 1) + 1+ 1 + (– 1) + 1 + 1 + (–1 ) + 1 + 1 = 3. The representation is:⌫tot9−113This is, of course, a reducible representation. Reduce it now withthe reduction formula (see Chapter 4):5.1. Normal Modes223a A1 = (1/4)[1 · 9 · 1 + 1 · (−1) · 1 + 1 · 1 · 1 + 1 · 3 · 1]= (1/4)(9 − 1 + 1 + 3) = 3a A2 = (1/4)[1 · 9 · 1 + 1 · (−1) · 1 + 1 · 1 · (−1) + 1 · 3 · (−1)]= (1/4)(9 − 1 − 1 − 3) = 1a B1 = (1/4)[1 · 9 · 1 + 1 · (−1) · (−1) + 1 · 1 · 1 + 1 · 3 · (−1)]= (1/4)(9 + 1 + 1 − 3) = 2a B2 = (1/4)[1 · 9 · 1 + 1 · (−1) · (−1) + 1 · 1 · (−1) + 1 · 3 · 1]= (1/4)(9 + 1 − 1 + 3) = 3The representation reduces to:⌫tot = 3A1 + A2 + 2B1 + 3B2 .These nine irreducible representations correspond to the ninemotional degrees of freedom of the triatomic water molecule.
Toobtain the symmetry of the genuine vibrations, the irreducible representations of the translational and rotational motion have to be separated. This can be done using some considerations described inChapter 4. The translational motion always belongs to those irreducible representations where the three coordinates, x, y, and z,belong. Rotations belong to the irreducible representations of the pointgroup indicated by Rx , Ry , and Rz in the third area of the charactertables.
In the C2v point group,⌫tran = A1 + B1 + B2and⌫rot = A2 + B1 + B2Subtracting these from the representation of the total motion, weget⌫tot = 3A1 + A2 + 2B1 + 3B2− (⌫tran = A1+ B1 + B2 )− (⌫rot =A2 + B1 + B2 )⌫vib = 2A1+ B22245 Molecular VibrationsThus, of the three normal modes of water, two will have A1 and oneB2 symmetry. Let us stress again: this information could be derivedpurely from the molecular point-group symmetry.5.1.3. Their TypesThe normal modes can usually—though not always—be associatedwith a certain kind of motion. Those connected mainly with changesin bond lengths are the stretching modes.
The ones connected mainlywith changes of bond angles are the deformation modes. These maybe mainly either in-plane or out-of-plane deformation modes. Thesimplest deformation mode is the bending mode.Examine now the symmetries of these different types of vibration.For this purpose a new type of basis set is used. Since we are interested in the changes of the geometrical parameters, these changes arean obvious choice for basis set. The geometrical parameters are alsocalled internal coordinates, and the basis is the displacement of theseinternal coordinates.Let us continue with the water molecule and determine thesymmetry of its stretching modes. The molecule has two O–H bonds,so the basis will be the changes of these O–H bonds.
The representation of this basis set is⌫str2 0 0 2and with inspection of the C2v character table we see that it reduces toA1 + B2 . This means that the stretching of the O–H bonds contributesto the normal modes of A1 and B2 symmetry. (We shall later see thatthese are the symmetric and antisymmetric stretches, respectively.)The third internal coordinate which can be considered in the watermolecule is the bond angle, H–O–H. Its change will be the bendingmode.
All symmetry operations leave this basis unchanged, so therepresentation is:⌫bend1 1 1 1and it belongs to the totally symmetric representation, A1 . What canwe conclude? B2 appears only in the stretching mode, so the B2normal mode will be a pure stretching mode. The A1 symmetry mode,however, appears in both the stretching and the bending mode. At5.2.
Symmetry Coordinates225this point we cannot say whether one of the A1 normal modes willbe purely stretch and the other purely bend or they will be a mixture.This depends on the energy of these vibrations. If they are energetically close, they can mix extensively. If they are separated by a largeenergy difference, they will not mix. In the case of H2 O, for example,the two A1 symmetry modes are quite well separated, while in Cl2 Othey are completely mixed.Modes of different symmetry never mix, even if they are close inenergy.
(This is a general rule which will have its analogous versionfor the transitions among electronic states as will be seen later inChapters 6 and 7.)The above analysis of the types of normal modes brings us to thelimit where simple symmetry considerations can take us. Nothing yethas been said about the pictorial manifestation of the various normalmodes. Above we deduced, for example, that the B2 normal mode ofthe water molecule is a pure stretch. The question may also be asked,how does it look? This question can be answered with the help ofsymmetry coordinates.5.2. Symmetry CoordinatesThe symmetry coordinates are symmetry-adapted linear combinationsof the internal coordinates.
They always transform as one or anotherirreducible representation of the molecular point group.Symmetry coordinates can be generated from the internal coordinates by the use of the projection operator introduced in Chapter 4.Both the symmetry coordinates and the normal modes of vibrationbelong to an irreducible representation of the point group of themolecule. A symmetry coordinate is always associated with one oranother type of internal coordinate—that is pure stretch, pure bend,etc.—whereas a normal mode can be a mixture of different internalcoordinate changes of the same symmetry.
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