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Lecture 17 Computational Spectroscopy Predicting real observables ©2013, Jordan, Schmidt & Kable Lecture 17 Copyright Notice Some images used in these lectures are taken, with permission, from “Physical Chemistry”, T. Engel and P. Reid, (Pearson, Sydney, 2006); denoted “ER” throughout the lectures and other sources as indicated, in accordance with the Australian copyright regulations. ©2013, Jordan, Schmidt & Kable Lecture 17 Learning outcomes 17.1 Calculating the geometry tells you something about the rotational spectrum 17.2 Calculating force constants allows calculation of the normal modes of vibration 17.3 HF overestimates frequencies due to incorrect dissociation 17.4 DFT can get band positions pretty well by multiplication by a factor of 0.97 17.5 The normal modes of vibration can be mixtures of the various coordinates 17.6 Most accurate way is to solve Schrödinger equation on an accurate potential energy surface! ©2013, Jordan, Schmidt & Kable Lecture 17 How to be chemically accurate? For chemistry, we would like energies to be accurate to within about 4 kJ/mol. So we need to use a very good method, but it would be very expensive to do everything with configuration interaction at the complete basis set limit. Geometry optimizations and frequency calculations (for zero point energies) take a lot of effort, and cannot be done with high levels of theory. We could calculate geometries at one level, and energies at another, e.g. CCSD/6-31G(d)//B3LYP/6-31G(d). Here we get the density function theory geometry and then use a so-called coupled-cluster calculation to get the energy. But, smart people (Curtiss, Pople) have developed combinations of methods which end up being chemically accurate, with more modest cost. ©2013, Jordan, Schmidt & Kable Lecture 17 Level of correlation (# of configurations) The Quantum Chemistry Landscape The answer pointless Sensible compromise Hartree-Fock limit 1st year chemistry Quality of basis set ©2013, Jordan, Schmidt & Kable Lecture 17 Level of correlation (# of configurations) The Quantum Chemistry Landscape pointless QCISD/6-311G(d,p) DE(QCI) The answer Diffuse DE(+) MP4/6-311G(d,p)// MP2/6-31G(d) more polarizations DE(2df) MP2/6-31G(d) Hartree-Fock limit 1st year chemistry Quality of basis set ©2013, Jordan, Schmidt & Kable Lecture 17 Level of correlation (# of configurations) G1 Theory pointless QCISD/6-311G(d,p) DE(QCI) DE(+) The answer DE(2df) G1 (chemically accurate) MP4/6-311G(d,p)// MP2/6-31G(d) The corrections are added to approximate performing QCISD calculation with large basis MP2/6-31G(d) Hartree-Fock limit 1st year chemistry Quality of basis set ©2013, Jordan, Schmidt & Kable Lecture 17 Predicting Molecular Spectra Spectra represent transitions between energy levels of a molecule. Generally speaking, we can separate motions due to translations, rotations and vibrations, with electronic transitions being separated again. Etrans<Erot<Evib<Eelec Generally, we do not consider quantized translational energy levels. Only in exceptional circumstances would one bother (e.g. confined molecule). If we know the equilibrium structure of a molecule, we can predict the rotational energy levels. ©2013, Jordan, Schmidt & Kable Lecture 17 Predicting Rotational Spectra If we know the equilibrium structure of a molecule, we can predict the rotational energy levels. This is easy for a diatomic. The energy levels are given by ( ) E = BJ J +1 Where J is the angular momentum quantum number and B is the rotational constant. This energy level expression comes from solving the Schrödinger equation. m is the reduced mass. B= 2 2m R2 ©2013, Jordan, Schmidt & Kable m1m2 m= m1 + m2 Lecture 17 Predicting Rotational Spectra – carbon monoxide OK, so I need to know R, the equilibrium bond length. Do an energy minimization. Do MP2/6-311G(d) 1 ***** EQUILIBRIUM GEOMETRY LOCATED ***** comment (A single descriptive title card) COORDINATES OF SYMMETRY UNIQUE ATOMS (ANGS) ATOM CHARGE X Y Z -----------------------------------------------------------O 8.0 0.0000000000 0.0000000000 -0.0190160600 C 6.0 0.0000000000 0.0000000000 1.1190160600 COORDINATES OF ALL ATOMS ARE (ANGS) ATOM CHARGE X Y Z -----------------------------------------------------------O 8.0 0.0000000000 0.0000000000 -0.0190160600 C 6.0 0.0000000000 0.0000000000 1.1190160600 INTERNUCLEAR DISTANCES (ANGS.) -----------------------------O 1 O 2 C C 0.0000000 1.1380321 * 1.1380321 * 0.0000000 ©2013, Jordan, Schmidt & Kable m1m2 12 ´16 m= = = 6.857amu m1 + m2 12 +16 B= 2 2m R2 = 3.772 ´10-23 J = 1.899cm-1 R Lecture 17 Predicting Rotational Spectra – carbon monoxide OK, so I need to know R, the equilibrium bond length. Do an energy minimization. -1 B = 1.899cm = 56.9 GHz 2B = 114 GHz So, there should be radiation at about 114 GHz, and multiples thereof emanating from CO in interstellar space. In this case, we actually have an experimental transition of 115 GHz. Let’s go and look for CO in space… ©2013, Jordan, Schmidt & Kable ( ) E = BJ J +1 6B J=3 Transitions with DJ=±1 4B J=2 2B J=1 0 J=0 Lecture 17 Predicting Rotational Spectra – carbon monoxide ©2013, Jordan, Schmidt & Kable 2B J=1 0 J=0 Lecture 17 CO maps the universe ©2013, Jordan, Schmidt & Kable Lecture 17 Spectroscopy in the Daily Mail June 2011 ©2013, Jordan, Schmidt & Kable Lecture 17 Predicting Rotational Spectra – larger species For larger species we have three rotational constants, A, B and C which are obtained by finding the principle inertial axes of the molecule. These axes are such that the moment of inertia tensor is diagonal. c E ωT Iω m y 2 z 2 I mxy mxz mxy mx z myz 2 2 mxz myz 2 2 m x y b a ©2013, Jordan, Schmidt & Kable Lecture 17 Vibrational Spectra For diatomics, there is only one vibrational mode, so calculation of the vibrational energy levels is simple. Harmonic Approximation: The oscillator is assumed to be parabolic, and the energy levels are given by 1 2 V k R Re 2 E e v 1 / 2 1 e 2c k m v=3 7we/2 v=2 5we/2 v=1 3we/2 we/2 ©2013, Jordan, Schmidt & Kable v=0 Re Lecture 17 Vibrational Spectra – carbon dioxide We can calculate the force constant, k, by looking for the second derivative of the potential energy curve. E e v 1 / 2 -112.56 1 e 2c -112.60 m Nm-1 m = 6.857 amu e = 2510 cm-1 RHF/6-311G(d) -112.58 -112.62 Energy (hartree) k = 2544 k -112.64 -112.66 -112.68 -112.70 -112.72 -112.74 -112.76 -112.78 0.8 experiment is 2169 cm-1 ©2013, Jordan, Schmidt & Kable 1.0 1.2 1.4 C-O distance (Å) Lecture 17 Vibrational Spectra – carbon dioxide Even when we do the most careful differentiation, Hartree-Fock is giving a frequency that is quite high. DFT does better. Method e (cm-1) RHF/6-311G(d) 2435 B3LYP/6-311G(d) 2210 Experiment 2169 The reason that Hartree-Fock overestimates the frequency is due to neglect of correlation. HF cannot describe dissociation, so the well is too steep, and the frequency is too high. If you use HF frequencies, better think about scaling by about 0.89! ©2013, Jordan, Schmidt & Kable Lecture 17 Vibrational Spectra – carbon dioxide Normally, we want to compare to band positions, which also need to account for anharmonicity which is missing in the harmonic approximation. As it turns out, if you multiply DFT frequencies by the magic number 0.97, you get great results. J=3 0.97×2210 = 2143cm-1. v=1 R P J=2 J=1 J=0 J=3 v=0 J=2 J=1 J=0 ©2013, Jordan, Schmidt & Kable Lecture 17 How do complicated molecules vibrate? More complicated molecules than diatomics have many vibrational modes (3N-6 for a non-linear molecule). q1 q2 In this orientation, we can see that if we roll a ball vertically or horizontally, these vibrations will repeat forever. These are the normal modes of the molecule. ©2013, Jordan, Schmidt & Kable Lecture 17 How do complicated molecules vibrate? How about in this orientation? q1 q2 Here, the coordinates q1 and q2 are not the normal modes. Vibrating in q1 will cause movements in q2. ©2013, Jordan, Schmidt & Kable Lecture 17 How do complicated molecules vibrate? The normal modes are a rotation of the original coordinates. q1 q’1 q’2 q2 The mathematical procedure which calculates the normal modes of the system is known as normal mode analysis. ©2013, Jordan, Schmidt & Kable Lecture 17 E.g. Carbon dioxide The normal modes of CO2 are not the individual CO stretches, but linear combinations which are called symmetric and antisymmetric (also asymmetric) ©2013, Jordan, Schmidt & Kable Lecture 17 Doing the job properly To calculate spectra accurately, you just can’t beat solving the nuclear Schrödinger equation on the various electronic potential energy surfaces. How many quanta of the “normal modes” are in each wavefunction? ©2013, Jordan, Schmidt & Kable Lecture 17 Crazy 2d vibrational wavefunctions ©2013, Jordan, Schmidt & Kable Lecture 17