Chemistry 43 Lecture 7 Vibrational Spectroscopy NC State University
The Dipole Moment Expansion The permanent dipole moment of a molecule oscillates about an equilibrium value as the molecule vibrates. Thus, the dipole moment depends on the nuclear coordinate Q. μ Q = μ 0 + μ Q Q + where μ is the dipole operator.
Rotational Transitions Rotational transitions arise from the rotation of the permanent dipole moment that can interact with an electromagnetic field in the microwave region of the spectrum. μ Q = μ 0 + μ Q Q +
Interaction with radiation An oscillating electromagnetic field enters as E 0 cos(ωt) such that the angular frequency hω is equal to a vibrational energy level difference and the transition moment is M rot = μ 0 0 2π 0 π * Y J+,M cos θ Y JM sin θ dθdφ
Interaction with radiation The choice of cos(θ) means that we consider z-polarized microwave light. In general we could consider x- or y-polarized as well. x sin(θ)cos(φ) μ 0 = μ X i + μ Y j + μ Z k y sin(θ)sin(φ) μ 0 = μ 0 sinθcosφi + sinθsinφ j +cosθk z cos(θ)
The total wave function The total wave function can be factored into an electronic, a vibrational and a rotational wave function. Ψ = ψ el χ v Y JM M rot,vib = χ * v Y J M ψ * el μψ el dτ el χ v Y JM dτ nuc = χ * v Y J M μ(q)χ v Y JM sinθdθdφ where μ(q) is the dipole moment as a function of nuclear coordinates.
The total wave function The total wave function can be factored into an electronic, a vibrational and a rotational wave function. Ψ = ψ el χ v Y JM M = χ * v Y J M μ(q)χ v Y JM sinθdθdφ M rot = Y J M μ 0 Y JM sinθdθdφ and M vib = μ Q χ * v Qχ v dq
Selection rules for the μ = μ o + dμ + dq Q= 0Q 2 So, harmonic oscillator The dipole moment can be expressed as a Taylor expansion, v' μ v = v' v μ 0 + v' Qv Ordinarily, dμ dq Q =0 >> d 2 μ dq 2 dμ dq Q =0 d 2 μ Q 2 + dq 2 6 Q =0 dμ dq Q = 0 Q =0 >>> d 3 μ dq 3 d 3 μ dq 3 Q = 0 + v' Q 2 v d 2 μ dq 2 Q = 0 >>> = 0 for symmetric bonds Q 3 + Q = 0 <v' Q n v> 0 only when v' = v ± n +
Vibrational Transition
Vibrational Transition
Vibrational Selection Rule v = v + v = v - v = 2 v = v = 0 Q
Fundamentals and Overtones The selection rule for the harmonic oscillator is usually given as Δv =, but higher order terms are possible. The Δv = transition is known as the fundamental. The higher order terms are called overtones. The resonance condition for the harmonic oscillator is, ΔE = E v+ -E v = (v + 3/2)hν - (v + /2)hν = hν all of the Δv = transitions would occur at the same energy, the classical frequency, and the overtones would occur at 2hν for the first overtone and at 3n for the second overtone How many vibrations does a Li-Li molecule undergo during ps? (ν = 35.4 cm - )
Vibrational transitions Vibrational transitions arise because of the oscillation of the molecule about its equilibrium bond configuration. As the molecule oscillates infrared radiation can interact to alter the quantum state. μ Q = μ 0 + μ Q Q + M vib = μ Q * χ v + Qχ v dq
Vibrational transitions Infrared absorption As an example we can calculate the transition moment between the state v = 0 and v =. χ 0 = α π /4 e αq2 /2, χ = α π /4 2αQe αq2 /2 M vib = μ Q = μ Q α π /2 α π /2 2α 2α e αq2 /2 Q 2 e αq2 /2 dq π 2α 3/2 = μ Q 2α
Cartesian basis for molecular vibrations There are 3N-5 vibrational degrees of freedom for linear molecules and 3N-6 for nonlinear molecules CO 2 (linear) SO 2 (nonlinear) For the 3N Cartesian vector basis, {x i i =, 2, 3,, 3N} V = 2 i j b ij x i x j where the elements of the matrix, b ij are the Cartesian force constants and represent the restoring force experienced along x i when x j is changed. However, the Cartesian basis is not convenient for molecules.
