Chemical Bonding CHEM 6277-10 Lecture 21 Molecular Vibrational Spectroscopy Lecturer: Hanning Chen, Ph.D. 11/06/2014 Quiz 20 10 minutes Please stop writing when the timer stops ! Vibrational Degrees of Freedom For a system with N atoms, (N > 1) Linear Molecules Non-Linear Molecules Translational 3 3 Rotational 2 3 Vibrational 3N-5 3N-6 Total 3N 3N Nv = 4 Nv = 3 Nv = 1 Geometry Optimization How to find a molecular geometry with the lowest energy? Let us start from an arbitrary geometry denoted by { ! ! ! R1, R2 ,..., RN } step by step ! ! ! E R1, R2 ,..., RN ({ Assumption: { }) { ! ! ! R1', R2' ,..., RN ' ! ! ! > E ' R1', R2' ,..., RN ' ({ ! ! ! R1', R2' ,..., RN ' mandatory condition: } }) } { ! ! ! R1, R2 ,..., RN R2 R1 } 2 atomic coordinates 3 R3 1 is the optimized structure! (I am feeling lucky!) ∂E ({ R}) ∂({ R}) ! ! ! R= R1', R2' ,..., RN ' { = 0 } quasi-Newton Method { Taylor expansion of energy at the initial structure ∂E ( R ) E ( R') = E ( R ) + ∑ ( R − Ri ) ∂Ri i=1 3N ' i 3N 3N R={R} + ∑ ∑ ( R − Ri ) ! Energy gradients evaluated at R ∂E ( R ) Ei = ∂Ri ' i i=1 j=1 ( ∂ E ( R) R − Rj ∂Ri ∂R j ∂ E ( R) Eij = ∂Ri ∂R j R={R} 2 ∂ E ( R) 1 ' + ∑ ( Ri − Ri ) R={R} 2 2 i=1 ∂Ri 2 3N ∂ E ( R) Eii = 2 ∂ Ri R={R} 2 R={R} R={R} second-order derivative 3N 3N E ( R') = E ( R ) + ∑ ( R − Ri )Ei + ∑ ∑ ( R − Ri ) no mandatory condition 2 } 2 ' i i=1 ) ' j first-order derivative 3N ! ! ! ! R1, R2 ,..., RN = R ' i i=1 j≠i ( ) 3N 2 2 1 ' R − R j Eij + ∑ ( Ri − Ri ) Eii 2 i=1 ' j The energy expression is NOT quite useful. Further Partial Differentiation Derivative with respect to the atomic coordinate components: ∂E ( R') ∂E ( R ) = +∑ ∂Ri ' ∂Ri ' i=1 3N ∂( R − Ri ) ' i ∂Ri ' 3N 3N Ei + ∑ ∑ ∂( R − Ri ) ' i ∂Ri ' i=1 j≠i 3N ( ) ( ∂ R − R ( 1 i) ' 2 R j − R j Eij + ∑ Eii 2 i=1 ∂Ri ' ) ' i 3N 2 0 = 0 + Ei + ∑ R − R j Eij + ( R − Ri ) E j≠i ' j 3N ( ' i 2 ii ) Ei = ∑ R j − R Eij j=1 For all coordinate components: Hessian Matrix second derivative ⎛ E11 ⎜ ⎜ E21 ⎜ ... ⎜ ⎜⎝ E(3N )1 E12 ... E1(3N ) E22 ... E2(3N ) ... ... ... ... E(3N )(3N ) E(3N )2 ' j ⎞⎛ R −R 1 1' ⎟⎜ ⎟ ⎜ R2 − R2' ⎟⎜ ... ⎟⎜ ⎟⎠ ⎝ R3N − R3N ' ⎞ ⎛ E1 ⎟ ⎜ ⎟ = ⎜ E2 ⎟ ⎜ ... ⎟ ⎜ ⎠ ⎝ E3N ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Gradient Vector first derivative Feeling Lucky? Taylor expansion is truncated at the second order ! 2 3N 3N 3N ∂E ( R ) ∂ E ( R) ' ' ' E ( R') = E ( R ) + ∑ ( Ri − Ri ) + ∑ ∑ ( Ri − Ri ) ( R j − R j ) R={R} i=1 ∂Ri i=1 j=1 2 ∂ E ( R) 1 ' + ∑ ( Ri − Ri ) R={R} 2 ∂Ri ∂R j 2 i=1 ∂Ri 3N 2 R={R} Energy The potential energy surface is NOT necessarily a perfect harmonic well. global minimum local minimum R Quasi-Newton method does NOT help the system escape a local minimum ! Molecular Mechanics of a Vibrating System one-dimensional 3-atom model system: equilibrium positions: instantaneous positions: Hooke’s Law: F = k(x − x0 ) Newton’s Law: 2 d x F = −m 2 dt 1 2 3 x10 x1 x20 