5.3 Rotational Raman Spectroscopy General Introduction

Transcription

1 5.3 Rotational Raman Spectroscopy General Introduction When EM radiation falls on atoms or molecules, it may be absorbed or scattered. If λis unchanged, the process is referred as Rayleigh scattering. In 1871, Lord Rayleigh showed Scattered intensity I s λ -4 Eq. (5.41) In a scattering process, if λ is changed, the process is referred As Raman scattering. If the wavenumber decreases, it is Stokes Raman scattering. If the wavenumber increases, it is anti-stokes Raman scattering. Electronic, vibrational, and rotational transitions may be involved in Raman scattering. wbt 1

2 5.3.2 Theory of Rotational Raman Spectroscopy Molecular polarizability α = a measure of the degree to which the electrons in the molecule can be displaced relative to the nuclei. In general, the polarizability of a molecule is anisotropic. This Means that α may have different magnitudes when measured in different directions. When monochromatic radiation falls on a molecule, the E of the radiation induces in the molecule an electric dipole μ, μ =αe Eq. (5.43) The magnitude E of the vector can be written E = A sin 2πc ~ ν t Eq. (5.44) where A is the magnitude and ~ ν is the wavenumber of the monochromatic radiation. wbt 2

3 A surface drawn so that the distance from the origin to a point on the surface has a length of α -1/2, where α is the polarizability in that direction, forms an ellipsoid. As the polarizability ellipsoid rotates with the molecule at a frequency ν rot, the radiation sees the polarizability Changing at a a frequency of 2ν rot. The variation of the polarizability α with rotation is given by α = α 0,r + α 1,r sin 2πc(2 ~ ν rot )t Eq. (5.45) where α 0,r is the average polarizability and α 1,r is the magnitude of the change of polarizability during rotation. Substitute Eqs. (5.44) and (5.45) into Eq. (5.43), the magnitude of the induced dipole moment becomes μ = α 0,r A sin 2πc ~ ν t - ½α 1,r Acos 2πc ( ~ ν + 2 ~ rot )t + ½α 1,r A cos 2πc( ~ ν ν - 2 ~ ν rot )t Eq. (5.46) wbt 3

4 All three terms in Eq. (5.46) represent scattering of the radiation. The first term corresponds to Rayleigh scattering, where the wavenumber ~ ν remains unchanged. The second and third correspond to the anti-stokes and Stokes Raman scattering with wavenumbers of ( ~ ν + 2 ~ ν rot ) and ( ~ ν - 2 ~ ν rot ), respectively Rotational Raman Spectra of Diatomic and Linear Polyatomic Molecules In a diatomic or linear polyatomic molecule, rotational Raman Scattering obeys the selection rule: ΔJ = 0, ± 2 Eq. (5.47) wbt 4

5 Rotational Raman spectrum of a diatomic or linear polyatomic Molecule. Selection rule: ΔJ = 0, ± 2 wbt 5

6 Recall: When molecules initially in the ΔJ = 0 state encounter intense, monochromatic EM radiation of wavenumber ~ ν, they may absorb the radiation and undergo transition to the upper electronic, vibrational, or rotational state. In another situation, the EM radiation is not absorbed but undgoes a scattering process. The EM radiation will cause a change in the charge density of the molecule, producing an induced dipole in the molecule as expressed in Eq. (5.43). The molecule is said to be in a virtual state. When a molecule undergoes a virtual state V 0, it may return, according to the selection rules (Eq. 5.47), to J = 0 (Rayleigh), or J = 2 (Stokes). If a molecule initially in the J = 2 state, it may go to the virtual state V 1 and return to J = 2 (Rayleigh), J = 4 (Stokes), or or J = 0 (anti-stokes). wbt 6

7 Raman and Rayleigh scattering process involving virtual states V 0 and V 1. Selection rule: ΔJ = 0, ± 2 wbt 7

8 Conventionally, ΔJ = J(upper) J(lower), one usually consider only ΔJ = +2. The magnitude of a Raman displacement from the incident radiation is Δ ~ ν = F(J + 2) F(J) Eq. (5.48) where, Δ ~ ν = ~ ν ~ ν L ( ~ ν L is the wavenumber of the incident laser radiation) is positive for anti-stokes and negative for Stokes lines. When centrifugal distortion is neglected, the rotational term values F(J) are given by Eq. (5.12) and Eq. (5.48) becomes Δ ~ ν = 4B 0 J + 6B 0 Eq. (5.49) When centrifugal distortion is taken into account, the rotational term values F(J) are given by Eq. (5.19) and Eq. (5.48) gives Δ ~ ν = (4B 0 6D 0 )(J + 3/2) 8D 0 (J + 3/2) 3 Eq. (5.50) wbt 8

9 Series of rotational transitions are referred to as branches and they are labeled with a letter according to the values of ΔJ as follows: ΔJ Branch O P Q R S The intensities along the branches show a maximum since the population of the initial levels of the transitions, given by Eq. (5.15), show similar behavior. wbt 9

10 Rotational Raman spectrum of 15 N 2. (The lines marked with a cross are grating ghosts and not part of the spectrum) wbt 10

11 Previous viewgraph shows the rotational Raman spectrum of 15 N 2 recorded with nm radiation from an argon ion laser. Analysis gives a B 0 value of ± cm -1, yielding the bond length r 0 of ± Ả. A feature of the 15 N 2 spectrum shows an alteration of 1:3 for the J value of the initial level of the transition even:odd. This is an effect due to the nuclear spin of the 15 N nuclei. wbt 11

