Report

Chemistry 2 Lecture 6 Vibrational Spectroscopy Learning outcomes from lecture 5 • Be able to draw the wavefunctions for the first few solutions to the Schrödinger equation for the harmonic oscillator • Be able to calculate the energy separation between the vibrational levels for the harmonic oscillator • Be able to explain the dependence of this separation on the masses of the atoms and the strength of the bond • Be aware of and be able to calculate the zero point energy of a vibrating molecule • Be able to draw the potential energy curve for a real molecule and explain why it is different to the harmonic potential Assumed knowledge Light behaves like an oscillating electromagnetic field. The electric field interacts with charges. Two charges separated in space represent a dipole moment which can interact with an electric field. Energy can only be taken or added to the electric field in units of hn (photons). Revision: Light as a EM field Revision: The Electromagnetic Spectrum Spectral range l (nm) n (Hz) Radio ~1 x 109 ~108 ~0.03 ~10-8 NMR/ESR Microwave ~100,000 ~1012 ~30 ~10-2 Rotational Infrared ~1000 ~1014 ~3,000 ~103 Vibrational Visible 400-750 4-6 x 1014 14,00025,000 1 – 3x105 Electronic Ultraviolet 100-400 ~1015 ~40,000 ~5x105 Electronic <100 >1016 >100,000 >106 Core electronic X ray n~ (cm-1) Energy (kJ/mol) Spectroscopy Revision: Fundamental equations Quantity E hn n c ~ n 1 l n Energy l c Symbol SI Unit Common Unit E J kJ/mol Frequency n s-1 or Hz s-1 or Hz Wavelength l m nm or mm Wavenumber n~ m-1 cm-1 Constant Symbol Value Speed of light c 3.00 x 108 m/s Planck constant h 6.626 x 10-34 Js Classical absorption of light Because light is an oscillating electromagnetic field, it can cause charges to oscillate. If the charge can oscillate in resonance with the field then energy can be absorbed. Alternatively, an oscillating charge can emit radiation with frequency in resonance with the original oscillation. For example, in a TV antenna, the oscillating EM field broadcast by the transmitter causes the electrons in your antenna to oscillate at the same frequency. Classical absorption of light What about a molecule as an antenna? How can we get oscillating charges in a molecule? + - + - Classical absorption of light + - + Rotating a permanent dipole causes an oscillation of charge Classical absorption of light + + - Vibrating a permanent dipole causes an oscillation of charge The Quantum Harmonic Oscillator Solution of Schrödinger equation: n h 2 k 1 n m 2 But: n 1 k 2 m n = 0,1,2,… Oscillation (vibrational) frequency n (n Need to know! 1 2 G (v) (v )hn 1 2 ) S.I. units G(v) & in cm-1 G(v) is the “energy” from the bottom of the well and is the “harmonic frequency” “Anharmonic oscillator” (A.H.O.) harmonic anharmonic Molecules dissociates The Morse Anharmonic Oscillator all in wavenumber (cm-1) units: G (v) = ( G (v) = ( v + 12 ) e v + 12 ) e 2 1 v + 2 ( ) e 2 4 De 2 1 v + 2 e xe ( ) This one you DO have to know how to use In spectroscopy, we tend to use the letter “vee” to indicate the quantum number for vibration. The vibration frequency is indicated by e in cm-1. When solving the general quantum mechanical problem we used the letter n, to minimize confusion with “nu”, the vibrational frequency in s-1. Harmonic frequency G (v) v 1 Anharmonicity constant V (r) The Morse energy levels 2 1 - v 2 e xe 2 e Harmonic term Anharmonic term wexe is usually positive, so vibrational levels get closer together The force constant for the AHO is the same as for the harmonic case: 1 k 1 k or = n 2 c m 2 m r Different levels of approximation… • General AHO: G (v) v 1 • Morse oscillator: G (v) v 1 2 3 1 1 - v 2 e x e v 2 e y e - ... 2 e 2 1 - v 2 e xe 2 e • Harmonic oscillator: G ( v ) v 1 2 The level of approximation to use depends on: i) the information you have ii) the information you need The important equations! • In wavenumber (cm-1) units: G (v) = ( G (v) = ( v + 12 ) e v + 12 ) e 2 1 v + 2 ( ) e2 4 De 2 1 v + 2 e xe ( The ones you DO have to know how to use ) De e 2 4 e x e De e 2 4 e x e V (r) Dissociation energy D0 This is the energy from the bottom of the well to the dissociation limit. De But we know about zero-point energy, therefore slightly less energy is required to break the bond. D0 = De – G(0) We can estimate the bond dissociation energy from spectroscopic measurements! G(0) (Zero-point energy) r Selection rules • All forms of spectroscopy have a set of selection rules that limit the number of allowed transitions. • Selection rules arise from the resonance condition, which may be expressed as a transition dipole moment: μ 21 ( r ) μ ( r ) 1 ( r ) dr 0 * 2 Upper state molecular lower vibrational state coordinate dipole • Selection rules tell us when this integral is zero Harmonic Oscillator Selection Rules selection rule: Dv=±1 Dv=+1: absorption Dv=-1: emission If an oscillator has only one frequency associated with it, then it can only interact with radiation of that frequency. Selection rules limit the number of allowed transitions Thermal population At normal temperatures, only the lowest vibrational state (v =0 ) is usually populated, therefore, only the first transition is typically seen. Transitions arising from v0 are called “hot bands” (Their intensity is strongly temp. dependent) Much of IR spectroscopy can understood from just these two results of the quantum harmonic oscillator: E = (v+½)hn and Dv = ±1 For example: v=10 More problems with harmonic model…. -1 CO 6352 cm 0 .8 (first overtone) -1 4260 cm 1 .0 -1 A bsorbance 1 .2 2143 cm , (fundam ental) CO 1 .4 What are these? 