x2 x30 x3 Forces acting on atom 1: F1 = k11 ( x1 − x10 ) + k12 ( x2 − x20 ) + k13 ( x3 − x30 ) Each atom vibrates at a fundamental angular frequency x1 = Asin (ω t ) + x10 2 d x1 2 F1 = −m1 2 = −m1ω ( x1 − x10 ) dt ω : Atomic Forces For atom 1: −m1ω 2 ( x1 − x10 ) = k11 ( x1 − x10 ) + k12 ( x2 − x20 ) + k13 ( x3 − x30 ) mass-weighted displacements: x = m1 ( x1 − x10 ) x = m2 ( x2 − x20 ) x = m3 ( x3 − x30 ) ' 2 ' 1 ' 3 mass-weighted force constants: k = ' 11 k11 m1 m1 k = ' 12 k12 m1 m2 k = ' 13 k13 m1 m3 Relation between the atomic displacements: −ω x = k x + k x + k x 2 ' 1 ' ' 11 1 ' ' 12 2 ' ' 13 3 Matrix Representation For a molecule with N vibrational degrees of freedom: ⎛ mass-weighted ⎜ Hessian ⎜ ⎜ Matrix ⎜ ⎜⎝ ' 11 ' 12 ... k k ' 22 ... ... ' kN1 ... ' kN 2 ... ... k k ' 12 ⎞ ⎛ k ⎟ ⎜ ' k2 N ⎟ ⎜ ⎟⎜ ... ⎟ ⎜ ' kNN ⎟⎠ ⎜⎝ ' 1N ⎞ ⎛ x ⎟ ⎜ ' x2 ⎟ 2⎜ = − ω ⎟ ⎜ ... ⎟ ⎜ ' ⎜⎝ x N ⎟⎠ ' 1 ⎞ x ⎟ ' x2 ⎟ ⎟ ... ⎟ ' x N ⎟⎠ ' 1 Standard Eigenvalue-Eigenfunction Problem Eigenvalues: Normal mode analysis: Eigenfunctions: k = ' ij kij mi m j ω vibrational frequencies {x } = ' i normal mode 1 mi ∂ E m j ∂ri ∂rj 2 energy perturbation atomic displacements Some Examples Water (H2O) molecule: wavenumber −1 1 cm = 3.0 × 10 Hz 10 3657.1 cm -1 1594.7 cm -1 3755.9 cm Carbon Dioxide (CO2) molecule: 1388 cm 677 cm -1 -1 2349 cm 677 cm -1 -1 -1 Infrared Spectroscopy How can we detect the vibrational modes? Infrared: 100 → 10000 wavenumbers Unfortunately, the molecular vibrations are INVISIBLE, because the corresponding wavelengths are longer than RED light. Theory of Quantum Harmonic Oscillator Hamiltonian: 2 ˆ p 1 2 ˆ H= + mω xˆ 2m 2 Wavefunction: ϕ n (x) = Energy levels: 1⎞ ⎛ En = !ω ⎜ n + ⎟ ⎝ 2⎠ 1 4 ⎛ mω ⎞ e ⎜ ⎟ n ⎝ ⎠ π ! 2 n! 1 mω x 2 − 2! mω x Hn( ) ! wavefunction orthogonality: ϕ n ' ϕ n = δ nn ' Principles of Infrared Spectroscopy Fermi’s Golden Rule: transition rate: Ri→ f 2π ˆ = Ψi H ' Ψ f ! 2 ρf coupling energy Born-Oppenheimer Approximation: vibronic wavefunction: Ψ(R, r) = ψ vib (R)ϕ ele ( R, r ) Under light irradiation: ψ vib (R) : vibrational ϕ elec (R) : electronic ! ! ˆ H ' (ω ) = E (ω )ir Separation of vibrational and electronic degrees of freedom: Ri→ f 2π " 2 = ψ i (R) ϕ elec (R, r) r ϕ elec (R, r) ψ f (R) ρ f E ! ! µ (R, r) : dipole moment Dipole Moment Perturbation Taylor expansion of the dipole moment: µ0 ⎛ ∂µ ⎞ µ ( R, r ) = µ ( R0 , r ) + ( R − R0 ) ⎜ ⎟ ⎝ ∂R ⎠ static dipole moment R0 derivate of the dipole moment atomic displacement Transition rate: 2 2 2 2π 2π ⎛ ∂µ ⎞ 2 2 Ri→ f = ψ i (R) µ0 ψ f (R) ρ f E + ⎜⎝ ⎟⎠ ψ i (R) ( R − R0 ) ψ f (R) ρ f E ! ! ∂R δ if Ri→ f 2 2 2π 2 2π ⎛ ∂µ ⎞ 2 2 = µ0 ψ i (R) ψ f (R) ρ f E + ⎜⎝ ⎟⎠ ψ i (R) ( R − R0 ) ψ f (R) ρ f E ! ! ∂R =0 Ri→ f 2 2 2π 2 2 2π ⎛ ∂µ ⎞ 2 2 = µ0 δ if ρ f E + ⎜⎝ ⎟⎠ ψ i (R) ( R − R0 ) ψ f (R) ρ f E ! ! ∂R 2 Vibrational Transition Rate 2 2π ⎛ ∂µ ⎞ 2 = ⎜⎝ ⎟⎠ ψ i (R) ( R − R0 ) ψ f (R) ρ f E ! ∂R 