GE 140a 2019 Lecture 7 ISOTOPIC EFFECTS ON ELECTRON AND NUCLEAR MOTIONS IN ATOMS AND MOLECULES

1. GE 140a 2019 Lecture 7 ISOTOPIC EFFECTS ON ELECTRON AND NUCLEAR MOTIONS IN ATOMS AND MOLECULES

2. Review: Plank’s postulate E = nhv Energy of a ‘microscopic’ oscillator Frequency of oscillation Constant Cardinal number Max Planck Put forward as an assumption regarding the accessible vibrational energies in the oscillators (interatomic bonds) present in a solid body, in an effort to explain the black-body emission spectra of hot objects

3. Review: Modeling the emission spectrum as e- ‘jumps’ between shells The Rydberg formula: • l = wavelength emitted; R = Rydberg constant; Z = nuclear charge; nf and ni are integer quantum numbers of the final and initial ‘energy levels’ • Each Rydberg ‘constant’ is specific to each nuclide (element or isotope)

4. Isotope effects on electronic emission lines The energy associated with an atomic electron orbital: En = (-1/2) Z2a2 µ c2(1/n2)  Where Z = nuclear charge (+1 for hydrogen), a is the fine-structure constant, c is the speed of light, n is the quantum energy level of the bound electron, and µ is the reduced mass of the nucleus-electron system: µ = memN/(me + mN) Where me is the electron mass and mN is the nuclear mass For the hydrogen atom, me = 9.10939E-31, mN = 1.67262E-27 and µ = 9.10443E-31 Note En is measured relative to the free electron, and so is negative for a bound electron and rises (toward zero) as n goes up.

5. Notice the relationship to the Rydberg equation = n/c = ∆E/ch Ei = (-1/2) Z2a2 µ c2(1/ni2)  Ef = (-1/2) Z2a2 µ c2(1/nf2)  ∆E = Ef - Ei =(-1/2)Z2a2µc2 (1/nf2–1/ni2) So R = -(a2µc)/2h i.e., one of the reasons why the Rydberg constant is specific to each nuclide is that they differ in reduced masses of their various nucleus-electron orbits. I.e., the energies of electron transitions are influenced by isotopic mass.

6. Working through the isotope effect on the Balmer emission lines For the Hydrogen atom, me = 9.10939E-31, mN = 1.67262E-27 and µ = 9.10443E-31 For the Deuterium atom, me = 9.10939E-31, mN = 3.34358E-27 and µ = 9.10691E-31 All other terms being constant, each Ei term, and thus each n, in deuterium is shifted by a factor of 1.000272 (1 part in 3675) Thus the Balmer ‘red’ line (H-a) is shifted from 656.45377 nm wavelength for hydrogen to 656.27520 nm for deuterium —a small but observable shift This was the method Urey first used to discovery deuterium

7. Review: Taking these concepts to molecular motions The Fraunhofer lines Caused by absorption of sunlight by cool gas in outer solar atmosphere Most of the classic ones are electronic transitions and are relatively insensitive to mass (like the Balmerlines or less). But the long wavelength ones (IR) correspond to molecular vibrations (esp., H2O) The fact that they are discrete and sharp has same importance for understanding molecular vibrational energies as the H emission spectrum had for understanding electron orbital energies.

8. The physics of molecular absorption Absorption of high energy photons excites electronic energies Excited state Light couples (is absorbed by and excites) the molecule when the energy/frequency of a photon overlaps that of the electrons or nuclear motions (the ‘resonance principle’) …or by exciting nuclear (interatomic) motions (heat) Energy Ground state Absorption of low energy photons that overlap the frequencies of vibrations and rotations Releases of excess electronic energy as light (fluorescence) The highest energy electron orbitals are ~10 eV – in the UV part of the electromagnetic spectrum The lowest energy molecular motions are ~0.1 eV – in the IR part of the electromagnetic spectrum

9. Vibrational absorption lines have a connection to the Debye-Einstein models of phonons in solids Specific heats of solids are fit by models that postulate the chemical energies of those materials are contained in a finite number of interatomic lattice vibrations, or ‘phonons’, which follow Plank’s rule (E = nhv) Snapshots of lattice deformations corresponding to the N = 1, 2, 3 states of a ‘phonon’

10. An important question about this process drew the attention of chemical physicists ca. 1910-1930 ~5-10 eV Atomic and molecular electronic transitions Clearly best described with quantum mechanics, and isotopic mass influences energy (e.g., Balmer lines) ~0.1-5 eV Molecular vibrations (and ‘ro-vibrations’) Energy Are these classical or quantum mechanical? Plank’s black-body model and Debye/Einstein theory say ‘yes’; lots of other people at the time said ‘no’ ~0.05 eV Atomic and molecular translations Seem best described with classical physics, which says isotopic mass changes velocity, but not energy

