Laser Molecular Spectroscopy CHE466 Fall 2009

1. Laser Molecular SpectroscopyCHE466Fall 2009 David L. Cedeño, Ph.D. Illinois State University Department of Chemistry Vibrational Spectroscopy

2. x axis re Vibrational Motion Diatomic Molecules: The Harmonic Oscillator Model Vibrational motion is that related to the displacement of nuclei relative to each other along a given axis (called a stretching) or a common plane (bending). Vibrational motions represent most of the nuclear motion in a molecule. Recall that there are (3N – 6) or (3N – 5) vibrations for non-linear and linear molecules respectively, with N = # atoms. In order to describe vibrational motions we assume a classical model in which nuclei are represented by spheres of various masses joined by springs of different thickness (force constants). For a diatomic molecule: If the motion is described using Newtonian physics (classical) then the motion of one sphere relative to the other along the dimension x (molecular axis) would be harmonic, i.e. the motion will be oscillatory around an equilibrium position (re when atoms are at rest).

3. Vibrational Motion Diatomic Molecules: The Harmonic Oscillator Model The kinetic (T) and potential energies (V) are: k is the force constant of the spring and m is the reduced mass. The potential energy is a function of the displacement relative to equilibrium x = r – re

4. Vibrational Motion Diatomic Molecules: The Harmonic Oscillator Model Of course, molecular vibrational motion cannot be described using classical physics. Thus a quantum mechanical solution is required. The Hamiltonian is given by The energy is obtained using the Schrodinger equation and appropriate wavefunctions. These have the form: and Hv are the Hermite polynomials:

5. Vibrational Motion Diatomic Molecules: The Harmonic Oscillator Model The solution of the Schrodinger equation for the energy of the vibrational motion is quantized. In other words, there are vibrational states (with v = 0, 1, 2, …) that have an specific energy Ev: Ev = hne(v + ½) Where is the so-called harmonic frequency.

6. Vibrational Motion Diatomic Molecules: The Harmonic Oscillator Model Note that a consequence of the quantum nature of vibrational motion is that a molecule will always have a vibrational energy (v = 0). This is called the zero point energy. For a diatomic molecule using the harmonic oscillator model: E0 = ½hne Note that the harmonic oscillator model predicts that the spacing between adjacent vibrational levels is the same and equal to hne. Example: Find the zero point energy and spacing between vibrational states of the HI molecule. Use k = 4.0277 x 104 N/m and assume a harmonic oscillator model.

7. Vibrational Motion Diatomic Molecules: The Transition dipole integral We learned that the transitions probability for a rotational transition is extremely small (for practical reasons zero) if the molecule does not have a permanent dipole moment. Is that also the case for vibrational motion? Since in a vibration there is a displacement of nuclei realtive to each other, the dipole moment may not be constant, it may vary with the displacement (x). We can represent that variation using the following equation: Where me is the permanent dipole moment and the other term is the first order variation of the dipole moment with displacement. It turns out that based on this, a molecule that does not have a permanent dipole moment (i.e. me = 0) could be excited vibrationally (i.e. the probability of transitions is not negligible) as long as the vibrational motion induces a transitory dipole moment and the transition rule related to changes in v number does not vanish the transition dipole integral:

8. Vibrational Motion Diatomic Molecules: Transition Rules under the Harmonic Oscillator Model Based on the previous equation, homonuclear diatomic molecules cannot be vibrationally excited with light, because they do not have a permanent dipole moment, and its vibrational motion cannot create a transitory one. On the contrary, heteronuclear diatomic molecule are able to absorb infrared radiation (strongly). The transition rule is: Dv = ± 1 Most molecules at room temperature are in the lowest vibrational state (v = 0), thus the transition that is more likely to be observed is the one corresponding to v” = 0 to v’ = 1. This is called the fundamental transition. An increase in temperature may populate other vibrational states (say v = 1 and 2), thus in adition to v = 0, thus the transitions v” = 1 to v’ = 2, and v” = 2 to v’ = 3 are also possible. Bands associated to this transitions are called hot bands.

