A Spectroscopy Primer

An Introduction to Atomic, Rotational, Vibrational, Raman,

Electronic, Photoelectron and NMR Spectroscopy

Robert J. Le Roy

Department of Chemistry, University of Waterloo Waterloo, Ontario N2L 3G1, Canada

c Robert J. Le Roy, 2003-2011

i ii Preface

Spectroscopy is the scientist’s window on the molecular world. As molecules are too small to be seen directly by the human eye, we rely on their interaction with light (or electromagnetic radiation), to determine their properties, how they are formed from their constituent atoms, and how they react. Interaction with light probes the molecules’ characteristic rotational and vibrational motions, and we then attempt to explain that behaviour in terms of theoretical models. This allows us to determine what atoms a particular molecule is composed of, the length and strength of its bonds, and more generally, the patterns in which atoms assemble to form the diverse and myriad molecules on which we rely for industrial applications, for modern drugs, and for life itself. These notes begin by examining the interaction between light and matter as predicted by models derived from quantum mechanics, and by outlining the principles of spectroscopy. We then study the types of spectra associated with several different regions of the electromagnetic spectrum, and see that absorption or emission of light in those distinct regions tends to be associated with different types of molecular motion. We will see how the structures of molecules found both in interstellar space and closer to home can be determined with rotational (or microwave) spectroscopy. We will then see how vibrational (or infrared) and Raman spectroscopy may be used to determine the strengths of bonds and to identify characteristic groups of atoms within molecules. We will also see that the electronic energy binding atoms together in molecules and molecular ions can be studied using electronic and photoelectron spectroscopy. Finally, we will see how nuclear magnetic resonance spectroscopy uses the magnetic properties of atomic nuclei within molecules to learn about the structures of complex molecules, such as proteins, and to image tissues within human beings. The credit for these notes rests not only with the author, but also with several colleagues who provided important input. In particular, Professor John Hepburn developed the first offerings of this material at the University of Waterloo, while Professor William Power’s renovated version of his course notes were a key template for the current document. In addition, Professors Fred McCourt and Peter Bernath have freely offered valuable and abundant suggestions and criticism throughout the development process, while Drs. Iain McNab and John Ogilvie have provided helpful comments and suggestions on early drafts of the manuscript. I am also particularly grateful to Professor Fred McCourt for a thorough, critical reading of the current version of this document. Finally, the curiosity and bafflement of several classes of Chemistry 129 and 209 students stimulated many revisions and improvements that now grace these pages. All surviving errors are, of course, mine. I invite all readers to help to improve these notes further by suggesting changes that they feel might be helpful.

Robert Le Roy Waterloo, August 2011

iii iv PREFACE Contents

Preface iii

Contents vii

List of Figures x

List of Tables xi

List of Symbols xiii

1 Light, Quantization, Atoms and Spectroscopy 1 1.1LightandtheElectromagneticSpectrum...... 1 1.1.1 Physics in 1900 ...... 1 1.1.2 WavePropertiesofLight...... 2 1.1.3 TheQuantumTheoryofLight...... 3 1.1.4 ABriefNoteonUnits...... 7 1.2QuantumTheoryofMatter...... 8 1.2.1 TheSpectrumoftheHydrogenAtom...... 8 1.2.2 TheBohrTheoryoftheAtom...... 9 1.2.3 deBroglieWavelengths...... 11 1.3 Wave Mechanics and the Schr¨odingerEquation...... 12 1.3.1 AParticleinaOne-DimensionalBox...... 12 1.3.2 Orbital or Rotational Motion: A Particle on a Ring ...... 17 1.4ElectronicStructureofAtomsandMolecules...... 18 1.4.1 HydrogenicAtomicOrbitals...... 18 1.4.2 Multi-ElectronAtomsandAtomicSpectroscopy...... 21 1.4.3 Molecular Energies and the Born-Oppenheimer Approximation ...... 22 1.5Spectroscopyatlast...... 24 1.6Problems...... 25

2 Rotational Spectroscopy 27 2.1ClassicalDescriptionofMolecularRotation...... 27 2.1.1 Why Does Light Cause Rotational Transitions? ...... 27 2.1.2 RelativeMotionandtheReducedMass...... 28 2.1.3 MotionofaRotatingBody...... 29 2.2QuantumMechanicsofMolecularRotation...... 31 2.2.1 TheBasics...... 31 2.2.2 EnergyLevels,SelectionRules,andTransitionEnergies...... 32 2.2.3 IllustrativeApplications...... 33 2.3Complications!...... 35 2.4DegeneraciesandIntensities...... 39 2.5RotationalSpectraofPolyatomicMolecules...... 42

v vi CONTENTS

2.5.1 LinearMoleculesare(Relatively)EasytoTreat!...... 42 2.5.2 IllustrativeApplications...... 44 2.5.3 Non-LinearPolyatomicMoleculesareMoreDiﬃcult...... 45 2.6ConcludingRemarks...... 47 2.7Problems...... 49

3 Vibrational Spectroscopy 51 3.1ClassicalDescriptionofMolecularVibrations...... 51 3.1.1 WhyDoesLightCauseVibrationalTransitions?...... 51 3.1.2 TheCentreofMassandRelativeMotion...... 52 3.1.3 The Classical Harmonic Oscillator ...... 53 3.2 Quantum Mechanics of the Harmonic Oscillator ...... 54 3.3 Anharmonic Vibrations and the Morse Oscillator ...... 57 3.3.1 EigenvaluesandPropertiesoftheMorsePotential...... 57 3.3.2 OvertonesandHotBands...... 59 3.3.3 Higher-Order Anharmonicity and the Dunham Expansion ...... 60 3.4DissociationEnergiesandBirge-SponerPlots...... 62 3.5VibrationsinPolyatomicMolecules...... 64 3.6RotationalStructureinVibrationalSpectraofDiatomics...... 67 3.7WhyAreVibrationalLevelSpacingssoLarge?...... 71 3.8Problems...... 72

4 Raman Spectroscopy 75 4.1“Light-As-A-Wave”DescriptionofRamanScattering...... 75 4.2“Light-As-A-Particle”DescriptionofRamanScattering...... 78 4.3RotationalRamanSpectra...... 79 4.4VibrationalRamanSpectra...... 80 4.5RamanSpectraofPolyatomicMolecules...... 81 4.6Problems...... 82

5 Electronic Spectroscopy 83 5.1WhyDoesLightCauseElectronicTransitions?...... 83 5.2Vibrational-RotationalStructureinElectronicSpectra...... 84 5.3VibrationalPropensityRulesinElectronicTransitions...... 88 5.4Problems...... 93

6 Photoelectron Spectroscopy 95 6.1 Photoelectron Spectroscopy: ThePhotoelectricEﬀectRevisited...... 95 6.2Koopmans’Theorem...... 97 6.3VibrationalFineStructureinPhotoelectronSpectra...... 97 6.4MolecularOrbitalsandPhotoelectronSpectra...... 100 6.5SomeComplicationsinPhotoelectronSpectra...... 103 6.6X-RayPhotoelectronSpectroscopy(XPS)...... 105 6.7AugerElectronSpectroscopy(AES)...... 107 6.8Problems...... 108

7NMRSpectroscopy 111 7.1BasicsofNMRSpectroscopy...... 111 7.1.1 AngularMomentumandNuclearSpin...... 111 7.1.2 Magnetic Moments and Nuclei in a Magnetic Field ...... 112 7.1.3 NMRSpectra...... 113 7.2ChemicalShifts...... 117 CONTENTS vii

7.2.1 ElectronicShieldingofNucleiand‘ChemicalShifts’...... 117 7.2.2 What Determines Chemical Shifts, and The Chemical Shift Scale ...... 117 7.2.3 WorkingWithChemicalShifts...... 119 7.3Spin-SpinCoupling...... 120 7.3.1 Basics:CouplingfromaSingleNeighbour ...... 120 7.3.2 ‘Equivalence’, and Coupling from Multiple Equivalent Nuclei ...... 121 7.3.3 Spin-Spin Coupling to More Than One Type of Neighbour ...... 123 7.4MolecularStructuresfromNMRSpectra...... 124 7.5Problems...... 125 viii CONTENTS List of Figures

1.1 The electric field of light oscillates in space and in time ...... 3 1.2 Black-body radiation: observed distributions and the Rayleigh-Jeans law prediction . . . . . 4 1.3 The photoelectric effect: A.Theexperiment; B.Theobservations...... 5 1.4Thehydrogenatomemissionspectrum...... 9 1.5Hydrogenatomenergylevelsandtransitions...... 10 1.6Propertiesofthreesquare-wellparticle-in-a-boxsystems...... 14 1.7 Level energies and wave functions for four particle-in-a-box systems ...... 16 1.8Particle-on-a-ringorbitsandwavefunctions...... 18 1.9Definitionofsphericalpolarcoordinates...... 19 1.10 Effective radial wave functions of H atom orbitals for n =1− 4 ...... 20 1.11 Angular behaviour of H atom orbitals for n =1− 3 ...... 20 1.12Atomicorbitalenergiesinsomemany-electronatoms...... 21 1.13 Potential energy curves for the ten lowest energy electronic states of Li2 ...... 23 1.14 Regions of the electromagnetic spectrum and associated types of spectroscopy ...... 24

2.1Componentofthedipoleﬁeldofarotatingpolardiatomicmolecule...... 28 2.2Centreofmassandrelativecoordinatesforatwo-bodysystem...... 28 2.3Rotationalenergiesandlevelspacingsforalinearrigidrotor...... 32 2.4MicrowaveabsorptionspectrumofCOgas...... 33 2.5MicrowaveemissionspectrumofgaseousHF...... 35 2.6 Microwave absorption spectrum of H–C≡C–C≡C–C≡N ...... 36 2.7 Graphical determination of B0 and D0 forCO...... 38 2.8 Angular momentum projections for J =2...... 40 2.9 Boltzmann rotational population distribution for CO at T =293K...... 41 2.10Fourtypesoflinearmolecules...... 42 2.11 Structure of H2O...... 46 2.12 Rotational moments of inertia for some symmetric non-linear polyatomic molecules ...... 47 2.13 Gas phase molecular structure of azulene determined from rotational spectroscopy ...... 48 2.14 The rotational emission spectrum of cyanodiacetylene in Sagittarius B2 ...... 48

3.1 Dipole moment of a vibrating polar diatomic molecule ...... 52 3.2 Vibrational modes of CO2 ...... 52 3.3 Harmonic oscillator eigenvalues and wavefunctions for three model systems ...... 55 3.4 Vibrational levels and transitions of a Morse potential energy function ...... 58 3.5VibrationalextrapolationsandtheBirge-Sponerplot...... 63 3.6 Vibrational normal modes of: A.waterH2Oand B. acetylene C2H2 ...... 65 3.7RoomtemperatureabsorptionspectrumofDCl...... 67 3.8Spectrumandenergylevelpatternforaninfraredband...... 68 3.9NaClemissionspectrumshowingbandheads...... 70 3.10 High temperature (1800 K) emission spectrum of GeO ...... 71

ix x LIST OF FIGURES

4.1 Oscillating induced dipole moment of a rotating non-polar molecule ...... 76 4.2 Oscillating induced dipole moment of a vibrating non-polar molecule ...... 77 4.3 Incident ν0 and scattered νs lightinRamanscattering...... 78 4.4 Schematic illustration of rotational and vibrational Raman spectra ...... 81

5.1 Schematic illustration of rotational and vibrational structure in electronic spectroscopy . . . . 84 5.2VibrationalbandsintheelectronicspectrumofSrS...... 85 5.3 Rotational structure near the (0,0) band head in the A 1Σ+ − X 1Σ+ spectrumofCuD.... 87 5.4 Deﬁnition of the “stationary point” for a particular (v,v) electronic transition...... 89 3 1 + 2 + − 5.5 Potential curves and turning points for the Br B( Π0u ) X( Σg )system...... 90 5.6 Classical prediction for the time a vibrating molecule spends at a particular radius ...... 91 5.7 Dependence of vibrational band intensities on the relative radial positions of upper- and lower- statepotentialenergyfunctions...... 92

6.1 Molecular orbital level-energy picture of the photoionization of H2 ...... 96 + 6.2 He(I) photoelectron spectrum of H2 and the associated H2 and H2 potential energy curves . 98 6.3 Molecular orbital diagram and He(I) photoelectron spectrum for HCl ...... 100 6.4 Molecular orbital diagram and He(I) photoelectron spectrum for N2 ...... 102 6.5 He(I) photoelectron spectrum and molecular orbital diagram for O2 ...... 104 6.6XPSspectraforCatomsindiﬀerentmolecularenvironments...... 106 6.7 Schematic level energy diagram illustrating XPS core-hole decay ...... 107 6.8He(I)photoelectronspectrumofCO...... 109

3 7.1 Space quantization of angular momentum P for R=2 ...... 112 1 7.2 Nuclear spin energy levels in a magnetic field for I = 2 ...... 114 7.3 NMR spectra of various atomic nuclei in 250 MHz and 600 MHz spectrometers...... 115 7.4 1HNMRspectrumofethanol...... 116 7.5 Chemical shifts of 1Hand13Cnucleiinvariousenvironments...... 118 7.6 Energy level diagram for a proton without and with coupling to a neighbouring proton of a differenttype...... 120 7.7 Simulated NMR spectrum for neighbouring single H atoms A and B with different chemical shifts...... 121 7.8 Energy level diagram for a nucleus without and with coupling to two identical neighbouring nuclei...... 122 7.9 Simulated NMR spectrum for proton A interacting with two equivalent neighbouring protons B...... 122 7.10 Pascal’s triangle and the calculation of weights for split NMR peaks...... 123 7.11 1H NMR spectra of ethyl chloride, n-propyl iodide, iso-propyl iodide and tert-butyl alcohol . 124 7.1260MHzNMRspectrumofanunknownorganiccompound...... 126 List of Tables

1.1Conversionfactorsforenergyunitsencounteredinspectroscopy...... 7 1.2Wavefunctionsofhydrogenicorbitals...... 19

2.1PredictedandobservedmicrowavespectrumofCO...... 34 2.2 Experimental microwave transition energies for ground state (v =0)CO...... 37 2.3 Interstellar molecules detected by their rotationalspectra...... 49

3.1 Characteristic group vibrational energiesν ˜i for common chemical function groups ...... 66

4.1Labelsforvarioustypesofrotationaltransitions...... 79

+ + 6.1 Molecular parameters for some states of HCl, HCl ,N2 and N2 ...... 101 + 6.2 Molecular parameters for some states of O2 and O2 ...... 103 6.3 Compare calculated orbital binding energies and experimental ionization energies for CO . . . 105 6.4 Ionization energies of the 1s electronsoftheﬁrst-rowelements...... 105 6.5 Nitrogen 1s chemical shifts relative to the core ionization energy for N(1s) in gaseous N2 ...106

7.1TableoftheNMRpropertiesofvariousnuclearisotopes...... 113

xi xii LIST OF TABLES List of Symbols

Fundamental Constants

The current values of some of the more commonly used molecular constants are listed below. They are taken from a more comprehensive list reported in the Reviews of Modern Physics,Vol. 80, pp. 663–730 (2008), and can also be found online in the web page of the United States National Institute for Standards and Technology: http://physics.NIST.gov/cuu/Constants. That NIST web page will also present updated values of these constants, as they become available.

Speed of Light c =2.997 924 58×108 ms−1 Planck Constant h =6.626 068 96×10−34 joules s = h/2π =1.054 571 628×10−34 joules s −18 −1 Rydberg Constant R∞ =2.179 871 97×10 joules = 109 737.315 685 27 cm −18 −1 RH =2.178 685 42×10 joules = 109 677.583 41 cm −23 −1 −1 −1 Boltzmann Constant kB =1.380 650 4×10 joules K =0.695 035 6 cm K Molar Gas Constant R =8.314 472 joules mol−1 K−1 23 Avogadro Number NA =6.022 141 79×10 −11 Bohr Radius a0 =5.291 772 0859×10 m=0.529 177 208 59 A˚ −27 1 12 Atomic mass unit 1 u =1mu =1.660 538 782×10 kg ≡ 12 m( C) −27 Mass of the Proton mp =1.672 621 637×10 kg = 1.007 276 466 77 u −31 −4 Mass of Electron me =9.109 382 15×10 kg=5.485 799 0943×10 u Fundamental Charge e =1.602 176 487×10−19 coulombs −1 2 Inertial Constant Cu =16.857 629 u cm A˚

ν frequency (hertz, Hz) λ wavelength (metre, m) ν˜ wavenumber (cm−1) E energy(J=joule=kgm2 s−2,eV,cm−1) E energy in cm−1 F rotational energy in cm−1 Measures G vibrational energy in cm−1 p momentum (kg m s−1) v speed (m s−1) L angular momentum (kg m s−1) F force (newton, N = kg m s−2 or J m−1); Fin spectroscopists’ units (cm−1 A˚−1) T temperature (Kelvin, K)

xiii xiv LIST OF SYMBOLS

n principal orbital angular momentum m magnetic J rotational v vibrational Quantum Numbers S total electronic spin angular momentum

mS z-component of S I total nuclear spin angular momentum

mI z-component of I

V (r) molecular potential energy De dissociation energy re equilibrium bond length k, k˜ bond force constant in J/m2,cm−1/A˚2 μ reduced mass β Morse potential exponent I moment of inertia −1 Bv rotational constant in cm −1 Dv centrifugal distortion constant in cm νe vibrational frequency in Hz −1 ωe vibrational frequency in wavenumbers, cm −1 Molecular Parameters ωexe vibrational anharmonicity, in cm −1 Be inertial rotational constant in wavenumbers, cm α molecular polarizability M dipole moment IE ionization energy

Ke− photoelectron kinetic energy εorbital molecular orbital energy γ nuclear magnetogyric ratio σ chemical shielding constant δ chemical shift JAB nuclear spin-spin coupling constant

Atomic Masses Many of the numerical calculations encountered in doing problems associated with this course involve the use of atomic masses. A complete listing of the masses of all the stable isotopes of all elements may be found online from the NIST web page http://physics.NIST.gov/PhysRefData/Compositions/ as well as in §1 (on pp. 1-15 to 1-18 of the 82nd edition) of the Handbook of Chemistry and Physics. If by some strange oversight you do not own one yourself, copies of recent editions of this tome may be found in the reference section of any signiﬁcant library. From the University of Waterloo, the current edition of this invaluable sourcebook is available on-line at http://www.hbcpnetbase.com/. Chapter 1

Light, Quantization, Atoms and Spectroscopy

What Is It? Spectroscopy is the branch of science that uses light to probe the properties of atoms, molecules and materials. The discrete frequencies (or colours, or energies) of light absorbed or emitted by particular species tell us about the spacings between the quantized energy levels that are associated with various internal motions of the system. In atoms these are motions of the electrons, and the transitions tell us about changes in the electronic conﬁguration. In molecules the internal motions also include rotation, vibration and the orientations of the nuclear spins.

HowDoWeDoIt? We measure the discrete frequencies and intensities of the electromagnetic radiation that is absorbed, emitted or scattered by atoms or molecules.

WhyDoWeDoIt? By interpreting the observed spectra in terms of quantum-mechanical models for the distribution of discrete level energies of a molecule, we learn about its structure and properties, including the lengths and strengths of its bonds, the identity and relative positions of its constituent atoms, the energies and symmetries of its molecular orbitals, and the overall behaviour of the molecule. The observed patterns of level energies, combined with the mathematical tools of statistical thermodynamics, allow us to predict thermodynamic properties (enthalpy, entropy, heat capacity, etc.) of molecular gases, and equilibrium constants for chemical reactions. The spectrum of a given species also provides an absolutely unique molecular ﬁngerprint, an essential tool for environmental and analytical chemistry, and for astrophysics. The observed spectral transition intensities tell us about the populations of various chemical species in a given system, and more microscopically, about the relative populations of individual molecular energy levels. The “selection rules” that govern which levels may be coupled by allowed transitions tell us about the symmetry of the molecular electronic wave functions, and hence help us identify particular molecular states. More generally, spectroscopy is our most exquisite probe of the properties of matter. It stimulated the development of quantum mechanics, the theory that underlies our current understanding of the nature of atomic-scale matter, and it provides a powerful means for testing basic theories.

1.1 Light and the Electromagnetic Spectrum

1.1.1 Physics in 1900

At the beginning of the past century, the underlying basis of our modern view of the nature of the physical world had been discovered.

1 2 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

• The atomic theory was known. It was understood that all matter was made up of particles called atoms, and their approximate sizes and masses were known.

• The electron was known to be a small particle, and its mass and charge were fairly accurately known.

• The kinetic theory of gases, which explains the macroscopic properties of gaseous systems in terms of the classical motions and collisions of a myriad of tiny atomic or molecular particles was known, and it successfully explained the empirically determined ideal gas law PV = nRT .

• The Periodic Table of the elements had been developed empirically, although the reason for the peri- odicity, the fact that atoms in a given column of the table have similar chemical properties, was not yet understood. [This had to await the development of quantum mechanics!]

• Light was known to be a wave phenomenon. Virtually all properties of light – how it reflects from a plane surface, how it is refracted (changes direction) at an interface between two media (e.g., at the air-water interface), how its colours are dispersed on passing through a prism or reflecting from a ruled grating, and interference effects – were explained by Maxwell’s theory of electromagnetic radiation and the ordinary wave theory that governs water waves and the transmission of sound.

However, a handful of troubling observations could not be explained in terms of the existing world view, and in a few years this led to a revolution in our understanding of matter at the atomic and molecular scale.

1.1.2 Wave Properties of Light Electromagnetic radiation (or light) consists of mutually perpendicular, oscillating electric and magnetic fields traveling through space at a finite speed. The electric and magnetic fields are perpendicular both to one another and to the direction of propagation of the light. As with any wave phenomenon, its nature may be characterized by any one of three inter-related properties that provide a quantitative measure of what we commonly call the “colour” of light:

• Frequency (symbol: ν) – number of oscillations per unit time (units: cycles s−1 = hertz, Hz)

• Wavelength (symbol: λ) – distance spanned by one cycle (units: meters, m, or more commonly nanome- ters nm, where 1 nm = 10−9 m)

• Wavenumber (symbol:ν ˜) – number of cycles in 1 cm, or inverse of the wavelength in cm (units: cm−1)

As with all other wave phenomena, the product of the frequency and the wavelength is the speed of the traveling wave. In the case of light traveling in a vacuum, this speed is a fundamental constant of nature, and has the value c = νλ =2.997 924 58×108 ms−1 . (1.1) Note that frequency and wavelength are both expressed in SI units, while the wavenumber is not. This point requires particular notice, since when converting between frequency and wavelength using Eq. (1.1), one normally expresses wavelength in units m (or more commonly nm), and in this conversion the value of c should have units m s−1. In contrast, for conversions involving the wavenumber of light,ν ˜ [cm−1],

1 ν ν˜ = = , (1.2) 102 λ 102 c afactorof102 is required to convert the units of wavelength from m to cm. This mixture of units is unfortunate, but it is unavoidable because of the widespread use of the wavenumber in cm−1 to characterize light. Wavenumbers are not diﬃcult to work with, but one must be careful about the units. We see later that both the frequency and wavenumber of light are directly proportional to its energy (in J), and in common usage we often describe amounts of energy in terms of the associated value of the frequency or wavenumber. Although light consists of both oscillating electric and magnetic ﬁelds, the fact that molecules consist of negatively charged electrons distributed about positively charged nuclei makes them particularly susceptible 1.1. LIGHT AND THE ELECTROMAGNETIC SPECTRUM 3

oscillation period 1/ν + E0 field viewed at E(x,t) fixed point 0 in space

− E0 0 2 4 time/106 -15 s 8 10

wavelength λ E + 0 field E(x,t) viewed at fixed point 0 in time

− E0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 distance / μm Figure 1.1: The electric field of light oscillates in space and in time. to the effects of the electric fields. For plane polarized light traveling in the x direction, the the solution of Maxwell’s equations show that the oscillation of its electric field along the axis of polarization is described by the expression 2πν E(x, t)=E0 cos x − 2πν t + φ0 , (1.3) c 14 −1 in which t represents time and φ0 is a constant phase factor. For light of frequency ν =3×10 s , Fig. 1.1 shows how the electric field at a given point in space oscillates in time, and how the electric field at a given instant of time oscillates as a function of distance along the direction of propagation. For this case use of Eqs. (1.1) and (1.2) shows that λ = 999.308 nm andν ˜ = 10 007 cm−1. Visible light is only a small part of the entire range or spectrum of electromagnetic radiation, so we classify electromagnetic radiation in terms of its frequency or wavelength. The visible portion of the spectrum runs roughly from 400 to 700 nm. However, the electromagnetic spectrum that we use in spectroscopy stretches over 15 orders of magnitude, from large wavelengths of hundreds of meters (radio frequency waves or “rf”) to the very small wavelengths associated with γ rays. Some comments on the full electromagnetic spectrum are given at the end of this chapter.

1.1.3 The Quantum Theory of Light What was wrong with the wave theory of light? Although Maxwell’s classical electromagnetic wave theory of light explained many observations with great accuracy, two troubling experiments resolutely resisted explanation. Explaining them earned Max Planck and Albert Einstein Nobel Prizes in physics in 1918 and 1921, respectively, and gave birth to the theory we now call quantum mechanics.

Max Planck and ‘black-body’ radiation It had long been observed that a solid hot object emits light whose intensity distribution as a function of wavelength (or frequency, or colour) is independent of the nature of the hot material, and depends only on the temperature. This phenomenon is called black-body radiation, and it is associated with a host of familiar phenomena such as ﬁre, heating elements in an oven, tungsten ﬁlaments in incandescent light bulbs, and stars, including our sun. By the late 19th century those intensity distributions had been carefully measured 4 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

13 ν / Hz 14 15 10 10 10

T = 15 000 K 20 Rayleigh-Jeans law (15 000 K) Intensity Ι(ν,T) / 10-7 J m-2

10 T = 6 000 K ( × 10 ) 5 T = 300 K ( × 104 ) 0 infrared visible ultraviolet Figure 1.2: Black-body radiation: observed distributions and the Rayleigh-Jeans law prediction.