2 Ri→ f Vibrational transition dipole moment: 2 2 2π ⎛ ∂µ ⎞ 2π ⎛ ∂µ ⎞ 2 2 = ⎜⎝ ⎟⎠ ψ i (R) R ψ f (R) ρ f E − ⎜⎝ ⎟⎠ ψ i (R) R 0 ψ f (R) ρ f E ! ∂R ! ∂R 2 Ri→ f 2 2 2π ⎛ ∂µ ⎞ 2π ⎛ ∂µ ⎞ 2 2 2 2 = ⎜⎝ ⎟⎠ ψ i (R) R ψ f (R) ρ f E − ⎜⎝ ⎟⎠ R0 δ if ρ f E ! ∂R ! ∂R 2 Ri→ f µif ⎛ ∂µ ⎞ ⎜⎝ ⎟⎠ : ∂R derivative of electronic dipole moment 2 2π ⎛ ∂µ ⎞ 2 2 = ⎜⎝ ⎟⎠ µif ρ f E ! ∂R 2 Ri→ f =0 µif : vibrational transition dipole moment Selection Rules for Infrared Spectroscopy 1 ⎛ mω ⎞ ψ i (R) R ψ f (R) = n ⎜ ⎟ 2 n! ⎝ π ! ⎠ only when Rule 1: 1 2 e ∫ mω R 2 − ! f = i + 1 or f = i − 1 ΔE = ±!ω mω x mω x Hi ( )RH f ( )dR ! ! ψ i (R) R ψ f (R) ≠ 0 ψ i (R) R ψ f (R) = Vibrational transition can only occur between two neighboring vibrational states ! Rule 2: ⎛ ∂µ ⎞ ≠ 0 ⎜⎝ ⎟⎠ ∂R A normal mode that does not change electronic dipole moment is not IR-detectable! Infrared Active Normal Modes Water (H2O) molecule: IR active IR active IR active Carbon Dioxide (CO2) molecule: IR inactive IR active IR active IR active Raman Spectroscopy Nobel Prize Physics excited λsca λinc 1930 vibrational manifolds ground Rayleigh scattering Intensity: Chandrasekhara. V. Raman Stokes shift anti-Stokes shift I Rayleight ≫ I Stokes > I anti − Stokes Theory of Raman Spectroscopy induced dipole moment: − P + E Pi = α ij E j α ij : molecular polarizability tensor Electromagnetic field: E = E0 cos (ω 0t ) Taylor expansion of the molecular polarizability: ∂α ∂α α = α (R0 ) + dR = α (R0 ) + A cos ω vibt ∂R ∂R Atomic displacements: dR = A cos ω vibt Dipole Moment Fluctuation ∂α ⎛ ⎞ P ( t ) = α (t)E(t) = ⎜ α (R0 ) + A cos ω vibt ⎟ E0 cos (ω 0t ) ⎝ ⎠ ∂R AE0 ∂α P ( t ) = α (R0 )E0 cos (ω 0t ) + ( cos(ω 0 − ω vib )t + cos(ω 0 + ω vib )t ) 2 ∂R Rayleigh scattering Stokes scattering anti-Stokes scattering A vibrating dipole emits electromagnetic fields ! I Raman ⎛ ∂α ⎞ ∝⎜ ⎟ ⎝ ∂R ⎠ 2 Raman selection rule: ∂α ≠0 ∂R Raman Active Normal Modes Water (H2O) molecule: Raman active Raman active Raman active Carbon Dioxide (CO2) molecule: Raman active Raman inactive Raman inactive Raman inactive Review of Homework 20 12.16 State whether each of the following is a group. (a) All the integers (positive, negative and zero) with the rule of combination for forming the product of two elements being addition. (b) All positive integers with the rule of combination being multiplication. (c) All real numbers except zero, with the rule of combination being multiplication. (1) close (2) associative (3) identity (4) inversible (a) A+B=C, (A+B)+C=A+(B+C), I=0, A+(-A)=I=0 → YES (b) A*B=C, (A*B)*C=A*(B*C), I=1, A*(1/A)=I=1 → NO (c) A*B=C, (A*B)*C=A*(B*C), I=1, A*(1/A)=I=1 → YES Homework 21 Reading assignment: Homework assignment: Chapter 15.10, 15.11, 15.12, 15.13 Problems 15.34, 15.39 Homework assignments must be turned in by 5:00 PM, November 7th, Friday to my mailbox in the Department Main Office located at Room 107, Corcoran Hall