Internal coordinates For our vibrational basis, we will use four different types of internal coordinates (R): bond stretchings (ν) + in-plane angle bendings (δ) out-of-plane angle bendings (π) - + torsions (τ) + π ν + - δ - τ +
The Wilson FG Matrix Method The equations of motion can be written as, (FG - λe)q = 0 (here, F is the force constant matrix, G is the inverse mass matrix and E is the identity matrix) The secular equation for vibrations is, FG - λe = 0 where the λs are the vibrational eigenvalues, ν =30.3 λ
Symmetry Coordinates, S V = 2 f ij S i S j V must remain invariant under all operations. Therefore f ij = 0 if S i and S j belong to different irreps - the F-matrix is block diagonalized when symmetry coordinates are used. The F- and G-matrices are still set up in terms of internal coordinates but are then symmetrized (block diagonalized) by a similarity transformation, G sym = UGU t F sym = UFU t Each block of the resulting F sym G sym matrix is then treated as a separate problem.
Construction of Symmetry Coordinates To generate symmetry coordinates, S. Determine Γ cart [=(Γ x +Γ y +Γ z ) Γ unsh ] the 3N Cartesian displacement vectors and decompose it into irreps. 2. Subtract the external degrees of freedom (x, y, z, R x, R y, and R z ) from Γ cart. 3. Decide on a set of internal coordinates and determine the reducible representations of the symmetrically equivalent sets and decompose each. 4. Construct the symmetry coordinates, S, of each irrep as linear combinations of the appropriate internal coordinates as determined in 3.
The water molecule belongs to the C 2v point group. The 3N Cartesian vectors for water. C 2v E C 2 σ v Γ xyz = Γ x +Γ y +Γ z 3 - Γ unsh 3 3 Γ cart =Γ xyz Γ unsh 9-3 σ v '
Γ cart = 3a + a 2 + 3b + 2b 2 ; Γ trans = a + b + b 2 ; Γ rot = a 2 + b +b 2 Therefore, Γ vib = 2a + b Choose internal coordinates (R), C 2v E C 2 σ v σ v ' Α Α 2 - - B - - - - B 2 There are two symmetrically equivalent sets of R: Δr and Δα C 2v E C 2 σ v σ v ' Γ Δα Γ Δr 2 0 2 0 a a +b
SALCs: C 2v E C 2 σ v σ v ' Γ Δα Γ Δr 2 0 2 0 r r r 2 r r 2 Δα Δα Δα Δα Δα ν ν (a ) S = ( 2 Δr +Δr ) 2 Symmetric stretch δ δ 2 (a ) S 2 =Δα ν a ν 3 (b ) S 3 = 2 Δr Δr 2 ( ) Antisymmetric stretch
The symmetry coordinate definitions in the form S = UR, S S 2 = S 3 / 2 / 2 0 0 0 / 2 / 2 0 Δr Δr 2 Δα The force field for the internal coordinates is written as, Δr Δr 2 Δα F = Δr Δr 2 Δα f rr f rr; rf rrα f rr f rr rf rrα rf rrα rf rrα r 2 f αα where the r and r 2 terms are used to maintain the same units for all force constants since angle bendings are unitless and have force constants in J not J/m 2.
We can apply the U matrix that transforms to symmetry coordinates to obtain the F sym matrix. Fsym = UFU t F sym = / 2 / 2 0 0 0 / 2 / 2 0 f rr f rr; rf rrα f rr f rr rf rrα rf rrα rf rrα r 2 f αα / 2 0 / 2 / 2 0 / 2 0 0 F sym = / 2 / 2 0 0 0 / 2 / 2 0 / 2 f rr + f rr rf rrα / 2 f rr f rr / 2 f rr + f rr rf rrα / 2 f rr f rr 2/ 2rf rrα r 2 f αα 0 F sym = f rr + f rr 2rf rrα 0 2rf rrα r 2 f αα 0 0 0 f rr f rr
Infrared absorption of water Water is the main absorber of the sunlight in the atmosphere. The 3 million million tons of water in the atmosphere (~0.33% by weight) is responsible for about 70% of all atmospheric absorption of radiation, mainly in the infrared region where water shows strong absorption. It contributes significantly to the greenhouse effect ensuring a warm habitable planet, but operates a negative feedback effect, due to cloud formation reflecting the sunlight away, to attenuate global warming. The water content of the atmosphere varies about 00-fold between the hot and humid tropics and the cold and dry polar ice deserts.