(solid curves in Fig. 1.2), but the best available theory, the Rayleigh-Jeans law, predicted that the observed intensity per unit frequency was given by the expression

2πkT I(ν, T )= ν2 . (1.4) c2 As shown by the dashed curve in Fig. 1.2, this prediction sharply disagrees with experiment. In particular, although it is accurate at low frequencies, it predicts that the intensity distribution would increase to infinity as a function of frequency. This feature of the classical description was termed the ultraviolet catastrophe, because its predictions implied dire consequences for all forms of life in the universe if black-body radiation did indeed behave that way. Fortunately for us, it was well known that the radiation from hot objects behaves differently, with an intensity distribution function I(ν, T ) that passes through a maximum whose position and magnitude depend on temperature, and then dies off at higher frequencies, as shown by the solid curves in Fig. 1.2. A key assumption of the classical theory was that the energy could be emitted or absorbed by the hot object in increments of any possible size. However, in 1900 Max Planck showed that if, instead, one assumed that the energy could only be emitted or absorbed in finite increments or “quanta” whose size ε depended linearly on the frequency of the light according to the expression

ε = ε(ν)=hν = hc102 ν˜ (1.5) in which h is a tiny scaling factor, that same derivation gave the distribution law

2πhν3 1 I(ν, T )= . (1.6) c2 ehν/kT − 1 This function has the correct qualitative behaviour shown by the solid curves in Fig. 1.2: it increases as ν2 at small frequencies, passes through a maximum, and dies oﬀ exponentially at high frequencies. Planck also showed that if his scaling factor was given the value h =6.626×10−34 J/Hz (now called the Planck constant),1 Eq. (1.6) yielded essentially exact agreement with experiment! This result was truly remarkable, but for a number of years many people (initially including Planck himself) were reluctant to accept the full implications of the quantization postulate of Eq. (1.5). In particular, although people were compelled to accept the results of his derivation, since the agreement with experiment was so good, many refused to accept the hypothesis of “quantization” on which it was based, and kept trying

1 Since 1 Hz = 1 s−1, the units of h are more commonly expressed as joules×seconds, or J·s. 1.1. LIGHT AND THE ELECTROMAGNETIC SPECTRUM 5

incident positive light of anode to A frequency ν collect ↑ B fast e- maximum - electron e kinetic emitted energy electrons e- - metal e negative cathode with grid for 0 measuring frequency [s-1] voltage = 0 - ν → e kinetic ν0 energy - W0 A current meter Figure 1.3: The photoelectric effect: A.Theexperiment; B. The observations. to find alternative derivations that required no such assumption. This conflict was reflected in Planck’s statement that2 “A new scientific truth does not triumph by convincing its opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up which is familiar with it.” Indeed, in the early days, insofar as Planck’s theory was accepted at all, it was assumed that the quantization was a property of the material object emitting or absorbing the light. It was only somewhat later that it was recognized to be an intrinsic property of light.

Albert Einstein and the photoelectric effect A second troublesome phenomenon that 19th century physics failed to explain was the photoelectric effect, which is illustrated schematically in Fig. 1.3. In this experiment it was found that when light is shone on the surface of certain metals in a vacuum, electrons are emitted. A positive electrode was used to collect the electrons and measure the net current, while a negatively charged mesh of variable voltage was positioned between the anode and the cathode. The maximum kinetic energy of the emitted electrons was then measured by determining how large a negative voltage on that grid was required to completely shut off the current. This yielded the following observations:

• There is no time lag between the arrival of the light beam at the surface and the emission of the ﬁrst electrons.

• The number of emitted electrons increases with the intensity of the light, but their maximum kinetic energy is unaﬀected by it.

• The maximum kinetic energy of the emitted electrons increases with the frequency of the incident light, but does not depend on its intensity.

• For each metallic material there is a characteristic threshold frequency ν0 below which no electrons are emitted, independent of the intensity of the light.

These results were completely inconsistent with the accepted view of light as a wave phenomenon, according to which electron emission from a surface was an erosion phenomenon, like water waves wearing away a cliﬀ. In 1905, Albert Einstein showed that all observations associated with the photoelectric eﬀect were explained if one assumed that the energy associated with light of frequency ν was “quantized” in tiny bundles of size ε(ν)=hν, in which the constant h can be determined from the slope of the type of plot shown in Fig. 1.3 B. The fact that there is a threshold frequency below which no electrons are emitted merely indicates

2Quoted from The Quantum Physicists and an Introduction to Their Physics,byW.H.Cropper,OxfordUniversity Press, 1970. 6 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY that there exists a characteristic minimum energy required to tear an electron free from a particular metal. This threshold energy W0 = hν0 (1.7) is called the “work function” of the metal. Quanta of light of frequency ν0 have sufficient energy to dislodge the electrons, but no remaining energy to transfer to the electron as kinetic energy. In this theory, Einstein extended Planck’s idea of quantized energy packets emitted or absorbed by matter and predicted that the energy carried about by light was also quantized. When accurate data later became available, it was also found that the empirical constant h determined as the slope of the plot in Fig. 1.3 B was exactly the same as the empirical constant determined by fitting Eq. (1.6) to the observed intensity distributions of black-body radiation. This result showed that Planck’s energy quantum ε(ν)isinfacta property of light, and not of the emitting or absorbing material of the black-body. When it first appeared, Einstein’s proposal was rather unsettling, as it directly challenged the universally accepted view of light as a wave phenomenon. At the time, the data on which his conclusions were based were somewhat rough, so doubters had hope, and some expended considerable effort attempting to prove that the data on which his theory was based were unreliable. One of the most prominent of these was Robert Millikan, himself a later (1923) Nobel Laureate for his oil-drop experiment which determined the charge on the electron. However, in 1916 even he was compelled to say2 “I spent ten years of my life testing that 1905 equation of Einstein’s, and contrary to all my expectations, I was compelled in 1915 to assert its unambiguous experimental verification, in spite of its unreasonableness.” The skepticism seen at the end of this sentence reminds us of the remark by Planck quoted on p. 5. The fact that black-body radiation and the photoelectric effect are quantitatively explained using the same simple yet astounding assumption and the same value for a new fundamental constant heralded the dawn of quantum theory. In order to describe properly how electrons are dislodged from a metal and how the intensity of light emitted by a heated object varies with its “colour” (or wavelength), we conclude that electromagnetic radiation must consist of tiny bundles or “quanta” of energy whose magnitude is precisely determined by their frequency. At the same time, to describe the properties of propagation, reflection and refraction, electromagnetic radiation must be described as waves. This apparent dichotomy is known as the wave-particle duality of light, according to which light possesses the characteristics of both waves and particles.

Arthur H. Compton and “bouncing” photons

The final evidence that terminated arguments about whether or not light could show particle-like properties was provided by a set of experiments performed in 1922-23 by Arthur H. Compton at Washington Univer- sity in Saint Louis Missouri.3 He found that when monochromatic (single-wavelength) light of very short wavelength (X-rays) passed through thin films of solid material, the scattered light had two components: (i) intense scattered light with exactly the same wavelength as the incident X-rays, and (ii) low intensity scattered light with slightly longer wavelengths, where the magnitude of the wavelength shift varied with the angle of deflection from the direction of the incident beam. Compton showed that his observations were quantitatively explained if both the electrons in the material and the quanta or ‘photons’ of light behaved like classical billiard balls undergoing collisions subject to the normal energy and momentum conservation laws of classical mechanics. However, this evidence also required him to devise some definition for the momentum of a photon. This was done by combining Einstein’s famous special relativity mass–energy relationship, E = mc2 , with the light-energy expression of Eq. (1.5), while making use of the conventional classical definition of the momentum of an object as the product of its mass with its velocity:4

2 ε(ν)(∼ mλ c )=pλ c = hν . (1.8)

3 Not all of the key work establishing quantum mechanics was done in the great universities of Europe! 4 Because a photon has no rest mass, the symbol “mλ” in Eq. (1.8) represents a ﬁctitious quantity; it is the fact that light travels at the relativistic speed c which allows it to have a ﬁnite momentum. 1.1. LIGHT AND THE ELECTROMAGNETIC SPECTRUM 7

Rearranging this expression and making use of the usual frequency/wavelength relationship of Eq. (1.1) yields Compton’s expression for the momentum of a photon of light of wavelength λ:

pλ = h/λ . (1.9)

Thus, while Planck and Einstein showed that the energy associated with light of a given frequency (or wavelength, or colour) was quantized in minute packets of magnitude ε = hν = hc/λ, Compton showed that these quanta, commonly called photons, also “bounced” like classical rigid objects, with momenta given by Eq. (1.9). An exciting modern application of this particle-like property of photons was its use in the ﬁrst experiments to produce an ultra-cold atomic gas at temperatures in the milli-kelvin to micro-kelvin range, work which earned Steven Chu, Claude Cohen-Tannoudji and William Phillips the 1997 Nobel Prize in Physics (see http://www.nobel.se/physics/laureates/1997). The earliest of these experiments was simply based on the fact that when an atom moving towards a light source absorbs a photon, the momentum given up by the photon slows it down slightly. The average molecular speed in a gas is a direct measure of its temperature, and in an intense laser ﬁeld this process can occur an immense number of times per second, slowing the atoms to average speeds thousands of times smaller than they would have even in the intense cold of interstellar space.

1.1.4 A Brief Note on Units One potentially confusing issue in science is the wide variety of names and units that are used for seemingly identical quantities. Because of the Planck energy relation of Eq. (1.5), spectroscopists treat energies, frequencies and wavenumbers equivalently, jumping back and forth between J, Hz and cm−1 while talking all the time about “energy”. The Planck equation is the justiﬁcation for this, as it demonstrates the direct proportionality between the energy and frequency of light. Moreover, use of particular experimental techniques leads to the use of seemingly unrelated units such as the electron volt (eV) in certain types of spectroscopy (see Chapter 6). The table below [taken from Rev. Mod. Phys. 80, 633 (2008)] will facilitate conversions among these various “energy-like” units.

Table 1.1: Conversion factors for energy units encountered in spectroscopy.

joule (J) eV cm−1 Hz 1joule(J)= 1 6.241 509 65×1018 5.034 117 47×1022 1.509 190 45×1033 1eV= 1.602 176 487×10−19 1 8065.544 65 2.417 989 454×1014 1cm−1 =1.986 445 501×10−23 1.239 841 875×10−4 12.997 924 58×1010 1Hz= 6.626 068 86×10−34 4.135 667 33×10−15 3.335 640 951×10−11 1

Although all of the energy units appearing above are sometimes used in molecular spectroscopy, the most widely used unit is wavenumbers, with units cm−1, and in most cases it will be the unit used in this text. Moreover, although the SI unit of length is the meter, and the nanometer (1 nm = 10−9 m) is commonly used to characterize the wavelength of light, molecular dimensions are most commonly reported in units of A(1˚ A=10˚ −10 m), and this is the unit that will be use for molecular bond lengths. Similarly, although the SI unit for mass is kg, in discussing and performing calculations for molecules it is much more convenient 12 to use atomic mass units, mu =1u≡ m( C)/12 . In spite of the above, formal expressions for molecular level energies encountered in this course are normally derived and written down in SI units, with energy in joules, mass in kilograms, and length in meters. It would of course be quite tedious if we had to undertake detailed unit conversions in every calculation, but if we think ahead, this will not be necessary. The theoretical formulae for the energy associated with many phenomena considered in molecular physics contain a factor of the form 2/(2Md2)[J],inwhich = h/2π , h is the Planck constant, M is a mass in kg, and d is a length with units [m]. Because we prefer to input a mass with units [u], a ﬁrst step is to replace M [kg] by M [kg] = mu [kg]×M [u], where mu is the mass in kg of one atomic mass unit (see p. xiii). Similarly, 8 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY because we wish to quote distances in Angstr¨˚ oms, we substitute d [m] = 10−10 d [A].˚ Combining these terms with the factor 1/(102 hc) required to convert from [J] to [cm−1], the ubiquitous factor 2/(2Md2) becomes 2 2 20 ( [J · s]) 10 1 Cu 16.857 629 2 = 2 2 = 2 = 2 2(M [kg]) (d [m]) 2(mu [kg]) 10 hc (M [u]) d [A]˚ (M [u]) d [A]˚ (M [u]) d [A]˚ in which the numerical value of the constant Cu =16.857 629 056, which we call the “inertial constant”, is obtained on substituting values of the various physical constants into the initial versions of the above 2 2 2 −1 expression. This conversion of / 2Md [J] to Cu/ (M [u]) d [A]˚ [cm ] appears repeatedly in the following chapters.

1.2 Quantum Theory of Matter

1.2.1 The Spectrum of the Hydrogen Atom In 1900 it was known that atoms were roughly 10−10 m=0.1 nm in diameter, but it was not clear what their structure was or where the electrons were located. Then in 1911 Ernest Rutherford proposed the “nuclear” model of the atom, according to which all positive charge is located in a tiny nucleus of diameter ∼ 10−14 m, while the electrons move about it in orbits of diameter ∼ 10−10 m that deﬁne the eﬀective atomic size.5 However, particles moving in a circular orbit are constantly accelerating towards the center, and classical electromagnetic theory predicts that charged particles that are accelerating spontaneously emit light. This prediction is indeed obeyed by atomic-scale particles, and it is the basis for very intense tunable light source machines known as “synchrotrons”, such as the “Canadian Light Source” facility in Saskatoon, Saskatchewan (see http://www.lightsource.ca). However, within an atom this would spell disaster: if the orbiting electrons emitted light they would lose energy, slow down, and eventually spiral into the nucleus.6 This “collapsing atom” problem appeared to raise serious questions about the validity of the Rutherford model. Black-bodies were well known to emit light over a continuous range of frequencies, and the distribution of their intensities was explained by Planck, as discussed above. However, by the early 1900’s experimental spectroscopy had also shown that individual types of atoms and molecules absorbed or emitted light at certain discrete frequencies (or ‘colours’). The simplest atom, hydrogen, was the most intensively studied, since it should be the easiest to understand. Indeed, around 1885 the Swiss schoolteacher Johannes Balmer had shown that the lines of the emission spectrum of gaseous H atoms in the visible region could be exactly explained by the formula 2 n1 λ = A 2 in which n1 =3, 4, 5, 6, ... (1.10) n1 − 4 and A is a constant. Today it is more common describe this series of lines using the expression obtained on inverting the left- and right-hand sides of Eq. (1.10): R 1 − 1 ν˜ = H 2 2 , (1.11) (2) (n1) in which (recall Eqs. (1.1) and (1.2))ν ˜ is the wavenumber of the emitted spectral line, and the constant −1 RH = 109 677.583 41 cm is known as the ‘Rydberg constant’ for hydrogen. Outside the narrow frequency range known as the visible region, several other hydrogen atom emission series were also observed (see Figs. 1.4 and 1.5), and Swedish physicist Janne Rydberg showed that a generalization of the reciprocal version of Balmer’s equation allowed all of these series to be exactly explained by the expression R 1 − 1 ν˜ = H 2 2 , (1.12) (n2) (n1)

5 These conclusions were based largely on experimental work done at McGill University in Montreal, before his 1907 move to the University of Manchester in England, and they won Rutherford the 1908 Nobel Prize in Chemistry. 6 This slowing down does not occur in a synchrotron because energy is continuously infused into the particle beam to compensate for energy lost by emission of radiation. 1.2. QUANTUM THEORY OF MATTER 9

wavelength λ / nm 4000 1000 400 200 100 ⏐ ⏐ ⏐ ⏐ ⏐ ⏐

0 20000 40000 60000 80000 100000 120000 wavenumber ν− / cm-1 Figure 1.4: The hydrogen atom emission spectrum. If the emitted light is dispersed by a prism, photons of diﬀerent frequency cause blackening at diﬀerent locations on a photographic plate.

in which a particular value of n2 = 1, 2, 3, . . . characterizes each series, and for a given series n1 has the 2 values n1 = n2 +1,n2 +2,n2 +3, ..., etc. Because the quantity 1/n decreases rapidly as n increases, 1 1 1 1 1 1 , , , , , , ... = {1, 0.25, 0.111111, 0.0625, 0.040, 0.027777, ...} 1 4 9 16 25 36 2 each of these series converges to a characteristic limitν ˜∞ = RH/(n2) . Except for the n2=2 s e r i e s t h a t w a s named after Balmer, each of these series is named after the person who ﬁrst measured it.

n2 series region 1 Lyman far ultraviolet 2 Balmer visible 3 Ritz-Paschen near-infrared 4 Brackett mid-infrared 5 Pfund far-infrared

For a long time the perfect agreement of this beautifully simple equation with the observed spectra was viewed, as Neils Bohr wrote,2 “ ... as the lovely patterns on the wings of butterflies; their beauty can be admired, but they are not supposed to reveal any fundamental physical laws.” However, it was Bohr himself, in work published over the years 1913–1915, who definitively proved that in the case of atoms these patterns do indeed directly reflect fundamental physical laws.

1.2.2 The Bohr Theory of the Atom Drawing upon the nuclear model of the atom proposed by Rutherford in 1911 and the Planck/Einstein energy quantization of radiation, Bohr rationalized the atomic emission spectra of hydrogen in terms of a model which combined conventional classical mechanics with an ad hoc quantization postulate. He started from a mechanical picture in which the electron moved in a circular orbit with the classical centrifugal force away from the nucleus exactly balanced by the Coulomb attraction between the two opposite charges. He took care of the “collapsing atom” problem by simply ignoring it (a nice way to treat problems, if you can get away with it!), and assuming that the electron was in some sort of ‘stationary state’ in which the classical electrodynamics rules governing radiation by a moving charge simply did not apply. He then added a critical quantization postulate, that the allowed or stationary states were associated with integer multiples of the quantity h θ/˙ 2π,whereθ˙ is the classical angular speed of the orbiting electron (with units radians per second). As a ﬁnal step, he then introduced his famous correspondence principle, which asserted that for orbits with very large radii, and hence very small angular frequencies θ˙, quantum results must merge and agree with the results of classical electrodynamics. This constraint yielded a value for the proportionality constant relating the energies of the stationary states to the quantity h θ/˙ 2π, and then led to the level energy −1 expression (in cm ): 4 − e μH 1 − R 1 En = 2 2 2 2 2 = H 2 (1.13) 32π 0 10 hc n n 10 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

0 n=∞

energy n=4 1875. nm 1282. nm 1094. nm -1 /cm n=3 Ritz-Paschen

-20000 656. 5 nm 486. 2 nm 410. 3 nm 434. 2 nm Balmer n=2

-40000 Lyman

Coulomb potential -60000 − e 2 V(r) = ⎯⎯⎯⎯⎯⎯ (4πε 102 hc) r

97.25 nm 94.98 nm 93.78 nm 0 121.56 nm 102.57 nm -80000

-100000

n=1

-120000 0 1 2 3 4 5 6 7 r/Å Figure 1.5: Hydrogen atom energy levels and transitions.

in which e is the electron charge, 0 is the permittivity of vacuum (a known constant which arises in electrostatics), and μH = me mp/(me + mp) is what we call the “reduced mass” of this two-particle system (see § 2.1.2). Note that the “tilde” () over a symbol for energy indicates that its units are those of the wavenumber, cm−1, while the factor of 102 in the denominator of Eq. (1.13) converts from the SI unit of inverse length (m−1) to the common “spectroscopists’ unit” of cm−1. Substituting the known values of the various physical constants into Bohr’s expression for RH yields exactly the same value of this constant that Rydberg had determined empirically by ﬁtting the observed positions of the lines in the H-atom spectrum to Eq. (1.12)! Another result yielded by this derivation is that the magnitude of the angular momentum of the electron (L) is an integer multiple of , L = n ,wheren (= 1, 2, 3, 4, . . . ) is a positive integer, and = h/2π =1.054 571 628×10−34 J s. We shall see later that this quantization of angular momentum is central to our understanding of the spectra associated with molecular rotation. If Bohr’s quantized energy levels do indeed describe the only possible allowed energy states of the atom, by conservation of energy, the light emitted by an excited H atom carries the energy associated with a transition between a pair of such levels, as illustrated in Fig. 1.5. It is also clear that the diﬀerence between the energies of two of Bohr’s levels

− R 1 − 1 ΔE(n1,n2)=En1 En2 = H 2 2 =˜ν (1.14) (n2) (n1) agrees exactly with the empirical Rydberg expression of Eq. (1.12). Thus, radiation with wavenumber ΔE(n1,n2) is emitted or absorbed when the electron undergoes a transition between quantum states n1 2 and n2 . The theory also predicts that the radius of the orbit associated with quantum number n is n ×a0, 1 where a0 =0.529 177 208 59 A˚ is the radius of the Bohr orbit in the ground ( n = 1 ) level of an Hatom. Thus, Bohr explained that each series of lines in the atomic hydrogen emission spectrum is due to electrons “falling” from large-radius high-energy orbits (designated by n1) into a particular smaller-radius lower-energy orbit characterized by a particular n2 value. A straightforward extension of the basic derivation shows that for a general one-electron atom or ion “A” 1.2. QUANTUM THEORY OF MATTER 11

A consisting of a nucleus of mass mnuc and charge +Ze, the energy levels are given by the formula A − R 1 − 2 μA R 1 En = A 2 = Z H 2 , (1.15) n μH n

A A 7 in which μA = me mnuc/(me + mnuc) is the reduced mass of this system. This generalization allowed Bohr to explain the differences between the transition energies of the 1Hand2H atoms, and to predict accurately the transition energies of all one-electron atomic ions, such as He+,Li+2,Be+3,B+4, ... , etc. Because the electron mass is much smaller than any nuclear mass, the correction factor (μA/μH) is always close to unity. However, it must be included if we are to account for the differences between the observed transition energies for different isotopes of a given one-electron atom, such as those for H and D, or those for 7Li+2 and 6Li+2. For example, consider the one-electron 7Li+2 ion for which the nuclear mass is given by m(7Li+3)=7.016 004 55 − 3(0.000 548 579 909 43) = 7.014 358 81 u. The electronic reduced mass for this species is then 7Li+3 7Li+3 −4 μ7Li+2 = mnuc me/ mnuc + me =5.485 370 093×10 u ,

−4 which is only 0.0466% larger than the value of μH=5 .482 813 061×10 u, and only 0.0013% larger than −4 μ6Li+2=5 .485 298 698×10 u. Thus, although these differences are small, they are not negligible; for example, the transition energy of the first “Lyman-type” line (the 2p ← 1s transition) of a Li+2 ion is 9.640 cm−1 larger for 7Li+2 than for 6Li+2, a difference that is more than four orders of magnitude larger than the limits of experimental precision Bohr’s theory represented a huge step towards a practical quantum theory for matter, but it turned out that it was only able to provide an accurate description of the properties of one-electron atoms or ions, and a decade later it was superceded by what today is called quantum mechanics. However, because of the revolutionary implications of Bohr’s result regarding the properties of atoms and molecules, Eq. (1.14) (or the conventional SI units version of it, ΔE(n1,n2)=hν ) has become known as the Bohr resonance condition.

1.2.3 de Broglie Wavelengths Since light can be described as a wave with characteristics of a particle, shouldn’t matter (made up of particles) also possess wave characteristics? This question was posed by the French aristocrat and scholar Prince Louis-Victor de Broglie, and in his 1924 PhD thesis he argued convincingly that it must be true, even though there was at the time no evidence to support it. Surprisingly, this arbitrary and seemingly crazy hypothesis received an almost tolerant reception, at least in part because it was quickly picked up by Einstein who announced2 that de Broglie “had lifted the corner of a great veil”. Using intuitive reasoning based on the (by then) accepted fundamental wave-particle dual nature of light, he argued that a particle of mass m moving with velocity v would show wave-like properties, and be characterized by what is known now as its “de Broglie wavelength” h h λ = λ = ≡ (1.16) p mv p in which p = mv is the particle momentum. This prediction implies that beams of particles should display the full range of properties predicted by the classical theory of wave motion, including reflection, diffraction, and constructive and destructive interference. This hypothesis was confirmed experimentally in 1927, when George Thomson in the UK and Clinton Davisson and Lester Germer in the USA reported experiments that showed that beams of fast electrons reflecting off metal crystals did indeed show exactly the same diffraction properties as X-rays. Today it is well known that all atomic particles, electrons, neutrons, and protons, as well as whole atoms and molecules, possess wave-like as well as ordinary particle properties (such as mass). Indeed, in principle trucks moving in a column on a highway would have wavelike properties! However, the theory of wave-particle duality also

7 A Ignoring the very small relativistic mass correction, mnuc = MA − Zme ,inwhichMA is the standard atomic mass of atomic isotope A. 12 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY predicts that in the limit of extremely large quantum numbers or extremely small de Broglie wavelength, the wavelike properties of matter become indistinguishable from our normal classical mechanical experience for material objects. Thus, people should not expect their wavelike properties to mitigate the eﬀects of standing in the middle of the road in front of a stream of oncoming transport trucks.

1.3 Wave Mechanics and the Schr¨odinger Equation

Bohr’s successful combination of a classical description of the motion of the electron and nucleus with his seemingly arbitrary postulate regarding quantization had yielded a successful ‘quantum-plus-mechanics’ description of one-electron atoms. However, this model did not work for more complicated systems, and after a decade of efforts to extend or generalize it, it became clear that it was a dead end, and that an entirely new approach was required. The first successful more general theory to appear was the “matrix mechanics” proposed by Werner Heisenberg in 1925. However, because of its mathematical complexity and sophistication it is not normally considered to be a good starting point for a beginner’s approach to quantum mechanics. A somewhat more accessible approach was the differential equation formulation that the Austrian physicist Erwin Schrödinger developed and published in a series of papers in 1926. Conceptually, mathematically, and in a sense physically, Schrödinger’s mechanics was altogether different from Heisenberg’s mechanics. While Heisenberg’s method is algebraic, Schrödinger’s method begins with a differential equation. Heisenberg’s method builds on discrete and discontinuous quantities, whereas Schrödinger’s mechanics is based on a quantity that is continuous. On the wave-particle duality question, the Heisenberg procedure seems to side with the particle viewpoint; in contrast, Schrödinger’s differential equation is very explicitly a “wave equation”. As might be expected, Heisenberg and Schrödinger initially had difficulty accepting the validity of each other’s theories. In a letter to theoretical physicist Wolfgang Pauli, Heisenberg wrote:2 “The more I ponder about the physical part of Schrödinger’s theory, themoredisgustingitappearstome.” Similarly, although he published a formal mathematical proof of the equivalence of the two theories later in 1926, Schrödinger had doubts about the physical content of Heisenberg’s matrix mechanics, and wrote:2 “I was discouraged, if not repelled, by what appears to me a rather difficult method of transcendental algebra, defying any (intuitive) visualization.” Erwin Schrödinger took de Broglie’s proposed matter waves to mean that a proper description of particle behaviour would be given by a ‘wave function’ – a mathematical description of its wave nature – one of whose properties was that the particle could exist only in certain quantized states, as in the Bohr model. Schrödinger’s description also provided a rationale for these discrete states: they were states in which the electron waves would not destructively interfere. No rigorous derivation is possible for Schrödinger’s differential equation (nor for Heisenberg’s method); he simply wrote it down, based on his deep understanding of the mathematics of classical wave theory and a profound intuitive acceptance of the wave nature of particles. However, what he dreamed up provides the basis for all of our current understanding of atomic matter and of chemistry. We next describe the application of his method to two key illustrative cases: (i) the “particle-in-a-box” problem, a simple model system whose description illustrates how “quantization” arises, and (ii) the hydrogenic atom, which provides the basis for our description of atomic orbitals and of the chemist’s periodic table.

