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936 C. GROUP THEORY IN SPECTROSCOPY Now only those ψ l are allowed in optical transitions (from the ground state A g ) that are labelled by B 3u , because only the direct product B 3u × B 3u may contain the fully symmetric irreducible representation A g . Thus, the transitions A g ⇒B 3u as well as B 3u ⇒A g are allowed, if the light is polarized along x,i.e.perpendicular to the ring of the molecule. Now let us take light polarized along y, i.e. within the molecular plane, per- polarization y pendicularly to the N–N line. This time we are interested in the irreducible repre- sentations that arise from A g ×B 2u ,becausey transforms according to B 2u .Very similarly [by analyzing a(α)]wefindthat A g ×B 2u =B 2u This means that the allowed states are now of the B 2u type. Similarly, for polarization along z (z belongs to B 1u ), i.e. along the nitrogen-polarization z nitrogen direction, we have A g ×B 1u =B 1u Thus for polarization parallel to the NN axis of the molecule, absorption may occur from the ground state to any state of the B 1u type (and vice versa). Nothing more can be said when relying solely on group theory. We will not get any information about the energies of the transitions, or about the corresponding intensi- ties. To get this additional (and important) information we have work hard to solve the Schrödinger equation, rather than count on some easy profits obtained by the primitive multiplication of integers (as in group theory). To obtain the intensities, we have to calculate the transition moment integrals μ kl . However, group theory, by excluding from the spectrum many transitions (forbidden ones), provides a lot transition moment of important information on the molecule. Table C.7 collects the calculated light frequencies 41 (¯ν in wavenumbers, or cm −1 , ν =c¯ν,whereν is the usual frequency), the oscillator strengths f kl (in a.u.) f kl = 4πc 3 ν|μ kl | 2 (C.28) as well as the polarization of light for excitations from the electronic ground state for pyrazine and the pyrazine monocation. It is seen that the left-hand side of Ta- ble C.7 is consistent with the selection rules derived above. Indeed, a large f kl only corresponds to those transitions from the ground state of the pyrazine that have been predicted as allowed (B 1u ,B 2u and B 3u ). The predicted polarization also agrees with the observed polarization. oscillator strength Excitations from an excited state Calculations for absorption from the ground-state were particularly simple. Now let us see whether anything will be more complicated for the transitions from an excited state of the B 2g type of symmetry. We are going to calculate a(α) (for every α) for the following representations: 41 J. Koput, unpublished results. 4 Integrals important in spectroscopy 937 Table C.7. Wav e n umber s ( ¯ν), oscillator strengths (f kl ) and light polarization (in parentheses) Pyrazine Pyrazine monocation Excited state ¯νf kl Excited state ¯νf kl B 3u 28960 0015(x) B 1 27440 0007(x) B 2u 36890 0194(y) B 2 34130 0280(y) B 2g 38890 00A 2 45100 00 A u 41710 00A 1 49720 0126(z) B 1u 49800 0183(z) B 1 57380 0012(x) B 1g 57070 00A 2 57710 00 B 1u 57420 0426(z) A 1 58210 0625(z) A u 60170 00A 2 59830 00 B 2g 60970 00B 2 60370 0010(y) for polarization along x:B 2g ×B 3u for polarization along y:B 2g ×B 2u for polarization along z:B 2g ×B 1u . The characters of the representation B 2g ×B 3u are the following (Table C.6, the first finger goes along B 2g , the second – along B 3u ,etc.) 1 −1 −1 −1 −111 and are identical with the characters of B 1u . Hence, even without any calculation of a(α),wehaveB 2g ×B 3u =B 1u . Thus the transitions (for the polarization along x) are allowed only for the states labelled by B 1u because otherwise there is no chance of obtaining a fully symmetric integrand. Similarly, by multiplying B 2g and B 2u we obtain the following characters of B 2g ×B 2u : 1111−1 −1 −1 −1 and these are identical to the characters of A u , therefore B 2g ×B 2u = A u .Ifthe polarization of light is along y, the only excitations (or deexcitations) possible are for states belonging to A u . Finally, for polarization along z, we find the characters of B 2g ×B 1u : 1 −1 −11−111−1 that turn out to be those of B 3u . This means that B 2g × B 1u = B 3u and that the transitions are possible only for states belonging to B 3u . Example 14. Pyrazine monocation. As to the selection rules, nothing was said so far about the pyrazine monocation. We will be interested in excitations from the electronic ground state (as in Table C.7). The pyrazine monocation corresponds to symmetry group C 2v (Table C.8). The ground state belongs to the fully symmetric irreducible representation A 1 . Since (as before) we begin by excitations from the ground state, let us see which irreducible representations