11. A mechanical model for the kinematics of chemical bonds Chemical bonds are (sort of) like springs The ‘springiness’ of vibration/bending of chemical bonds can be approximated as an harmonic oscillator: F = -kx • In this case the ‘spring’ is the potential energy well created by the electronic structure • If the electronic energies are independent of the motions of the nuclear masses (the ‘Born-Oppenheimer approximation’) then we can assume the spring has the same k in all isotopic versions of a given molecule. This is wrong but a useful starting point. Potential energy ~1/2kx2

12. Behaviors of harmonic oscillators k m 1/2 1/2 k µ 1 2π 1 2π Frequency of an ‘anchored’ spring with a test mass: n = n = The ‘two-body’ problem can be analyzed with the same expression by replacing m with the ‘reduced mass’, which deals with the fact that the oscillation involves both bodies moving relative to the center of mass. m1m2 m1+m2 µ = • k is the spring constant (in Kg.s-2) on the bond. Intra-molecular bonds generally have spring constants of ca. 10’s to 100 Kg.s-2. Intermolecular bonds (e.g., van der Waal’s forces) generally have spring constants of ca. 1-10 Kg.s-2). A slinky is 1 Kg.s-2 • n is the frequency in s-1; note spectroscopic data for n may be represented in cm-1

13. What about modes of motion that involve relative movements of 3 or more atoms? Smimj Smk for all i-j bonds for all k atoms that move µ = Consider fundamental ‘modes’ (independent motions) of CO2 These two ’degenerate’ – equivalent motions oriented along orthogonal spatial dimensions. Degenerate modes are both real, but you won’t see them as separate in an absorption spectrum. Their frequencies are the same. 16.16 16+16 (16.12)+(16.12) 2.16+12 for symmetric stretch of CO2 µ = = for other modes

14. How many fundamental vibrational modes do molecules possess? For a molecule with n atoms: • 3n-5 independent fundamental modes for linear polyatomic molecules • 3n-6 for non-linear polyatomic molecules CO: 1 CO2: 4 (2 bending modes are degenerate) C2H2: 7 (2 sets of 2 are degenerate) Bigger molecules are rarely truly linear H2O: 3 (no degeneracies) CH4: 9 (2 sets of triply degenerate; 1 set of doubly degenerate) C6H6 (benzene): 30 (11 sets of doubly degenerate)

15. Note measured absorption spectra can look much more complex Absorption spectrum of CO2 Fundamental modes of CO2 All in wavenumbers, or cycles per cm-1 1 wavenumber = n/c, where c is in cm/s 2349 (asymmetric stretch) 1388 (symmetric stretch) 667x2 (bending – 2 are degenerate) Messiness arises from rotation, rotation/vibration coupling, overtones, combination modes; note also each line will have a width that depends on T and P. If you are a spectroscopist this is all terribly interesting; from our perspective it is just a complicating detail that arises in the determination of fundamental modes.

16. Our first-order model for the kinematics of molecular vibrations is straighforward, but what does this say about vibrational energies? Should I model this as a ‘classical’ spring? Or as a ’quantum mechanical’ spring? Potential energy = 1/2kx2 Vibrational energies are quantized and take on values that are multiples of fundamental mode frequencies Vibration involves an exchange of potential and kinetic energy over cycle Isotopic mass changes frequency of oscillation, but has no effect on the potential or kinetic energy Isotopic mass changes µ, n and thus E

17. This discussion was the motivation for studying vapor pressure isotope effects • A purely chemical, thermodynamic isotope effect arising from the change in free energy of the liquid with isotopic substitution (at fixed T, P, gravitational potential energy, etc.) • Confirms ~10-15 year speculation that vibrational energies should be treated as we do electron orbitals in a ‘Bohr-like’ atom: as quantized, following Plank’s law 1930-32: Vapor pressure isotope effects observed, T (K) 20 100 25 50 36Ar 40Ar liquid ice 20Ne 22Ne liquid acond.-vapor ice T-1

18. Our core description of quantum mechanical molecular vibrations • A quantized spring has access only to discrete energy levels, and can never reach the exact minimum. This is a result of the uncertainty principle (location and momentum cannot be simultaneously known). • The lowest accessible state is the ‘zero point’ energy, which is half of one of Plank’s hypothesized energy levels Ei = (ni+1/2)hn Energy of each state: Where Ei is the energy of quantum state ni = 0, 1, 2, 3…, h is Plank’s constant and n is the vibration frequency of the harmonic oscillator.