1.3.1 A Particle in a One-Dimensional Box For the special case of a particle of mass m with total energy E moving in one dimension along the x coordinate subject to a potential energy field V (x), Schrödinger’s differential equation for the particle/wave’s displacement y(x) has the form

2 d2 y(x) − + V (x) y(x)=Ey(x) . (1.17) 2m dx2 1.3. WAVE MECHANICS AND THE SCHRODINGER¨ EQUATION 13

The prototype “particle-in-a-box” problem is one in which the potential energy imposed on the particle traps it in a box with impenetrable walls at x =0 and x = L , but allows it to move absolutely freely inside the box (i.e., V (x)=0 for0

y(x)= sin(kx)andy(x)= cos(kx) . (1.19)

However, before proceeding further it is useful to summarize the rules of wave mechanics as they apply to the range of systems discussed in this course.

Rules of Schr¨odinger’s Wave Mechanics 1. For every molecular system there exists a “wave function” ψ , which contains all the information that we can possibly know about the system.

Note that although we wrote Eqs. (1.17) – (1.19) using the familiar calculus name y(x)for the dependent variable, it is a virtually universal convention to use the Greek letter psi , written as ψ , to represent the solution of Schr¨odinger’s diﬀerential equation in quantum theory. Unless stated otherwise, this practice shall be adopted from here on.

2. This wave function ψ is the solution of Schr¨odinger’s diﬀerential equation for the system, and it depends on all spatial coordinates (x, y, z) of all particles comprising the system.

Of course the total wave function also depends on time, but these notes only consider its time-independent part. Note too that while any real particle moves in three dimensions, it is often possible to describe that behaviour mathematically in terms of three separate one- dimensional problems, such as the particle-in-a-box problem that we have just discussed.

3. The wave function ψ is a continuous mathematical function of all coordinates of the system.

4. The probability of ﬁnding the system with a particular conﬁguration (i.e., with a particular set of spatial coordinates) is proportional to |ψ|2 .

The wave equation solution ψ is a mathematical function that can have positive or negative (or complex number) values, but the square of its absolute value is always non-negative: |ψ|2 ≥ 0 . Since the particle must be somewhere, the probabilities must all add up to unity. For our one-dimensional system this requirement is expressed mathematically by the “normalization” condition +∞ |ψ|2 dx = 1 (1.20) −∞ This shows that |ψ|2 is actually a “probability density”, or in our one-dimensional case, the probability per unit distance. 14 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

V(x) n ↑ energy ⎯ 7 m=1.0 m=1.0 m=2.25

n=4 6

5 n=3 4

n=2 3 2 n =1 1 00x → L=1.0 x → L=1.5 0x → L=1.0 Figure 1.6: Potential energy function V (x) (solid curves), discrete level energies (dashed lines) and wave functions (dotted curves) for three “square-well” particle-in-a-box systems.

For the case of our particle conﬁned to a rigid one-dimensional box, we can see that the general function

ψ(x)=A sin(kx)+B cos(kx) (1.21) in which A and B are arbitrary numerical constants, satisfies the first three rules. It is a continuous function of all of the coordinates of the system (here, x), and it is easy to verify that if sin(kx)andcos(kx)are solutions of Eq. (1.18), then so is any linear combination of them. The intriguing point then is the final rule: we will see that it is the source of the requirement that the energy of the system be quantized.

How does quantization arise? Recall that in our simple one-dimensional system the particle is trapped in the interval between x =0 and L. Since Rule #4 tells us that the probability of ﬁnding the particle at a given position x is proportional to |ψ(x)|2 , this conﬁnement means that ψ(x) must be identically zero everywhere outside the box (i.e., for x<0andx>L). If that is true, the wave function continuity requirement of Rule #3 means that the wave function inside the box must go to zero at the walls (or boundaries) where it meets the wave function solution outside the box. In mathematical language this means that

ψ(x =0)= 0 and ψ(x = L)=0. (1.22)

Applying the ﬁrst of these matching or “boundary conditions” to the general wave function solution of Eq. (1.21) shows that

ψ(x =0)= A sin(k × 0) + Bcos(k × 0) = A × 0+B × 1=B =0, (1.23) so the general solution of Eq. (1.21) is reduced to the form ψ(x)=A sin(kx) . Applying the second boundary condition in Eq. (1.22) then yields the condition

ψ(x = L)=A sin(kL)=0. (1.24)

Our knowledge of trigonometric functions tells us that Eq. (1.24) is satisfied whenever kL= nπ where n is an integer, or if A = 0 . However, if either n =0 orA = 0 , the function ψ(x) would be zero everywhere: inside as well as outside the box. Since |ψ|2 is proportional to the probability of finding the particle at a particular location, and since it must be located somewhere, these “trivial solutions” are unacceptable, because they could not satisfy the normalization condition of Eq. (1.20). Thus, we reach the inescapable conclusion that the only distinct “non-trivial” solutions of the Schrödinger equation for this system which satisfy our four Rules of wave mechanics are those associated with discrete values of the 1.3. WAVE MECHANICS AND THE SCHRODINGER¨ EQUATION 15 constant k corresponding to positive integer values of n :8 2mE nπ k = = for n =1, 2, 3, 4, ... (1.25) 2 L

Rearranging this expression shows that these allowed solutions can only occur if the system energy has a discrete value given by the equation 2 π2 E = E = n2 (1.26) n 2 m L2

In “spectroscopists’ units”, with mass in u, length in Angstr¨˚ oms and energy in cm−1, this yields the expression that we commonly use for calculations: 2 π 2 −1 En = Cu n [cm ] (1.27) m[u] (L[A])˚ 2

−1 2 in which Cu =16.857 629 056 [u cm A˚ ] is the ubiquitous numerical factor introduced in the discussion of units in §1.1.4. It is easy to show that the normalization condition of Eq. (1.20) means that the scalar factor in Eq. (1.24) must have the value A = 2/L (recall that our ﬁrst boundary condition required that B =0).Thus,it is the coupling of the requirements that the wave function must be a solution of the Schr¨odinger equation and must be continuous with the interpretation of |ψ(x)|2 as a probability density that gives rise to the phenomenon of quantization of energy in molecular systems. Equation (1.26) shows that the energy of our particle is only allowed to have the discrete values En = 2 2 2 2 2 2 2 2 2 2 2 2 π /2mL ,4π /2mL ,9π /2mL ,16π /2mL , ... etc., and the associated (normalized) wave functions (or “eigenfunctions”) are ψn(x)= 2/L sin(nπx/L) . For three related “square-well” particle-in-a-box models, Fig. 1.6 shows the allowed energy levels (horizontal dashed lines) and the associated wave functions (dotted curves).9 They illustrate properties that apply to virtually all quantum systems.

• For every allowed “eigenstate” (or distinct solution of the Schr¨odinger equation), the “eigenfunction” (or wave function) has an integer number of positive and/or negative loops, separated by “nodes”.10 [A “node” is a point where the wavefunction vanishes: ψ = 0 . In two-dimensions we would have nodal lines or curves, and in three dimensions we would have two-dimensional nodal surfaces, as seen later in Fig. 1.11.]

• The energy of the system increases with the number of nodes. In particular, the wave function for the nth level has n loops (or extrema) and n − 1 “internal” (i.e., not at a boundary) nodes.

• When the width of the interval in which a particle is trapped increases, the levels all shift to lower energies, as shown by the a comparison between the ﬁrst and second panels of Fig. 1.6. 2 [For this particular ‘particle-in-a-box’ case of a square well potential: En ∝ 1/L .]

• For a given potential function, increasing the mass of the particle decreases the level energies, as shown by a comparison between the ﬁrst and third panels of Fig. 1.6. [For this particular ‘particle-in-a-box’ case of a square well potential: En ∝ 1/m.]

8While negative integer value of n would also satisfy Eq. (1.24), the associated normalized solutions would not be ‘distinct’ (linearly independent) from those for positive n values. 9 Each diagram is a combination of two types of plot: (i) an energy vs. distance plot showing the allowed level energies (dashed lines) and how the potential energy function (solid line segments) varies with distance, and (ii) plots of a wave function ψn(x) vs. distance x, for each of which the zero of the y–axis is placed at the energy of the corresponding level. This type of combined energy/wave-function plot is used quite frequently in spectroscopy and quantum mechanics. 10Our use of the German adjective eigen,asineigenvalue, eigenfunction, and eigenenergy. The word “eigen”, which translates literally as “proper”, comes from the mathematics of matrices (which have eigenvectors and eigenvalues), and is widely used in quantum mechanics. 16 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

∞∞ r → 0 V(x) n =4 n =6 n =3 n =2

n =5 − e2 V(r)= ⎯⎯⎯ n =4 4πε0 r n =3 A B n =2 n =1 n =1 − RH 0Lx →

V(r) =½k(r−r )2 e υ =5 υ =5 De → υ =4 υ =4 υ =3 υ =3 υ =2 υ =2 C υ =1 D υ =1 υ =0 υ =0

re r → re r → Figure 1.7: Level energies and wave functions for particle-in-a-box problems associated with differently shaped potential energy functions. A – square-well potential with infinitely rigid walls; B –Coulomb potential and effective radial wave functions for an electron in an H atom; C – harmonic-oscillator potential for molecular vibration; D – Morse potential for molecular vibration.

The above properties also apply to the wave functions for the orbital motion of an electron in an atom and the rotation of a molecule (see §1.3.2), for the radial motion of an electron in an atom, and for vibrational motions in a molecule. For the two latter cases the differential equation governing the radial or stretching motion can be written in precisely the same form as Eq. (1.17), the only difference being that the potential energy function differs from one case to another. In the various cases, the precise mathematical dependence of the level energy on the trap size (L in Eq. (1.26) or Fig. 1.6) or particle mass m differ, but qualitative features remain the same. For example, Fig. 1.7 shows the energies (horizontal dashed lines) and wave functions (dotted curves) for the lowest few levels of a particle of mass m trapped in potential energy wells (generalized boxes) with four representative shapes. Case A is the conventional “square-well” problem discussed above, with V (x) = 0 inside the box; case B is for the radial motion of an electron in an H atom (c.f. Fig. 1.5); case C is the harmonic-oscillator model for molecular vibration, which is discussed in Chapter 3, and case D shows the solutions of the radial Schrödinger equation for molecular vibration obtained using the more realistic Morse function model for molecular vibration, also discussed in Chapter 3. Following the spectroscopic convention for labeling eigenstates of vibrational motion, the levels in segments C and D of Fig. 1.7 are labeled v =0,1,2,3,... , rather than n =1,2,3,4,... etc. However, this naming convention has no effect on the properties listed above. It is immediately clear that although the four properties listed above apply to all cases illustrated in Fig. 1.7, the pattern of level energy spacings depends very strongly on the shape of the box. In particular, 1.3. WAVE MECHANICS AND THE SCHRODINGER¨ EQUATION 17 in cases B–D the width of the well (effectively, the width of the box) increases with energy, and the rate at which it does so increases from C to D to B. As a result, whereas the spacings between adjacent levels increase with energy for the square-well problem of Case A, this trend is halted for C and reversed for cases D and B, with the most extreme reversal being for the Coulomb potential (Case B), which is the case for which the width of the potential energy well (the effective box length) increases most rapidly with increasing energy. We will see in Chapter 3 that this dependence of level spacings on well width can be used to determine the properties of the potential-energy function governing molecular vibration. A final observation about the results shown in Figs. 1.6 and 1.7 is the fact that the lowest allowed energy level never lies at the bottom or energy zero of the potential energy function. Except for the Coulomb case, for which the potential function goes to minus infinity (Fig. 1.7 B), the gap between the energy of the lowest allowed level and the potential minimum is called the “zero-point energy” of the system. This existence of a finite zero-point energy is a special property of some quantum systems that has no analog in classical mechanics. However, as we shall see in the next section, it does not appear in the quantum description of rotational or orbital motion.

1.3.2 Orbital or Rotational Motion: A Particle on a Ring Orbital motion of an electron or rotational motion of a molecule occur in three-dimensional space and require more sophisticated mathematical treatments than are appropriate here. However, the simple one-dimensional problem of a particle of mass m moving in a flat circular orbit with radius r provides a realistic illustrative model for these motions. Your first course in quantum mechanics will teach you that the Schrödinger equation for this model problem has the form 2 2 − d ψ(φ) 2 2 = Eψ(φ) 2mr dφ d2 ψ(φ) 2mr2E or = − ψ(φ)= − b2 ψ(φ) (1.28) dφ2 2 in which φ is the polar angle characterizing the position of the particle on the ring. As this equation has exactly the same form as Eq. (1.18), its general solution is also given by Eq. (1.21) except that the distance variable x is replaced by the angle φ, and the constant b becomes defined by the expression b2 =2mr2E/2 . In this case, it is the third Rule of wave mechanics (see p. 13) that imposes quantization on the system. A central feature of orbital motion is that it is never ending; the particle keeps on going around and around, and its wave function must reflect this fact. In particular, when the particle completes a full orbit of 2π radians or 360◦, it returns to the location from which it started. Thus, if the wave function is to be continuous everywhere, then necessarily ψ(φ+2π)=ψ(φ)forall possible values of φ. This circular boundary condition can be satisfied only if the sinusoidal wave function undergoes precisely an integer number of full oscillations when the particle makes a full circuit around the ring. Stated mathematically, this means that b×2π =2π where must be an integer. The definition of b means that this condition yields the energy “eigenvalue” equation: 2 2 π2 E = 2 = 2 (1.29) 2mr2 2m (πr)2 in which =0,1,2,3,..., and the associated wave functions can be written as ψ(φ)= 1/2π sin(φ+δ), where δ is an arbitrary phase constant. For practical work we wish, of course, to have distances in Aand˚ massesinuandtogenerateenergiesincm−1, in which case Eq. (1.29) becomes Cu 2 −1 E = 2 [cm ] (1.30) m[u] r[A]˚

−1 2 in which Cu =16.857 629 056 [u cm A˚ ] is the numerical constant introduced earlier. Figure 1.8 illustrates the wave functions (oscillating curves) for the eight lowest-energy particle-on-a-ring states. Comparing Eq. (1.29) with Eq. (1.26) and Fig. 1.7 with Fig. 1.8, we see that our particle-on-a-ring is eﬀectively a particle-in-a-square-well-box for a box length of L = πr. One important diﬀerence, however, 18 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

l =0 l =1 l =2

l =3 l =4 l =5

Figure 1.8: Particle-on-a-ring orbits (simple circles) and the wave functions (oscillating curves) associated with the eight lowest-energy levels. is that in the present case the = 0 state with a total energy of zero is allowed; i.e., there is no zero-point energy for rotation. This reflects the fact that in the square-well problem, if the energy lay at the potential minimum the wavefunction inside the box could only join continuously to that outside if its amplitude was everywhere zero, a result that would contradict the normalization condition of Eq. (1.20). In contrast, for continuity to be satisfied in the present case, all the wavefunction must do is join to itself smoothly whenever the angle φ increases by 2π, and this is possible for E =0 ifψ(φ) is a constant (equal to 1/2π ). We will see in Chapters 2 and 3 that these different eigenvalue properties are reflected in the nature of the lowest allowed levels associated with rotational and vibrational motion.

1.4 Electronic Structure of Atoms and Molecules

The simple one-dimensional problems described above are as far as we are going to proceed with actually solving the Schrödinger equation. However, note that the time-independent Schrödinger equation for a general system is written symbolically as H ψ = Eψ (1.31) in which H is the Hamiltonian operator for the system, a generalization to three dimensions (and multiple H − 2 d2 particles, if appropriate) of the operator 1D = 2m dx2 + V (x) appearing in Eq. (1.17). As with the one-dimensional problems discussed in §1.3, Eq. (1.31) always has many different solutions corresponding to discrete allowed energy eigenvalues. This section reviews the properties of some familiar atomic and molecular orbital solutions of the Schrödinger equation that you encountered in your introductory chemistry course(s) in the light of the quantum-mechanical language introduced in the preceding section.

1.4.1 Hydrogenic Atomic Orbitals The Schrödinger equation for the hydrogen atom may be written as a three-dimensional version of Eq. (1.17) for a pseudo-particle with effective mass μH = me mp/(me + mp), where me and mp are the masses of the electron and proton, respectively (note that since mp me , μH ≈ me ). Because an atom is inherently spherically symmetric, it is most convenient to describe this system using the spherical polar coordinates r, θ and φ in place of the conventional Cartesian coordinates x, y and z (see Fig. 1.9). Although mathematically somewhat more complicated than Eq. (1.17), this differential equation can also be solved exactly in closed form. This solution yields precisely the same energy level expression obtained from Bohr’s “old quantum 1.4. ELECTRONIC STRUCTURE OF ATOMS AND MOLECULES 19 theory”, Eqs. (1.13) and (1.15), and the associated wave functions z→ define the familiar hydrogenic atomic orbitals presented in introductory Chemistry courses. As the electron in the atom is free (x,y,z) to move in three dimensions, it should be no surprise that these + θ (r,θ,φ) x = r sinθ cosφ solutions are characterized by three quantum numbers, n, and r→ y = r sinθ sinφ m . For the generalized case of an electron orbiting around a bare z = r cosθ → nucleus with charge +Ze,wheree is the magnitude of the elec- y tron charge (i.e., for the one-electron atom or ions H, He+1,Li+2, φ Be+3,...etc.),theeigenfunctionsofthelowestfewlevelsarelisted x→ in Table 1.2. 11 Figure 1.9: Definition of spherical po- It is easy to see that each of these hydrogenic functions may lar coordinates in terms of rectangular be written as the product of a function of r times a function of θ Cartesian coordinates. times a function of φ times a constant factor which imposes the three-dimensional analog of the normalization condition of Eq. (1.20). Moreover, if we associate the hydrogenic quantum number |m| with the one-dimensional orbital quantum number of §1.3.2, we see that the φ–dependent parts of these hydrogenic wavefunctions have the same form as the simple flat-orbit wavefunctions of §1.3.2.

The radial part of each of these eigenfunctions, Rn,l(r) , is itself the product of an exponential term times a member of the family of “Laguerre polynomials”, a class of functions whose properties have been thoroughly studied by mathematicians. However, a little mathematical manipulation of the diﬀerential equation that yields these radial functions converts it to the familiar form of Eq. (1.17) with x replaced by r and V (x)by 2 −C1/r, with C1 = Ze /4π0 , the Coulomb coeﬃcient for an electron interacting with a nucleus of charge

11 The word “hydrogenic” labels atomic systems consisting of a single electron interacting with a nucleus of charge +Ze (e.g., H, 2HorD,He+,Li+2,Be+3, ..., Fe+25, . . . etc.), for all of which the mathematical description is identical.

Table 1.2: Wave functions of hydrogenic orbitals expressed as real functions of r,for n =1,2and3.Here Z is the atomic number of the nucleus, and ao =0.519 177 2083 A˚ is the Bohr radius.

3/2 1 Z −Zr/ao n =1,=0,m =0; ψ1 = √ e s π ao 3/2 1 Z Zr −Zr/2ao n =2,=0,m =0; ψ2 = √ 2 − e s 4 2π ao ao 3/2 1 Z Zr −Zr/2ao =1,m =0; ψ2 = √ e cos θ pz 4 2π ao ao 3/2 1 Z Zr −Zr/2ao =1,m = ±1; ψ2 = √ e sin θ cos φ px 4 2π ao ao 3/2 √1 Z Zr −Zr/2ao ψ2py = e sin θ sin φ 4 2π ao ao 3/2 2 2 √1 Z Zr Z r −Zr/3ao n =3,=0,m =0; ψ3s = 27 − 18 +2 2 e 81 3π ao ao ao √ 3/2 2 2 √2 Z Zr − Z r −Zr/3ao =1,m =0; ψ3pz = 6 2 e cos θ 81 π ao ao ao √ 3/2 2 2 ± √2 Z Zr − Z r −Zr/3ao =1,m = 1; ψ3px = 6 2 e sin θ cos φ 81 π ao ao ao √ 3/2 2 2 √2 Z Zr − Z r −Zr/3ao ψ3py = 81 6 2 e sin θ sin φ π a o ao ao 3/2 2 2 √1 Z Z r −Zr/3ao 2 − =2,m =0; ψ3d 2 = 2 e 3cos θ 1 z 81 6π ao ao √ 3/2 2 2 ± √2 Z Z r −Zr/3ao =2,m = 1; ψ3dxz = 2 e sin θ cos θ cos φ 81 π ao ao √ 3/2 2 2 √2 Z Z r −Zr/3ao ψ3dyz = 2 e sin θ cos θ sin φ 81 π ao ao 3 2 / 2 2 2 ± √1 Z Z r −Zr/3ao =2,m = 2; ψ3d 2 2 = 2 e sin θ cos 2φ x −y 81 2π ao a o 3/2 2 2 √1 Z Z r −Zr/3ao 2 ψ3dxy = 2 e sin θ sin 2φ 81 2π ao ao 20 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

3 2 rR1s(r) 1 0 2 rR (r) rR (r) 2p 1 2s

-1

1 rR3d(r) rR3s(r) rR3p(r)

-1

rR4s(r) rR4p(r) rR (r) 0 4f

rR4d(r) -1 0 5 10 15 20 0 5 10 15 20 0 5 10 15 20 r/Å r/Å r/Å Figure 1.10: Eﬀective radial wave functions of H atom orbitals for n =1− 4.

+Ze. The eigenfunctions of this particular version of Eq. (1.17) are simply rRn(r) . Thus, the radial part of the wavefunction for an electron in a hydrogenic atom is effectively a solution of the one-dimensional Schrödinger equation for a particle of mass μH trapped in a “box” defined by a Coulomb potential energy function. Because of this relationship to our familiar particle-in-a-box problem (see also Fig. 1.7), we have chosen to consider the product rRn(r) rather than Rn(r) itself when discussing the radial behaviour of hydrogenic wave functions. The radial and angular behaviour of some of the orbitals of Table 1.2 are shown in Figs. 1.10 and 1.11, respectively. As with the one-dimensional eigenfunctions discussed in § 1.3, these wave functions oscillate between positive and negative values along any one of the three polar coordinates r, θ or φ, and the value of the associated energy eigenvalue depends on the total number of nodes or nodal surfaces. For example, the

ψ2s(r, θ, φ) eigenfunction has one internal radial node and no angular nodes, whereas the three ψ2p(r, θ, φ) functions have no internal radial nodes and one angular node; thus, each of these n = 2 wavefunctions has a total of one nodal surface. Similarly, each of the three types of n = 3 solutions has a total of two nodal surfaces, while each solution for n =4 has three. Thus, the familiar “principal” quantum number for hydrogenic atomic orbitals may be written as

n = {no. radial nodes} + {no. angular nodes} +1 in which (as in §1.3.2), the total number of angular nodes is given by the orbital quantum number . Thus, the three- dimensional hydrogen atomic orbitals may also be considered to be solutions of a particle-in-a-box problem. At this point it is important to recall what the atomic wave function ψ really is: it is a mathematical description of the properties of the electron, and it generally has oscillatory behaviour of the type commonly associated with classical wave motion. Because it is an ordinary mathemati- Figure 1.11: Angular behaviour of H atom cal function, it can have positive and negative values, as seen orbitals for n =1− 3. 1.4. ELECTRONIC STRUCTURE OF ATOMS AND MOLECULES 21

Figure 1.12: Atomic orbital energies in some many-electron atoms. in Figs. 1.6 − 1.8, 1.10 and 1.11. The algebraic sign of the wave function is important when we consider how orbitals overlap to hybridize (e.g., to yield sp or sp3 orbitals) or to yield bonding or anti-bonding molecular orbitals. However, insofar as the wavefunction represents a physical property, it is its square |ψ|2,the “probability density” of ﬁnding a system in a particular conﬁguration, that matters.

1.4.2 Multi-Electron Atoms and Atomic Spectroscopy The orbital wave functions that are commonly used to describe electronic structure in atoms or molecules are solutions to the simple one-electron hydrogenic atom Schrödinger equation described above. However, although orbitals in multi-electron atoms have the same types of symmetry seen there, and many similar properties, they are different for two reasons. The first is simply the fact that the larger nuclear charge +Ze gives rise to an electron–nucleus attraction which is Z times stronger than that in a simple H-atom; this observation also applies to our general hydrogenic (one-electron) atoms, as seen in the exact wave function and eigenvalue expressions of Table 1.2 and Eq. (1.15). More serious complications arise from the effects of electron–electron repulsion and the fact that inner-shell electrons partially shield the nucleus from the outer ones, so that they effectively feel only a portion of its actual charge. On the one hand, these considerations greatly increase the complexity of the Schrödinger equation so that it cannot be solved exactly in closed form, not even for the two-electron He atom. On the other hand, numerical quantum-chemistry computer programs can solve the resulting Schrödinger equation to very high precision for a wide range of cases, and for simple atoms those computational results can be made almost as exact as one could desire. For elements in the first rows of the periodic table, Fig. 1.12 shows the accurate calculated energies of the valence (outer shell) electrons. On the extreme left side of this figure we see that the He+(1s) binding energy is four times larger than that for H(1s), a straightforward manifestation of the effect of the Z2 factor in Eq. (1.15). However, the analogous binding energy of a 1s electron in a neutral He atom is only approximately twice (instead of four times) as large as that for an H atom, partly because of the electron–electron repulsion energy, and partly because each of those two 1s electrons partially shields the nucleus from the other. For the two 1s electrons of a neutral Li atom the same considerations apply, making their binding energy far smaller than the factor of nine times stronger than that for H that is predicted by Eq. (1.15) for a one-electron Li+2 system. Moreover, the fact that those two 1s electrons are so tightly bound to the nucleus means that they shield it very efficiently from the outermost 2s electron whose binding energy, as a result, is only about twice 22 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

(instead of nine times!) as strong as that for the 2s orbital of an H atom. These same considerations explain most of the trends among the level energies as one continues across this diagram. The most significant additional effect, seen for the 2p electrons for elements from C to F, is that when the p subshell has between 2 and 5 electrons, the p orbitals in a given atom are no longer all energetically equivalent to one another. The trends in behaviour seen in Fig. 1.12 are certainly interesting in their own right. However, the main objective of this discussion is to point out that as in the simple H atom, complex many-electron atoms also have ladders of energy levels, and that the associated wave functions also oscillate increasingly rapidly in space and have more nodal surfaces as the energy increases. For a multi-electron atom the total electronic energy is the sum of the energies of all its electrons, and as with the H atom, there always exist an infinite number of empty allowed eigenstates lying above the highest occupied ones. In atomic spectroscopy, light whose photon energy matches one of the level spacings in the atom promotes one or more of the electrons from one of the occupied lower-energy orbitals into one of the unfilled higher- energy orbitals. For hydrogenic atoms those orbital energies are accurately given by Eq. (1.15), but for multi-electron atoms there are no analogous explicit expressions for the orbital energies. On the other hand, accurate measurements have been made of the allowed spectroscopic transition energies of all of the atoms in the periodic table, and those results pose a persistent challenge to theoreticians as the latter pursue the development of ever better and more accurate computational methods. However, the most important practical application of atomic spectroscopy is the fact that the unique spectroscopic fingerprints of allowed transitions for each species provides a marvelously general technique for identifying the atomic composition of unknown molecules and materials.