arise from A 1 × B 1 (for the x polarization of light, see Table C.8, x transforms according to B 1 ), A 1 × B 2 (for the y polarization) 938 C. GROUP THEORY IN SPECTROSCOPY Table C.8. C 2v group characters C 2v E C 2 σ v (xz) σ v (yz) A 1 11 1 1 zx 2 y 2 z 2 A 2 11 −1 −1 R z xy B 1 1 −11−1 x R y xz B 2 1 −1 −11yR x yz and A 1 × A 1 (for the z polarization). We calculate the characters of A 1 × B 1 by multiplying 1 by 1 −11−1 and checking in Table C.8 that these correspond to B 1 (it has to be like this, because the characters of A 1 are all equal to 1), i.e. A 1 ×B 1 =B 1 . Similarly, even without immediate checking, we see that A 1 ×B 2 =B 2 and A 1 ×A 1 =A 1 .Inthiswaythe following allowed transitions from the ground state (A 1 ) have been predicted: for polarization along x:A 1 →B 1 for polarization along y:A 1 →B 2 for polarization along z:A 1 →A 1 . This agrees with f kl =0 values of Table C.7. Now we are able to compare the spectrum for pyrazine and for its monocation, Table C.7. Attaching a proton to the pyrazine (creating its monocation) does not look like something that would ruin the UV-VIS spectrum. We might expect that the frequencies of the bands, even their intensities should be somehow similar in both molecules. As we can see from the Table, the frequencies are similar indeed. For both molecules there are forbidden (f kl =0) and allowed (f kl =0) transitions. Note that what is allowed for the pyrazine is also allowed for its cation, the light po- larization coincides, even the values of f kl are similar (we have taken into account that the transition to B 1u in pyrazine with frequency 49800 cm −1 corresponds to the transition to A 1 in the monocation with frequency 49720 cm −1 ). In the mono- cation there are some additional transitions allowed: to B 1 and to B 2 .Thisisquite understandable, because the number of symmetry operations for the monocation is smaller, and the higher molecular symmetry the more numerous are forbidden transitions. If a molecule had no symmetry operations at all (except of course the identity symmetry), then all transitions would be allowed. Thus, practically with zero effort, we find the selection rules in UV-VIS for any molecule we want. Selection rules in IR and Raman spectra The selection rules derived above pertain to electronic transitions, when the posi- tions of the nuclei are fixed in space. Now a vibrational excitation of the molecule will be considered, while the electronic state is assumed to be unchanged. The vi- brations of a molecule are related to its vibrational levels (each corresponding to an irreducible representation) and the corresponding vibrational wave functions, 4 Integrals important in spectroscopy 939 Fig. C.6. Small amplitude harmonic vibrations of a molecule (N atoms) are described by 3N − 6in- dependent harmonic oscillators (normal modes). Each normal mode is characterized by an irreducible representation. A diagram shows the vibrational energy levels of three normal modes corresponding to the irreducible representations 1 2 3 The modes have different frequencies, hence the interlevel separations are different for all of them (but equal for a given mode due to the harmonic potential). On the right-hand side all these levels are shown together. and the IR spectrum results from transitions between such levels. Fig. C.6 shows the energy levels of three normal modes. In the harmonic approximation the problem of small amplitude vibrations (Chapters 6 and 7) reduces to the 3N − 6 normal modes (N is the number of atoms in the molecule). Each of the normal modes may be treated as an indepen- dent harmonic oscillator. A normal mode moves all the atoms with a certain fre- quency about their equilibrium positions in a concerted motion (the same phase). The relative deviations (i.e. the ratios of the amplitudes) of the vibrating atoms from equilibrium are characteristic for the mode, while the deviation itself is ob- tained from them by multiplication by the corresponding normal mode coordinate Q ∈(−∞ ∞).ThevalueQ = 0 corresponds to the equilibrium positions of all the atoms, Q and −Q correspond to two opposite deviations of any atom from its equilibrium position. Each normal mode belongs to an irreducible representation of the symmetry group of the molecule. What does it really mean? In any mode the displacements of the equivalent atoms from equilibrium have the same absolute value, although they may differ in sign. 940 C. GROUP THEORY IN SPECTROSCOPY We assume that small atomic deviations satisfy the symmetry requirements of the symmetry group of the molecule (valid for all atoms in equilibrium po- sitions) and transform according to the irreducible representation, to which the normal mode belongs. Squaring the deviations destroys information about their signs, i.e. the absolute values of the deviations of the