9. Vibrational Motion Diatomic Molecules: Anharmonicity It is obvious that a harmonic potential energy does not represent vibrational motion accurately, especially at large values of v (i.e. highly excited vibrational states). Pictorially, putting a large amount of energy in a vibration does not translate into infinite elongation (x) of the bond. Instead, the coulombic and covalent attraction of the nuclei will be overcome and the molecule will dissociate. Also, compression of the bond at a very large displacement will induce repulsive forces that increase the potential energy at a larger rate than predicted by a harmonic oscillator model.

10. Vibrational Motion Diatomic Molecules: Anharmonicity Morse proposed a potential that represents diatomic molecules very well: De is the dissociation energy of the molecule (relative to V(x) = 0) and b is the Morse parameter for the molecule (it defines the width of the potential curve). A solution to the Schrodinger equation is still possible upon using wavefunctions that are slightly different from the ones obtained using the harmonic oscillator model. The vibrational energy is given by: Ev = hne(v + ½) – hnexe(v + ½)2 + hneye(v + ½)3 + … The parameters xe and ye are the first and second order anharmonicity constants. In most cases (for routine work) the energy is expressed as a two term equation (i.e. depending up to (v + ½)2), but more terms may be necessary.

11. Vibrational Motion Diatomic Molecules: Anharmonicity A direct results of anharmonicity is that the vibrational states of a given molecule are not evenly spaced. Indeed the energy difference between adjacent levels decreases as v increases.

12. Vibrational Motion Diatomic Molecules: Anharmonicity The energy spacing between adjacent levels is given by: DE = Ev’ – Ev” = hne – 2henexe (v”+ 1) Which can be written in wavenumbers as: w = we – 2wexe (v”+ 1) Note then that if have the wavenumbers for Dv = 1 transitions, a linear plot of w vs. v’ would give we and wexe from the intercept and the slope.

13. Vibrational Motion Diatomic Molecules: Anharmonicity, zero point energy and dissociation energy The zero point energy corrected for anharmonicity is equal to: E0 = ½ hne – ¼ henexe The dissociation energy can be defined either in terms of the bottom of the potential curve (De) or more realistically from the zero point energy (D0 from v = 0)

14. Vibrational Motion Diatomic Molecules: Anharmonicity, zero point energy and dissociation energy D0 can be obtained form the DE = Ev’ – Ev” = hne – 2henexe (v”+ 1) plot as an integral or summation of all the spaces between adjacent levels (i.e. DE) extrapolated to the maximum v possible, assuming that the extrapolation is valid. De can be calculated as: De = D0 + E0

15. Vibrational Motion Diatomic Molecules: Transition rules The anharmonicity of the potential perturbs the wavefunctions and breaks the harmonic transition rule. Due to anharmonicity, it is possible to reach higher vibrational states via light absorption because allowed transitions correspond to: Dv = ±1, ±2, ±3, … Most common transitions are those originating from v” = 0, thus it is possible to observe bands for: v” = 0 → v’ = 1 (Fundamental band) v” = 0 → v’ = 2 (first overtone band) v” = 0 → v’ = 3 (second overtone band) et cetera. Usually the intensities of the overtone bands are reduced dramatically (about 10v’-fold as v’ increases)

16. Vibrational Motion Diatomic Molecules: Transition rules Example: CO FTIR spectra of fundamental, and first and second overtone transitions. Mina-Camilde et al., J. Chem. Ed., 1996, 73, 804-807.

17. Vibrational-Rotational Spectroscopy Diatomic Molecules So far, vibrational motion has been described ignoring the effect of rotational motion on the spectra. Since the photons exciting a vibration contain enough energy to excite the molecule rotationally, one cannot ignore rotational motion. The figure shows a typical high resolution spectrum of a fundamental vibrational transition for a diatomic molecule

18. Vibrational-Rotational Spectroscopy Diatomic Molecules The main band consists of two groups of narrow bands, separated by a gap at about the center of the main band w (cm-1)

19. 4 3 2 J’ = 1 v’ = 1 J’ = 0 4 3 2 J” = 1 v” = 0 J” = 0 Vibrational-Rotational Spectroscopy Diatomic Molecules Each band corresponds to a rotational transition occurring simultaneously with the vibrational transition. The rotational transitions correspond to DJ = ± 1 (transition rule) Transitions with DJ = 1 occur at larger frequencies than those with DJ = –1 (see red arrows vs. blue arrows). Thus they group apart from each other in the spectrum. The set of bands at higher frequency is called the R branch (DJ = 1), while the other set is the P branch (DJ = -1). The gap in between is a consequence of the absence of DJ = 0 transitions in diatomic molecules. The lowest frequency R band corresponds to J” = 0 → J’= 1, while the highest energy P band corresponds to J” = 1 → J’= 0. They are denoted wR(0) and wP(1) respectively (in wavenumbers).