Infrared absorption of water Main vibrations of water isotopomers v, cm - v 2, cm - v 3, cm - H 2 6 O 3657.05 594.75 3755.93 H 2 7 O 3653.5 59.32 3748.32 H 2 8 O 3649.69 588.26 374.57 HD 6 O 2723.68 403.48 3707.47 D 2 6 O 2669.40 78.38 2787.92 T 2 6 O 2233.9 995.37 2366.6
Overtones of water Even in water vapor ν ν 3, but symmetries are different, Γ Γ 3. However, the third overtone of mode has the same symmetry as the combination band Γ Γ Γ = Γ Γ 3 Γ 3. Strong anharmonic coupling leads to strong overtones at,032 and 0,63 cm -. These intense bands give water and ice their blue color. ν symmetric stretch 3825 cm - ν 2 bend 654 cm - ν 3 asymmetric stretch 3935 cm -
Frequency shift due to molecular interactions Hydrogen bonding lowers O-H force constant and H-O-H bending force constant. vapor liquid ν 3825 3657 ν 2 654 595 ν 3 3935 3756
Raman scattering The Raman spectrum of a compound with vibrational modes at 300 and 800 cm - observed with a 9,500 cm - exciting line at 300K. The very strong line at n o is due to Rayleigh scattering. The anti-stokes component of the Raman scattering is weaker than the Stokes component by exp(-e vib /kt). In normal Raman, the scattering is not from a stationary state but from a virtual state. Virtual states can be considered as stationary states that have been spread out by the uncertainty principle, ΔE Δt h where ΔE is the amount by which the photon's energy fails to be in resonance with the closest lying excited state and Δt is the time the system can spend in this state.
Stokes and anti-stokes E h/ t The two types of normal Raman scattering. Since Δt (c) -, if the exciting line, n 0 is 20,000 cm - and the lowest excited state is 30,000 cm -, then ΔE = x0 4 cm - and Δt (x0 4 x 3x0 0 ) - 3 fs or about a vibrational period - normal Raman scattering is a very fast process.
Molecular polarizability What fraction of the molecules will undergo scattering will depend upon how the electrons are affected by the electric field, i.e., on the molecule's polarizability (α). One can view the process as the photon initiating an oscillating dipole (μ) in the molecule as the electrons oscillate with the electric field; it is then the oscillating dipole that radiates the scattered photon (much like a radio transmitter). The strength of this induced dipole is proportional to the electric field of the photon. The proportionality constant is the molecular polarizability, α. Since the dipole and the electric field are actually vectors, the polarizability is a tensor. μ ind = αε
The polarizability tensor μ x μ y μ z α xx α xy α xz = α yx α yy α yz α zx α zy α zz where α xz ( 0) is a measure of how strongly the z-component of the electric field (z-polarization) induces a dipole in the x- direction: this implies that the vibration interacts with the electric field in such a way so as to rotate the polarization of the electric field. Indeed, one of the important pieces of information available from Raman scattering is the polarization change that the incoming photon undergoes. ε x ε y ε z
Classical description of Raman scattering The molecular vibration at ω v alters the polarizability according to: α molecule = α M + α M Q ΔQcos(ω vt) A transition dipole moment from state i to f is created by Interaction of radiation at frequency ω 0. μ fi = α M Q ΔQcos(ω vt)e 0 cos(ω 0 t) = α M Q ΔQ 2 E 0 cos([ω 0 ω V ]t) + cos([ω 0 + ω V ]t)
Isotropic and anisotropic tensor components In solutions or pure liquids which involve rotationally averaged, randomly-oriented species, there are three irreducible tensors which remain invariant: the isotropy, G o = α xx2 + α yy2 + α zz2 = 3α, which transforms as the totally symmetric representation; an isotropic α-tensor is diagonal all off-diagonal elements are zero; the symmetric anisotropy, G s, which transforms like the d-orbitals; these are the off-diagonal elements of the a- tensor ; and
The depolarization ratio The intensity of scattered radiation which is polarized perpendicular to that of the incoming radiation is defined as I and that of the radiation parallel to the incoming radiation as I then the depolarization ratio, ρ is defined as, ρ = I /I When the polarizability tensor is diagonal, the vibration does not rotate the electric vector: all of the intensity will remain parallel to the incoming radiation and ρ = 0. This is isotropic scattering. However, if any of the off-diagonal elements of the polarizability tensor are non-zero, then there is intensity in I and ρ > 0.