1.4.3 Molecular Energies and the Born-Oppenheimer Approximation A molecule differs from a multi-electron atom in that instead of having a single positively charged nucleus, it has two or more positively charged nuclei about which the electrons are distributed. This makes the description of molecules much more complicated than for atoms because one must take account of the simultaneous motion of both the nuclei and the several electrons. Fortunately, a remarkably effective approximation method, the Born-Oppenheimer approximation, allows the nuclear and the electronic motion to be treated separately, and permits a relatively straightforward treatment of the nuclear motion – the molecular vibrations and overall rotation of the molecule. In 1927, Max Born and Robert Oppenheimer recognized that since nuclei are much more massive than electrons ( mp ≈ 1800 me ), if they had comparable energies the nuclei would move much more slowly than electrons. This situation suggested that for each instantaneous configuration of those slowly-moving nuclei one might ignore their motion and solve the Schrödinger equation for the electrons alone. The detailed properties of those solutions would, of course, vary as the nuclei slowly moved about, but taking account of such changes would be much simpler than trying to solve the Schrödinger equation for all electrons and nuclei at the same time. In the mathematical language of quantum mechanics, the Born-Oppenheimer approximation assumes that the total Hamiltonian for the system may be written as the sum of a Hamiltonian describing the behaviour of the electrons and one describing the nuclear motion Htotal = Helectrons + Hnuclei (1.32) and that the total wave function can be written as a product of functions characterizing the electronic and the nuclear motion ψtotal = ψelectrons × ψnuclei (1.33)

The electronic wave functions ψelectrons are the solutions of the electronic Schrödinger equation Helectrons ψelectrons = Eel ψelectrons (1.34) for the nuclei fixed in one particular configuration. However, both these wave functions and the electronic energy eigenvalues Eel will vary as the nuclei move. Just as an atom has many electronic energy levels, so does a molecule. For a molecule, however, those energies depend on the relative positions of the various nuclei. For the diatomic molecule Li2, Fig. 1.13 1.4. ELECTRONIC STRUCTURE OF ATOMS AND MOLECULES 23

Li 30000 2 energy 2 2 /cm-1 Li( S) + Li( P)

20000

10000 Li(2S) + Li(2S)

0 2 re 4 r/Å 6 8 Figure 1.13: Potential energy curves for the ten lowest energy electronic states of Li2.

shows how the energies Eel = Eel(r) calculated by solving the electronic Schrödinger equation vary with the 12 internuclear distance r for all molecular states that correlate with the two lowest Li2 dissociation limits. In particular, at each selected internuclear distance r, the electronic Schrödinger equation was solved numerically to determine the electronic energy levels shown by the stacks of square points at that value of r.Onecan perform such electronic structure calculations on as dense a mesh of distances as desired; doing so and joining adjacent points associated with electronic wave function solutions of the same symmetry yields the curves seen in Fig. 1.13. These curves of electronic energy vs. r turn out to be the effective potential energy functions, normally denoted V (r)=Eel(r) , that govern the vibrational motion and collisions of the atoms. Figure 1.13 shows that these molecular potential energy curves have a wide variety of shapes. Some, such as the very lowest curve, have the attractive single-well shape that is typical of the ground electronic state of most molecules. We will see in Chapter 3 that we can think of molecular vibrations as being the motion of a frictionless ball rolling back and forth in the well formed by this type of potential. In contrast, other curves (such as the second-lowest and the uppermost ones in Fig. 1.13) are mainly “repulsive”, so that if a ball were set free to roll along one of them, it would rapidly roll down and out to infinity, a process which corresponds to dissociation of the molecule. One also sometimes finds molecular potential energy curves with “humps” or with two or more minima. Although this diverse range of behaviour may seem somewhat intimidating, in this course we consider mainly molecular states with the simple single-minimum behaviour of the lowest curve in Fig. 1.13, which is characteristic of the ground (or lowest-energy) state of most molecules. Up to this point our discussion of the Born-Oppenheimer approximation has focussed on discussing the solution of the electronic Schrödinger equation, Eq. (1.34), and on the dependence of its energy eigenvalues on internuclear separation. However, an amazing feature of this approximation is the extremely simple form of the resulting differential equation governing the nuclear motion. As you will learn in your first quantum mechanics course, on substituting Eqs. (1.32) and (1.33) into Eq. (1.31) and then neglecting certain minor terms, one obtains an effective Schrödinger equation for vibrational motion that for a diatomic molecule has the familiar particle-in-a-box form

2 d2 ψ(r) − + V (r) ψ(r)=Eψ(r) (1.35) 2μ dr2 in which μ = mA mB/ (mA + mB) is the eﬀective or vibrational “reduced mass” for a diatomic molecule formed from atoms A and B with masses mA and mB, respectively. This equation clearly has exactly the same

12 I. Schmidt-Mink, W. M¨uller and W. Meyer, Chemical Physics 92, 263 (1985). 24 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

Figure 1.14: Regions of the electromagnetic spectrum and the type of molecular motion and spectroscopy associated with each. form as our generic Schrödinger equation Eq. (1.17) for the one-dimensional motion of a particle subject to the potential energy function V (x)! Hence, solution of the electronic Schrödinger equation Eq. (1.34) yields the potential-energy function V (r) that governs the dynamical behaviour of the nuclei, and the differential equation for the latter is essentially the same as that for a particle trapped in a one-dimensional box. Chapter 3 exploits this result further.

1.5 Spectroscopy at last . . .

By now you must realize that all the information that we need concerning a molecule and its quantized states can be obtained from the Schrödinger equation, Htotal ψtotal = Etotal ψtotal (1.36) in which Htotal incorporates all interactions within the system, Etotal is one of the observable energy eigenvalues of the system, and ψtotal is the wave function that describes the properties of that eigenstate. What we actually measure in spectroscopy are differences between the energies of pairs of quantized energy levels. This is done by observing the emission or absorption of light whose photon energies correspond to those level spacings. Analyzing such results could be extremely complicated if we had to solve the Schrödinger equation simultaneously for all of the types of motion that a molecule can have. In particular, a molecule can have electronic, vibrational, rotational and nuclear-spin-orientation degrees of freedom, each with their own characteristic sets of level energies. Fortunately, within an extended Born-Oppenheimer type of approximation, each type of motion can be treated independently, which allows us to treat molecular energies for the different types of motion separately. Moreover, as illustrated by Fig. 1.14, transitions between levels associated with the different types of motion tend to lie in distinctly different regions of the electromagnetic spectrum. (A fascinating survey of the properties of the different regions of the electromagnetic spectrum may be found on the NASA www site at http://missionscience.nasa.gov/ems/.) As a straightforward extension of Eq. (1.32), the Hamiltonian for nuclear motion Hnuclei can be written as a sum of terms representing four different types of nuclear behaviour, thereby allowing Htotal to be written as: Htotal = Helectrons + Hnuclei = Helectrons + Hvibration + Hrotation + Hnuclear−spin (1.37) 1.6. PROBLEMS 25

Similarly, the overall wave function can be written as a product of wave functions corresponding to the diﬀerent types of motion:

ψtotal = ψelectrons × ψvibration × ψrotation × ψnuclear−spin (1.38) and the total energy as the sum of the various types of energy:

Etotal = Eel + Evibration + Erotation + Enuclear−spin (1.39) In the following, Chapter 2 will discuss the spectra associated with Hrotation, Chapters 3 and 4 the spectra associated with Hvibration and its coupling with Hrotation, Chapters 5 and 6 the types of spectra arising from the combination of the eﬀects of Helectrons + Hvibration + Hrotation , while Chapter 7 discusses the manifestations of Hnuclear spin, which are completely uncoupled from the other types of motion.

1.6 Problems

1. Calculate the frequency and wavenumber of electromagnetic radiation with a wavelength of 4.0×10−10 m. What region of the electromagnetic spectrum does this fall into? What is the energy of the photons from this radiation in joules/photon and kilo-joules/mole?

2. Given that it takes a minimum energy of 4.40 eV to dislodge electrons from the surface of chromium metal, what is the maximum kinetic energy of electrons emitted from a chromium surface when it is irradiated with ultraviolet radiation of wavelength 200 nm? What is the de Broglie wavelength of the emitted electron?

3. Two energy levels in a molecule are separated by 7.50×10−22 J. What is the energy separation in −1 kJ·mol and cm−1? What are the frequency and wavelength of light that will drive a transition between these two energy levels?

4. The Ritz-Paschen series of the H atomic emission spectrum consists of a series of lines corresponding to n → n transitions into n =3 from n > 3 levels. What are the wavelengths, wavenumbers and frequencies of the four longest-wavelength transitions in this series? [Give your answers to 6 signiﬁcant digits.]

5. For each of the following transition frequencies, calculate the corresponding wavelength, wavenumber, and energy separation in both J and eV: (a) 95.3 [MHz] (b) 20 [GHz] (c) 6.4×1017 [Hz]

6. Spectroscopists often talk about energy separations in units other than standard SI energy units, e.g., using Hz, cm−1, or electron volts to characterize a change of energy. What is the reason for this?

7. In a Bose-Einstein condensate experiment, the atoms that were “condensed” were 87Rb, which have an atomic mass of 86.909 u. Given that their velocity in the condensate was 5.0×10−3 ms−1,whatwas their de Broglie wavelength?

8. Sketch the radial and angular portions of all atomic orbitals in the n = 4 shell of the hydrogen atom, indicating the algebraic sign (+ or −) in the various regions. What is the signiﬁcance of the mathematical signs in the various regions?

9. Calculate the energy per photon and the energy per mole of photons for radiation of wavelength: (a) 1.00 cm (microwave) (b) 600. nm (red) (c) 550. nm (yellow) (d) 400. nm (blue) (e) 200. nm (ultraviolet) (f) 150. pm (X-ray) 26 CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY

10. The Brackett series of H-atom emission lines consists of a series of lines corresponding to transitions from the n1 > 4tothen2 = 4 atomic shell. What is the long wavelength limit and the associated value of the quantum number n1 for these transitions? What is the short wavelength limit for this series? [give your answers in nm]

11. The hydrogenic level energy expression of Eq. (1.15) applies to any one-electron system. Using this equation, determine the wavenumbers for the two lowest-energy transitions, and name the associated spectral region: (a) for the “Balmer series” of 9Be+3 (b) for the “Pfund series” of 56Fe+25.

12. Eq. (1.15) shows that different isotopes of atomic hydrogen have different discrete level energies, and hence different level energy spacings. For the Lyman emission series of a deuterium (D) atom, what are the wavelengths and wavenumbers for the four transitions of lowest energy?

13. For an electron trapped in a 1-D box of length L =5.5 A,˚ assuming all possible transitions are ‘allowed’, what are the energies in cm−1 of the four lowest-energy transitions ? [Note: this is an approximate model for the energies of π electrons in linear conjugated hydrocarbons.]

14. Because a photon carries one unit of angular momentum, allowed rotational or angular motion transitions must correspond to changes of ±1 in the relevant angular momentum quantum number. Consider a system consisting of a particle moving on a ring (as discussed in § 1.3.2), for which three adjacent transitions are observed to occur with wavenumbers 349.8, 489.6 and 629.7 cm−1,

(a) What are the values of the quantum number for the upper and lower levels of these transitions? (b) If the particle is an H atom, what is the radius of the ring around which that H atom is moving? (c) If the H atom of part (b) were replaced by a 133Cs atom moving on a ring with the same radius, what would be the wavenumbers and frequencies of the transitions associated with the same quantum number changes? In what region of the electromagnetic spectrum would they lie?

15. Consider a particle trapped in a rigid box of length L =0.75 A.˚ For this type of system, the most strongly allowed spectroscopic transitions are those for which Δn = ±1. (a) If the particle is a 7Li atom, what are the wavenumbers (in cm−1) and wavelengths (in nm) of the four transitions of lowest-energy? Note that transitions are normally labeled by the quantum number of the lower level, and by the quantum number change in the transition; thus, in the present case the

wavenumber of a transition between nupper and nlower would be identiﬁed asν ˜Δn(nlower), where Δn = nupper − nlower . (b) If the 7Li atom were replaced by a 6Li atom, what would be the “isotope shift” Δ˜ν(n)=˜ν(6Li)− ν˜(7Li) for each of the transitions considered in part (a)? 6 (c) For the Li atom in this hard-wall box, what would be the wavenumbers of theν ˜2(3) andν ˜3(2) transitions?

16. The visible emission of Li, Na and K atoms occurs at 670.7 nm, 589.2 nm and 405.0 nm, respectively. Assuming that these emissions are all due to electronic transitions from the valence p orbitals to the valence s orbitals, determine the valence s − p energy spacings (in joules) for each of these atoms.

17. For long-chain conjugated hydrocarbon molecules, a surprisingly good approximation to the ﬁrst electronic excitation energy of the extended π bond is to treat it as a particle-in-a-box problem, with the electron being trapped in a rigid box whose length is the sum of the lengths of the bonds sharing the π electrons. Using appropriate bond lengths based on those of cyanoacetylene (see e.g., Chapter 9 of Handbook of Chemistry and Physics,athttp://www.hbcpnetbase.com), predict the energy (in cm−1)of the lowest energy electronic transition of the molecule H–C≡C–C≡C–C≡N. Chapter 2

Rotational Spectroscopy

What Is It? Rotational spectroscopy detects transitions between the quantized energy levels of a molecule rotating freely in space.

HowDoWeDoIt? Transitions are observed by measuring the frequency and amount of microwave radiation that is absorbed or emitted by rotating molecules in the gas phase.

WhyDoWeDoIt? A knowledge of the pattern of rotational energy level spacings gives us values of the moment(s) of inertia of a molecule, from which we can determine the geometry of that molecule – its bond lengths and bond angles.

2.1 Classical Description of Molecular Rotation

2.1.1 Why Does Light Cause Rotational Transitions?

Since molecules consist of positive nuclei surrounded by a distribution of negatively charged electrons, they will clearly interact with the oscillating electric field of incident light. One such type of interaction is the scattering of light, which we mentioned in §1.1.3 and will discuss further in Chapter 4. However, since the distribution of charge in a molecule tends to be fixed to its structural framework, those fields will also exert forces on that framework which can cause the kinds of rotational and vibrational excitations discussed in the next three chapters. Let us begin by considering a polar diatomic molecule, such as HF, that is rotating about a fixed point in space. If nothing interferes, it will rotate forever with some fixed angular velocity. As this occurs, the component of its dipole along a chosen axis in the plane of rotation will oscillate sinusoidally, as illustrated in Fig. 2.1. However, if the electric field of light incident on this molecule oscillates at exactly the same frequency as the natural rotational motion of the molecule, the molecule will receive a periodic “push” in phase with its motion. As a result, it will pick up energy from the field and rotate faster. This is the mechanism by which a molecule gains rotational energy from incident light. It behaves like a child on a swing – if she receives a periodic push exactly in phase with the natural motion of the swing, the amplitude of the swing increases. Similarly, the oscillating dipole associated with the rotation of the molecule can do exactly what the broadcast antenna of a radio station does – emit electromagnetic radiation at the frequency of the oscillation. This is why a molecule can spontaneously emit light. Of course, the emission by the radio station antenna is very intense, since it is constructed to have a length that is a large fraction (ideally one half, so that one end of the antenna will be electrically positive at the same instant the other is negative) of the wavelength of the emitted radiation. However, such emission is very weak for a molecule, because the charge separation is generally orders of magnitude smaller than the wavelength of light associated with the frequency of the natural motion of the molecule (see Fig. 1.14). However, spontaneous emission of light by a rotating polar molecule does occur.

27 28 CHAPTER 2. ROTATIONAL SPECTROSCOPY

+ orientation − − of + − − + + − molecule + − +

direction of dipole

↑ vertical component of time → molecular dipole

Figure 2.1: Behaviour of the vertical component of the dipole ﬁeld of a polar diatomic molecule rotating clockwise in the plane of the paper.

It is immediately clear that for a molecule to undergo absorption or emission transitions of this type, it must have a permanent dipole moment. Thus, molecules such as HF, H2O, C6H5OH, or indeed any molecule that is not “symmetric”, will produce rotational spectra, and will therefore be called “rotationally active” or “microwave active”, since pure rotational transitions occur in the microwave (MW) region of the 1 electromagnetic spectrum. On the other hand, symmetric molecules with no dipole moment, such as CO2, CH4,C6H6 or C60, will be “rotationally inactive”, in that while they do still rotate, that rotational motion cannot gain energy by absorbing photons from an incident light ﬁeld, or lose it by emitting radiation.

2.1.2 Relative Motion and the Reduced Mass m1 Consider a system consisting of two particles of mass m1 and m2 → z r located at positions specified by vectors r1 and r2, respectively, as il- r→ + lustrated in Fig. 2.2. In studying molecules we are normally interested 1 → m2 in the motion of the atoms relative to one another, and not in their Rcm absolute positions in space. It is therefore convenient to replace these r→ individual-particle coordinates r1 and r2 with coordinates specifying 2 the relative position of the two nuclei r = r2 − r1 and the position R cm of the centre of mass.Thecentre of mass of a system (indicated y by the symbol “+” in Fig. 2.2) is the “balance point”, defined by the fact that relative to an arbitrary origin, the total leverage about that origin of all the particles comprising the system is equal to that for a single “effective” particle with the mass of the whole system (here x simply m1 + m2) located at R cm : Figure 2.2: Position R cm of the centre of mass “+”, and relative coordi- (m1 + m2) R cm = m1 r1 + m2 r2 (2.1) nates r for a two-body system. Combining our definitions of r and R cm, we may then express the one-particle positions r1 and r2 in terms of the centre of mass and relative coordinates R cm and r : m2 m1 r1 = R cm − r and r2 = R cm + r (2.2) m1 + m2 m1 + m2 Using the usual equations of classical mechanics, we know that the total kinetic energy of our two-particle system may be written as

1 2 1 2 1 2 1 2 KEtot = m1|v1| + m2|v2| = m1|r˙1| + m2|r˙2| (2.3) 2 2 2 2 1 The reason that rotational transitions fall in the microwave region of the spectrum is discussed in §3.7. 2.1. CLASSICAL DESCRIPTION OF MOLECULAR ROTATION 29

where vi = r˙i = dri/dt is the velocity of particle–i. Substitution of Eq. (2.2) into Eq. (2.3) yields the result that 2 2 1 m2 1 m1 KEtot = m1 R˙ cm − r˙ + m2 R˙ cm + r˙ 2 m1 + m2 2 m1 + m2 2 2 2 1 2 m2 m2 = m1 R˙ cm − R˙ cm · r˙ + r˙ 2 m1 + m2 m1 + m2 2 2 2 1 2 m1 m1 + m2 R˙ cm + R˙ cm · r˙ + r˙ 2 m1 + m2 m1 + m2 2 2 1 1 m1 m2 = (m1 + m2) R˙ cm + r˙ 2 2 m1 + m2

=KEcm +KErel (2.4) where the subscript “cm” stands for ‘centre of mass’, and the subscript “rel” stands for ‘relative motion’. In other words, the total kinetic energy for our two-particle system is simply the sum of

• the kinetic energy of a particle located at the centre of mass R cm, whose mass is the sum of the masses of all of the particles in their system, plus • the kinetic energy of relative motion, which for our two-particle system is the kinetic energy of a particle having the effective mass m1 m2/(m1 + m2) that is located at position r2s relative to a fixed origin. This effective mass associated with the relative motion in a two-particle system occurs so ubiquitously that we give it a special name and symbol; the symbol is the Greek letter “mu” (written μ) and the name is the −1 reduced mass: m1 m2 1 1 μ = = + . (2.5) m1 + m2 m1 m2 Note that while our definition of the centre of mass readily generalizes to many-particle systems, there is no multiple-particle analog of the quantity μ ;itmaybeusedonly for two-particle (i.e., diatomic molecule) systems.2

2.1.3 Motion of a Rotating Body Let us continue our discussion of a general two-particle (diatomic molecule) system. Since we are interested in the rotation of the system, we shall ignore the overall translational kinetic energy represented by the centre-of-mass term KEcm in Eq. (2.4). The total internal kinetic energy is then that of a particle of mass μ moving about a fixed origin. Since we wish to consider only rotation, the magnitude of the relative position vector |r| = re is a constant. At this point it is convenient to replace the conventional rectangular Cartesian representation of r = (x, y, z) by the spherical polar coordinates (re,θ,φ) defined in Fig. 1.9. Taking derivatives of the Cartesian coordinates with respect to time (this differentiation being denoted by a “dot” over the variable name: x˙ ≡ dx/dt) while recalling that re is constant, we obtain

x˙ = re θ˙ cos θ cos φ − φ˙ sin θ sin φ

y˙ = re θ˙ cos θ sin φ + φ˙ sin θ cos φ

z˙ = −re θ˙ sin θ. The internal or rotational kinetic energy may then be written as 2 1 1 2 2 2 KErot = 2 μ r˙ = 2 μ x˙ +˙y +˙z = ......

1 2 ˙2 ˙2 2 = 2 μ (re) θ + φ sin θ (2.6)

2 Note that this derivation is also the origin of the reduced masses μH and μA that appeared in the discussion of the Bohr model for a hydrogenic atom in §1.2.2. 30 CHAPTER 2. ROTATIONAL SPECTROSCOPY

If we assume, for convenience, that the rotation occurs in the x-y plane, then θ is ﬁxed at θ =90◦ ,so θ˙ =0 2 and sin θ = 1 . Moreover, since the constant factor μ (re) is a property characteristic of this rigid rotating 2 system, it is convenient to give it a special name and symbol: the symbol is I = μ (re) and the name is the moment of inertia. In this notation we can therefore write

2 2 1 2 (I φ˙) |L | KErot = I φ˙ = = . (2.7) 2 2 I 2 I In Eq. (2.7) we have introduced the symbol L to represent the angular momentum of the system. In a more systematic three-dimensional derivation, we would ﬁnd that

|L |≡|r × p = |r × (mr˙)| = |I ω| = Iφ,˙ (2.8) which shows that L is perpendicular to the plane of rotations. Here, ω is a vector generalization of our φ˙ for the case of rotation in an arbitrary plane (rather than strictly in the x-y plane). It is immediately clear that our expression Eq. (2.7) for the rotational kinetic energy of our rigid two-particle system is precisely 1 2 2 analogous to the familiar classical expression 2 mv = p /2m for the translational kinetic energy of a particle with mass m, speed v, and linear momentum p = mv. ThemomentofinertiaI is the analog of the particle mass m, the angular velocity φ˙ the analog of the linear velocity v, and the angular momentum |L | the analog of the linear momentum p. For a rigid molecular system rotating freely in space, the potential energy is identically zero, so that Erot =KErot . While we will not attempt to derive it here, the ﬁnal result obtained above for the case of two particles, the fact that L2 Erot =KErot = , (2.9) 2 I in which L ≡|L | is the magnitude of the orbital angular momentum, also holds for any multi-particle system. In particular, for a system of N particles, the position of the centre of mass is deﬁned by a straightforward generalization of Eq. (2.1), namely, N N mi R cm = mi ri (2.10) i=1 i=1 and the moment of inertia about an axis through that centre of mass is given by the equation

N I ⊥ 2 = mi (di ) (2.11) i=1

⊥ in which di is the perpendicular distance from particle–i to that speciﬁc axis of rotation. For a linear ⊥ molecule, di is just the distance from that particle to the center of mass, while for non-linear molecules there will be diﬀerent moments of inertia for rotation about three orthogonal axes through the centre of mass. We have now completed our overview of the classical mechanics of a rotating system. The essential results we wish to carry forward are the following.

• The total energy of any N–particle system can be decomposed into the sum of the kinetic energy for its overall translational motion in space, KEcm (which we will ignore), plus the total internal energy for the relative motion of the particles within the system.

• For rotational motion of a rigid system of N particles, the potential energy is zero and the rotational energy is given by Eq. (2.9).

• For the special case of a diatomic molecule (or any two-particle system), the general deﬁnition of the 2 moment of inertia reduces to the simple expression I = Id = μ (re) ,where μ = m1 m2/(m1 + m2).