20. Vibrational-Rotational Spectroscopy Diatomic Molecules: Energy and Spectral Bands The energy of a molecule is then the sum of rotational and vibrational energies: Ev,J = Ev + EJ = hne(v + ½) – hnexe(v + ½)2 + BvJ(J +1) – DJJ2(J +1)2 With Bv = Be – a(v + ½), where a is the vibration-rotation coupling constant. Therefore the spectral bands correspond to DE = Ev’,J’ – Ev”,,J” , which can be separated into R and P bands (in wavenumbers). For the fundamental vibrational transition (v” = 0 to v’ = 1): Where w0,1 = we – 2wexe

21. Vibrational-Rotational Spectroscopy Diatomic Molecules: Energy and Spectral Bands If centrifugal distortion is neglected (i.e. DJ≈ 0), then: Where w0,1 = we – 2wexe. This is the so-called band center of the fundamental vibrational transition. It is possible to obtain all the spectroscopical constants B1, B0, we, wexe, Be, and a from the analysis of vibration-rotational spectra. This implies using a combination of the equations above in a way that other equations are created in a way that the dependence on some f the parameters is cancelled out. This methodology is called a Combination of Differences approach.

22. Vibrational-Rotational Spectroscopy • Diatomic Molecules: Spectroscopical Constants and the Combination of Differences Method • Obtaining B1and B0 and DJ • If one subtracts wP(J”) from wR(J”) the following is obtained: • Which can be written as: • Thus a linear plot can be obtained, in which the slope is -4DJ and the intercept is 2B1. • Similarly, if one subtracts wP(J”+1) from wR(J”-1) the following is obtained: • Which can be written as: • Thus a linear plot can be obtained, in which the slope is -4DJ and the intercept is 2B0.

23. Vibrational-Rotational Spectroscopy • Diatomic Molecules: Spectroscopical Constants and the Combination of Differences Method • Obtaining Beand a • Given that Bv = Be – a(v + ½), then at least two values of Bv are needed to obtain Be and a. • Obtaining weand wexe • If wP(J”) from wR(J”+1) are added the following is obtained: • Thus a linear plot can be obtained with an intercept equal to w0,1 = we – 2wexe. • Knowledge of other band centers, for instance the one for the first overtone band (v” = 0 to v’ = 2, w0,2) allows one to obtain we and wexe. It can be shown that any overtone, the band center has a wavenumber equal to:

24. Vibrational-Rotational Spectroscopy Diatomic Molecules: Intensities The intensities of the P and R bands are dictated by Boltzmann’s statistics, as is the case for the population of rotational levels of the initial state: Where EJ” is the rotational energy.

25. 4 3 2 J’ = 1 v’ = 1 J’ = 0 4 3 2 J” = 1 v” = 0 J” = 0 Vibrational-Rotational Raman Spectra Diatomic Molecules:transitions It is possible to obtain the Raman vibrational and rotational spectra of diatomic molecules. The vibrational transition rule still holds (i.e. Dv = ±1, ±2, ±3,…) as well as the rotational transition rule (DJ = 0, ±2). The interesting event is that a vibrational-rotational band in the Raman has three branches: S (for DJ = +2), Q (for DJ = 0) and O (for DJ = -2)

26. Vibrational-Rotational Raman Spectra Diatomic Molecules: Below is the Raman spectrum of the fundamental band of 12C16O. Notice that most Q bands overlap. Also, the S band with the lowest energy correspond to J” = 0 and denoted wS(0), while the O band with the highest energy correspond to J” = 2 and denoted wO(2) in wavenubers.

