Vibrational Spectroscopy Vibrational states and energies Vibrational spectroscopy Infrared Raman Normal Mode Analysis
Vibrational States and Energies
Quantum approach to vibration Solution is Gaussian Energy is quantized 2 2 h ∂ Ψ 2 k – Q Ψ = EΨ 2 + 2 2µ ∂Q
1 E = v + hν 2 Q
v is the quantum number Allowed transitions v’ -> v + 1, v’ -> v - 1
The bonding electronic state gives rise to a potential energy surface for the nuclear motion
χ
2
Harmonic approximation 2 1 V Q = kQ 2 1/4 α – αQ 2 χ0 Q = π e 1/4 α – αQ 2 χ 1 Q = π 2αQe
... E v = v + 1 hω 2
There is a potential energy surface that corresponds to each electronic state of the molecule The shift in the nuclear displacement arises from the fact that the bond length increases in the σ* state compared to the σ state. The overlap of the vibrational wave functions is key to understanding the shape of absorption bands.
There are 3N-6 vibrational degrees of freedom in a molecule with N atoms Three degrees of freedom are required for translation. Three degrees of freedom are required for rotation. For example, in H2O there are 9 total degrees of freedom and 3 vibrational degrees of freedom. In C6H6 there are 36 degrees of freedom and 30 Vibrational degrees of freedom. Exception: In linear molecules there are only 2 rotational degrees of freedom and therefore the number of vibrations is 3N - 5.
The vibrational degrees of freedom can be expressed as normal modes. All normal modes have the same form for the harmonic oscillator wavefunction and differ only in the force constant k and reduced mass µ. The total wavefunction is a product of normal modes. The total nuclear wavefunction for water is χ1χ2χ3. The normal mode wavefunctions of water correspond to the symmetric stretch, bend, and asymmetric stretch. These are linear combinations of the stretching and bending internal coordinates of H2O.
Normal modes of water
ν1 symmetric stretch 3825 cm-1
In water vapor ν1 ≈ ν3, but symmetries are different, Γ1 ≠ Γ3. (Γ is the symmetry) However, the third overtone of 1 has the same symmetry ν2 bend 1654 cm-1 as the combination band Γ1 Γ1 Γ1 = Γ1 Γ3 Γ3 . Strong anharmonic coupling leads to strong overtones at 11,032 and 10,613 cm-1. ν3 asymmetric stretch 3935 cm-1
Comparison of harmonic and anharmonic potentials
Frequency shift due to molecular interactions Hydrogen bonding lowers O-H force constant and H-O-H bending force constant.
vapor → liquid ν1 3825 → 3657 ν2 1654 → 1595 ν3 3935 → 3756 The intermolecular hydrogen bonding stretching mode is difficult to observe.
Vibrational Transitions
Transition dipoles In order for infrared light to be absorbed the polarization must be aligned with the direction of the transition moment. For a vibrational mode this is determined by the directional change in the dipole moment. This is shown below for the bending mode of H2O.
O
O H
H
H
H
Transition dipoles The change in ground state dipole moment during vibration interacts with light. ∂µ g µ = µg + Q+ ∂Q
The first term is static and does not contribute to the transition. Calling the vibrational wavefunctions χi the transition moment is:
∂µ g µ 10 = ∂Q
χ 1Qχ 0dQ,
Dipole derivatives The vibrational wavefunctions χi are Gaussians, thus the transition moment for transition from vibrational state 0 to vibrational state 1 is:
∂µ g µ 10 = ∂Q
∞
e –∞
– αQ 2/2
∂µ g 1 dQ = 2α ∂Q
2 – αQ 2/2
Qe
The transition dipole moment is proportional to the dipole derivative. This is true for any normal mode of vibration (i.e. harmonic).
Absorption of infrared radiation leads to vibrational transitions
v=0
Absorption of infrared radiation leads to vibrational transitions
v=1 v=0
The selection rule for vibrational transitions is ∆v = ±1
v=2 v=1 v=0
Two methods for calculating FC The overlap of nuclear wavefunctions in the ground and excited state known as the FC factor can be expressed in a sum-over-states ∞ ∞ 2
FC =
Σ Σ v′ = 0 v = 0
v|v′
or time correlator formalism.
1 FC = π
∞
i|i(t) e – (ωi + ω0)te – Γtdt 0
The sum-over-states method The absorption lineshape is determined by the displacement of the excited state potential energy surface. The FC factors can be calculated analytically in the linear coupling approx-imation (no frequency shift).
FC =
∞
Σ
v=0
S ve – S δ ε – vhω v!
The Franck-Condon factor determines the envelop of the absorption lineshape ∆
∆
∆
S = ∆2/2 S is electron-phonon coupling ∆ is nuclear displacement
Resonance Raman spectrum is obtained by a laser light scattering experiment Detector Lens Sample UV Laser
Spectrograph
Inelastic light scattering produces a frequency shift. There is exchange of energy between the vibrations of the molecule and the incident photon.
Raman scattering Raman scattering is an inelastic light scattering process. In the resonant picture it involves evolution in the excited state so it also depends on the FC factor and the transition dipole moment. On the left a sum-over-states picture is shown.
Kramer-Heisenberg-Dirac Raman scattering depends on the molecular polarizability. The electric field drives the system into excited state |n> and then from |n> it returns to the final state |f>. αρσ
if
1 = Σ – n = 1 ω0 + ωnf + iΓ n ω0 – ωnf – iΓ n h
Two electric field interactions are required. In the above expression ρ and σ are any two of the Cartesian coordinates x, y, or z. The energy differences ωnf include both electronic and vibrational contributions e.g. ωnf = ωeg - nωvib. We can consider both non-resonant and resonant cases for KHD.