equivalent atoms are the same. This means that the squares of deviations transform according to the fully symmetric representation of the group. To establish the vibrational selection rules, let us first define the vibrational states of 3N − 6 harmonic oscillators (normal modes). The ground state of the system is no doubt the state in which every normal mode i is in its ground state, ψ i0 . The ground-state wave function of the i-th normal mode reads as (p. 166) ψ i0 =N 0 exp −a i Q 2 i (C.29) where a i > 0 is a constant, and Q i is the normal mode coordinate. Whatever this normal mode is, the wave function contains the square of Q i ,i.e.thesignofthe deviations of the equivalent atoms is irrelevant. The squares of the deviations, and therefore function ψ i0 itself, transform independently of i. Let us denote this fully symmetric irreducible representation by A 1 .Thewave- function of the first excited state of a normal mode has the form (p. 166) ψ i1 =N 1 Q i exp −a i Q 2 i (C.30) and we see that ψ i1 transforms exactly as the coordinate Q i does, i.e. according to the irreducible representation to which the normal mode belongs (because Q 2 i in the exponent and therefore the exponent itself both belong to the fully symmetric representation). In the harmonic approximation the total vibrational wavefunction of the system of 3N −6 normal (i.e. independent) oscillators can be written as: ψ osc 0 =ψ 10 ψ 20 ψ 30 ψ 3N−60 (C.31) the zeros in the indices mean that all the modes are in their ground states. This means that ψ osc 0 transforms according to the representation being the direct prod- uct A 1 ×A 1 ×A 1 ×···×A 1 =A 1 (a banality, all the characters of A 1 are equal 1). Now let us focus on the excited states of the 3N −6 vibrational modes. The excited states may be quite complex, but the most important (and the simplest) are those with all the normal modes in their ground states, except a single mode that is in its first excited state. A transition from the many-oscillator ground state to such an excited state is called a fundamental transition. The intensities of the fundamental fundamental transition transitions are by at least one order of magnitude larger than the others. This is why we will focus on the selection rules for such transitions. Let us take one such 4 Integrals important in spectroscopy 941 singly excited state (with the first mode excited): ψ osc 1 =ψ 11 ψ 20 ψ 30 ψ 3N−60 (C.32) The function ψ 11 corresponding to the first excited state transforms according to the irreducible representation , to which the normal mode 1 belongs. Thus, ψ osc 1 transforms according to ×A 1 × A 1 × A 1 ×···×A 1 = , i.e. it belongs to the same irreducible representation as ψ 11 . Of course, if the only excited mode were the i-th one, then the many-oscillator wavefunction would belong to the same irreducible representation as the wavefunction of the particular oscillator does. We will need this result later. IR selection rules. Let us consider a molecule with a fixed position in a Cartesian coordinate system. To excite the molecule, IR light (because the separation of the vibrational levels corresponds to the infrared region) is used, which is polarized along the x axis. Theory of electromagnetism says the transition integral 42 decides the intensity of the absorption ψ osc 0 ˆμ x ψ osc 1 dτ (C.33) where ˆμ x stands for the dipole moment component x. The selection rules estab- lish which integrals of this kind will be zero for symmetry reasons. To this end we need information about the irreducible representations to which ψ osc 0 ˆμ x ψ osc 1 belong. 43 Since ψ osc 0 transforms according to A 1 , for the integral to survive, the function ψ osc 1 has to belong to the same irreducible representation as ˆμ x (and therefore x itself). We showed above that ψ osc 1 belongs to the same irreducible representation to which the normal mode 1 belongs. In other words, the rule is: SELECTION RULE IN IR the transition from the ground state is allowed for those normal modes that transform as x,wherex is the direction of light polarization, and similarly for light polarization along y and z. Raman selection rules. The physics of Raman spectra 44 is different: rather than direct absorption this is light scattering (in the UV-VIS region) on molecules. It turns out, that beside the light the source is emitting, we also detect quanta of energy lower or higher by hν,whereν is the vibrational frequency of the molecule. 42 The integration goes over the coordinates of the nuclei. 43 We are going to analyze the direct product of these three representations. If it contains the fully symmetric representation, the integral is not zero. 44 Chandrasekhar Venkata Raman (1888–1970), Indian physicist, professor at the University of Cal- cutta and at the Indian Scientific Institute in Bangalore. In 1928 Raman discovered light scattering that is accompanied by a change of frequency (by frequency of the molecular vibrations). Raman received the Nobel prize in 1930 “for his work on the scattering of light and for the discovery of the effect named after him”. 