Classical Normal Mode Analysis: Harmonic Approximation Each absorption in a vibrational spectrum corresponds to a normal mode. Characteristics of Normal Modes 1. Each normal mode acts like a simple harmonic oscillator. 2. A normal mode is a concerted motion of many atoms. 3. The Center of mass doesn’t move. 4. All atoms pass through their equilibrium positions at the same time. 5. Normal modes are independent; they don’t interact. Carbon Dioxide in Air 40 38 36 34 32 30 28 26 24 Bend: 22 -1 H2O 1595 cm Single Beam 20 18 16 Asymmetric stretch: -1 14 H2O 3756 cm 12 10 Symmetric stretch: Asymmetric stretch: -1 Bend: 8 H O 3652 cm -1 2 CO 2 2349 cm CO 667 cm -1 6 2 4000 3500 3000 2500 2000 1500 1000 500 Wavenumbers (cm-1) 1 dV d2 V V = k x 2 F = - = -kx k = (1) 2 dx dx 2 d2 x F = ma m = -kx (2) dt 2 d2 x x = A sin(2 πνt) = -4π2ν2 x -4π2ν2 m x = -kx (3) dt 2 Coordinates: Atom i: X i, Y i, Z i, Displacements: x i = Xi – Xi,eq , yi = Y i – Yi,eq Calculate the potential energy V(x 1, y 1, z 1, x 2, y 2, z 2, x 3, y 3, z 3,…., x N, y N, z N) ∂2V 11 2 = k 1 2 ∂ x1 xx change of the force on atom 1 in the x-direction when you move atom 1 in the x-direction Colby College ∂2 V 11 1 2 2 = k same atom same direction ∂ y1 yy ∂2V 11 1 2 ∂ ∂ = k same atom different directions x1 y1 xy ∂2V 12 = k different atom same direction 1 2 ∂x1∂x2 xx ∂2V 12 ∂ ∂ = kxy different atom and direction 1 2 x1 y2 change of the force on atom 1 in the x-direction when you move atom 2 in the y-direction π2ν2 11 11 11 12 12 1N -4 m1x1= -kxx x1 - kxy y1 - kxz z1 - kxxx2 - kxy y2 -…- kxz zN (4) π2ν2 11 11 11 12 12 1N -4 m1y1= -kyx x1 - kyy y1 - kyz z1 - kyx x2 - kyy y2 -…- kyz zN : π2ν2 21 21 21 22 22 2N -4 m2x2= -kxx x1 - kxy y1 - kxz z1 - kxx x2 - kxy y2 -…- kxz zN : N1 N1 N1 N2 N2 NN -4π2ν2 m z = -k x - k y - k z - k x - k y -…- k z N N zx 1 zy 1 zx 1 zx 2 zy 2 zz N Total of 3Nx3N terms on the right. For example: triatomic, only allow vibration along x-axis: O1 C2 O3 x π2ν2 11 12 13 -4 m1x1= -kxx x1 - kxx x2 - kxx x3 (5) 21 22 23 -4π2ν2 m x = -k x - k x -k x (6) 2 2 xx 1 xx 2 xx 3 31 32 33 -4π2ν2 m x = -k x - k x - k x (7) 3 3 xx 1 xx 2 xx 3 Use simultaneous equation solver, or….. Convert to mass weighted coordinates and mass weighted force constants: 12 k ~ ~ ~12 xx x1 = m1 x1 x2 = m2 x2 k = (8) xx m1 m2 Colby College then in the new coordinates: ~11 ~12 ~13 π2ν2 ~ ~ ~ ~ -4 x1 = - kxx x1 - kxx x2 - kxx x3 (9) ~21 ~22 ~23 π2ν2 ~ ~ ~ ~ -4 x2 = - kxx x1 - kxx x2 - kxx x3 (10) ~31 ~32 ~33 π2ν2 ~ ~ ~ ~ -4 x3 = - kxx x1 - kxx x2 - kxx x3 (11) 11 12 13 k k k 2 2 xx xx xx -4π ν m1 x1 = - m1 x1 - m2 x2 - m3 x3 (12) m1 m1 m1 m2 m1 m3 Mass weighted force constants and mass weighted displacements: 11 12 13 kxx kxx kxx m1 m1 m1 m2 m1 m3 ~ ~ x1 x1 21 22 23 k k k ~ 2 2 ~ xx xx xx = 4π ν (13) - x2 - x2 m2 m1 m2 m2 m2 m3 ~ ~ 31 32 33 x3 x3 kxx kxx kxx m3 m1 m3 m2 m3 m3 gives a symmetric matrix. This is an eigenvalue-eigenvector equation. The eigenvalues are the squared normal mode frequencies. The eigenvectors are the mass weighted normal coordinate displacements: ~ ~ ~ xi yi zi xi = sin(2 πνt) yi = sin(2 πνt) zi = sin(2 πνt) mi mi mi 1 k k Units: ν = or 4 π2ν2 = with k in N m -1, µ in kg molecule -1 2π m m ~ν 1 ν c ν~ To convert to wavenumbers: = λ or = λ = c with ν~ in cm -1, c in cm s -1, and m in g mol -1 : π2 2~ν2 4 c k ~2 k/m = or ν = -5 1000 g/kg N A m 5.8921x10 Carbon Dioxide Stretches: Sym 1340 cm -1 Asym 2349 cm -1 Colby College 11 33 12 23 By symmetry : k = k k = k xx xx xx xx 11 Guesses: k =1600 N m -1 xx 12 11 22 11 -1 13 kxx = -kxx kxx = 2 k xx = 3200 N m kxx = 0 O1 C2 O3 1600 1600 - 0 16 16 16 12 O1 -100 115.47 0 1600 3200 1600 C - - = 115.47 -266.67 115.47 2 - 12 16 12 12 12 16 0 115.47 -100 O 1600 1600 3 0 - 16 12 16 16 ≈ Eigenvector 1: E=0.00175039 0 ------------- Eigenvector 2: E= -100 Symmetric stretch: -0.707107 ~ 100 –1 ν = -5 = 1303 cm 0 5.892x10 0.707107 ------------- Eigenvector 3: E= -366.662 Asymmetric stretch: -0.369279 ~ 366.66 –1 ν = -5 = 2495 cm 0.852799 5.892x10 -0.369279 (for about 5% errors) Valence Force Field For m1 = m 3 q = r – r 1 12 o m q2= r23 – ro r12 2 r23 δ = θ – θo m1 m3 1 2 1 2 2 θ V = k q + k q + k δ δ 2 1 1 2 1 2 θ 2 2 2m 1 2 o k1 4π νasym = 1 + sin (14) m2 2 m1 θ θ 2 2 2 2m 1 2 o k1 2 2m 1 2 o kδ 4π (νsym + νbnd ) = 1 + cos + 1 + sin 2 (15) m2 2 m1 m1 m2 2 ro 4 2 2 2m 1 k1 kδ 16 π (νsym νbnd ) = 2 1 + 2 2 (16) m2 m1 ro G. Herzberg, “Molecular Spectra and Molecular Structure II. Infrared and Raman Spectra of Polyatomic Molecules,” Van Nostrand, Princeton, N. J., 1945. Colby College .