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TAMU CHEM 327 - PPt5x-327

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Chapter 5 – MOLECULAR SPECTROSCOPYWhen the Born-Oppenheimer approximation is applied, the electronic motion can be separated from the nuclear motion; Once ψeland E(R) have been determined as a function of R, the nuclear motion may be treated;where[])R(Eψ)R(ψE(R)Tnucnucnucrr=+)r(R,E(R)ψ)r(R,ψReHelel2elrr=⎥⎦⎤⎢⎣⎡+)r,R(EΨ)r,RΨ(ReHT2elnucrrrr=⎟⎠⎞⎜⎝⎛++The Schrödinger equation for H2+⎟⎟⎠⎞⎜⎜⎝⎛+−∇−=BeAe22e2elr1r1e)r(2mHrhTransitions in moleculesElectronic: ΔE ~ 5 eVλ ~ 250 nm~ 40,000 cm-1Vibrational: ΔE ~ 0.2 eVλ ~ 6.2 μm~ 1600 cm-1Rotational: ΔE ~ 4x10-4eVλ ~ 3 mm~ 3 cm-1UV-visibleInfra-redMicrowaveννν1cm1wavenumber1cνλ1ν −===Schematic diagram showing the relative ordering of electronic,vibrational, and rotational energy levelsJ111091v’=0v’=1vibrationalenergy levelsrotationalenergy levelsv’=2excited electronic energy levelJ111091v=0v=1v=2ground state electronic energy levelrotationalenergy levelsvibrationalenergy levelsThe electromagnetic spectrumROY G BIVwavelength (m)radiowavesmicro-wavesinfraredvisibleultra-violetx rayswavenumber (103cm−1)103110−310−610−91214161820 22 24 26Transition rates and selection rulesThree general types of transitions:stimulatedabsorptionstimulatedemissionspontaneousemissionRadiation fieldrequiredStimulated transitions:E(t)The time dependent potential energy acting on a moleculeIs primarily associated with its dipole moment;tcos2πE(t)E0νrr=t),r(Eμ-t),rV(rrrr⋅=∑−=iirμrrewhere μ is the dipole moment;= transition dipole moment integralFor a transition from state m to state n, the transition rate is2mnmn2μ)ρ(ν32πλh=Where ρ(νmn) is the time averaged energy density at the transition frequency, and∫=τψμψμdmnmnr*mnTotal transition rates and Einstein coefficientsmnThe total rate of transition m→nisRmn= λmnNm= NmBmnρ(νmn) – stimulated absorptionThe total rate of transition n→misRnm= λnmNn= NnBnmρ(νnm) – stimulated emissionplus the rate of spontaneous emission from m→n;R’nm= NnAnmAt equilibrium, Rmn= Rnm+ R’mnwhich results in Bnm= BmnandAnm= [(8πhν3nm)/c3]Bnm.2mn2mnμ32πBh=whereThe selection rules for a transition are determined by the transitionDipole moment integral. Selection rules for rotational and vibrational transitions= transition dipole moment integral∫=τψμψμdmnmnr*mnRotational transitions:a). Molecule must have a permanent dipole momentb). ΔJ = ±1c). ΔM = 0, ±1Vibrational transitions:a). Dipole moment must change as molecule vibratesb). Δv = ±1O C OSymmetric stretchO C OAsymmetric stretchO C OBending(one in xz plane &one in yz plane)zVibrations of the CO2moleculeMolecule Vibrational Rotational O2CO2H2OCH4Exercise: Indicate whether the following molecules display vibrationaland rotational spectra.No. Symmetric stretching does not change the dipole moment.No. Homonuclear diatomic molecules do not have permanent dipole moments.Yes, for the asymmetric stretch and bending vibrations.No. CO2is a linear symmetric molecule and so its dipole moment is zero .Yes.Yes.Yes. No.Types of rotorsRotational spectra may be classified according to the symmetry of the Molecule as defined