942 C. GROUP THEORY IN SPECTROSCOPY For the Raman scattering to be non-zero, at least one of the following integrals should be non-zero ψ osc 0 ˆα qq ψ osc 1 dτ (C.34) where ˆα qq with qq = x yz is a component of the polarizability tensor which transforms as one of the following (cf. eq. (12.40), p. 636): qq = x 2 , y 2 , z 2 , xy, xz, yz or their linear combinations (this information is available in the tables of characters). Identical reasoning leads to the conclusion that the normal mode excited in a fundamental transition has to belong to the same irreducible representation as the product qq . It remains to be seen to which irreducible representations the normal modes be- long. The procedure consists of two stages. Stage 1. First, the global Cartesian coordinate system is chosen, Fig. C.7. In this system we draw the equilibrium configuration of the molecule, with numbered atoms. A local Cartesian coordinate system is located on each atom with axes par- allel to the axes of the global coordinate system. For each atom, we draw the arrows of its displacements along x, y and z, oriented towards the positive values (alto- gether 3N displacements), assuming that the displacements of equivalent atoms have to be the same. When symmetry operations are applied, these displacements transform into themselves 45 and therefore form a basis set of a (reducible) repre- sentation of the symmetry group of the molecule (in its equilibrium position). This representation will be decomposed into the irreducible representations. Stage 2. The reducible representation describes the genuine (internal) vibra- tions as well as the six apparent vibrations (three translations and three rotations). The apparent vibrations correspond to those irreducible representations that are associated to x, y, z (translations) and R x ,R y ,R z (rotations). We know from the corresponding table of characters what the later ones are. Summing up: the re- ducible representation mentioned above has to be decomposed into irreducible representations. The decomposition yields = a( 1 ) 1 + a( 2 ) 2 + a( 3 ) 3 . From this decomposition we have to subtract (in order to eliminate the appar- ent vibrations) all the irreducible representations the x y, z,R x R y and R z be- long. After these two stages we are left with the number of the irreducible represen- tations which pertain to the genuine vibrations. 46 Only after this can we establish the vibrational selection rules according to the procedure used before. All this will be shown by a simple example of the carbonate anion CO 2− 3 that in its equilibrium configuration corresponds to the D 3h symmetry group, Fig. C.7. 45 For example, a displacement of an atom along x under a symmetry operation turns out to be a displacement of another equivalent atom. 46 Rather internal motions. Note that some of these genuine vibrations may correspond to rotations of the functional groups in the molecule. 4 Integrals important in spectroscopy 943 Fig. C.7. The carbonate anion CO 2− 3 ,the coordinate system used and the versors de- scribing the displacements of the atoms. Example 15. IR and Raman spectra of the carbonate anion. To decompose a re- ducible representation into irreducible ones, we do not need the reducible rep- resentation be given in full details. It is sufficient to know its characters (p. 920). These characters are easy to deduce by considering what happens to the displace- ment vectors along x i , y i , z i (for atom i) under all the symmetry operations. What will greatly simplify our task is that only the diagonal elements of the matrices of the reducible representation contribute to the characters. How it looks in practice is shown in Table C.9. Thus, the characters of the reducible representation have been found. To de- compose the representation, we have to know the table of characters for the D 3h symmetry group, Table C.10. Let us write (in the same order as in Table C.10) the characters of the reducible representation just found: 12 0 −24−22 Now, let us find (p. 920) how many times [a(α)] the irreducible representa- tion α is present in the reducible representation (the sum over classes: number of operations in class × the calculated character × the character of irreducible representation): a A = 1 12 [1 ×12 ×1 +2 ×0 ×1 +3 ×(−2) ×1 +1 ×4 ×1 +2 ×(−2) ×1 +3 ×2 ×1]=1 Similarly, we find (only needing to know how to multiply such numbers as 12 3) that a(A) =1 a E =3 a A 1 =0 a A 2 =2 a(E) =1 This means that the reducible representation in question decomposes into =A 1 +A 2 +3E +2A 2 +E (C.35) 944 C. GROUP THEORY IN SPECTROSCOPY Table C.9. Class The