While we will not attempt to derive it here, these same points hold in the exact quantum mechanical description of an N–particle system, and they provide the basis for our description of rotational spectra. 2.2. QUANTUM MECHANICS OF MOLECULAR ROTATION 31

2.2 Quantum Mechanics of Molecular Rotation

2.2.1 The Basics In Bohr’s theory of the H atom, the central postulate required to make it work (unjustifiable at the time!) was that the orbital angular momentum of the electron was only allowed to have discrete values which were integer multiples of = h/2π =1.054 571 628 × 10−34 J s. However, the discussion of §1.3.2 shows that for the case of a particle of mass μ rotating in a plane at a fixed radius r = re , solution of the Schrödinger equation gives the allowed energies as

( )2 ( )2 Erot = 2 = (2.12) 2μ(re) 2 Id

Comparing this expression with the classical rotational energy expression of Eq. (2.9) shows that for the case of rotation in a plane, the allowed values of the angular momentum are L = L2D = , for any (non- negative) integer value of . More generally, quantum mechanics tells us that all forms of angular momentum are quantized such that the component of the angular momentum along any space-ﬁxed axis has values that diﬀer by increments of . Of course, molecular rotation actually occurs in 3 dimensions. A straightforward extension of the discussion of §1.3.2 shows that for orbital or rotational motion in three-dimensional space, the allowed values of the angular momentum are |L | = L = L3D = J(J +1) for J =0, 1, 2, 3, ... etc. (2.13)

Note that while the symbol is commonly used to represent the total orbital angular momentum quantum number of an electron in an atom, it is a near-universal convention in spectroscopy to use the diﬀerent symbol J to represent the total angular momentum of a molecule, even though the mathematical descriptions of the large J two types of motion are very similar. Note too that since J(J +1)=J (1+1/J) −−−→ J , this angular momentum quantization is only slightly diﬀerent from our result for orbital motion in two dimensions. Coupling this three-dimensional angular momentum result with Eq. (2.9) then yields the quantum mechanical expression for rotational energy of a rigid molecular system: 2 Erot(J)= [J(J + 1)] [J] . (2.14) 2I However, spectroscopists usually express energies in cm−1, and it is customary to use the separate symbol “F ” to denote rotational energies in those units: 2 F (J) ≡ Erot(J)=Erot(J)/ 10 hc = B [J(J + 1)] [cm−1] , (2.15) where as before, J =0,1,2,3,...,etc.SinceI has units (mass)×(distance)2 , we can write

2 20 10 Cu −1 B = 2 = 2 [cm ] , (2.16) 2I 10 hc I [u A˚ ]

−1 2 2 where as usual Cu=1 6 .857 629 [u cm A˚ ], and I = μ(re) for diatomic molecules and is given by Eq. (2.11) for larger systems (see §2.5).3 This quantity B is called the rotation constant, or more explicitly the inertial rotational constant of the molecule. Since it is virtually always quoted in spectroscopists’ energy units of cm−1, it is written without the ‘tilde’ which we use to indicate when the energy F is in cm−1.Itsvalue(or values plural for non-linear molecules) is (or are) clearly determined by the molecular structure – the bond length for the case of a diatomic – and we will see that its experimental determination is the key to accurate determination of molecular structures. 3 Note that Cu would be the numerical value of the inertial rotational constant B for a diatomic molecule with a reduced mass of 1 u (approximately true for D2) and a bond length of 1 A;˚ this is the reason for calling it the “inertial constant”. 32 CHAPTER 2. ROTATIONAL SPECTROSCOPY

2.2.2 Energy Levels, Selection Rules, and Transition Energies From Eq. (2.15) we see that the allowed rotational levels of a molecule have energies of 0, 2B,6B,12B, 20B, . . . , etc., as shown in Fig. 2.3. The systematic increase in the level spacings would appear to make it absolutely trivial to make rotational assignments, i.e., to identify the upper and lower level quantum numbers associated with observed transitions. However, this is not quite as simple as it might seem. As discussed in Chapter 1, a photon is a particle with zero rest 2 56B J=7 mass and a momentum of pλ = h/λ =10 hν˜ . However, an important additional property is that it has an intrinsic angular momentum 4 (the property we call spin ) of exactly 1, in the usual angular mo- ∼ν = 14B mentum unit of . We also know from classical mechanics and from our everyday experience that in any multi-particle collision, the total → amounts of linear and angular momenta are conserved (e.g., imagine 42B J=6 two cars colliding while sliding and spinning on a frictionless sheet of ∼ ice). Since the total angular momentum of a system (e.g., molecule ν = 12B plus photon) must always be conserved, whenever a molecule absorbs or emits a photon of light, the rotational quantum number J 30B J=5 must change by ΔJ = Jupper − Jlower = ±1,whereJupper and Jlower rotational energy ∼ are the values of the rotational quantum number for the upper and ν = 10B lower levels of the transition. The ± sign appears here because angu- 20B J=4 lar momentum is a vector property, and the result of vector addition depends on the relative orientations of the two vectors: in this case, ν∼ =8B quantum mechanics allows only these two choices for the relative 12B J=3 alignment. ν∼ =6B The essential implication of the above selection rule is that only 6B J=2 rotational transitions between adjacent J levels are allowed. For our ∼ν =4B rigid-rotor system this means that the allowed transition energies are 2B J=1 0 J=0 ν˜rot =ΔF (J)=F (J +1)− F (J) J Figure 2.3: Rotational energies and − = B [(J +1)(J +2) J(J +1)] = 2B (J + 1) (2.17) level spacings for a linear rigid rotor. It is interesting to note that the frequency of the light absorbed or rot ≈ I ≈ I emitted in such a transition, νJ J /(2π ) L/(2π ) [Hz], is equal to the classical rotation frequency θ/˙ (2π)=L/(2πI) for a molecule with angular momentum of magnitude L = J (see Eq. (2.7)). Thus the classical picture presented in §2.1.1 is consistent with the quantum mechanical ΔJ = ±1 selection rule that gives rise to Eq. (2.17). Note too that Eq. (2.17) introduces an important spectroscopic convention which we will encounter again and again. It is the fact that we always use the rotational quantum number of the lower energy level to label a transition. This is true independent of whether the transition is absorption (upper ← lower) or emission (upper → lower), and independent of which quantum number is the larger (in vibrational and electronic spectroscopy, the higher energy level can have the smaller rotational quantum number). rot As illustrated in Fig. 2.3, the allowed transition energies for a linear rigid rotor areν ˜J =2B ,4B, 6B,8B, . . . , etc. In other words, this pure rotational spectrum will consist of a set of spectroscopic lines whose energy (or wavenumber) increases linearly with J, and hence those lines will be equally spaced with a separation of rot ≡ rot − rot − − Δ˜νJ ν˜J ν˜J−1 =ΔF (J) ΔF (J 1) =2B[(J +1)− J]=2B (2.18)

Examples of spectra of this type are presented in Figs. 2.4 − 2.6. In summary, we see that spectroscopic rotational transitions can only occur when the energy of the absorbed or emitted light exactly equals the spacing between the initial and ﬁnal levels, and if the following two selection rules are obeyed: 4 We will encounter spin again in Chapter 7, as it is central to nuclear magnetic resonance (NMR) spectroscopy. 2.2. QUANTUM MECHANICS OF MOLECULAR ROTATION 33

Figure 2.4: Microwave absorption spectrum of CO gas. Note the weak transitions due to the less abundant isotopologues of CO.

Rotational Selection Rule 1: ΔJ = ±1 , since the photon has an angular momentum of 1 that must be added to or subtracted from the angular momentum of the molecule when a spectroscopic transition occurs.

Rotational Selection Rule 2: The molecule must possess a permanent electric dipole moment, in order to give the oscillating electric ﬁeld of the light something to which it can apply a torque.

2.2.3 Illustrative Applications

Exercise (i): Predict the Microwave Spectrum of CO Consider the diatomic molecule 12C16O, for which a fragment of the microwave spectrum is seen in Fig. 2.4. Because it is a heteronuclear molecule it will have a permanent electric dipole, so we expect that it will undergo rotational transitions when it is exposed to radiation of the appropriate frequency. Assuming that we know that its bond length is exactly re =1.128 322 A,˚ let us predict its rotational level energies and the “frequency” (i.e., the energy or “colour”)5 of the that which will induce rotational transitions. Solution. Firstly, we need to know the inertial rotational constant B for CO. From a standard table of atomic masses,6 we obtain7

12 16 m( C) = 12.000 000 u and m( O) = 15.994 915 u .

It is then straightforward to calculate

m1 m2 12.000 000 × 15.994 915 μ = = u=6.856 209 u , m1 + m2 12.000 000 + 15.994 915 and hence to obtain B as

Cu 16.857 629 −1 −1 B = = cm =1.931 285 cm . 2 . × . 2 μ [u] re [A]˚ 6 856 209 (1 128 322) Now that the rotational constant is known, it is straightforward to calculate the energies of the rotational levels using Eq. (2.15), and the allowed microwave transition frequenciesν ˜J =ΔF (J) from Eq. (2.17); the results of these calculations are shown in Table 2.1. 5 Note that in sloppy common usage we often speak of the “frequency” of a transition when we are actually referring to its energy in cm−1 or other units. 6 See, e.g., the NIST web page http://physics.NIST.gov/PhysRefData/Compositions/ or §1 of the online edition of the Handbook of Chemistry and Physics,athttp://www.hbcpnetbase.com. 7Recall that 1 u = 1 atomic mass unit = 1.660 538 782 × 10−27 kg. 34 CHAPTER 2. ROTATIONAL SPECTROSCOPY

Table 2.1: Predicted and observed microwave spectrum of CO.

rot rot JJ(J +1) F (J)=B[J(J +1)]ν ˜J =2B(J +1)ν ˜J (obs) 00 0.0 3.845 034 3.845 033 12 3.845 034 7.690 068 7.689 919 26 11.535 102 11.535 102 11.534 510 31223.070 204 15.380 136 15.378 662 42038.450 340 19.225 170 19.222 223 53057.675 510 ......

The above calculation seems very straightforward, and the predicted J =1← 0 transition energy seen in rot −1 Table 2.1 is in very good agreement with the best experimental valueν ˜J (pbs) = 3.845 0335 cm .However, we see in Table 2.1 that the agreement becomes systematically worse with increasing J, and for the J =5← 4 −1 transition the discrepancyν ˜calc − ν˜obs =0.002 947 cm is orders of magnitude larger than the experimental uncertainty for this transition. We shall see below that the discrepancies between our calculated values and experiment can be explained if we allow for rotational stretching (centrifugal distortion) of the CO molecule. Exercise (ii): Determine the Bond Length of HF While it is all very nice to be able to predict a spectrum from a known molecular bond length, in the real world we normally wish to solve the inverse problem, that is, to determine a molecular bond length from an experimentally measured spectrum. Consider the case of HF. Low temperature experimental measurements of its pure rotational spectrum contained two adjacent lines with energies ofν ˜ = 123.129 67 and 163.936 16 cm−1. What are the rotational assignments for these two lines, and what is the molecular bond length? Solution. To begin, we must ﬁrst use the experimental data to determine the rotational constant B.Equa- tion (2.18) indicates that the separation between the two spectroscopic lines is equal to 2B,soweobtain B 1 . − . . −1 . = 2 [163 936 16 123 129 67] = 20 403 245 cm To determine the HF bond length, we then simply re-arrange Eq. (2.16) while utilizing the special expression 2 for the moment of inertia of a diatomic molecule (I = μre ),

2 2 I = Cu/B =0.826 222 94 [u A˚ ]=μ (re) .

The known atomic masses6 yield a reduced mass of −1 1 1 μ(HF) = + =0.957 055 278 u , 1.007 825 032 18.998 403 20 so we then obtain re = I/μ = 0.826 222 94/0.957 055 278 A=0˚ .929 137 8 A˚ .

Finally, to determine the rotational assignments of these two lines, we see that utilizing our experimental value of B in Eq. (2.17) yields

rot J +1 =ν ˜J /2B = 123.129 67/(2 × 20.403 245)=3.017 40 ≈ 3 , or J =2, (2.19)

since J must be an integer. This shows that our lower energy line corresponds to the transition J =3← 2, which indicates in turn that the higher energy line corresponds to J =4← 3 . Note that in this line-labeling we have used the standard spectroscopic convention that when both appear, the quantum number label for the upper level of a transition is written ﬁrst. One troublesome point about the above discussion is that the quantity actually obtained in the calculation of Eq. (2.19) is not precisely an integer, and the deviation from the nearest integer is orders of magnitude larger than what could be due to experimental uncertainties. Similarly, using the larger of the two transition energies yields J + 1 = 163.936 16/2 × 20.403 245 =4.017 40 ≈ 4 . However, as with the discrepancy mentioned at the end of the previous example, it turns out that this apparent irregularity is also due rotational stretching (centrifugal distortion) of the molecule, which is discussed in detail in the following section. 2.3. COMPLICATIONS ! 35

Figure 2.5: Microwave emission spectrum of gaseous HF showing rotational assignments for v =0 and1.

2.3 Complications ! The preceding numerical examples and the spectra shown in Figs. 2.4−2.6 point to the existence of a number of complications that arise in experimental rotational spectra.

Complication #1: Isotopologues In the calculations associated with the two illustrative examples presented above, we were always careful to use the precise atomic mass of one particular isotope of each atomic species. This was necessary, since calculating B from a knowledge of the bond length, or the inverse problem of determining re from an experimental B value, both involve the value of the reduced mass μ, and the result will clearly be different when masses of different isotopes of a given element are used. Note that since transitions are discrete properties of individual molecules, the abundance-averaged atomic mass (the standard “atomic weight”) should never be used in this type of calculation; when in doubt, one should use the mass of the most abundant isotope. At the same time, we expect that the electronic energy or chemical binding for different isotopic forms of a given chemical species (different isotopologues) should be the same. While that is not precisely true, the deviations are usually extremely small – well below the resolution we consider in this course. Thus, we make a fundamental assumption that different isotopic forms of a given chemical species have exactly the same electronic potential energy curves, and hence exactly the same equilibrium bond length re. In view of the above, in Exercise (i) of the previous section we can see that different B values, and hence different rotational energies and rotational line spectra, would be predicted for minor isotopologues of CO, such as 13C 16O, 12C 18Oor12C 17O. Indeed, the most abundant of these minor isotopologues are responsible for the minor peaks seen in Fig. 2.4. Thus, unless special isotopically pure samples are used, the various isotopologues of a given species present in a normal sample will each give rise to its own set of equally-spaced lines, with a relative intensity determined by the relative abundance of that isotopologue. This does not present much difficulty for CO, where the major isotopologue is overwhelmingly more abundant than the others (natural carbon is only 1.1% 13C), but it can make the spectra quite complicated for other cases. For example, Ge has three major isotopes with abundances of 20 − 36% and two more with abundances of about 7% (the complexity of the resulting infrared spectrum of GeO is illustrated by Fig. 3.10 on p. 71), while Mo, Ru, and Sn all have seven or more isotopes with significant abundance, so molecules formed from normal samples of these atoms will have quite congested spectra. On the other hand, this complexity can be a very useful tool for identifying the chemical species giving rise to a given spectrum, since the fact that isotope shifts may be accurately predicted from the precisely known reduced mass ratios is a very sharp diagnostic 36 CHAPTER 2. ROTATIONAL SPECTROSCOPY

Figure 2.6: Microwave absorption spectrum of H–C≡C–C≡C–C≡N, showing vibrational satellites. for identiﬁcation of a particular chemical species.

Complication #2: Vibrational Stretching and Vibrational Satellites A second type of complication is illustrated by the presence of two sets of (roughly) equally spaced lines in the HF emission spectrum of Fig. 2.5. It is to be expected that when a molecule is excited into higher vibrational energy levels, the amplitude of its vibrational motion will increase. Because of the asymmetry of typical intermolecular potential energy curves, this means that the average bond length will increase with the degree of vibrational excitation. For example, in Fig. 1.7-D on p. 16 it is clear that the average bond length in the v = 5 vibrational level is much greater than that in the v = 0 “ground” level. This in turn means that the B value in this v = 5 level will be much smaller, and hence that the spacings (of roughly 2B, see Eq. (2.18)) between the lines in its pure rotational spectrum will be distinctly smaller than those for the lower vibrational levels. Because of this dependence of the average bond length on vibrational level, it is customary to introduce the subscript “v” and use the symbol Bv to represent the inertial rotational constant. For a diatomic molecule, this means that Eq. (2.16) may be re-written as

Cu Bv = 2 (2.20) μ(rv) in which rv is the effective average bond length for a moleule in vibrational level v. For HF, this vibrational stretching causes the J =18← 17 transition to shift from 692.481 cm−1 for v = 0 to 665.937 cm−1 for v = 1 . From Eq. (2.20) we can see that this corresponds to a 3.8% decrease in Bv from v = 0 to 1, which in turn corresponds to an increase of 1.9% in the effective bond length rv. For a molecule (such as HF) that has a small moment of inertia, both the energy spacings between vibrational levels and the v–dependence of Bv (and hence also of rv) tend to be relatively large. The size of the vibrational level spacing means that excited vibrational levels will not have significant populations at normal temperatures, and this means that microwave spectra due to pure rotational transitions within higher vibrational levels will be very weak. Indeed, the only reason that they can be seen in the emission spectrum of Fig. 2.5 is that the sample was very hot. On the other hand, for polyatomic molecules with large moments of inertia, many of the several possible vibrational modes (see §3.5) have small vibrational energy spacings, and this allows their excited vibrational levels to have substantial populations at room temperature. An example of this latter type of species is the linear molecule cyanodiacetylene HC5N, for which a segment of the microwave spectrum is shown in Fig. 2.6. In this case the spacing between adjacent pure −1 −1 rotational lines is approximately 2.54 GHz or 0.0847 cm , which implies that the B value is 0.04236 cm , 2 8 and hence that the moment of inertia is I(HC5N) = Cu/0.04236 = 390 u A˚ . However, we can see that for each J+1 ← J rotational transition there is a cluster of closely spaced lines in Fig. 2.6. They are the pure rotational transitions within the ground level and within each of the several thermally populated excited vibrational levels of this molecule; the latter are called vibrational satellites. 8 Note that since this is not a diatomic molecule, there is no analog of μ, and the moment of inertia is defined by Eq. (2.11). 2.3. COMPLICATIONS ! 37

Complication #3: Rotational Stretching or Centrifugal Distortion: the Non-Rigid Rotor The preceding subsection introduced the idea that molecules are non-rigid, and that vibrational excitation can stretch bonds, and hence change moments of inertia and Bv values. Similarly, we know from personal experience that when any object rotates it “feels” an outward centrifugal force pulling it away from the centre of rotation. This also applies to molecules. Up to now, our model of a molecule has been balls (atoms) attached to the ends of rigid sticks (chemical bonds). However, chemical bonds behave more like springs than sticks, as they can stretch or compress when forces are applied, so a more accurate model for rotation must take into account the centrifugal stretching of chemical bonds. As a molecule rotates, the nuclei will be pulled apart by centrifugal forces, and those forces will increase as the rate of rotation (and hence the rotational energy) increases. For a diatomic molecule this implies that as the value of the rotational quantum number J increases, the eﬀective bond length rv = rv(J) will tend 2 to increase. This in turn leads to an increase in the moment of inertia I = I(J)=μ (rv(J)) , and since the spacing between rotational lines is 2Bv(J)=2(Cu/I) , at higher values of J the rotational line spacings will rot − rot become progressively smaller. This behaviour is evident in Fig. 2.5, where theν ˜J=14 ν˜J=13 line spacing −1 rot rot −1 of 36.0 cm is visibly larger than theν ˜25 − ν˜24 line separation of 26.4 cm , and it explains why these rot −1 rotational transition energies (e.g.,ν ˜24 = 904.385 cm ) are not simple 2(J+1) multiples of the rotational −1 constant B = B0 =20.403 cm determined in example (ii) on p. 34. It also explains the discrepancies referred to at the end of each of the illustrative examples of §2.2.3. This rotational stretching, or centrifugal distortion, is normally accounted for by including an additional term in the rotational energy expression of Eq. (2.15), to yield

2 Fv(J)=Bv[J(J +1)]− Dv [J(J +1)] , (2.21) where as usual Bv ≡ Cu/Iv(J=0 ) , a n d Dv is a positive quantity called the centrifugal distortion constant, whose magnitude depends inversely on strength of the bond. In general Dv Bv . However, weak bonds are expected to distort more than strong bonds, and hence the former will have relatively larger values of Dv/Bv. As implied by our discussion of Fig. 2.5, both the rotational transition energies

ν˜rot =ΔF (J) ≡ F (J +1)− F (J) J v 2 2 = Bv[(J +1)(J +2)]− Dv[(J +1)(J +2)] − Bv[J(J +1)]− Dv[J(J +1)] 3 =2Bv(J +1)− 4Dv(J +1) (2.22) and the rotational line spacings

rot rot − rot − 2 Δ˜νJ =˜νJ ν˜J−1 =2Bv 4Dv(3J +3J + 1) (2.23) will also be modiﬁed by the eﬀect of centrifugal distortion.

Although Dv is usually relatively small (Dv ∼ 0.0001Bv), it cannot readily be ignored when dealing with high precision data, not even at low values of J. For example, consider the experimental data for 12C16O rot listed in Table 2.2. The fact that the line spacings Δ˜νJ are not constant shows that even for CO, which has a very strong or “stiﬀ” triple bond, the eﬀects of centrifugal distortion are quite evident in high precision measurements.

Table 2.2: Experimental microwave transition energies for ground state (v =0)CO.

rot rot rot − 2 J transitionν ˜J Δ˜νJ =˜νJ ν˜J−1 (3J +3J +1) 01← 03.845 033 — — 12← 17.689 919 3.844 886 7 23← 211.534 510 3.844 591 19 34← 315.378 662 3.844 152 37 45← 419.222 223 3.843 561 61 56← 523.065 043 3.842 820 91 38 CHAPTER 2. ROTATIONAL SPECTROSCOPY

intercept =2B =3.845 059 3.845 0

∼ ΔνJ /cm−1 −5 slope = − 4 D0 = − 2.459×10

3.844

3.843

0 25 50 75 100 (J2+3J+1) Figure 2.7: Graphical determination of B0 and D0 for CO.

Exercise (iii): Using the experimental data of Table 2.2, determine the values of B0 = Bv=0 and D0 for CO. Solution. For the 1 ← 0 transition, J = 0 , and hence

rot 3 ν˜0 =2B0(J +1)− 4D0(J +1)

=2B0 − 4D0 =3.845 033

or D0 =[2B0 − 3.845 033] /4 (2.24)

Similarly, for the 6 ← 5 transition, J = 5 , and hence

rot ν˜5 =12B0 − 864D0 =23.065 043

Using Eq. (2.24) for D0,weobtain

rot ν˜5 =23.065043 = 12B0 − 864 [2B0 − 3.845 033] /4 (2.25)

−1 Solving Eq. (2.25) for B0 and substituting the result into Eq. (2.24) then yields B0 =1.922 529 cm and −6 −1 D0 =6.138 × 10 cm . An alternate, and somewhat better approach to the determination of B0 and D0 is to utilize all of the data in Table 2.2, instead of only the first and last frequency differences. In particular, Eq. (2.23) shows that a plot rot 2 of Δ˜νJ vs. (3J +3J + 1) should be linear, with intercept 2Bv and slope −4Dv . As shown in Fig. 2.7, this −6 −1 approach yields the same B0 value, but a D0 value of 6.147×10 cm that differs slightly from the one obtained above. In general, performing least-squares fits to full data sets while taking proper account of data uncertainties is the optimum way of determining molecular constants from experimental data.

Another way to think about centrifugal distortion is to use the line spacings to approximate the derivative of the energy with respect to [J(J+1)] in order to obtain a value for the centrifugally distorted effective bond length, rv(J). In particular, let us define C eff ≡ dE(v, J) − ≡ u Bv (J) = Bv 2Dv[J(J +1)] 2 (2.26) d[J(J +1)] μ [rv(J)]

Rearranging this result to solve for rv(J), and then multiplying the numerator and denominator by rv(J=0 ) yields the expression 2 1/2 Cu/μ [rv(J=0 ) ] rv(J)=rv(J =0) (2.27) Bv − 2Dv[J(J +1)] 2.4. DEGENERACIES AND INTENSITIES 39

However, rv(J=0) is just the bond length that the molecule would have if there was no rotational stretching, so the numerator on the right hand side is our deﬁnition of Bv, and this equation yields 1/2 Dv rv(J)=rv(J =0) 1 − 2[J(J +1)] (2.28) Bv

−6 For the ground vibrational level of CO, the results in Fig. 2.7 show us that D0/B0 =3.30×10 , so it will require a relatively large value of J (≈ 55) to cause even a 1% increase in the CO bond length. In contrast, −4 for the ground vibrational level of HF the ratio D0/B0 =1.0×10 , so 1% stretching occurs for J=9 a n d 10% stretching for J=30 ; thus, centrifugal stretching can be quite substantial. However, the essential point of this discussion is not to worry about precisely how fast the bond stretches with increasing J, but rather to appreciate the fact that in a more accurate description, molecules are actually non-rigid rotors whose rotational level energies are best represented either by Eq. (2.21), or by the even more general expression

2 3 4 Fv(J)=Bv[J(J +1)]− Dv [J(J +1)] + Hv [J(J +1)] + Lv [J(J +1)] + ... (2.29) in which Hv and Lv are known as higher-order centrifugal distortion constants. It is the fact that the rotational energy actually depends on the set of rotational constants {Bv,Dv,Hv,Lv, ... etc.} that led us to label Bv as the inertial rotational constant.

2.4 Degeneracies and Intensities

Rotational spectroscopy is normally performed by passing microwave radiation through a sample chamber containing the gas phase molecules of interest. Transitions may be detected in either absorption or emission mode. In absorption (the more common), microwave radiation of known intensity passes through the sample and transitions are detected as a reduction of its intensity at specific frequencies. In emission, the light emitted by a sample (which is usually very hot) is dispersed by a spectrometer. Since there is zero signal except at the particular frequencies emitted by the molecule, this latter approach can often detect weak transitions that would be very difficult to discern in an absorption experiment. [A lit match can be seen from a great distance in the dark.] For either absorption or emission, there is a single fundamental principle governing the relative strengths of the different lines in the spectrum.

All else being equal, the intensities of absorption and emission lines will be proportional to the populations of the initial levels.9

For a system in thermal equilibrium, two properties govern the relative population of a given level:

• the “degeneracy” of the level; i.e., the number of distinct molecular quantum states with exactly the same energy, and

• the Boltzmann thermal population distribution for the system.

The ﬁrst of these factors is a property of the molecule itself, and the second is a property of the ensemble of all of the molecules in the system. Let us consider each of these factors in turn.

Degeneracy of Molecular Rotational Levels

We have seen earlier that rotational angular momentum is a vector quantity whose magnitude is allowed by quantum mechanics only to have one of the discrete values L = |L | = J(J +1) , for integer values of the total angular momentum quantum number J (= 0, 1, 2, 3, . . . etc.). Since L is a vector quantity, it has components Lx, Ly and Lz pointing along the x, y and z axes in space. However, the Heisenberg uncertainty principle of quantum mechanics forbids us from simultaneously “knowing” (i.e., being able to

9 As usual in science, “all else” is never truly equal. However, we will initially overlook complicating niceties and consider only the effect of level population on intensities. 40 CHAPTER 2. ROTATIONAL SPECTROSCOPY determine experimentally) more than one of these components, in addition to L itself. Quantum mechanics also tells us that for a system with a given value of J, the one component we may know can only have one of the 2J+1 discrete values: −J, −(J − 1), −(J − 2),...,(J − 2),(J − 1) or J. This result leads to the introduction of another quantum number for specifying the state of the system, the angular momentum projection quantum number MJ , which is allowed to have one of the 2J + 1 integer values: −J, −J +1, −J +2,... , J − 2, J − 1andJ. Since the value of this angular momentum component only tells us about the direction in which L is pointing, and not about its magnitude (i.e., not about the magnitude of the rotational speed), these 2J+1 different projection states are said to be degenerate,inthatforagivenJ,the allowed MJ states all have exactly the same energy. Note that MJ can never be higher than +J or lower than −J. While any one of the three Cartesian components of L could be the one that is “known”, for mathematical convenience we almost always choose the space-fixed z-axis to be the axis of quantization. As a consequence of this choice, to identify the rotational state of a molecule it is necessary to specify both the total angular mom- | | entum quantum number J that defines the magnitude of L , and the quan- z tum number MJ that defines the value of Lz . For a rotating molecule with J = 2 , Fig. 2.8 illustrates the allowed projections of L onto the laboratory − Lz= 2 h → (or space-fixed) z axis. For present purposes, we are only concerned with L knowing that the number of degenerate sublevels associated with any given 1−h value of J is gJ=2 J+1; however, we will see in Chapter 7 that the whole phenomenon of NMR spectroscopy depends on the properties of the spatial projection of the spin angular momentum vectors of nuclei. 0−h This concept of degenerate MJ levels should be quite familiar, as it is the same property encountered in the discussion of the quantum numbers − specifying the orbital motion of an electron in a hydrogen atom. The -1 h only difference is that in discussing the electron orbits, the total angular momentum quantum number is given the label (instead of J), and the -2 −h projection quantum number was m (rather than MJ ). In that case the (2+1)-fold degeneracy associated with a given value of gave rise to the three p (for =1), the five d (for =2 ) , a n d t h e s e v e n f (for =3) degenerate Figure 2.8: Angular momen- substates. tum projections for J=2 .