27. Vibrational-Rotational Raman Spectra Diatomic Molecules: Neglecting the effect of centrifugal distortion: A combination of differences method can be used to determine B0, and B1: w0,1 is obtained from the origin of the Q band (i..e wQ(0)): wQ(0) = w0,1

28. A1 B2 A1 Vibrational Spectroscopy Polyatomic Molecules Normal Modes As seen previously, the number of vibrational modes of motion increases proportionally to the number of atoms in the molecule (as 3N – 5 in linear molecules, or 3N – 6 in non-linear ones). Each vibrational motion may be, however, involve the combination of the motions of various atoms in terms of both stretching and bending. It is possible to break down these complicated motion into a set of harmonic motions that move at the same frequency, and in phase. Such a vibrational mode is called a normal mode. For example in water, the vibrational motion of the molecule is expressed in terms of 3 normal modes, which all of you already are familiar with:

29. n1 n3 n2 Vibrational Spectroscopy Polyatomic Molecules It is possible to solve the Schrodinger equation for the molecules in terms of the energy of each individual normal mode. It turns out that each mode has a vibrational quantum number associated to it (vi). Thus the vibrational energy of each mode assuming a harmonic motion is given by: Ei = hni(vi + ½) If a normal mode is degenerate, then: Ei = hni(vi + di/2) Where di is the degeneragy. The total vibrational energy of the molecule is the summation of the normal mode energies:

30. Vibrational Spectroscopy Polyatomic Molecules In the water molecule, the 3 normal modes have associated quantum numbers v1, v2, and v3. In triatomic molecules (like water), the bending mode is denoted with the label 2, while the symmetric stretch is label 1. The transition rules are the same as for diatomic molecules: Dv = ±1, ±2, ±3…, which means that there is a possibility to have as many as 3N – 6 fundamental, and overtone bands. Fortunately, there are some restrictions based on symmetry to the probability of a transitions. Besides the fundamental and overtone bands, there is the possibility of having combination bands, which originate from the excitation of two normal modes simultaneously. Thus for water, the following is possible: Three fundamental bands: vi” = 0 → vi’ = 1 (i = 1, 2 or 3) Various first overtones: vi” = 0 → vi’ = 2, 3, 4, etc., (i = 1, 2, or 3) Various combination bands, see some examples: v1” = 0, v2” = 0, v3” = 0 → v1’ = 0, v2’ = 1, v3’ = 1 v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 1 v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 1, v3’ = 0

31. Vibrational Spectroscopy Polyatomic Molecules: Notation Let us introduce some notation to account for the transitions. There are two notations, which are used by spectroscopists to distinguish a particular transition. Let us see some examples: Sym. Stretch of water: v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 0. Notation: Notation: |0,0,0> → |1,0,0> 1st overtone of the bending mode of water: v1” = 0, v2” = 0, v3” = 0 → v1’ = 0, v2’ = 2, v3’ = 0. Notation: Notation: |0,0,0> → |0,2,0> Combination of sym. stretch and asym. stretch v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 1. Notation: Notation: |0,0,0> → |1,0,1>

32. Vibrational Spectroscopy Polyatomic Molecules: Notation Let us introduce some notation to account for the transitions. There are two notations, which are used by spectroscopists to distinguish a particular transition. Let us see some examples: Sym. Stretch of water: v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 0. Notation: Notation: |0,0,0> → |1,0,0> 1st overtone of the bending mode of water: v1” = 0, v2” = 0, v3” = 0 → v1’ = 0, v2’ = 2, v3’ = 0. Notation: Notation: |0,0,0> → |0,2,0> Combination of sym. stretch and asym. stretch v1” = 0, v2” = 0, v3” = 0 → v1’ = 1, v2’ = 0, v3’ = 1. Notation: Notation: |0,0,0> → |1,0,1>

33. Vibrational Spectroscopy Polyatomic Molecules: An IR spectrum of CO2 (gas phase) Bands: 201 : 667.5 cm-1 301 : 2349 cm-1 n2 n3

34. n2+ n3 2n2 2n1 n1+ n3 2n2+ n1 n2 n1 n3 Vibrational Spectroscopy Polyatomic Molecules: An IR spectrum of N2O (gas phase) Bands: 201 : 588.8 cm-1 202 : 1167.0 cm-1 101 : 1285.0 cm-1 301 : 2223.5 cm-1 101 202 : 2461.5 cm-1 102 : 2563.5 cm-1 201 301 : 2798.5 cm-1 101 301 : 3498.1 cm-1