character of the corresponding matrix Eχ(E)=12 Justification: each versor transforms into itself. Hence each diagonal element is equal to 1, and the number of them is equal to 3 times the number of atoms =12 2C 3 χ(C 3 ) =0 Justification: 0 from the oxygens, because they transform into other oxygens +1(from z 4 ) + cos120 ◦ (from x 4 ) +cos120 ◦ (from y 4 ) =0 3C 2 χ(C 2 ) =−2 Justification: it is sufficient to consider only one of the operations of the class – others will have the same character. Let us take the rotation about the C 2 axis going through O 1 and C. Then the only unit vectors that transform into themselves (eventually changing sign – then the contribution to the character is −1) are those related to O 1 and C. We have χ(C 2 ) = −1(from z 4 ) +(−1)(from z 1 ) −1(from x 1 ) −1(from x 4 ) +1(from y 1 ) +1(from y 4 ) =−2 σ h χ(σ h ) =4 Justification: the contribution from each atom will be the same, i.e. χ will be equal to 4 times the contribution from a single atom, the latter equals: −1(from z) + 1(from x) + 1(from y) =1 2S 3 χ(S 3 ) =−2 Justification: only C gives a contribution, which is equal to: −1(from z 4 ) − 1 2 (from x 4 ) − 1 2 (from y 4 ) =−2 3σ v χ(σ v ) =2 Justification: Let us take only a single operation from the class, the one, which represents the reflection in the plane going through O 1 and C 4 . Then the contributions to χ are the same for both atoms, and one gives: −1(from x) +1(from z) +1(from y) =1. Table C.10. Characters of the irreducible representations of symmetry group D 3h D 3h E 2C 3 3C 2 σ h 2S 3 3σ v A 1 111111 x 2 +y 2 z 2 A 2 11 −111−1 R z E 2 −102−10x y x 2 −y 2 xy A 1 11 1 −1 −1 −1 A 2 11 −1 −1 −11z E 2 −10−210R x R y xz yz From the table of characters, we see that the apparent vibrations (see the irre- ducible representations corresponding to x y z R x R y R z ) belong to A 2 ,E , A 2 ,E . After subtracting them from , we obtain the irreducible representations that correspond to the genuine vibrations: A 1 A 2 2E i.e. one vibration of symmetry A 1 (and a certain frequency ν 1 ), two vibrations (each doubly degenerate) of symmetry E (they differ by frequency ν 3 = ν 4 )andone vibration of A 2 symmetry (corresponding to frequency ν 2 ). 4 Integrals important in spectroscopy 945 SELECTION RULES FOR IR: Therefore, we expect the following selection rules for the fundamental tran- sitions in the IR spectrum for the CO 2− 3 anion: 1. x and y belong to representation E , and therefore frequencies ν 3 and ν 4 areactiveinIR; 2. z belongs to representation A 2 , and therefore frequency ν 2 is active in IR. SELECTION RULES FOR RAMAN SPECTRA For the Raman spectra we expect the following selection rules. The vibra- tions with the frequency will be active: 1. ν 1 ,becausex 2 +y 2 and z 2 belong to A 1 ; 2. ν 3 and ν 4 , because x 2 −y 2 and xy belong to E , while the vibration of the frequency ν 2 will be inactive in Raman spectroscopy, because none of the polarizability components (symbolized by x 2 y 2 ,etc.)belongs to A 2 . The results are collected in Table C.11 (sign “+” = active vibration, sign “–” = inactive vibration, the polarization of the light is shown in parentheses). As seen from Table C.11, in case of the carbonate anion the vibration ν 1 is inac- tive in IR, but active in Raman spectroscopy, while the opposite is true for ν 2 .The vibrations with frequencies ν 3 and ν 4 are active both in IR and in Raman spectra. EXCLUSION RULE If the molecule under study has a centre of symmetry, the exclusion rule is valid, i.e. the vibrations that are active in IR are inactive in the Raman spectrum, and vice versa. This follows from the fact that, in this case, x y z belong to different irreducible representations than x 2 y 2 z 2 xyxzyz.Indeed,thex y z are antisymmet- ric with respect to the inversion operation, whereas x 2 y 2 z 2 xyxzyzor their combinations are symmetric with respect to inversion. This guarantees that they belong to different irreducible representations, therefore for a molecule with Table C.11. Transitions in CO 2− 3 active (+) and inactive (−) in IR and in Raman spectra Representation ν IR (polarization) Raman A 1 ν 1 −+ A 2 ν 2 + (z) − E ν 3 + (circular) + E ν 4 + (circular) + . contains the square of Q i ,i.e.thesignofthe deviations of the equivalent atoms is irrelevant. The squares of the deviations, and therefore function ψ i0 itself, transform independently of i. Let us. the absolute values of the deviations of the equivalent atoms are the same. This means that the squares of deviations transform according to the fully symmetric representation of the group. To establish. positions of all the atoms, Q and −Q correspond to two opposite deviations of any atom from its equilibrium position. Each normal mode belongs to an irreducible representation of the symmetry group of
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