19. Our core description of quantum mechanical molecular vibrations All members of a population of molecules will be in quantum state 0 when temperature is absolute 0; at higher temperatures, the population will contain members having a range of quantum states, with a distribution described by the Maxwell-Boltzmann law: exp(-Ei/kT) Sjexp(-Ei/kT) Ni N = Where Niis the number of molecules in quantum state i, N is the total number of molecules k is the Boltzmann constant and T is temperature (Kelvin)

20. Note that if I increase reduced mass, frequency drops and the energy of each accessible state will decrease: µ µ’ µ µ’ n’ n 1/2 1/2 (n - n’) = n(1 - ) = Ei - Ei’ = [(ni+1/2)h].(n - n’) For order-of-magnitude estimates or very low temperature systems, we can ignore the high-energy states and simply calculate the difference in energy between the ni= 0 states, called the ‘Zero Point Energies’, or ZPE’s. ZPE - ZPE’ = 1/2h(n - n’)

21. How do I sum energies over a population (and consider terms beyond vibration)? The partition function (Q) (where degenerate states get counted multiple times) Q = Snie-Ei/kT where ni = the number of particles in the population of interest that have energy Ei. So, Q is a weighted sum of the energies of all molecules in a population Components to the partition functions of gaseous molecules Vibration Translation Rotation Qtot = Qtrans x Qrot x Qvib Quantum theory predicts isotope effects for all of these energies

22. The translational partition function Volume of system Debroglie wavelength (2πMkT)3/2 h3 Qtrans = = V Translation The distribution of possible locations at a given momentum imposed by the uncertainty principle. This will return at the end of term as a key term in ‘quantum tunneling’ fractionations. Where M is molecular weight, k is Boltzmann’s constant, h is Plank’s constant, and V is the system volume. If we only care about the change in partition function on isotopic substitution, this simplifies: 3/2 Qtrans’ Qtrans M’ M = Note classical translational energies of gas particles depend on temperature but not mass whereas quantum mechanical translational energies depend on mass. This is one of those ‘science is hard’ times of this class. I won’t (and can’t) explain it well, but I encourage you to enter your Darth-Vader meditation room and contemplate it.

23. The rotational partition function Skipping all extraneous derivation steps… Where s is the symmetry number and I the moment of inertia (this term is more complex for tri-atomic and higher-order molecules) Qrot’ Qrot sI’ s’I Similarly: = Don’t spend a lot of time contemplating this; the difficult part (estimating I) will disappear before we get much further

24. The vibrational partition function Qvib = Snie-Ei/kT Where each Ei is the vibrational energy of one member of the population having a frequency nand a Boltzmann-like distribution of states, ni’s. A solution to this summation and canceling of terms shared by isotopic forms leads to: Mostly the ZPE contribution Mostly contributions from higher-energy quantum states of each vibration Qvib’ Qvib e-Ui’/2 1-e-Ui e-Ui/2 1-e-Ui’ ∏ = Where Ui = hni/kbT for each vibrational fundamental mode, i

25. Here’s a bit of the maths behind the vibrational term (or all modes j and quantum energy levels n: ∏ Qvib= Se-Ej,n/kT j n Ei = (ni+1/2)hn (hnj/kT) = Uj ∏ e-5Uj/2 e-3Uj/2 + Qvib= + [ e-Uj/2 … ] (series expansion of sum) j ∏ [ e-Uj e-2Uj + Qvib= 1 + e-Uj/2 … ] j 1 + x+ x2… = 1/(1+x) Qvib = ∏ e-Uj/2/(1-e-Uj) j Qvib’ Qvib e-Uj’/21-e-Uj e-Uj/2 1-e-Uj’ ∏ = j

26. A serious complication An anharmonic classical system: the freely swinging pendulum An anharmonic description of a quantized vibrational mode Bonds are not really harmonic oscillators, so our model for how frequency, mass and energy interrelate all contains an error that is only a decent approximation at low temperatures, when most bonds are in the n=0,1 vibrational states, where the energy ‘well’ is pretty close to parabolic. This can be addressed with ‘anharmonicity corrections’, though these are usually as inaccurate as the original problem. ‘Path integral’ first-principle methods are a better solution, but beyond our portfolio here. Go talk to Tom Miller.