35. Vibrational Spectroscopy Polyatomic Molecules: Selection Rules and Symmetry Why is that all fundamental bands of N2O, and H2O (both bent molecules) are observed in the IR spectrum, but for CO2 only two of them are observed? Because of the symmetry of the molecule. Recall that IR intensities are proportional to the strength of the dipole moment that is induced by the vibrational motion. In other words the transition dipole integral Rv’v” must be valued (i.e. ≠ 0). There is a symmetry requirement that implies that the quantity to be integrated must be totally symmetric or contain a term that is totally symmetric: Г(Yv’) x Г(m) x Г(Yv”) ⊇ A Since the dipole moment follow the symmetry of translational motion then the equation above can be expressed as: Г(Yv’) x Г(Ti) x Г(Yv”) ⊇ A (with i = x, y, or z) If the lower state is the ground state (i.e. vi” = 0), then the wavefunction is totally symmetric and Г(Yv’) = A. Thus the equation reduces to: Г(Yv’) x Г(Ti) ⊇ A (with i = x, y, or z) Meaning that allowed transitions will be those with wavefunctions that include or follow the symmetry representation of the dipole moment (i.e. translational motion) in any direction: Г(Yv’) ⊇Г(Ti) (with i = x, y, or z)

36. A1 A1 B2 Vibrational Spectroscopy • Polyatomic Molecules: Selection Rules and Symmetry • EXAMPLE 1: Fundamental Transitions of Water • Determine symmetry group: C2v • Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: For water these are A1, B1 and B2. • Allowed transitions that originate from the ground state are those that correspond to normal modes that correspond to A1, B1 and B2 symmetry. In the case of water: Гvib = 2A1 + B2, so all fundamental bands are IR active. Sym Stretch Asym Stretch Bending

37. + + A1g A2u E1u Vibrational Spectroscopy • Polyatomic Molecules: Selection Rules and Symmetry • EXAMPLE 2: Fundamental Transitions of CO2 • Determine symmetry group: D∞h • Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: For CO2 these are A2u and E1u. • Allowed transitions that originate from the ground state are those that correspond to normal modes that correspond to A2u and E1u symmetry. In the case of CO2: Гvib = A1g + A2u + E1u, so only A2u (asym. stretch) and E1u (two degenerate bending) are IR active. Sym Stretch Asym Stretch Bending Note: For D∞h: A1g is also notated as Sg+, A2u as Su+ and E1u as Pu.

38. Vibrational Spectroscopy Polyatomic Molecules: Selection Rules and Symmetry Cross Product Symmetry Rules: This are required to find the symmetry representation of overtones and combination transitions. 1. Single degeneracy: A x A = A A x B = B B x B = A 1 x 1 = 1 1 x 2 = 2 2 x 2 = 1 g x g = g g x u = u u x u = g Example: A1g x A2u = A2u A1u x A2u = A2g 2. Degenerate: Depends on symmetry group, but overall must multiply characters and then reduce the result. Example: Determine cross product of E x E for group C3v: 3. Tables of cross products rules are available in the course website. Some worth noting are the one for the linear molecules. For instance the notation is different The reducible representation is: 4 in E, 1 in 2C3, and 0 in 3sv. Reduction yields: E x E = A1 + A2 + E

39. Vibrational Spectroscopy • Polyatomic Molecules: Selection Rules and Symmetry • EXAMPLE 3: Overtone Transitions of CO2 • Determine symmetry group: D∞h • Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: For CO2 these are A2u and E1u. • 1st overtone of asym. stretch (i.e. transition 302): In this case the cross product of the symmetry representation is required. The fundamental (301)corresponds to symmetry A2u, thus the overtone corresponds to symmetry A2u x A2u = A1g. Therefore, the overtone 302 has a symmetry representation that DOES NOT contain one of the translational representations, thus the band is IR inactive. • 1st overtone of bending (i.e. transition 202): In this case the cross product of the symmetry representation is required. The fundamental (201)corresponds to symmetry E1u, thus the overtone corresponds to symmetry E1u x E1u = A1g + A2g + E2g (Also notated as: Pu x Pu = Sg+ + Sg– + Dg). Then, this overtone (202) has a symmetry representation that DOES NOT contain one of the translational representations, thus the band is IR inactive.