0 .6 (second overtone) 1 .6 0 .4 x 10 0 .2 x 100 0 .0 2000 4000 6000 -1 W a v e n u m b e r (c m ) What are the new peaks? • Three peaks… i. 2143 cm-1 ii. 4260 cm-1 iii. 6352 cm-1 These are almost 1 : 2 : 3 which suggests transitions might be 01 02 03 v=4 v=3 v=2 v=1 v=0 Anharmonic oscillator (A.H.O.) selection rules: There are none! But!... Harmonic and anharmonic models are very similar at low energy, so selection rules of AHO converge on HO as the anharmonicity becomes less: H.O. A.H.O. A.H.O. selection rule: Dv=±1,±2, ±3 Intensity gets weaker and weaker (typically 10× weaker for each) Anharmonic oscillator (A.H.O.) A.H.O. selection rule: Dv= ±1,±2, ±3 Dv = 1 : fundamental Dv = 2 : first overtone Dv = 3 : second overtone, etc Typical Exam Question • Consider the infrared absorption spectrum of CO below. a) From the wavenumber measurements on the spectrum, assign the spectrum, hence determine the harmonic frequency, e (in cm-1) and the anharmonicity constant exe (in cm-1). b) Estimate the bond dissociation, D0, for this molecule. There is no absorption below 2000 cm-1. Fundamental: 2143 = G(1) – G(0), Overtone: 4260 = G(2) – G(0) Using the spectra to get information… G ( v ) v 1 2 e - v e xe 2 1 2 G(1)-G(0) = [(1.5)e – (1.5)2 exe] - [(0.5)e – (0.5)2 exe] 2143 = e – 2exe …(1) G(2)-G(0) = [(2.5)e – (2.5)2 exe] - [(0.5)we – (0.5)2 exe] 4260 = 2e – 6exe …(2) Two simultaneous equations (simple to solve) → e = 2169 cm-1, and exe = 13 cm-1 De e 2 De 4 e x e ( 2169 ) 4 13 2 90 , 500 cm -1 V (r) Using the spectra to get information… D0 De - G (0) 90 ,500 - 1080 89 , 400 cm D0 -1 De 89,400 cm-1 = 1069 kJ/mol c.f. exp. value: 1080 kJ/mol Why the difference? Remember Morse is still an approx. to the true intermolecular potential. Still 2% error is pretty good for just 2 measurements! G(0) (Zero-point energy) r Equations to know how to use… m1 × m2 mm +m 1 2 n 1 k 2 m G ( v ) ( v 1 2 ) G (v) v 1 2 1 - v 2 e xe 2 e D0 De - G (0) De e 2 4 e x e Learning outcomes • Be able to manipulate and use the key equations given in the green box at the end of the lecture. •Utilize the harmonic oscillator and anharmonic oscillator as a model for the energy level structure of a vibrating diatomic molecule. Next lecture • The vibrational spectroscopy of polyatomic molecules. Week 11 homework • Work through the tutorial worksheet and the practice problems at the end of the lectures and check your answers with those available online • Play with the “IR Tutor” in the 3rd floor computer lab and with the online simulations: http://assign3.chem.usyd.edu.au/spectroscopy/index.php Practice Questions Which of the following diatomic molecules will exhibit an infrared spectrum? a) HBr b) H2 c) CO d) I2 2. An unknown diatomic oxide has a harmonic vibrational frequency of ω = 1904 cm−1 and a force constant of 1607 N m −1. Identify the molecule. a) CO b) BrO c) NO d) 13CO 3. As the energy increases, the vibrational level spacing for a harmonic oscillator is: a) increases b) decreases c) stays constant 4. As the energy increases, the vibrational level spacings for a Morse oscillator usually: a) increase b) decrease c) stay constant 5. As the energy increases, the vibrational level spacings for an anharmonic oscillator usually: a) increase b) decrease c) stay constant 1. Practice Questions 6. For a Morse oscillator the observed dissociation energy, D0, is related to the equilibrium vibrational frequency and the anharmonicity by the following expression: a) ωe2/4ωexe b) [ωe2/4ωexe]-G(0) c) (v+½)ωe+(v+½)2ωexe d) (v+½)ωe 7. Which of the following statements about the classical and quantum harmonic oscillator (HO) are true (more than one possible answer here)? a) The classical HO frequency is continuous, whereas the quantum frequency is discrete. b) The classical HO has continuous energy levels, whereas the quantum HO levels are discrete. c) The classical HO depends on the force constant, but the quantum HO does not. d) The classical HO may have zero energy, but the quantum HO may not. e) The classical HO does allow the bond to break, whereas the quantum HO does. Practice Questions 8. Which of the following statements correctly describe features of the quantum Morse oscillator and energy levels associated with it? a) The vibrational energy levels get more closely spaced with increasing v. b) The vibrational energy levels approach a continuum as the dissociation energy is approached. c) The Morse oscillator and HO are nearly the same at very low v d) The Morse potential is steeper than the HO for r < re e) The Morse potential exactly describes the interatomic potential.