Regimes of Raman polarization Τhe intensities of the two components can then be expressed as I 0G o + 4G s and I 3G s. The total intensity of the scattered radiation is given as I tot = I + I 0G o + 7G s. The depolarization ratio is then, ρ = 3G s 0G o + 4G s Thus, we see that isotropic Raman (G s = 0) is polarized and ρ = 0 and anisotropic Raman (G o = 0) is polarized and hence ρ = 3/4
In normal Raman, activity is due to non-zero values for G o and/or G s. G o is always totally symmetric so totally symmetric vibrations are always Raman active. G s transforms like the d- orbitals so modes which have the same symmetry as one of the d-orbitals will also be Raman active. Furthermore, since only non-zero values of G o can result in ρ < 3/4, only totally symmetric modes are polarized. As one example, consider the C 2v point group, C 2v Raman selection rules A z x 2, y 2, z 2 G o + G s 0 ρ 0.75 A 2 R z xy G s + G a ρ = 0.75 B x, R y xz G s + G a ρ = 0.75 B 2 y, R x yz G s + G a ρ = 0.75 More than one entry here means that a modes scatter by G 0 and G s and 0 ρ 0.75
Molecules with a center of symmetry Molecules that have a center of symmetry (i.e. an inversion center) have vibrational and Raman selection rules that are mutually exclusive. As a consequence one can determine those selection rules using the character table. Irreps that transform as x,y,z are infrared active. Irreps that transform as xy,xz,yz,or x 2,y 2,z 2 are Raman active If a molecule does not have a center of symmetry then a mode can be both infrared and Raman active.
Character table for D h point group E 2C σ v i 2S C' 2 linear functions, rotations quadratic A g =Σ + g x 2 +y 2, z 2 A 2g =Σ - g - - R z E g =Π g 2 2cos(φ) 0 2-2cos(φ) 0 (R x, R y ) (xz, yz) E 2g =Δ g 2 2cos(2φ) 0 2 2cos(2φ) 0 (x 2 -y 2, xy) E 3g =Φ g 2 2cos(3φ) 0 2-2cos(3φ) 0 A u =Σ + u - - - z A 2u =Σ - u - - - E u =Π u 2 2cos(φ) 0-2 2cos(φ) 0 (x, y) E 2u =Δ u 2 2cos(2φ) 0-2 -2cos(2φ) 0 E 3u =Φ u 2 2cos(3φ) 0-2 2cos(3φ) 0
Carbon dioxide motion along z-axis E 2C σ v i 2S C' 2 linear functions, rotations quadratic A g =Σ + g x 2 +y 2, z 2 A u =Σ + u - - - z Γ 3 3 3 - - - E i linear functions, rotations quadratic A g =Σ + g x 2 +y 2, z 2 A u =Σ + u - z Γ 3 - The reducible representation can be decomposed into A g + 2A u
Normal modes - CO 2 Symmetric stretch (infrared ν 2289 inactive) cm - Symmetric stretch (Raman active). +. Asymmetric stretch 3 ν - Bends 3 2349 cm - ν 2 667 cm - (IR active) (infrared active) ν 2 667 cm - (IR active) There are 4 normal modes (3N - 5). Three of them are infrared active since they show a dipole moment change in their motion.