The Thermal Population Distribution A central result of statistical thermodynamics is the fact that For a system in thermal equilibrium at temperature T , the probability for finding a molecule in a −Ei/kB T particular quantum state i with energy Ei is proportional to e , −23 −1 −1 −1 10 in which kB =1.380 650 4×10 [J K ]=0.695 035 6 [cm K ] is known as the Boltzmann constant. It is important to note, however, that this statement refers to “a particular quantum state” of the system. In order to specify the probability of finding the system in a particular energy level, we must sum over the populations of all distinct quantum states with that energy. Thus, an alternate formulation of the above result is the statement that For a system in thermal equilibrium at temperature T , the probability of finding a molecule in a −Ei/kB T particular energy level Ei is proportional to gi e , in which gi is the total degeneracy of level Ei. Since the sum of the probabilities for all possible levels must add up to 1, the fraction of all molecules (of a given species) with energy Ei is −Ei/kB T fi(T )= gi e Q(T ) , (2.30) −Ei/kB T in which the quantity Q(T )= i gi e , with the sum running over all possible distinct energy levels Ei, is called the molecular partition function. 10 Note that the Boltzmann constant is simply the per-molecule value of the ideal gas law “gas constant” −1 −1 23 R =8.314 472 J mol K =NA kB ,where NA =6.022 141 79 × 10 is the Avogadro number. 2.4. DEGENERACIES AND INTENSITIES 41

gJ = (2J+1) pop 10 Jmax(T)

−BJυ (J+1)/kB T gJ e

−BJ(J+1)/k T 10 × e υ B

0 0 10 J 20 Figure 2.9: Boltzmann rotational population distribution for CO at T = 293 K.

The molecular partition function is a rough measure of the number of quantum states of that molecule having signiﬁcant equilibrium populations at the given temperature. It is a very important property, since it determines the macroscopic thermodynamic behaviour of a system. We won’t discuss the partition function further in this course, but you will encounter it in upper-year chemistry courses, where it is employed to connect microscopic and macroscopic properties of matter. You will see there that our ability to make reliable predictions of the equilibrium thermodynamic properties of many systems depends critically on the knowledge of the patterns of energy levels {Ei} determined from spectroscopic measurements. For a rotating linear molecule the energy levels are speciﬁed by the total angular momentum quantum number J, and their energies are given by Eq. (2.15), (2.21) or (2.29) (depending on how precisely the pattern of level energies is known), with the level degeneracies being gJ=2 J+1. The fraction of molecules in the level with energy Fv(J) is therefore given by −Fv(J)/kBT fJ (T )= (2J +1)e Q(T ) . (2.31)

For simplicity, let us consider the rigid rotor case, for which Fv(J)=J(J+1 ) Bv . For such a system, it is easy to see that fJ (T ) is the product of the term (2J+1), which increases linearly with J,andtheterm e−Bv J(J+1)/kBT , which decreases exponentially as J increases. The competition between these two terms gives the overall behaviour shown by the solid curve in Fig. 2.9, while the inﬂuence of the initial-state population on experimental line intensities may be seen in Figs. 2.4 and 2.5 (a cleaner example will be seen in the vibration-rotation spectra discussion of Chapter 3). Since Eq. (2.31) provides us with a simple analytic expression for the (fractional) equilibrium population of any given rotational level, it is a straightforward matter to address the following question. For a system in thermal equilibrium at temperature T , what is the value of J for the most highly populated rotational level ? pop Calculus tells us that the location of this maximum will be the value of J, denoted Jmax(T ), for which the derivative of fJ (T ) with respect to J is zero. Hence, for a rigid rotor: d 2 −B (2J +1)e−Bv J(J+1)/kBT =0=e−Bv(J +J)/kBT 2+(2J +1) v (2J +1) (2.32) dJ kBT

pop Removing the exponential common factor and solving for J = Jmax(T ) yields the expression pop kBT 1 J = Jmax(T )= − (2.33) 2 Bv 2 for the most populated level at the given temperature. As mentioned earlier,9 however, “all else” is never truly equal, and theory tells us that in addition to the initial-state population, absorption intensity depends on a linear power of the transition energyν ˜, while the 42 CHAPTER 2. ROTATIONAL SPECTROSCOPY full emission intensity expression includes a factor ofν ˜4. These factors have little eﬀect on the rotational line intensities in vibrational or electronic spectra (see Chapters 3 and 5), but they can be quite important for pure rotational spectra. In particular, using the rigid rotor expression of Eq. (2.17), a slight generalization of the above discussion shows that the value of J for the most intense line in a pure rotational absorption or emission spectrum, respectively, is given by11 √ √ abs pop em pop J max(T ) ≈ 3 J max(T )andJ max(T ) ≈ 5 J max(T ) (2.34)

2.5 Rotational Spectra of Polyatomic Molecules

2.5.1 Linear Molecules are (Relatively) Easy to Treat! For a general polyatomic molecule consisting of N atoms distributed in space, introducing centre-of-mass and relative coordinates in the d same manner as in our diatomic molecule derivation in §2.1.2 yields A m m the same separation of the total energy into the sum of the kinetic 1 2 energy of the centre of mass plus the internal energy. That internal d d energy in turn separates into a sum of the internal vibrational energy 12 12 plus the overall rotational energy of the molecule. For any linear B molecule, that rotational energy is also given by Eq. (2.9), where m2 m1 m2 again the allowed values of |L | are given by Eq. (2.13). Thus, up to this point the treatment of an arbitrary linear molecule (such as d12 d22 d12 HC5N of Fig. 2.6) is identical to that for a diatomic molecule. In C fact, the only diﬀerence between the treatment of a rigid diatomic m1 m2 m2 m1 molecule and a rigid linear polyatomic is in the way we describe the I moment of inertia . d12 d23 As mentioned in §2.1.3, for an arbitrary multi-particle system D the position of the centre of mass is given by Eq. (2.10) and the m1 m2 m3 moment of inertia by Eq. (2.11). For the particular case of a diatomic 2 molecule, the latter happens to simplify to the expression Id=μ(re) Figure 2.10: Four types of linear used in the early parts of this Chapter, but that expression cannot molecules. be used for a molecule consisting of more than two particles. We therefore begin by demonstrating the use of Eqs. (2.10) and (2.11) for the four types of linear molecule illustrated in Fig. 2.10. Since these molecules are all linear, the three-dimensional centre-of-mass position vector R cm =(xcm,ycm,zcm) may be replaced by a one-dimensional coordinate, and for convenience, we assume that the atoms lie on the x axis, with the leftmost atom being at the coordinate origin. Note, however, that precisely the same results for the position of the centre of mass in the molecule and the resulting value of I are obtained using any other choice for this origin.

Case A: ADiatomicMolecule. Applying Eq. (2.10) yields the centre-of-mass position xcm =(0×m1 + d×m2) /(m1 + m1)=d[m2/(m1 + m2)] , a point that lies a distance x2 = d − xcm = d[m1/(m1 + m2)] from atom m2. The moment of inertia is then readily calculated from Eq. (2.11):

2 2 I = m1 (−xcm) + m2 (d − xcm) 2 2 2 2 m2 d m1 d = m1 2 + m2 2 (m1 + m2) (m1 + m2) m1 m2 2 2 = d = μd = Id . (2.35) m1 + m2

Thus, the general deﬁnition (of Eq. (2.11)) for the moment of inertia also yields our familiar diatomic molecule result. 11 R.J. Le Roy, Journal of Molecular Spectroscopy 192, 237 (1998). 2.5. ROTATIONAL SPECTRA OF POLYATOMIC MOLECULES 43

Case B: A Symmetric Triatomic Molecule. In cases such as this, one can see by inspection that the centre of mass must lie on the middle atom, at a distance d12 from the origin and from each of the other atoms. However, a straightforward application of Eq. (2.10) yields the result:

xcm =(m2 ×0+m1 d12 + m2(d12 + d12)) /(m2 + m1 + m2)

= d12(m1 +2m2)/(m1 +2m2)=d12 . (2.36)

Since the middle atom m1 lies at the centre-of-mass position, it makes no direct contribution to the value of the moment of inertia, and hence

2 2 2 2 I = m2(−d12) + m1(0) + m2(d12) =2m2 (d12) . (2.37)

Case C: A Symmetric Tetra-atomic Molecule. This is clearly another case where we know by inspection that the centre of mass lies precisely in the middle of the molecule. Once again, however, our formal deﬁnition readily gives the result:

xcm =[m1 ×0+m2 d12 + m2(d12 + d22)+m1(d12 + d22 + d12)] /(m1 + m2 + m2 + m1)

m2(2d12 + d22)+m1(2d12 + d22) 1 = = d12 + 2 d22 , (2.38) 2(m1 + m2) which conﬁrms our intuitive prediction. Combining this result with our formal deﬁnition of the moment of inertia then yields

1 2 1 2 1 2 1 2 I = m1(−d12 − 2 d22) + m2(− 2 d22) + m2( 2 d22) + m1(d12 + 2 d22) 1 2 1 2 =2m1 (d12 + 2 d22) +2m2( 2 d22) . (2.39)

Case D: An Asymmetric Triatomic Molecule. This case is a little more interesting, because while the centre of mass must lie on the molecular axis, one cannot locate it by inspection. However, our general deﬁnitions still apply. In particular, recalling that the leftmost atom, m1, is located at the coordinate origin, we obtain

xcm =[m1 ×0+m2 d12 + m3(d12 + d23)]/ (m1 + m2 + m3) . (2.40)

Relative to this centre of mass, atom m1 is located at x1 = −xcm ,atomm2 at x2 =(d12 − xcm), and atom m3 at x3 =(d12 + d23 − xcm) . Our deﬁnition of the moment of inertia then yields

2 2 2 I = m1 (−xcm) + m2 (d12 − xcm) + m3 (d12 + d23 − xcm) { ···} = some tedious algebra 2 2 2 = m1(d12) + m3(d23) − (m1 d12 − m3 d23) / (m1 + m2 + m3) . (2.41)

Note that if m1=m3 and d12=d23 , this expression collapses to the symmetric triatomic result of Eq. (2.37).

Case E: Other Types of Linear Molecules. For larger non-symmetric multiple-atom linear molecules, one can, of course, apply the general deﬁni- tions of Eqs. (2.10) and (2.11) and obtain closed-form algebraic expressions for the moment of inertia as functions of the relevant atomic masses and bond lengths. Such expressions become ever clumsier, and little is gained by writing them down. However, when the bond lengths are all known, numerical application of our equations for xcm and I is remarkably straightforward, and readily gives accurate results. For larger molecules that have a centre of symmetry, such as N≡C–C≡C–C≡N (cyanodiacetylene), it is quite straightforward to locate the centre of mass by inspection (in this case, in the middle of the carbon–carbon triple bond), and then simply write down an expression for the moment of inertia. At 44 CHAPTER 2. ROTATIONAL SPECTROSCOPY

ﬁrst glance this would seem to be rather uninteresting, since such symmetric molecules would have no permanent dipole moment, and hence no normal (i.e., dipole allowed) rotational spectra. However, as we shall see in Chapters 3 and 4, the rotational level spacings of such species may be observed as ﬁne structure in vibrational or electronic spectra, or more directly by Raman spectroscopy. Thus, it is still quite relevant to understand how their moments of inertia, that are obtained from the experimentally determined Bv values, are related to the molecular bond lengths.

2.5.2 Illustrative Applications Exercise (iv): Predict the microwave spectrum of cyanodiacetylene H–C≡C–C≡C–C≡N. The microwave spectrum of cyanodiacetylene was shown in Fig. 2.6. If we had first observed that spectrum in a gaseous mixture in chemical effluent from an industrial plant, how would we know what it was due to? One way to identify unknown species in spectra is to use predictions based on general chemical knowledge to predict the structure, and hence the moment(s) of inertia and the spectra of various possible culprits. We will now do that for cyanodiacetylene, H–C≡C–C≡C–C≡N. From the data table in Chapter 9 of the Handbook of Chemistry and Physics, we find that for the simpler molecule cyanoacetylene (H–C≡C–C≡N), the bond lengths are given as rC−H=1 .058 A,˚ rC≡C=1 .205 A,˚ rC−C=1 .378 Aand˚ rC≡N=1 .159 A.˚ Assuming that these bond lengths are essentially unchanged in cyanodiacetylene, if we choose the origin of our coordinate system to lie at the H atom, the position of the centre of 1 12 14 mass is (using the masses of H, Cand NformH, mC and mO, respectively)

xcm =[mH ×0+mC rC−H + mC (rC−H + rC≡C)+mC (rC−H + rC≡C + rC−C)

+ mC (rC−H +2rC≡C + rC−C)+mC (rC−H +2rC≡C +2rC−C)

+ mN (rC−H +2rC≡C +2rC−C + rC≡N)] /[mH +5mC + mN]=······ = 319.7687/75.01090 = 4.2630 [A]˚ .

This shows that the centre of mass lies roughly in the middle of the second C≡C bond, and relative to that point, the seven atoms in the molecular chain running from H to N, are located at x = −4.2630, −3.2050, −2.0000, −0.6220, +0.5830, +1.9610 and 3.1200 A,˚ respectively. From this knowledge of each atom’s distance ⊥ di to the centre of mass, we can readily calculate

2 2 2 2 I(HC5N) = 1.007 825(−4.2630) +12.000 000 (−3.2050) +(−2.0000) +(−0.6220) 2 2 2 +(0.5830) +(1.9610) +14.003 074(3.1200) 2 = 380.759 [u A˚ ] .

This result then gives us a prediction for the rotational constant of cyanodiacetylene:

−1 B(HC5N) ≈ Cu/I =0.04427 [cm ]

The similarity between this prediction and the experimental value of B=0 .0444 [cm−1] , obtained as half of the spacings between the clusters of lines in Fig. 2.6, gives us confidence that the molecule giving rise to that spectrum was actually HC5N. Further confirmation could be obtained by predicting the isotopic shifts of the moment of inertia (and hence of the transition frequencies) obtained on replacing 14Nby15N (0.37% abundance) or on replacing one of the 12Catomsby13C (1.1% abundance), and looking for weak shifted lines in the experimental spectrum at the corresponding predicted frequencies. Exercise (v): Determining the Bond Lengths of Carbonyl Sulfide O=C=S . As we have seen for diatomic molecules, one of the key applications of rotational spectroscopy is for determining molecular bond lengths from the experimentally determined moment(s) of inertia. Let us use this approach to determine the lengths rCO and rCS of the C=O and C=S bonds in the linear triatomic molecule O=C=S. One thing that is immediately clear is that we cannot determine these two independent bond lengths from a single experimental observable. However, consideration of Eq. (2.41) suggests that if we had experimental rotational constants (and hence experimental I values) for two different isotopic forms of this molecule, we would be able to use the resulting two equations in two unknowns to determine the two bond lengths. Let us assume, then, that experiment has given us B0 values for two different OCS isotopologues:

16 12 32 −1 B0( O C S) = 0.202 864 [cm ] 16 12 34 −1 B0( O C S) = 0.197 910 [cm ] . 2.5. ROTATIONAL SPECTRA OF POLYATOMIC MOLECULES 45

We know that for any molecule, Bv = Cu/I ,sothetwomomentsofinertiaare

16 12 32 2 Ia ≡ I( O C S) = Cu/0.202 864 = 83.098 18 [u A˚ ] 16 12 34 2 Ib ≡ I( O C S) = Cu/0.197 910 = 85.178 26 [u A˚ ] .

16 12 32 To simplify the following expressions, we deﬁne mO = m( O), mC = m( C), mS = m( S) and ∗ 34 mS = m( S), all in units u. If we explicitly write out Eq. (2.41) for our two isotopologues, we obtain

2 2 2 (mO rCO − mS rCS) Ia = mO(rCO) + mS(rCS) − mO + mC + mS m r − m∗ r 2 I m r 2 m∗ r 2 − ( O CO S CS) . b = O( CO) + S( CS) ∗ mO + mC + mS

After expanding the squares in the numerators on the right hand side of these expressions, cross multiplying to eliminate the denominators, and collecting terms, we obtain

2 Ia (mO,mC,mS)=(mO mC + mO mS)(rCO) (2.42) 2 +(mO mS + mC mS)(rCS) +2mO mS rCO rCS ∗ ∗ 2 Ib (mO,mC,mS)=(mO mC + mO mS)(rCO) (2.43) ∗ ∗ 2 ∗ +(mO mS + mC mS)(rCS) +2mO mS rCO rCS .

∗ Multiplication of Eq. (2.42) by the ratio mS/mS then yields

∗ ∗ mS mS 2 Ia (mO,mC,mS) =(mO mC + mO mS) (rCO) (2.44) mS mS ∗ ∗ 2 ∗ +(mO mS + mC mS)(rCS) +2mO mS rCO rCS .

Because the two ﬁnal terms in Eqs. (2.43) and (2.44) are identical, subtraction of the left- and right-hand sides of Eq. (2.44) from the corresponding sides of Eq. (2.43) yields ∗ ∗ ∗ mS mS 2 Ib (mO,mC,mS) − Ia (mO,mC,mS) = mO mC 1 − (rCO) mS mS

Since IA and IB and all of the atomic masses are known, we can readily solve this equation to obtain rCO =1.167 415 A.˚ Substitution of that value back into Eq. (2.42) then yields the quadratic equation

2 895.055 3899 (rCS) + 1194.010 013 (rCS) − 4024.609179 = 0

whose solution yields rCS=1 .555 92 A˚ (the other root of this quadratic is a negative number, and has no physical signiﬁcance).

2.5.3 Non-LinearPolyatomicMoleculesareMoreDiﬃcult......

A polyatomic molecule is in general a three-dimensional object, and the full three-dimensional version of Eq. (2.10) is required to locate its centre of mass. In this case, a proper description of the system requires the introduction of three moments of inertia associated with rotation about three unique orthogonal axes through the centre of mass, called the “principal axes”. However, for molecules with any signiﬁcant degree of symmetry, those axes are generally aligned in an intuitively natural way relative to the structure of the molecule, and for molecules with a high degree of symmetry, two or more of the moments of inertia may be equal. For example, for any linear molecule one of these principal axes is the molecular axis, and the two others are perpendicular to it. From the deﬁnition of Eq. (2.11) it is clear that rotation about the axis running through those atoms will have a moment of inertia of zero, and hence cannot contribute to the energy of rotation, while the moments of inertia about the other two axes must be identical. Hence, the rotational energy depends only on that one value of I. For non-linear molecules things are more complicated, and while a proper treatment of them is beyond the scope of this course, we will consider some special cases which are relatively tractable. 46 CHAPTER 2. ROTATIONAL SPECTROSCOPY y Consider the case of the water molecule, whose structure is shown in Fig. 2.11. Because of its symmetry, it is intuitively obvious that the centre of mass and two of our principal axes of rotation will lie in the 0 .9 x × 58 plane of the molecule, and that the centre of mass must lie on the y Å axis which passes through the O nucleus and bisects the H–O–H bond angle. Relative to an (arbitrarily chosen) origin at the centre of the 16 1 O atom, the two H atoms are located at the coordinate positions θ = 104.5°

(xH,yH)=(±rOH sin(θ/2), −rOH cos(θ/2)) Figure 2.11: Structure of H2O. =(±0.7575, −0.5865) .

Applying Eq. (2.10) for the x and y coordinates of the centre of mass, we obtain

mH ×(−0.7575) + mO ×0+mH ×(+0.7575) xcm = =0 2 mH + mO mH ×(−0.5865) + mO ×0+mH ×(−0.5865) ycm = = − 0.06564 , 2 mH + mO while symmetry tells us that zcm = 0 . The centre of mass is shown as an “×”inFig.2.11;thefactthatitis located so close to the centre of the 16O atom is due to the large diﬀerences between the masses of the two types of atoms. With the centre of mass located, a little additional arithmetic yields the values of the three moments of inertia as:

2 2 2 2 Ix = mH(−0.5865 + 0.0656) + mO(0.0656) + mH(−0.5865 + 0.0656) =0.6158 [u A˚ ] ≡ IA 2 2 2 2 Iy = mH(−0.7575) + mO(0) + mH(0.7575) =1.1566 [u A˚ ] ≡ IB 2 Iz = Ix + Iy =1.7724 [u A˚ ] ≡ IC . I ⊥ This definition of z reflects the simple Pythagoras theorem result that the distance (in the plane) from 2 2 atom i to the z axis (which is ⊥ to the plane of the paper) is simply (xi − xcm) +(yi − ycm) ,andthe associated relation Iz = Ix + Iy holds true for any planar molecule. Note that the last term in each of these equations introduces another spectroscopic convention; while our choice of the x, y and z Cartesian axes is arbitrary, the three moments of inertia of a molecule are always labeled such that IA ≤ IB ≤ IC . In conclusion, it is clear that the water molecule may rotate about any of the A, B,orC axes, and that there will be inertial rotational constants Bv associated with each. However, another viewpoint is that the molecular symmetry axis (here the y axis) may rotate in space like a linear molecule, while at the same time the molecule can internally rotate about that body-fixed axis. This second type of behaviour means that yet another rotational quantum number, conventionally labeled K, is required to describe fully the state of a non-linear polyatomic molecule. With a little algebra, it may be shown that the moments of inertia for a few specially symmetric types of non-linear polyatomic molecules can be written in closed form. Those results for spherical rotors (or “spherical tops”), species for which all three moments of inertial are identical, and for symmetric rotors (or “symmetric tops”), molecules for which two moments of inertia are equal to each other and differ from the third, are shown in Fig. 2.12. For the former, I and I⊥, respectively, are defined as moments of inertia about axes parallel and perpendicular to the symmetry axis of the “top”. Ignoring centrifugal distortion, a little additional dose of quantum mechanics shows that for a spherical top, sph Cu F (J)=Bsph[J(J +1)] = I [J(J +1)] , (2.45)

2 and that the degeneracy of the rotational energy levels is now gJ =(2J +1) . Similarly, for a symmetric top sym sym 2 F = F (J, K)=B⊥[J(J +1)]+ B − B⊥ K Cu Cu Cu = [J(J +1)]+ − K2 (2.46) I⊥ I I⊥ 2.6. CONCLUDING REMARKS 47

Figure 2.12: Rotational moments of inertia for some symmetric non-linear polyatomic molecules; mtot is the total mass of the molecule. in which J =0,1,2,3,...etc.(asusual),and K =0,1,2,..., J (i.e., 0 ≤ K ≤ J ). In this case the energy level degeneracy is 2(2J +1)for K>0andsimply(2J +1)when K =0. For more general polyatomic molecules that lack one of these two special types of symmetry (i.e., in which IA, IB and IC are all diﬀerent), there are no general closed-form expressions for the level energies, not even for our simple little water molecule. On the other hand, it is still a straightforward matter to treat them numerically, so computers save the day.

2.6 Concluding Remarks

In conclusion, we have seen in this chapter how light causes molecules to become rotationally excited, and how their rotational energy levels and the associated spectra can be described within the laws of quantum mechanics. We have also seen how the properties of pure rotational spectra reflect the values of the moment(s) of inertia, and how the latter may in turn be used empirically to determine the structure of molecules – the bond lengths and (although not discussed here) the bond angles in non-linear species. This capability of rotational spectra is extremely important, as the most accurate information we have about molecular structure comes from analyses of this type. For diatomic molecules such structure determinations are quite straightforward (and you are expected to become quite proficient at it!). For linear triatomics and larger symmetric linear molecules it is also reasonably feasible to determine all of the bond lengths, should data for enough different isotopologues be available. Up to a point, this is also true for complicated non-linear molecules, although the number of 48 CHAPTER 2. ROTATIONAL SPECTROSCOPY

Figure 2.13: Gas phase molecular structure of azulene determined from rotational spectroscopy. independent isotopologue measurements required will clearly grow rapidly. In particular, one needs one independent inertial constant for each independent geometric parameter (bond length or bond angle) of the system. This clearly must have been quite challenging for a molecule such as azulene C6H5NH2,whose structure and spectroscopically determined structural parameters are shown in Fig. 2.13. By systematically varying the composition of isotopes forming the molecule (i.e., 1Hvs.2Hand12Cvs.13C in various locations, and 14Nvs.15N), all of the pertinent bond lengths and bond angles were deduced from the changes in the moments of inertia, which were in turn determined from changes in the rotational line spacings. One other important application of microwave spectroscopy is identifying and quantifying the amounts of particular chemical species present in both normal and unusual environments. A particularly important example of the latter is its use in ﬁnding and identifying molecules in interstellar space and remote plane- tary environments. Pure rotational spectroscopy has proved to be particularly useful for this, because the relatively high transparency of our atmosphere in the microwave region has facilitated very extensive studies by ground-based observatories. For many years one of the world’s most productive observatories of this type was the “Algonquin Radio Observatory” in Algonquin Park, run by the National Research Council of Canada. Molecules in outer space are found in very cold clouds of gas, but because of molecular collisions they are distinctly “warmer” than the surrounding background, so they are very well suited for observation by emission. Since they are very cold, only the lowest levels are populated. An example of a spectrum obtained in this way is shown in Fig. 2.14. This is the same species for which a conventional room temperature spectrum was shown in Fig. 2.6. In Fig. 2.14 the very low temperature freezes out all of the vibrational satellites, and what we see here is some of the hyperﬁne structure of the J =1→ 0 line in the ground state.12 Many

12 Hyperﬁne structure is the splitting of individual rotational lines due to level splittings caused by non-zero nuclear spins –

Figure 2.14: The rotational emission spectrum of cyanodiacetylene in Sagittarius B2. 2.7. PROBLEMS 49 types of novel molecules have been discovered in this way, and spectroscopists have reproduced their spectra in laboratory experiments to conﬁrm the frequencies of the transitions. A list of some of the molecules, ions and free radicals identiﬁed in this fashion is presented in Table 2.3.

Table 2.3: Interstellar molecules detected by their rotational spectra. Diatomics OH,CO,CN,CS,SiO,SO,SiS,NO,NS,CH,CH+ + + + Triatomics H2O, HCN, HNC, OCS, H2S, N2H ,SO2, HNO, C2H, HCO, HCO ,HCS + Tetratomics NH3,H2CO, HNCO, H2CS, HNCS, N≡C–C≡C, H3O ,C3H(linear), C3H(cyclic) 5-Atomics N≡C–C≡C-H, HCOOH, CH2=NH, H-C≡C–C≡C, NH2CN 6-Atomics CH3OH, CH3CN, NH2CHO, CH3SH 7-Atomics CH3–C≡C-H, CH3CHO, CH3NH2,CH2=CHCN, N≡C–C≡C–C≡C-H 8-Atomics HCOOCH3,CH3–C≡C–C≡N 9-Atomics CH3OCH3,CH3CH2OH, N≡C–C≡C–C≡C–C≡C-H 11-Atomics N≡C–C≡C–C≡C–C≡C–C≡C-H

2.7 Problems

12 −1 1. The rotational constant B for C2 has been determined to be 1.8198 cm . What is the C–C bond length? Could a rotational spectrum of this molecule be observed using microwave spectroscopy?

2. Chlorine ﬂuoride, 35Cl19F, exhibits a microwave spectrum where transitions are spaced 0.516 cm−1 apart, with no apparent change in the rotational spacing as J increases. Determine the Cl–F bond length of this molecule.