Analysis of isotope effects Vibrational spectra are analyzed within the harmonic approximation. Reduced mass m 1m 2 k µ= k ω = µ m1 + m2 m m 1
2
For a diatomic we can see the effect directly m 1m 2′ µ ω ′ k µ′ = = ω′ = ′ µ′ m1 + m2 µ′ ω
The same principle applies to polyatomics But first we must analyze the normal modes.
Normal Modes of Vibration Polyatomic molecules can be considered as a set of coupled harmonic oscillators. Although this is a classical model we shall see that it can used to interpret spectra using the quantum-mechanical harmonic oscillator wave functions. The collective motions of the atoms in a molecule are decomposed into normal modes of vibration within the harmonic approximation. The normal modes are mutually orthogonal. That is they represent linearly independent motions of the nuclei about the center-of-mass of the molecule.
Cartesian equations of motion N 1 T = Σ m i x i 2 + y i 2 + zi 2 2i = 1
V = V0 + +
+
N
Σ
i=1
N
Σ i, j = 1
N
Σ i=1
∂V x + ∂V y + ∂V z ∂x i i ∂y i i ∂zi i
∂ 2V x 2 + ∂ 2 V y 2 + ∂ 2 V z 2 ∂x i2 i ∂y i2 i ∂zi2 i ∂V x y + ∂V y z + ∂V z x ∂x i∂y j i j ∂y i∂z j i j ∂zi∂x j i j
T is the kinetic energy and V is the potential energy expressed as a Taylor’s series in x, y and z.
Mass-weighted coordinates 3N 1 T = Σ η 2i 2i = 1 3N 1 V = Σ a ijη iη j 2 i, j = 1
η 1 = m 1 x 1 , η 2 = m 1 y 1 , η 3 = m 1 z1 , ...η 3N = m N zN a ij =
∂V ∂η i∂η j
The equations of motion are: 3N
η i + Σ a ijη j = 0 j=1
Solutions Trial solutions have the form:
η i = η isin 0
λt + δ
η i = λ η icos
λt + δ
η i = – λη isin
λt + δ
0
0
These give a linear system of coupled equations:
– λη i + 0
3N
a ijη 0j = 0 Σ j=1
Which is equivalent to the matrix equation:
A – λI η 0 = 0 A is the matrix of coefficients. I is the identity matrix. λ are the eigevalues.
Diagonalization of the matrix The general form of the matrix equations is
a 11 a 12 a 13... a 21 a 22 a 23... a 31 a 32 a 33... ...
λ 1 0 0... 0 λ 2 0... – 0 0 λ 3... ...
η 01 η 02 =0 0 η3 ..
There is a trivial solution in which all of the terms in the η0 column vector are zero. The interesting solution, however, is the solution for which the determinant of the matrix |A - λI| is equal to zero. V = 1 η TAη 2
a 11 a 12 a 13... a 21 a 22 a 23... 1 V = η 1 η 2 η 3 ... a a a ... 2 31 32 33 ...
η1 η2 η3 ..
Definition of normal coordinates In fact, the procedure of finding det |A - λI| is a matrix diagonalization of A. To perform this diagonalization we transform to normal coordinates Qi where:
Qi =
3N
l kiη k Σ k=1
In matrix form Q = LTη, which can also be written:
Q1 l 11 l 21 l 31... l 12 l 22 l 32... Q2 = l 13 l 23 l 33... Q3 ... .
η1 η2 η3 ..
The Q are normal coordinates
L is a unitary matrix; its inverse is equal to its transpose L-1 = LT. The matrix L will diagonalize A. L TAL = Λ =
λ 1 0 0... 0 λ 2 0... 0 0 λ 3... ...
Diagonalization of the matrix The eigenvalues are
λ i = ωi = 2πν i
The kinetic energy is: 3N 2 T T T 1 1 1 T = Q L LQ = Q Q = Σ Qi 2 2 2i = 1 3N T T T 1 1 1 V = Q L ALQ = Q ΛQ = Σ λ iQi2 2 2 2i = 1
The uncoupled equations of motion are now represented by
Qi = Qi sin 0
λi t + δ
These solutions represent collective motions of the nuclei. Each Q contains simultaneous displacements of many nuclei.
Internal Coordinates Cartesian coordinates are less convenient than a coordinate system defined in terms of the bonds, angles, etc. of the molecule. Such a coordinate system is called an internal coordinate system. Only motions relative to the center-of-mass are included and thus there are 3N - 6 internal coordinates for a non-linear polyatomic with N atoms. For linear polyatomic molecules there are 3N - 5 internal coordinates. The internal coordinates are Stretch ∆r Bend ∆θ Torsion ∆τ 4 Wag ∆ω
Example: CO2 For example, for CO2 we have the following internal coordinates. ∆r 1 ∆r 2 R= ∆θ 1 ∆θ 2
∆θ1 ∆r2
∆r1
∆θ2
Example: CO2 For example, for CO2 we have the following internal coordinates.
V = 1 k r ∆r 12 + k r ∆r 22 + k θ∆θ 21 + k θ∆θ 22 + k rθ∆r 1∆θ 1 + 2 1 k ∆r ∆θ + k ∆r ∆θ + k ∆r ∆θ + k ∆r ∆r + ... rθ 2 2 rθ 1 2 rr 1 2 2 rθ 2 1 T
2T = η η = R BB T
T –1
T
–1
R=R G R
det FG – λI = 0
Normal modes - CO2
Asymmetric stretch Asymmetricstretch stretch Symmetricstretch stretch Asymmetric Symmetric -1 -1 2349cm cm-1 ν1 2289inactive) cm-1 (infrared 2349 cm νν3ν332349 (IRactive) active) (Raman active) (IR
.
+
.
Bends Bends -1-1 νν2 2667 cm 667 cm
(infrared active) (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.