40. Vibrational Spectroscopy • Polyatomic Molecules: Selection Rules and Symmetry • EXAMPLE 4: Overtone Transitions of Water • Determine symmetry group: C2v • Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: these are A1, B1 and B2. • 1st overtone of asym. stretch (i.e. transition 302): In this case the cross product of the symmetry representation is required. The fundamental (301)corresponds to symmetry B2, thus the overtone corresponds to symmetry B2 x B2 = A1. In other words, the first overtone has a symmetry representation that contains one of the translational representations, thus the band is IR active. • 2nd overtone of asym. stretch (i.e. transition 303): In this case the cross product of the symmetry representation is required. The fundamental (301)corresponds to symmetry B2, thus the overtone corresponds to symmetry B2 x B2 x B2 = B2. Then, the second overtone has a symmetry representation that contains one of the translational representations, thus the band is IR active.

41. Vibrational Spectroscopy • Polyatomic Molecules: Selection Rules and Symmetry • EXAMPLE 5: Combination Transitions of NH3 • Determine symmetry group: C3v • Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: these are A1, E. • The representation for vibrational motions is Гvib = 2A1 + 2E. Sym. Stretch (v1) and a bending (v2) are A1, Asym. stretch (v3) and another bending (v4) are E. • Combination (101301): In this case the cross product of the symmetry representation is required. Two fundamental are combined. They (101 and 301)correspond to symmetry representations A1 and E respectively. Thus the combination corresponds to symmetry A1 x E = E. Thus, the combination has a symmetry representation that contains one of the translational representations, thus the band is IR active. • Combination (101201): In this case the cross product of the symmetry representation is required. Two fundamental are combined. Both (101 and 201) correspond to symmetry representation A1. Thus the combination corresponds to symmetry A1 x A1 = A1 . Thus, the combination has a symmetry representation that contains one of the translational representations, thus the band is IR active.

42. v2, A1g v1, A1g v3, A2u v4, E1g v5, E1u + - + - - - + + Vibrational Spectroscopy • Polyatomic Molecules: Selection Rules and Symmetry • EXAMPLE 6: Combination Transitions of acetylene (C2H2) • Determine symmetry group: D∞h • Look up the Г(Tx), Г(Ty) and Г(Tz) in the character table: these are A2u and E1u. • The representation for vibrational motions is Гvib = 2A1g + A2u + E1g + E1u. Sym C-H and C≡C stretchs (v1 and v2) are A1, Asym. C-H stretch (v3) is A2u, the trans bending (v4) is E1g and the cis bending is E1u Note that out of this 5 fundamental transitions, only 301 and 501 are IR active.

43. Vibrational Spectroscopy • Polyatomic Molecules: Selection Rules and Symmetry • EXAMPLE 6: Combination Transitions of acetylene (C2H2) • Combination (101301): In this case the cross product of the symmetry representation is required. Two fundamental are combined. They (101 and 301)correspond to symmetry representations A1g and A2u respectively. Thus the combination corresponds to symmetry • A1g x A2u = A2u. Thus, the combination has a symmetry representation that contains one of the translational representations, thus the band is IR active. • 2. Combination (201302): In this case the cross product of the symmetry representation is required. One fundamental and one overtone are combined. The fundamental (201 and 301) correspond to symmetry representations A1g and A2u respectively. Thus the combination corresponds to symmetry A1g x (A2u x A2u) = A1g . Thus, the combination has a symmetry representation that DOES NOT contain one of the translational representations, thus the band is IR inactive.

44. Vibrational Spectroscopy Polyatomic Molecules: Selection Rules and Symmetry for Raman Given that the Raman effect is associated to an induced dipole proportional to the polarizability change of the molecule as it vibrates, then the symmetry of the molecule also determines the probability of a Raman vibrational transition. In this case, the selection rule is given by: Г(Yv’) x Г(ai,j) x Г(Yv”) ⊇ A In this case, the symmetry representation of the upper state must include the symmetry components of the polarizability tensor, which are also represented in the right hand side of the character tables. Example: water has polarizability representations with A1, A2, B1 and B2 symmetry, meaning that all fundamental transitions (2A1 and B2) are Raman active. Example: the molecule trans-diazene (HN=NH) is C2h and has Гvib = 3Ag + Au + 2Bu. Three fundamental modes are IR active (Au + 2Bu) and 3 IR inactive. Interestingly, there are 3 fundamental modes (3Ag) that are Raman active, and three (Au + 2Bu) that are Raman inactive. Molecules with inversion centers (centrosymmetric) behave this way.