3. The 14N1H radical has a bond length of 1.037 A.˚ Determine the moment of inertia I, the rotational constant B, and the wavenumber (cm−1)ofthe J =3← 2 transition for this species.

4. The ﬁrst rotational transition J =1← 0of12C16O has been measured at 115 271.204 MHz. Determine the C–O bond length.

5. Using the information given and/or results obtained in question #4, ﬁnd the frequency (in MHz) of the ﬁrst rotational transition for 13C17O. [Shortcuts are permitted if you can justify them.]

12 16 14 16 16 6. Given that C O, N Oand O2 have approximately the same bond lengths (1.16±0.04 A),˚ predict the relative magnitudes of their rotational constants B.

7. Three adjacent rotational transition lines for the ground vibrational level of 1H79Br are observed at 84.58, 101.42 and 118.21 cm−1. Determine which transitions these correspond to (i.e., determine the rotational energy levels involved). Determine B0, D0 and the H–Br bond length.

8. A space probe was designed to look for CO in the atmosphere of Saturn, and it was decided to use a microwave technique from an orbiting satellite. Given that the bond length of the molecules is 1.1282 A,˚ at what frequencies (in Hz) will the ﬁrst four transitions of 12C16O lie? What precision is needed in order to distinguish the ﬁrst transition of 12C16Oand13C16O in order to determine the relative abundances of the two carbon isotopes?

9. Predict the relative intensities of the ﬁrst 6 rotational transitions of 12C16O for a spectrum obtained at a temperature of T = 1000 K. Use the value for the inertial rotational constant obtained in Exercise (i) (on p. 33), and assume the rigid rotor model is valid.

10. The microwave spectrum of 12C14N exhibits a series of lines with spacings of 3.7992 cm−1. Determine the C–N bond length. in this case I(14N) = 1 . 50 CHAPTER 2. ROTATIONAL SPECTROSCOPY

11. The equilibrium bond length of LiH is 1.595 A.˚ Determine the inertial rotational constants for 7Li1H, 6Li1Hand7Li2H, assuming the bond lengths are identical. If these isotopologues were all present at their naturally-occurring levels, what would be the relative intensities of the three sets of spectral lines?

12. Hydrogen deuteride, 1H2H, has a bond length of 0.741 42 A,˚ and a centrifugal distortion constant of 0.0463 cm−1. What is the value of J for which the contribution of centrifugal distortion to the rotational level spacing matches that due to the inertial rotational constant?

13. We are told that the lowest-energy rotational level spacings in three unknown diatomic molecules, A2, AB and AC, are 2.89, 3.41 and 3.86 cm−1, respectively. Given that their bond lengths were determined independently to be 1.2074, 1.1508 and 1.1281 A,˚ respectively, identify the atoms A, B and C. Which of these molecules would not yield a microwave spectrum (justify your answer)?

127 35 −1 14. The rotational constant Bv of I Cl is 0.1142 cm . Determine the bond length in this molecule. 127 37 Determine the rotational constant Bv for I Cl, assuming the bond length does not change from one isotopologue to another.

15. Using the data from question #14, calculate the energies of the first five rotational energy levels and the first four rotational transitions of 127I35Cl, Will the rotational transitions for 127I37Cl occur at higher or lower energies?

16. The rotational spectrum of 1H127I consists of a series of lines separated by 13.10 cm−1. Determine the bond length of HI.

17. The ﬁrst seven lines of the microwave spectrum of 1H19F occur at: 41.10, 82.19, 123.15, 164.00, 204.62, 244.93 and 285.01 cm−1. Determine the HF bond length and the centrifugal distortion constant.

18. Using rotational spectroscopy, the comet Hyakutake that passed close to Earth in Spring 1996 was found to contain large concentrations of hydrogen isocyanide, HNC, a linear triatomic molecule. Given that the C≡N bond length is 1.180 Aand˚ the H–N bond length is 0.976 A,˚ at what frequency (in MHz) would the lowest energy rotational transition for 1H14N12C occur?

19. The vibrational level dependence of inertial rotational constant of H35Cl is given by the expression 1 1 2 −1 Bv =10.5933 − 0.3070(v + 2 )+0.00163(v + 2 ) cm . What are the average bond lengths rv in each of the three lowest vibrational levels, v =0,1and2?

20. Show that for a two-particle system the general deﬁnition of Eq. (2.11) reduces to our simple two- 2 particle or diatomic molecule result Id = μ (re) .

21. If the 12C16O molecule has a bond length of 1.131 A,˚ where does its centre of mass lie?

22. Show how we get from the ﬁrst line of Eq. (2.6) to the second.

23. From Eq. (2.34) and the HF rotational constant determined in Exercise (ii) (see p. 34), what temperature would be associated with the emission spectrum of Fig. 2.5? What would your result have been if we had considered only the thermal population factor? [Note that the surface temperature of the sun is around 6000 K.]

24. Compared to 12C16O, what is the relative abundance of 13C18O? Following up on the discussion of Exercise (i) (on p. 33), what would you predict for the frequencies ν (in GHz) of the three lowest-energy lines of its pure rotational spectrum?

25. For the linear molecule cyanoacetylene 1H−12C≡ 12C−12C≡ 14N, r(H−C) = 1.058 A,˚ r(C≡C) = 1.205 A,˚ r(C−C) = 1.378 Aand˚ r(C≡N) = 1.159 A.˚ (a) Where is its centre of mass? (b) What is its moment of inertia? (c) What is its inertial rotation constant? Chapter 3

Vibrational Spectroscopy

What Is It? Vibrational spectroscopy detects transitions between the quantized vibrational energy levels associated with bond stretching or bond angle bending.

HowDoWeDoIt?Transitions are observed by measuring the amount of infrared radiation that is absorbed or emitted by vibrating molecules in solid, liquid, or gas phases.

WhyDoWeDoIt?A knowledge of the vibrational level spacings gives us the value of the stretching (or bending) force constant which characterizes the stiﬀness of a bond, allows us to estimate the bond dissociation energy, and gives us a means of identifying characteristic functional groups of atoms within a large molecule.

3.1 Classical Description of Molecular Vibrations

3.1.1 Why Does Light Cause Vibrational Transitions?

As discussed at the beginning of Chapter 2, the fact that molecules consist of distributions of positive and negative charges means that they will be affected by external electric fields. This is true both for static fields applied in the laboratory, and for the rapidly oscillating electric field associated with light. We begin this chapter by examining how the latter can influence or change the vibrational motion of a molecule. Consider a polar diatomic molecule such as HF, which is fixed in space and vibrating with its (slightly) elastic bond regularly stretching and compressing. Since there is no dissipative friction inside a molecule, left to its own devices it will vibrate forever with some fixed period or frequency. As this occurs, the separation of positive and negative charges rhythmically increases and decreases, and hence so does the magnitude of the molecular dipole moment, as illustrated in Fig. 3.1. However, if the molecule is subjected to light whose “colour” (or wavelength, or wavenumber) is tuned so that its frequency exactly matches the natural vibrational oscillation of the dipole moment, that stretching motion will receive a periodic “push” in phase with its motion, which will cause it to pick up energy from the light field and become vibrationally excited. Once again (as for rotation), it is like a child being pushed on a swing – if subjected to a periodic push exactly in phase with its natural motion, the amplitude of the swinging increases. Similarly, the arguments presented in §2.1.1 show that the oscillating charge associated with this changing dipole can give rise to emission of light at the frequency of the vibration. While the above discussion is very similar to that presented at the beginning of Chapter 2, there is a critical difference. What is important here is not whether the molecule actually has a dipole moment, but rather whether that dipole changes with time as the molecule vibrates! It is this regular oscillatory change in the dipole, and not merely its existence, that gives the electric field of the light something to push against in order to excite the molecule vibrationally. It is immediately clear that virtually any molecule that has a permanent dipole moment will have a “dipole allowed” vibrational spectrum, since the vibration of its

51 52 CHAPTER 3. VIBRATIONAL SPECTROSCOPY

+ + + + + + length + + of molecule − − − − − − − − strength of dipole

equilibrium ↑ vertical dipole moment component of molecular time → dipole

Figure 3.1: Dipole moment of a vibrating polar diatomic molecule which is fixed and aligned in space. bond(s) will cause that dipole moment to oscillate. In particular, every chemically heteronuclear1 diatomic molecule (such as HCl, NaBr or OH) will be vibrationally or “infrared” active, since the differences in atomic polarizabilities will ensure that there is always some non-zero permanent dipole moment whose magnitude will oscillate when the bond stretches. This same argument tells us that (chemically) homonuclear diatomics such as H2,N2 or Rb2 will be vibrationally or infrared inactive. For polyatomic molecules the situation is more complicated, since they have more than one type of vibrational motion (or vibrational “mode”) that must be considered. As an illustrative example, consider the molecule O=C=O . Since it is a symmetric linear species, it has no permanent dipole symmetric stretch moment, and hence will be microwave inactive (i.e., it has no pure rotational spectrum). As will be discussed further in §3.5, CO2 has the three different types of vibrational motion illustrated in Fig. 3.2. These are the “symmetric stretch” mode in which both C=O bonds stretch and compress in phase with antisymmetric stretch each other, the “antisymmetric stretch” mode in which one bond compresses while the other stretches, and vice versa, and the bending mode. It is imme- bending mode diately clear that the symmetry of the molecule is maintained throughout the course of the symmetric stretching vibration, so the vibrational motion does not give rise to any temporary or “instantaneous” dipole moments. In con- Figure 3.2: trast, for both the antisymmetric stretch and bending modes, the molecular Vibrational modes of CO2. distortion that occurs during the vibrational cycle will give rise to a temporary dipole moment which oscillates in magnitude and direction (from left to right for the antisymmetric stretch, and up-and-down or in-and-out for bending) as the motion proceeds. Thus, these two latter modes will be vibrationally (or infrared) active, while the symmetric stretch mode is not. In the context of Fig. 3.1, these IR active modes correspond to a case in which the oscillating dipole moment allows transitions to occur even though the average or equilibrium dipole moment is zero.

3.1.2 The Centre of Mass and Relative Motion

As illustrated by our discussion of the classical mechanics of a two-particle system in §2.1, the total energy of a multi-particle system may always be separated into the kinetic energy of motion of the overall system, plus 2 the internal energy. For a rigid molecule that internal energy is simply the rotational energy Erot = L /2I of Eq. (2.9). However, if the bonds are somewhat elastic, both the vibrational kinetic energy and the potential energy governing that vibration also contribute to the internal energy. We also saw in Chapter 2 that for

1 Note that this argument does not apply to diatomic molecules that are only isotopically heteronuclear, such as 7Li6Li or HD, since the difference in nuclear masses does not give rise to any asymmetry of the overall charge distribution. 3.1. CLASSICAL DESCRIPTION OF MOLECULAR VIBRATIONS 53 a diatomic molecule, the internal energy (rotation plus vibration) is described in terms of the motion of a single pseudo-particle of mass μ = m1 m2/(m1+m2) located at the relative coordinate vector r. Rotation was concerned with the changing orientation of r in space, while our discussion of vibration will focus on the one-dimensional stretching of r. In other words, the vibration of a diatomic molecule can be described as the motion of particle of mass μ in one dimension, along the coordinate r. Near the end of Chapter 1 we discussed the Born-Oppenheimer approximation, which allowed us to separate the overall Schrödinger equation for the system into a Schrödinger equation for the motion of the electrons plus one for the motion of the nuclei. We also saw there that the r-dependence of the energy eigenvalues of the electronic Schrödinger equation yielded the potential energy curves V (r)=Eel(r) governing the vibrational motion of the nuclei, which appears in the effective radial Schrödinger equation of Eq. (1.35). Thus, the description of vibrational motion for a diatomic molecule, in either classical or quantum mechanics, is concerned with a particle of mass μ moving in one dimension along the radial coordinate r, subject to a potential energy function V (r). For a polyatomic molecule the situation is similar, except that we need to take account of multiple internal coordinates for relative motion, and we need special techniques (beyond the scope of this course) to determine the particular effective mass associated with each of those different modes of motion.

3.1.3 The Classical Harmonic Oscillator The basic starting point for any description of vibrational motion in classical or quantum mechanics is the harmonic oscillator model. It is based on Hooke’s Law, which states that when a system is displaced from equilibrium, the restoring force increases linearly with the displacement. For a diatomic molecule stretching (or compressing) away from its equilibrium distance r = re , this restoring force F can be written as

dV (r) F = − k(r − re)= − , (3.1) dr where the minus sign reminds us that F points in the opposite direction to the displacement, and k is the “force constant” that defines the stiffness of the vibrational motion. The last segment of this equation reminds us that in classical mechanics, any force can be written as the derivative of a potential energy function. Integrating this simple differential equation [ dy/dx = −kx,where x =(r − re)] yields the conventional quadratic expression for a harmonic oscillator potential energy function:

1 2 VHO(r)= 2 k(r − re) . (3.2)

Combining Eq. (3.1) with Newton’s famous second law, force = mass×acceleration , we obtain a diﬀeren- tial equation with a very familiar form (see Eq. (1.18)):

d2r F = μ = − k (r − re) (3.3) dt2 that may be rewritten as 2 d (r − re) k 2 = − (r − re)= − κ (r − re) , (3.4) dt2 μ where in this case κ = k/μ. As discussed in §1.3.1, this is a diﬀerential equation that we can readily solve to obtain the general solution

r(t) − re = A sin(κt+ δ)=A sin(2πνet + δ) (3.5) in which νe =(1/2π) k/μ is the frequency of the oscillatory vibrational motion in Hz, δ is an indeterminate phase constant, and A is the amplitude of the motion.2

2 Note that as often occurs in mathematics and science, we have changed the labels for the dependent and independent variables; the independent variable x in Eq. (1.18)-(1.21) is replaced here by the time t,andy is replaced by (r − re)asthe name of the dependent variable; however, the underlying mathematics is exactly the same. 54 CHAPTER 3. VIBRATIONAL SPECTROSCOPY

Our classical mechanics equation of motion Eq. (3.5) tells us that once the system starts vibrating, it oscillates back and forth forever. If we ignore the dissipative eﬀects of friction, a myriad of familiar macroscopic physical phenomena are described by this expression, including swings and pendulums, as well as the motion of a rigid ball rolling back and forth in a one-dimensional parabolic well. This last example is an exact analog of vibrational motion, since both correspond to having a particle (of mass μ) oscillate back and forth in a quadratic potential energy well. Note too that during this motion, the total energy is always a constant: at any instant the radial speed of the particle isr ˙(t) ≡ dr(t)/dt =2πνe A cos(2πνet + δ), and 1 2 since the instantaneous radial kinetic energy is 2 μ(˙r) , the sum of the potential energy and kinetic energy is constant:

1 2 1 2 Etot =KEradial + VHO(r(t)) = 2 μ(˙r) + 2 k (r(t) − re) 1 2 1 2 1 2 = 2 μ [2πνe A cos(2πνe t + δ] + 2 k [A sin(2πνe t + δ)] = 2 kA (3.6)

3.2 Quantum Mechanics of Molecular Vibration, and the Harmonic Oscillator

As indicated above, the vibrational motion in a diatomic molecule is described exactly by the one-dimensional motion of a particle of eﬀective mass μ subject to a potential energy function V (r). From §1.3 and §1.4 we see that this behaviour is governed by the particle-in-a-box Schr¨odinger equation

2 d2 ψ(r) Hˆvib ψ(r)= − + V (r) ψ(r)=Evib ψ(r) (3.7) 2μ dr2

We also saw in §1.3 and §1.4 that if V (r) is the square-well potential energy function of Fig. 1.7-A or the Coulomb potential of Fig. 1.7-B, we can solve this equation exactly in closed form to obtain the eigenvalue expressions of Eq. (1.27) and (1.13), respectively, together with exact analytic eigenfunctions. However, those potential energy functions are not very good models for vibrational motion. Fortunately, the diﬀerential equation obtained on substituting the harmonic oscillator potential of Eq. (3.2) into Eq. (3.7) is another one of those special cases that can be solved exactly. Its allowed energy eigenvalues are given by the simple expression HO 1 1 Evib = Evib (v)= k/μ (v + /2)=hνe (v + /2) (3.8) in which νe is exactly the same frequency (in Hz) appearing in the classical mechanical harmonic oscillator solution of Eq. (3.5), and the vibrational quantum number v has allowed values of v =0,1,2,3,.... As is the case for rotation, spectroscopists usually express vibrational energies in units cm−1, and it is customary to use the letter ‘G’ to represent vibrational energies in those units: HO HO 2 1 −1 G (v) ≡ Evib (v)/ 10 hc = ωe(v + /2)[cm ] (3.9) in which ! ! 2 −1 −2 1 2 Cu k˜ [cm A˚ ] −1 ωe = 2k = [cm ] (3.10) 2μ 102 hc μ [u]

−1 2 and Cu =16.857 629 [u cm A˚ ] is the ubiquitous numerical factor introduced in §1.1.4. Note that we choosetowritetheforceconstantask˜ when it is in “spectroscopists’ units”. In SI units k would normally have the units newtons/meter (or N m−1). However, it is equally valid to multiply numerator and denominator by 1 m and write it with units J m−2, or more generally, in units energy/length2 . Thus, expressing k˜ in units [cm−1 A˚−2] is quite appropriate when working with spectroscopic properties to describe molecules. Figure 3.3 show the quantum mechanical level energies and wavefunctions for a few of the lowest levels of three harmonic oscillator systems. Note that the naming convention used for molecular vibration problems diﬀers from that for the square-well or Coulomb potential (v =0,1,2,...,etc.,ratherthan n =1,2,3, . . . , etc.; see Fig. 1.7). However, the level counting and qualitative properties of the wave functions (the 3.2. QUANTUM MECHANICS OF THE HARMONIC OSCILLATOR 55

1 2 1 V(r) =− k (r-r ) ; μ = μ V(r) =− k r-r 2 = V(r) = k r-r 2 = . 2 e 0 2 ′( e ) ; μ μ0 ′( e ) ; μ 0 9 μ0 υ =5 {k ′= 2 k } {k ′= 2 k } υ =3 υ =3 υ =4 ω e υ =2 υ =3 υ =2

υ =2 ωe =1 υ =1 υ

υ =1

ωe υ =0 υ =0 υ =0 A B C

ωe /2 -0.20.0 0.2 -0.20.0 0.2 -0.20.0 0.2 (r-re ) (r-re ) (r-re ) Figure 3.3: Harmonic oscillator eigenvalues and wavefunctions for three model systems. Cases A and B use the same reduced mass μ, but diﬀerent potential energy force constants. Cases B and C have the same potential energy function, but diﬀerent μ values. nth level having n loops or extrema) is exactly the same. Note too that as for most other particle-in-a-box problems, the lowest harmonic oscillator level allowed by quantum mechanics does not have zero vibrational HO 1 energy, but rather has a zero point energy of G (0) = 2 ωe . We saw in §3.1.1 that vibrational transitions are governed by the following “physical” selection rule:

Vibrational Selection Rule 1: Vibrational transitions can only occur if the dipole moment of the molecule oscillates in the course of the vibrational motion.

In addition, the mathematical properties of the vibrational wavefunctions give rise to an “orthogonality” selection rule, which for the case of a harmonic oscillator is

Vibrational Selection Rule 2: If the molecular dipole moment varies linearly with bond stretching, a harmonic oscillator can only have Δv = ±1 transitions.

In other words, quantum mechanics only allows transitions between adjacent vibrational levels of a harmonic oscillator, so the observable transition energies are given by3

HO HO HO ΔG 1 ≡ G (v +1)− G (v) v+ /2 = ωe (v +1)+1/2 − ωe(v + 1/2)=ωe (3.11)

Note that in SI units (i.e., using Eq. (3.8)), these vibrational transition energies would be HO ΔEv+1/2(v)=hνe [J]; this tells us that the frequency of the light driving (or being emitted in) such a transition is exactly equal to the frequency of this vibration motion predicted by classical mechanics, a nice confirmation of the arguments presented in §3.1.1 For a diatomic molecule, vibrational transitions typically occur in the energy range 100 − 4 000 cm−1, which is in the infrared (IR) region of the electromagnetic spectrum. For example, for the ground electronic −1 states of HgI and IBr, ωe = 125.6 and 268.7cm , respectively, while" the ground electronic states of HgH −1 and HF have ωe=1387.1 and 4 138.5 cm , respectively. Since ωe ∝ k/μ˜ , these differences reflect differences in both the bond stiffness force constant k˜ and the reduced mass μ, and some surprising cancellations may occur. For example, while ωe(HgH)/ωe(HgI) ≈ 11.0 , most of that difference is due to the difference in

3 The subscript “v+1/2” on the vibrational spacing ΔGHO is the mid-point between v and v+1, reminding us that this is the spacing between those two levels. 56 CHAPTER 3. VIBRATIONAL SPECTROSCOPY reduced mass, μ(HgI)/μ(HgH) ≈ 77.5 , and the force constant k˜ for HgH is only about 1.6 times larger than that for HgI. This dependence of vibrational level energies and spacings on k˜ and μ is illustrated by Fig. 3.3. Panels A and B show the effect of doubling the force constant k on the level energies and wavefunctions for a given molecular species (i.e., for a given value of μ = μ0 ). This demonstrates why a given molecule in different electronic states is expected to have different vibrational spectra, since the potential energy functions differ −1 1 + from one state to another (e.g., see Fig. 1.13). For example, ωe = 1 405.498 cm for the X Σ ground 7 1 + −1 electronic state of LiH, while for the electronically excited A Σ state of this same species, ωe = 234.4cm ; i.e., the force constant k˜ is 36 times smaller in the excited state. Similarly, panels B and C of Fig. 3.3 show the effect of changing the reduced mass on the level energies and spacings for a given potential energy function. This illustrates the nature of isotope effects in vibrational spectra. For example, for the three minor LiH isotopologues 6LiH, 7LiD and 6LiD (with natural abundances of 7.499%, 0.014% and 0.001%, respectively) 1 + −1 4 in the X Σ ground electronic state, ωe=1420.048, 1 054.940 and 1 074.309 cm , respectively. Note that 1 the zero point energy of 2 ωe shown in Fig. 3.3-B does not correspond to a molecular transition, since there is no allowed level at the potential minimum. One key result illustrated by Fig. 3.3 and Eq. (3.11) is that the pure vibrational spectrum of a harmonic oscillator would be very boring! In particular, since all the vibrational spacings are the same, all allowed transitions within a given electronic state would be piled up on one another at exactly the same frequency ν˜vib = ωe . In practice, however, three additional factors make vibrational spectra much more interesting. (i) While a dipole moment does vary linearly for very small amplitude excursions from equilibrium, in general quadratic, cubic and higher-power terms are required to describe it accurately. For a harmonic oscillator those terms allow |Δv| > 1 transitions at energiesν ˜vib =2ωe ,3ωe , . . . etc. While those overtone transitions rapidly become much weaker as |Δv| increases, they could be observed at frequencies corresponding to multiples of ωe. (ii) The harmonic-oscillator function of Eq. (3.2) is not a very good overall model for a real potential energy curve. While almost all intermolecular potentials are approximately quadratic near their minima, a harmonic oscillator function goes to infinity at large |r − re|, so it could only describe bonds that never break. This is clearly unrealistic, as a real potential function must approach infinity as the nuclei are pushed together so that r → 0 , while as r →∞ it must approach an asymptote at the bond dissociation energy (see Fig. 1.13). Because of this shape asymmetry, the vibrational level spacings of a realistic potential are not constant (as they are for a harmonic oscillator), but become smaller with increasing energy, and the simple Δv = ±1 orthogonality selection rule is no longer strictly true (not even if the dipole moment is a strictly linear function of r). (iii) There is no angular momentum associated with radial vibrational motion. Thus, since angular momentum must be conserved when a photon is absorbed or emitted, all vibrational spectra must really be vibration-rotation spectra, with a simultaneous change in rotational quantum number J taking account of the photon angular momentum. These three factors mean that vibrational spectra are much richer and more interesting than would be the case for a pure harmonic oscillator, and allow such spectra to yield a wealth of detailed information about molecules. Subsequent sections discuss how factors (ii) and (iii) affect IR (i.e., vibrational) spectra.

Exercise (i): If an ab initio quantum chemistry calculation predicts a bond stretching force constant of k=1902 N m−1 for the ground electronic state of CO, what do we predict for the v=1 ← 0 pure vibrational transition energy of 12C16O, in cm−1? Answer: Since the given value of k is given in SI units, one way to address this question is to use Eq. (3.8). We begin by looking up the atomic masses, calculating the reduced mass, and converting to SI units: −1 −3 1 1 6.856 392 × 10 [kg/mole] −26 μ = + [u] = =1.138 531 × 10 [kg/molec] . 12.0 15.994 915 6.022 136 7 × 1023 [molec/mole] 4 7 A careful student will note that these ωe values are not precisely those predicted by multiplying the value for LiH by the ratios of the relevant μ values (which would yield 1 420.109, 1 054.721 and 1 074.115, respectively). The small discrepancies are due to “Born-Oppenheimer breakdown” eﬀects, which are currently a hot research topic. 3.3. ANHARMONIC VIBRATIONS AND THE MORSE OSCILLATOR 57

Substituting this value into Eq. (3.8), we obtain a vibrational energy spacing of

−1 k −34 1 902 [N m ] −20 hνe = =1.054 572 66 × 10 =4.310 317 × 10 [J] , μ 1.138 531 × 10−26 [kg]

−20 −34 13 or νe =4.310 317×10 /6.626 075 5×10 =6.505 083×10 [Hz] = 65.060 83 [THz]. Our familiar frequency to wavenumber conversion then yields

2 13 2 8 −1 ωe = νe/10 c = 6.505 083 × 10 /10 × 2.997 924 58 × 10 = 2 169.9[cm ]

An alternative approach to this problem is ﬁrst to convert the given value of k into “spectroscopists’ units”, and then to apply Eqs. (3.9) and (3.10). Since k=1902 [N m−1]=1902 [J m−2], we can either look up the J → cm−1 energy conversion factor in Table 1.1 on p. 7, or apply the Planck equation conversion factors:

−20 −2 10 5 −1 −2 k˜ = 1 902 [J m ] =9.574 892 × 10 [cm A˚ ] . 102 hc Substituting this result into Eq. (3.10) then yields the same result obtained before: 5 −1 ωe = 2×16.857 629 × 9.574 892×10 /6.856 392 = 2 169.9[cm ] .

Exercise (ii): A more common problem is that we wish to use experimental spectroscopic data to determine the properties of a molecule. Consider for example the case of LiH. Recent measurements showed that the v=1−0 7 −1 vibrational spacing of LiH was ΔG1/2 = 1 359.708 cm . What is the associated bond stretching force constant? Answer: This question may be readily addressed using Eqs. (3.11) and (3.10). From a table of atomic masses we determine that for this isotologue the reduced mass is

−1 1 1 μ = + =0.881 238 16 [u] . 7.016 003 0 1.007 825 035

Since we are given only one vibrational spacing, we must treat the system as a harmonic oscillator and assume −1 that ωe = 1 359.708 cm . Substituting this value and the reduced mass into Eq. (3.10) and solving for k˜ then yields 2 −1 −2 k˜ = (1 359.708) ×0.881 238 16/(2×16.857 629) = 48 323.47 [cm A˚ ] .