27. Putting it all together Qtot = Qtrans x Qrot x Qvib Combining terms using the Teller-Redlich spectrometric theorem, and simplifying: 3r/2 s s’ Q’ Q ni’ ni m’ m e-Ui’/2 1-e-Ui e-Ui/2 1-e-Ui’ = ∏ Where m is the mass of the isotope exchanged (not the whole molecule), and r is the number of atoms of the element of interest in the molecule (e.g., 1 for O in CO; 2 for O in CO2) The Teller-Redlich spectroscopic theorem is above my pay grade; go find a proper scientist if you want to learn how it did this.

28. An example: The 18O/16O partition function ratio of CO at 0˚ C CO: asymmetric dimer; one stretching vibration with w = 2140.8 cm-1 3r/2 s s’ Q’ Q ni’ ni m’ m e-Ui’/2 1-e-Ui e-Ui/2 1-e-Ui’ = ∏ Where Ui = hni/kT for each vibrational mode, i • Get all frequencies of all modes into units of s-1 • e.g., w = 2140.8 cm-1 x c (3x1010cm/s) gives n= 2.141x1013s-1 • Calculate frequencies of heavy isotopologues using harmonic approximation, e.g. 12x17.9991 12+17.9991 12x15.9949 12+15.9949 for 12C18O µ’= for 12C16O µ = µ µ’ n’ n 1/2 =

29. Keep chugging… 3r/2 s s’ Q’ Q ni’ ni m’ m e-Ui’/2 1-e-Ui e-Ui/2 1-e-Ui’ = ∏ Where Ui = hni/kbT for each vibrational mode, i • Evaluate s/s’and r • s/s’=1 (both isotopic species are asymetric ‘dumbells’) • r = 1 for CO • 4. Pull together data for m’s, h, kb and choose a T • m’/m ~ 17.9991/15.9949; T chosen to be 273 K • Calculate U’s • If you got n’s in right units, these should always be between ~1 and ~10 • 6. Plug’n play …

30. Finishing off the calculation… 1.1937 0.9758 1.000004 3r/2 s s’ Q’ Q ni’ ni m’ m e-Ui’/2 1-e-Ui e-Ui/2 1-e-Ui’ = ∏ CO: 1 1.3349 1.1459 This exercise shows that something about the energetics, and therefore stability and reactivity, of molecules changes on isotopic substitution, but how does it relate to familiar statements of chemical energies given to us by classical thermodynamics? This is actually a deep problem because Q values are the weighted sums of ‘ensembles’ of microstates that differ from one another in energy, whereas free energies, enthalpies, heat capacities, etc. are bulk properties of what are presumed to be homogeneous populations. Some useful points of contact between these approaches: Atom–basis expressions Molar–basis expressions Gi = -kBTln(Q) µi = dGi/dn = µi0+RTln(a) Note kB = R/NA Thus, Q is like an activity, and Q’/Q relates to the free energy change on isotopic substitution as: ∆G’ - ∆G = -kTln(Q’/Q)

31. Primary controls on Q’/Q values 3r/2 s s’ Q’ Q ni’ ni m’ m e-Ui’/2 1-e-Ui e-Ui/2 1-e-Ui’ = ∏ Ui = hni/kT µ µ’ n’ n 1/2 = • As isotopic mass ratios approach 1, Q’/Q approaches 1 (chemical isotope effects are strong for low m and weak for high m) • As reduced mass ratios approach 1, n’/n, U’/U and thus Q’/Q approach 1 (isotope effects are strong when the reduced mass of the oscillator changes a lot on isotopic substution; this is the case when all masses are low, and when the substituted atom is lighter than what it’s bound to) • U values fall and Q’/Q approaches 1 as temperature rises. We will explore this in more mathematical depth when we discuss isotope thermometers

32. How do I do this in condensed materials? There are alternate paths to the partition function for solids and liquids, but a way that follows simply from our discussion so far is to make sure you adjust the fundamental modes for intramolecular motions to reflect those in the condensed material, and add to your list of fundamental modes those that correspond to molecule-molecule motions or ’lattice modes’ This is actually tricky to do well by spectroscopic observation; often estimates come from first-principles molecular dynamics models.

33. What about our ‘special property’ descriptions of isotopic distribution? • Site preference: reduced masses of molecular vibrations are sensitive to the site of substitution. Thus, Q’/Q ratios are intrinsically site-specific • Mass laws: There is an explicit implication of this theory for the differences in Q’/Q values for rare isotopes of the same element having different masses (through the m’/m and µ’/µ terms). We’ll explore this more in lecture 8. • ’Clumping’: There is an explicit implication of this theory for the differences in Q’/Q values between singly and multiply substituted isotopologues (through the µ/µ’ terms). We’ll explore this more in lecture 10.