12 5.3.4 Nuclear Spin Statistical Weights When nuclear spin is included, the total wave function ψ is expressed as ψ = ψ e ψ v ψ r ψ ns Eq. (5.52) For a symmetrical (D h ) diatomic or linear polyatomic molecule, nuclear spin quantum number I = n + ½, where n is zero of an integer. Exchange of the two equivalent nuclei results in a change of sign of ψ which is said to be antisymmetric to nuclear exchange. The nuclei are referred to as Fermi particles (or fermions) and obey Fermi-Dirac statistics. On the other hand, if I = n, ψ is symmetric to nuclear exchange and the nuclei are referred to as Bose particles (or Bosons) and obey Bose-Einstein statistics. wbt 12

13 For 1 H 2 in the ground electronic state, both are symmetric to nuclear exchange. Since I = 1/2 for 1 H (a fermion), thus ψ and ψ ns which is said to be antisymmetric to nuclear exchange. Thus, must be antisymmetric For J = even, ψ r is symmetric (s) to nuclear exchange; for J = odd, ψ r is antisymmetric (a) to nuclear exchange. Some values of the nuclear spin quantum number I Nuc. I Nuc. I Nuc. I Nuc. I 1 H 1/2 12 C 0 16 O 0 30 Si 0 2 H 1 13 C 1/2 19 F 1/2 31 P 1/2 10 B 3 14 N 1 28 Si 0 35 Cl 3/2 11 B 3/2 15 N 1/2 29 Si 1/2 37 Cl 3/2 wbt 13

14 The nuclear spin wave function ψ ns is usually written as αor β, corresponding to M I = ½ or -1/2. Thus, there are four possible forms of ψ ns for 1 H 2 : ψ ns = α(1)α(2); β(1)β(2); α(1)β(2); β(1)α(2); The α(1)α(2) and β(1)β(2) wave functions are symmetric to interchange of the nuclei 1 and 2, whereas the other two are neither symmetric nor antisymmetric. Thus, it is necessary to take linear combination of these two forms to obtain symmetric or antisymmetric nuclear spin wave functions. Therefore, the three possible nuclear spin wave functions are α(1)α(2), β(1)β(2), and 2-1/2 [α(1)β(2) + β(1)α(2)]. The only antisymmetric nuclear spin wave function is 2-1/2 [α(1)β(2) - β(1)α(2)]. wbt 14

15 In general, for a homonuclear diatomic molecule, such as H 2, N 2, O 2, and F 2, there are (2I + 1)(I + 1) symmetric and (2I + 1)I antisymmetric nuclear spin wave functions, thus S/A = (I + 1)/I Eq. (5.56) For 1 H 2 and 15 N 2, each nucleus has I = ½. If the rotational quantum number J = odd, it requires a symmetric nuclear spin wave function to fulfill Eq. (5.52). Thus, there are 3 of them. In case of J = even, it requires an antisymmetric nuclear spin wave function. There is only 1 type of it. Therefore, the rotational Raman spectrum of 15 N 2 shows an alteration of 1:3 for the J value of the initial level of the transition even:odd. wbt 15

16 5.4 Structure Determination from Rotational Constants Measurement and assignment of the rotational spectrum of a diatomic or linear molecule result in a rotational constant B 0, which relates to the zero-point vibrational state. If the rotational constant of the excited states can be determined then B e (which relates to the molecule in the unattainable equilibrium configuration) can be obtained using Eq. (5.25). For a diatomic molecule B 0 and B e can be converted to moment of inertia using Eq. (5.11) or (5.12), and thence to bond lengths r 0 and r e by using I = µr 2. The major difference between r e and r 0 is that the former is independent of the isotopic species whereas the latter is. For a non-linear polyatomic molecule, its zero-point structure (in terms of bond lengths and angles) is isotope-dependent but its equilibrium structure is not. wbt 16

17 In a non-linear but planar molecule, e.g. H 2 CO, the out-ofplane principal moment of inertia I c is related to the other two by I c = I a + I b Eq. (5.58) H H C O b so there are only two independent rotational constants. Even for a small molecule such as H 2 CO, determination of the rotational constants in the v = 1 levels of all vibrations are considerably difficult. In larger molecules, it may be possible to determine A 0, B 0, and C 0. When making sufficient isotopic substitutions, one can obtain so-called r 0 structure. Ab initio and density functional theory calculations can predict various molecular structure parameters. wbt 17 a

18 Sample calculation 5.2 (a) From the value for B 0 of ± cm -1, obtained from the rotational Raman spectrum of 14 N 15 N, calculate the bond length r 0. (b) Why does it differ from r 0 for 14 N 2? (c) Would the values of r e differ? (d) Would there be an intensity alternation in the spectrum of 14 N 15 N? (e) Would 14 N 15 N show a rotational infrared spectrum? Possible solutions: (a) For 14 N 15 V, the reduced mass µ = ( x ) / ( ) g mol -1 x (10-3 kg g -1 / x mol -1 ) = x kg wbt 18

19 r 02 = h/(8π 2 cµb 0 ) = x J s / (8π 2 x x cm s -1 x x x 1.93 cm -1 ) = x m 2 r 0 = x m = pm = Å (bond length of 14 N 15 N at zero-point vibrational state) (b) The calculated value of r 0 differs from that of r 0 = Å for 14 N 2 because the zero-point level is lower, within the same anharmonic potential for the heavier isotopic species. (c) The values of r e would not differ because they refer to the minimum of the isotope-independent potential energy (d) No, because there is no exchange of identical nuclei on rotation about an axis perpendicular to the bond. (e) No, because the dipole moment of 14 N 15 N is zero. (In fact, like all such isotopically unsymmetrical molecules, it has an extremely small dipole moment, but too small to be detected by an infrared rotational spectroscopy. wbt 19