45. Vibrational Spectroscopy Polyatomic Molecules: Selection Rules and Symmetry for Raman Given that the Raman effect is associated to an induced dipole proportional to the polarizability change of the molecule as it vibrates, then the symmetry of the molecule also determines the probability of a Raman vibrational transition. In this case, the selection rule is given by: Г(Yv’) x Г(ai,j) x Г(Yv”) ⊇ A In this case, the symmetry representation of the upper state must include the symmetry components of the polarizability tensor, which are also represented in the right hand side of the character tables. Example: water has polarizability representations with A1, A2, B1 and B2 symmetry, meaning that all fundamental transitions (2A1 and B2) are Raman active. Example: the molecule trans-diazene (HN=NH) is C2h and has Гvib = 3Ag + Au + 2Bu. Three fundamental modes are IR active (Au + 2Bu) and 3 IR inactive. Interestingly, there are 3 fundamental modes (3Ag) that are Raman active, and three (Au + 2Bu) that are Raman inactive. Molecules with inversion centers usually behave this way.

46. v3, A2u v5, E1u - - + + Vibrational Spectroscopy Polyatomic Molecules: Vibrational Rotational Spectra Linear Molecules Based on symmetry considerations, the allowed (i.e. IR active) transitions for linear molecules can be classified as: A1g (Sg+) → A2u (Su+) and A1g (Sg+) → E1u (Pu) (symmetric molecules: D∞h) A1 (S+) → A2 (S+) and A1 (S+) → E1 (P) (asymmetric molecules: C∞v)

47. J’ = 3 A2u (Su+) J’ = 0 J” = 3 A1g (Sg+) J” = 0 v3, A2u Vibrational Spectroscopy Polyatomic Molecules: Vibrational Rotational Spectra Linear Molecules Sigma Type Bands: A1g (Sg+) → A2u (Su+) or A1 (S+) → A2 (S+) • Presence of P and R bands, absence of Q band: DJ = ± 1 • For symmetric molecules: alternation in the intensity due to spin statistics • Data is treated as done for a linear diatomic molecule (i.e. Bv and DJ can be obtained)

48. – J’ = 4 + + – E1u (Pu) – + + – – J” = 3 A1g (Sg+) + – + J” = 0 v5, E1u - - + + Vibrational Spectroscopy Polyatomic Molecules: Vibrational Rotational Spectra Linear Molecules Pi Type Bands: A1g (Sg+) → E1u (Pu) or A1 (S+) → E1 (P) J’ = 1 • Presence of P, Q and R bands: DJ = 0, ± 1 • Parity (related to symmetry of nuclear exchange upon rotation) must be taken into account: + ↔ – is allowed. • There is no J’ = 0, because degenerate (E or P in this case) vibrational states do not allow for total zero angular momentum • Rotational levels in degenerate states are split due to Coriolis forces. The effect of Coriolis is included in the energy equation: • EJ = BvJ(J + 1) – Bv ± ζi/2 J(J + 1) • For symmetric molecules: alternation in the intensity due to spin statistics

49. Vibrational Spectroscopy Polyatomic Molecules: Vibrational Rotational Spectra Symmetric Rotors Allowed transitions depend on the symmetry of the molecule. Many symmetric rotors are C3v, so let us use this as an example. According to the character table allowed vibrational transitions are: A1→ A1 and A1→ E Bands associated to the A1 symmetry are called parallel bands, while those associated to the E symmetry are called perpendicular bands

50. Vibrational Spectroscopy • Polyatomic Molecules: Vibrational Rotational Spectra • Symmetric Rotors (C3v case) • Parallel Bands: A1→ A1 • Presence of P, Q and R bands: DJ = 0, ± 1 and DK = 0 allowed for K ≠ 0 • DJ = ± 1 and DK = 0 allowed for K = 0 • Band is similar to that of a A1→ E band of linear molecule, but splitting is due to different K values.