below.TypeSymmetry ExamplesSpherical rotor Ix= Iy=IzCH4, SF6Symmetric rotor Ix= Iy≠ IzNH3Linear rotor Ix= Iy; Iz= 0 CO2Asymmetric rotor Ix ≠ Iy≠ IzH2OExercise: Show that the moment of inertia for a diatomic molecule reduces to I = μR2, where μ is the reduced mass.Conclusion: The rotation of a diatomic molecule is equivalent to a particle of mass μ moving on a sphere of radius R.Energy levels of a rigid rotor2I1)J(JE2roth+=Define the rotational constant B such that ,with B in units of cm−1. Then2IhcB2h=1)hcBJ(JErot+=The wavenumber of a rotational transition ishcEEνLUULJJJ,J−=Define the term value F(J):Then1)BJ(JhcEF(J)J+==)F(J)F(JνLUJ,JUL−=JUJLνhcΔE =Rotational energy level diagram for a linear rotorJ=4J=3J=2J=1J=0E=J(J+1)hcB20hcB12hcB6hcB2hcB0F(J)=E/hcΔFν=20B12B6B2B08B6B4B2BJ=0-1 J=1-2 J=2-3 J=3-42B2B4B6B8Bwave numberIntensitySpectral line spacing2B2BRotational spectrum of a linear rotor01122378wavenumberI = μRe2= 1.4495x10−46kg-m2EJ= J(J + 1)hcBExercise: Given that Re= 1.12832 Å and μ = 1.1385x10−26kg for CO,calculate (a) the energies and (b) the relative populations of the first eight rotational levels at 298 K. Then (c) determine the wavenumbersof the transitions between them.12222cm1.9307Ic8πhB,I8πh2IhcB−====h12Jg;egNNΔFνJ)/kTE-(EJ0J0J+===−JLEJ(10−4eV) ν (cm−1)gJNJ / N000 1 13.861 4.78 3 2.9457.722 14.35 5 4.72811.58 3 28.70 7 6.26015.454 47.84 9 7.47019.315 71.76 11 8.31823.176 100.46 13 8.79127.037 133.95 15 8.90430.898 172.22 17 8.69434.75Rotational Raman spectraincident photonscattered photonmoleculeMolecule is excited – Stokes process(photon energy decreases)Molecule is de-excited – anti-Stokes process(photon energy increases)Selection rules: a). Molecule must be anisotropically polarizable.+−+−≠b). ΔJ = ± 2μ = αERotational Raman spectrum of a linear rotor0→22→0νS= νinc− [F(JU) − F(JL)]νAS= νinc+ [F(JU) − F(JL)]Ground state potential energy diagram of HClMorse potentialHarmonic oscillatorpotentialMorse potential:2)Ra(Re]e[1DV(R)e−−−=a = [k/(2De)]1/2Exercise: Neglecting the second anharmonic term in the expressionfor Evib, calculate the following for CO at 298 K:a). The wavenumber for the fundamental transition from v = 1 to v = 0.b). The relative population of the v = 1 state.Given: = 2169.8 cm−1and Xe= 6.124x10−3.eνSelected vibrational group wavenumbers (in cm−1)Groupwavenumber Group wavenumberO−H stretch 3425 C=O stretch 1775N−H stretch 3350 C=C stretch 1650C−H stretch 2950 C−C stretch 1200C−H bend 1465 C−Cl stretch 700Vibration–rotation spectrav, Jv+1, J−1v+1, J+1v+1, JR-branch(ΔJ=1)P-branch(ΔJ=−1)Q-branch(ΔJ=0)A high resolution vibration-rotation spectrum of HClP-branchR-branch1−0 0−12−1 1−2ν = νvib+ [F(JU) − F(JL)] = νvib+ BUJU(JU + 1) − BLJL(JL+1)B = Be− αe(v + ½)Exercise: Calculate the wavenumber of the R-branch vibration-rotationtransition from v = 0, J = 1 in CO. Take into account anharmonicity and vibration-rotation interaction. Given: = 2169.8 cm−1, Xe= 6.124x10−3,Be= 1.9313 cm−1, and αe= 1.7504x10−2 cm−1.eνRaman vibration-rotation spectrav, Jv+1, J−2v+1, J+2v+1, JO-branch(ΔJ=−2)Q-branch(ΔJ=0)S-branch(ΔJ=2)Only Stokes (excitation) transitions


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