3.3 Anharmonic Vibrations and the Morse Oscillator

3.3.1 Eigenvalues and Properties of the Morse Potential We saw in our discussion of particle-in-a-box problems in §1.3 that the pattern of level energies in a one- dimensional potential energy well depended on the shape of that potential well – and in particular, on how rapidly the well gets wider with increasing energy. Let us consider the four particle-in-a-box problems pictured in Fig. 1.7 from this point of view. The preceding section shows us that for the exactly quadratic harmonic oscillator potential function, for which the width of the well increases as the square root of the energy (width =2 V (r)/k ), the level spacings are constant. We know from our particle-in-a-box discussion that making a box narrower increases the level spacings, and compared to the harmonic oscillator potential of Fig. 1.7-C, the (constant) width of the square well potential of Fig. 1.7-A becomes increasingly narrower with increasing energy. Thus, it is no surprise that the level spacings of the square-well potential increase with energy, since those for the harmonic oscillator are constant. Conversely, with increasing energy the potentials shown in Figs. 1.7-B and 1.7-D (see p. 16) get wide faster than does the harmonic oscillator function of Fig. 1.7-C, so their level spacings actually become smaller with increasing energy. Indeed, at their asymptotes the potential well becomes inﬁnitely wide, so we expect the level spacings to go to zero there. A realistic and widely used model for molecular potential energy functions is the Morse function

2 −β(r−re) −2β(r−re) −β(r−re) V (r)=De 1 − e = De e − 2 e +1 , (3.12) 58 CHAPTER 3. VIBRATIONAL SPECTROSCOPY

VMorse(r)

υ =7 υ =6 υ =5

υ =4 } hot bands third overtone D0 υ =3

De second overtone υ =2

first overtone υ =1

fundamental

υ =0 zero point energy ↑ r → re Figure 3.4: Vibrational levels and transitions of a Morse potential energy function.

in which De is the bond dissociation energy, the depth of the potential energy well, re is the equilibrium bond length, the position of the minimum of the potential, and β is a parameter (related to the force constant k) defining the stiffness of small amplitude vibrations near the potential minimum. The characteristic shape of a Morse potential and some of its properties are illustrated by Fig. 3.4. In the second version of Eq. (3.12) we can see that the (squared) positive first exponential term dominates the short-range repulsive behaviour, the negative middle term is responsible for the attractive potential well and long-range behaviour, while (constant) the third term defines the potential asymptote. Although this form is not ideal,5 the radial Schrödinger equation cannot be solved analytically for most more sophisticated potentials, and the Morse function is sufficiently realistic for the purpose of the following discussion. As was implied above, the Morse potential is another one of those accidentally special functions for which the radial Schrödinger equation may be solved analytically to give an exact closed-form expression for its vibrational level energies: Morse 2 G (v)=ωe(v + 1/2) − ωexe(v + 1/2) , (3.13) in which6 # ! $ 2 $ −1 −1 2 %4 Cu De [cm ] β [A˚ ] 2 De β 1 −1 ωe = = [cm ] (3.14) μ 102 hc μ [u] 2 −1 2 2 1 Cu (β [A˚ ]) −1 ωexe = β = [cm ] . (3.15) 2μ 102 hc μ [u]

2 Squaring both sides of Eq. (3.14) and dividing by Eq. (3.15) cancels out the factor Cu β and yields the relation Morse 2 De =(ωe) /4 ωexe . (3.16)

Comparing the harmonic oscillator and Morse energy level expressions Eqs. (3.9) and (3.13) suggests that the second term in the latter may be thought of as a correction to the basic harmonic oscillator result which eﬀectively accounts for bond breaking. To expand on this viewpoint, let us examine the algebraic form of

5 6 In particular, real intermolecular potentials have attractive inverse-power long-range tails such as V (r) ∼ D − C6/r ,and −β(r−r ) not the rapidly-dying exponential tail of a Morse function V (r) ∼ D − 2De e e . 6 Note that the constant “ωexe” should always be treated as a single symbol, and not as the product ωe ×xe . 3.3. ANHARMONIC VIBRATIONS AND THE MORSE OSCILLATOR 59 the Morse function as an expansion about the point r=re . In particular, using a Taylor series to expand the exponential function(s) of Eq. (3.12),

∞ (−x)n 1 1 1 e−x =1+ =1− x + x2 − x3 + x4 + ... , n! 2 6 24 n=1 our Morse potential energy function may be expanded as 2 3 7 4 1 5 V (r) ≈ De [β(r − re)] − [β(r − re)] + [β(r − re)] − [β(r − re)] + ... . (3.17) 12 4

For very small |r − re|, the cubic and higher-power terms will be much smaller than the leading quadratic Morse 2 term, and this potential collapses to the harmonic oscillator function of Eq. (3.2), with k˜ = k˜ =2De β . Thus, at small |r − re| our Morse potential can be thought of as being a harmonic oscillator function with cubic and higher power “anharmonic” terms added to make the shape more realistic. However, to allow the potential to dissociate properly would require use of the full inﬁnite series associated with the exponential functions. Because of the anharmonicity/asymmetry of the Morse potential energy function, the “orthogonality” selection rule that restricts harmonic oscillator transitions to Δv = ±1 is no longer precisely valid. However, there is still a very strong propensity to prefer small |Δv| values, so we replace “Vibrational Selection Rule 2” on p. 55 with the following:

Vibrational Selection Rule 2: Transitions in which the vibrational quantum number changes by one, Δv=±1 , are strongly allowed; transitions with Δv=±2, ±3, ... become much weaker with increasing |Δv|.

In view of the above, the strongest observed IR transitions still correspond to |Δv|=1, and the associated transition energies are

Morse Morse Morse ΔG 1 = G (v +1)− G (v) v+ /2 1 1 2 1 1 2 = ωe(v +1+ /2) − ωexe(v +1+ /2) − ωe(v + /2) − ωexe(v + /2)

= ωe − 2 ωexe(v +1) . (3.18)

This expression shows that as the vibrational quantum number v (and hence the energy) increases, the level spacings become systematically smaller, and go to zero at the dissociation limit. Another way of thinking about this result is to recognize that the energy level expression Eq. (3.13) is a parabolic function of v whose maximum lies at De. Taking the ﬁrst derivative of Eq. (3.13) with respect to v and setting it equal to zero, yields

dG(v) = ωe − 2 ωexe(v + 1/2)=0. (3.19) dv Solving this equation yields the eﬀective vibrational index associated with the dissociation limit

vD = ωe/2ωexe − 1/2 . (3.20)

We call this quantity an “eﬀective” vibrational index, because except for the extremely unlikely case that it was accidentally precisely an integer, there would be no vibrational level lying exactly at De. However, rounding vD down to the nearest integer yields vD =int{vD} , the quantum number of the highest bound vibrational level supported by this potential. Substituting the expression Eq. (3.20) for vD into Eq. (3.13) readily conﬁrms that this quantum number corresponds to the dissociation energy.

3.3.2 Overtones and Hot Bands According to Selection Rule 2, vibrational transitions with |Δv| > 1 are allowed for a Morse oscillator. The labeling in Fig. 3.4 introduces the special names used for some of these transitions. In particular, the 60 CHAPTER 3. VIBRATIONAL SPECTROSCOPY v=1 ← 0 transition is known as the vibrational “fundamental”, since at normal temperatures v=0 is the only level with substantial population, and since |Δv|=1 transitions are the most intense, it should be the dominant feature of a vibrational spectrum. The weaker Δv=2, 3, . . . , etc. transitions originating in v=0 are then known as the “first overtone”,“second overtone”,“third overtone”, . . . , etc. Finally, all transitions among higher vibrational levels (for which both vupper and vlower are > 0 ) are known as “hot bands”, independent of the values of |Δv|, since the lower levels of such transitions would only have significant populations at relatively high temperatures. Note that vibrational transitions in IR spectra are called bands, because their rotational fine structure causes them to consist of a number of lines spread over a finite range of frequencies, instead of being a single sharp line (see §3.6). While Δv=± 1 hot-band transition energies are given by Eq. (3.18), for overtones or |Δv| > 1 hot-band transitions, more general expressions must be used. While one could devise analogs of Eq. (3.18) for various possible values of |Δv|, in practice it is just as easy to substitute the appropriate integer values of vupper and vlower into Eq. (3.13), and take differences. It is important to note that Fig. 3.4 introduces a second definition for the bond dissociation energy. We might naturally think of the potential well depth De as the dissociation energy. However, since the lowest energy level allowed by quantum mechanics lies above the potential minimum by the “zero point energy” ZPE = G(v=0 ) ( w h i ch e q u a l s [ ωe/2 − ωexe/4] for a Morse oscillator), the minimum amount of energy required to dissociate a real molecule is D0 ≡ De−ZPE . Both De and D0 are sometimes referred to as the “dissociation energy”, and one must be careful to understand which one is being referred to in a given instance.

Exercise (iii): Determine the properties of the anharmonic LiH molecule. Exercise (ii) on p. 57 considered the case of 7LiH, assuming that only the energy of the fundamental vibrational −1 transition was known. However, given both that quantity (namely, ΔG1/2 = 1 359.708 cm ) and the energy −1 of the ﬁrst overtone transition ΔG(2 ← 0) = 2 674.560 cm , we may determine the molecular constants ωe and ωexe , and use the Morse model to estimate the dissociation energy De and the total number of bound vibrational levels of this molecule.

Answer: Using our Morse model expressions for ΔGv+1/2 and ΔG(2 ← 0)

ΔG0+1/2 =ΔG(1 ← 0) = 1 359.708 = ωe − 2ωexe(0 + 1) = ωe − 2ωexe (3.21) Morse Morse ΔG(2 ← 0) = 2 674.560 = G (v =2)− G (v =0)

1 1 2 1 1 2 = ωe(2 + /2) − ωexe(2 + /2) − ωe(0 + /2) − ωexe(0 + /2)

=2ωe − 6 ωexe . (3.22)

Multiplying Eq. (3.21) by 3 and subtracting Eq. (3.22) from it yields

−1 ωe =3×1 359.708 − 2 674.560 = 1 404.564 [cm ] ,

and substituting this back into Eq. (3.21) yields

−1 ωexe = (1 404.564 − 1 359.708)/2=22.428 [cm ] .

Finally, assuming a Morse model for the potential energy function, use of Eqs. (3.16) and (3.20) yield the values

2 −1 De = (1 404.564) /(4×22.428) = 21 990 [cm ] 1 vD = 1 404.564/(2×22.428) − /2 =30.813

7 for De and vD. The total number of bound vibrational levels of LiH (counting upward from v = 0 ) is therefore estimated to be vD +1=30+1=31.

3.3.3 Higher-Order Anharmonicity and the Dunham Expansion While the Morse potential function has a qualitatively realistic shape, and the analytic expression for its level energies is certainly simple to work with, it does not tell the whole truth. The observed vibrational energies and vibrational level spacings of real molecules are not precisely described by the simple quadratic 3.3. ANHARMONIC VIBRATIONS AND THE MORSE OSCILLATOR 61 function of Eq. (3.13) and the linear vibrational spacing function of Eq. (3.18). A common way of treating such data is to use the following straightforward empirical generalization of Eq. (3.13): 2 3 4 G(v)=ωe(v + 1/2) − ωexe(v + 1/2) + ωeye(v + 1/2) + ωeze(v + 1/2) + ... . (3.23) This in turn yields the vibrational spacing equation7 2 3 2 − 13 4 ΔGv+1/2 = ωe 2 ωexe(v +1)+ωeye 3v +6v + / + ωeze 4v +12v +13v +5 + ... . (3.24) It is clear that Eqs. (3.23) and (3.24) are becoming suﬃciently cluttered that using such expressions and solving multiple equations in multiple unknowns would rapidly become a rather unwieldy way of attempting to analyze experimental data. Moreover, that approach of determining Np parameters from Nd data essentially assumes that all of those input data are ‘perfect’, which is never really true. A more practical approach when one has data for multiple vibrational levels is to perform least-squares ﬁts to our polynomial expressions. This may readily be done using either a spreadsheet or a simple computer program. When

Dun 1 1 2 1 3 1 4 G (v)=Y1 0 (v + /2)+Y2 0 (v + /2) + Y3 0 (v + /2) + Y4 0 (v + /2) + ... , , , , 1 l = Yl,0 (v + /2) , (3.26) l=1 and he determined explicit analytic expressions for the various Yl,0 coefficients in terms of the potential expansion parameters {ai}: Y1,0 =2a0 Be (3.27) 3 2 Y2 0 = Be a2 − 5(a1) /4 (3.28) , 2 3/2 (Be) 2 2 4 Y3,0 = √ 10a4 − 35 a1 a3 − 17(a2) /2 + 225(a1) a2/4 − 705(a1) /9 (3.29) 4 a0 ... 2 in which Be=Cu/(re) (our ubiquitous inertial constant Cu appears again!). We note, of course, that this approach seems problematic, since each additional higher-order Yl,0 coefficient involves two additional ai potential coefficients. In practice, however, simultaneously taking account of the rotational energy level pattern resolves this problem (see §3.6), and one can determine the first Np {ai} potential parameters from empirically determined values of the Np leading Yl,0 and Yl,1 (see §3.6) coefficients. However, while it is appropriate that you be told about this more sophisticated theory, practical applications of this approach are beyond the scope of this text.8 So far, we have introduced three models for vibrational motion and the resulting transitions probed by spectroscopy. There is no set rule as to when each of these models should be applied. Obviously, the Morse or higher-order anharmonic oscillator provides a more realistic picture of molecular vibrations, and should be used whenever possible. However, if little or no information concerning the anharmonicity is available, the harmonic oscillator model is a convenient simplification. 7 6 Note that as for ωexe, thevariablenamesωeye and ωeze should each be treated as a single symbol, and not as a product. 8 It is clear that these Yl,0 coefficients correspond precisely to the empirical coefficients ωe, ωexe, ωeye, . . . etc., of Eq. (3.23), with the anomalous historical sign convention aberration that defines ωexe = −Y2,0 . 62 CHAPTER 3. VIBRATIONAL SPECTROSCOPY

3.4 Bond Dissociation Energies and Birge-Sponer Plots

While the Morse function is certainly more realistic than a harmonic oscillator model for representing a molecular potential energy curve, as the above discussion indicates, it is far from being exact. In addition to the reservation raised in footnote 5 on p. 58, its prediction that the vibrational level spacing decreases as an exactly linear function of v is not correct for real molecules. As a result, Eq. (3.16) may not provide us with particularly accurate estimates of the molecular dissociation energy. Moreover, while its shape is certainly more flexible, the Dunham potential of Eq. (3.25) goes to +∞ or −∞ (depending on the mathematical sign of the last ai coefficient) as r →∞, and for most more sophisticated potential energy functions, the radial Schrödinger equation cannot be solved analytically to give expressions for the level energies that can be inverted to yield a value of De or D0 (as we were able to do in Exercise (iii)). However, we still very much want to be able to determine accurate bond dissociation energies; thus, we must devise some alternative way of doing so. The nature of the problem is illustrated by panels A and B of Fig. 3.5. This example is a case in which experimental fundamental, overtone and hot band data have given us the vibrational energies for levels 9 v=0 − 6 . However, although we know that the curve in Fig. 3.5-B must extrapolate to an asymptote at D0, there would clearly be very large uncertainties associated with any such extrapolation on this plot. A better method of performing such extrapolations was introduced by Raymond Birge and Hertha Sponer in 1926. They recommended that the measured spacings between adjacent vibrational levels be plotted as shown in Fig. 3.5-C, with the point representing the spacing between levels v and v+1 being plotted at the half-integer ordinate value, v+1/2 . Since each of the small rectangles in that figure has a width of 1, its area is equal to the associated value of ΔGv+1/2 . Thus, the sum of the areas of the six rectangles shown is the energy difference {G(6) − G(0)} , which is essentially equal to the area under the smooth curve from v =0 to 6. It is therefore clear that if we could guess how to extrapolate this properly curve to the intercept, the area under the curve from v = 6 to the intercept would be the sum of all the missing rectangles – the quantity D − G(v=6) which is the binding energy of the highest observed level. The only remaining problem then is – how do we do this extrapolation? For almost 50 years there was no fundamental answer to the above question, and the most common approach was simply to use an empirical linear or polynomial extrapolation. In the sample problem of Fig. 3.5, −1 the six known experimental ΔGv+1/2 values are 476.00, 450.24, 423.19, 394.74, 364.90 and 333.74 cm .A straight line through the last two points has a slope of (333.74 − 364.90)/1=−31.16 , and yields the dot- dot-dash line seen in Fig. 3.5-C. The fact that it does not go through the points at small v shows that a

Morse oscillator model (which implied a strictly linear ΔGv+1/2 function) is not accurate for this system. Adding the areas of the six rectangles shown in Fig. 3.5-C is precisely equivalent to adding up the six given ΔGv+1/2 values, and yields 5 − −1 G(6) G(0) = ΔGv+1/2 = 2442.81 [cm ] . v=0 If we wish to use straight-line extrapolation to estimate the distance from level v+6 to the dissociation limit, it is ﬁrst necessary to determine the value of our linear function at the point v+6. Using the above slope we 1 ﬁnd that the value of our linear function there is 333.74 − 2 (31.16) = 318.16 . The distance from the point v=6 to the intercept is then simply 318.16/31.16 = 10.211 , and the area under that line in the extrapolation 1 −1 region is given by the formula for the area of a right-angle triangle, 2 ×318.16×10.211 = 1 624.29 cm .This is our linear Birge-Sponer extrapolation estimate of the distance from level v =6 to dissociation. Adding it to the directly measured value of {G(6) − G(0)} then yields

−1 D0 = 2442.81 + 1624.29 = 4067.10 cm .

As is suggested by the curved solid line extrapolation shown in Fig. 3.5-C, a somewhat more realistic extrapolation might be obtained by ﬁtting a polynomial to the experimental points. Doing this using a

9 This asymptote is at D0 rather than De, since experiment can only tell us directly about energies of one level relative to another, so information about the zero point energy De − D0 can only be obtained by extrapolation. 3.4. DISSOCIATION ENERGIES AND BIRGE-SPONER PLOTS 63

500 D 4000 D0 ? ∼ Birge-Sponer plot E (υ) 400 V(r) ? vib ⇑ 3000 ∼ ΔEυ + ½ υ = 6 ⇓ 300 5 4 2000 200 3

2 1000 A B 100 C 1

0 0 0 0 5 υ 10 15 0 5 υ 10 15 Figure 3.5: Vibrational extrapolations and the Birge-Sponer plot. spreadsheet or small computer program yields the polynomial expression 2 ΔGv+1/2 = 488.339 − 24.3609 v − 0.68152 v , and from the roots of this quadratic we find that the intercept of this curve is vD=1 4 .314 . From elementary calculus we may then determine the area under the curve for this quadratic extrapolation to be: 14.314 2 −1 D0 − G(v =6) = 488.339 − 24.3609 v − 0.68152 v dv = ... = 1439.98 [cm ] , 6 which in turn yields −1 D0 = 2442.81 + 1439.98 = 3882.79 cm as our quadratic-extrapolation estimate of D0 . −1 The 184 cm difference between our linear and quadratic extrapolation estimates for D0 may seem somewhat disconcerting. This is an example of the type of problem encountered when treating experimental data in the real world – there is often no obvious unique “right” answer. However, this difference is only 12% of the length of the energy extrapolation, and only approximately 5% of the size of these estimates for D0. Thus, on an absolute scale we have still obtained a pretty accurate value for the dissociation energy. It was mentioned above that for almost 50 years after 1926 there was no fundamental theoretical guidance regarding what shape to expect for a Birge-Sponer plot extrapolation. However, that situation changed in 1970 when Robert Le Roy and Richard Bernstein introduced a beautifully simple theory that gave an explicit analytic expression for the shape expected for a Birge-Sponer plot near its intercept at v=vD . While that theory is too complicated to present here, the essential result is that near its intercept, a Birge-Sponer plot has the shape − (n+2)/(n−2) ΔGv+1/2 X0 (vD v) , (3.30) in which n is the integer power defining the limiting long-range inverse-power behaviour of an intermolecular n 10 potential V (r) ∼ D − Cn/r , and has a value between 1 and 6. The proportionality constant X0 is a precisely known function of that long-range potential constant Cn and the reduced mass μ, and as in our Morse potential discussion, vD is the value of v at the intercept. [Note that for the Coulomb potential case of n = 1 , this theory yields the familiar H atom level energy expression of Eq. (1.15), with the caveat that for the special case of the Coulomb potential, vD becomes an integration constant with a value of −1.] For this more general theory, if n=5 the limiting extrapolation behaviour on Fig. 3.5-C would have 7/3 the functional form ΔGv+1/2 ∝ (vD − v) that yields the dotted curve extrapolation. It is interesting to note that although this optimal extrapolation yields vD =21.6 , which is almost 50% bigger than values obtained from our simple linear or quadratic extrapolation procedures, the corresponding true D0 value of 4134.37 cm−1 is only modestly different from those earlier values. While the real world is often not so kind regarding the accuracy of rough empirical Birge-Sponer plot extrapolations, it is still a very useful technique, and will be used in vibrational extrapolation problems discussed in this course. 10 The value of n for a given case is readily determined from a knowledge of the nature of the atomic fragments formed on dissociation. For most molecules dissociating to yield neutral atoms, n = 5 or 6. 64 CHAPTER 3. VIBRATIONAL SPECTROSCOPY

3.5 Vibrations in Polyatomic Molecules

Diatomic molecules are the simplest case we encounter in IR spectroscopy, since they possess only one type of vibrational motion, a bond stretch. Our “Vibrational Selection Rule 1” dictates that for its transitions to be allowed, a diatomic molecule must have a permanent dipole moment, as its electric dipole can only change during a stretch if it is present to begin with. The rules governing the IR spectra of polyatomic molecule are more complicated, because the fact that they are composed of more than two atoms means that they have more than one type of vibrational motion, and the fact that some of those modes of motion may distort a symmetric molecule into (temporarily) non-symmetric shapes means that they can have an oscillating dipole even if their equilibrium dipole moment is zero (see §3.1.1). Since each atom of a general N–atom molecule can move about in 3-dimensional space, to specify its conﬁguration fully requires 3N coordinates; in other words, such a system has a total of 3N degrees of freedom. However, as discussed in §2.1.2 and 3.1.2, we always introduce relative coordinates in order to allow us to describe the behaviour of the system in terms of the motion of the centre of mass (i.e., the overall translational of the molecule) plus the internal motion. Since three coordinates are required to deﬁne the position of the centre of mass, R cm =(xcm,ycm,zcm) , this leaves 3N−3 degrees of freedom for the internal motion. The overall rotation of the molecule will account for some of these 3N−3 internal degrees of freedom. The discussion of §2.1 indicated that for a diatomic molecule, or indeed any linear molecule, only two coordinates are required to characterize its orientation and rotational behaviour, the angular coordinates θ and φ of Fig. 1.9. Similarly, for a non-linear molecule, three angular coordinates are required (see §2.5.3). This leads us to the conclusion that

• a linear N–atom molecule has 3N − 5 vibrational degrees of freedom (the other ﬁve being three for translation plus two for overall rotation)

• a non-linear N–atom molecule has 3N − 6 vibrational degrees of freedom (the other six being three for translation plus three for overall rotation)

Each of these vibrational degrees of freedom corresponds to a separate type of co-ordinated overall motion of all of the atoms in the molecule, which is called a normal mode of vibration. Of course, all of the same considerations concerning harmonic and anharmonic vibrations discussed above apply to each of these vibrational modes. As a result, for a general polyatomic molecule the total vibrational energy is conventionally written as 1 1 1 G(v1,v2,v3, ...)= ωi(vi + /2)+ xi,j (vi + /2)(vj + /2)+ ... (3.31) i i j≥i where vi is the vibrational quantum number for mode i, ωi and xi,j are the harmonic and leading anharmonic vibrational constants. Equation (3.31) shows that in order to calculate the actual vibrational spacing associated with any particular mode it is necessary to know not only the harmonic and anharmonic constants ωi and xi,i associated with that mode (the analog of ωe and ωexe for a diatomic), but also all of the associated mixed anharmonicity constants, xi,j for j = i . Conversely, to obtain a knowledge of the complete set of harmonic and leading anharmonic constants for a species with Nm distinct vibrational modes would require Nm(Nm+3 ) /2 independent vibrational spacing measurements. It is usually quite diﬃcult to obtain that much information, so it is customary to characterize the energy associated with a given vibrational mode of a polyatomic molecule by the value of the observed fundamental vibrational spacing for that mode, denoted here asν ˜i, rather than by the value of the actual harmonic vibrational constant ωi. As a result, in Fig. 3.6 each of the vibrational modes is labelled with its name “νi” and with the value of this vibrational spacingν ˜i. Let us now consider a few particular cases.

(i) Adiatomicmolecule For a diatomic molecule, N=2, and since it is linear, it has 3N−5=1 vibrational mode, that mode being the usual radial stretching vibration. 3.5. VIBRATIONS IN POLYATOMIC MOLECULES 65

ν1 C − H symmetric stretch ∼ -1 ν1 = 3374 cm

ν2 C ≡ C stretch ∼ -1 ν1 ν2 = 1974 cm ∼ -1 ν1 = 3657 cm ν3 C − H asymmetric stretch ∼ -1 ν3 = 3287 cm

ν2 ν4 trans bend ∼ -1 ∼ -1 ν2 = 1595 cm ν4 = 612 cm A − + − + B

ν5 cis bend ∼ -1 ν3 ν5 = 729 cm ∼ -1 − + + − ν3 = 3756 cm

Figure 3.6: Vibrational normal modes of: A.waterH2Oand B. acetylene C2H2.

(ii) A linear triatomic molecule: O=C=O Since N=3 and the molecule is linear, it will have 3×3−5=4 vibrational degrees of freedom. Three of these were shown in Fig. 3.2, but where is the fourth? With a little thought we realize that since the molecule can move in three-dimensional space, it will have a bending mode out of the plane of the page, as well as the one shown in the ﬁgure. These two modes will be degenerate, in that they will have

exactly the same bending force constant and vibrational constants ω2 and {x2,j} (in this molecule, the bending mode is conventionally labelled mode 2). However, there will be two separate contributions to a b Eq. (3.31) associated with independent bending quantum numbers v2 and v2. Putting several quanta of vibrational energy into one of these bending modes and none into the other, and vice versa,would yield exactly the same total amount of vibrational energy. Similarly, if the bending could be accurately

described as a harmonic oscillator (i.e., all x2,j=0), putting one quantum of vibrational energy into each, or two into one and none into the other (or vice versa) would yield exactly the same vibrational energies. On the other hand, if the anharmonicity of this vibrational motion could not be ignored, the diﬀerence between the total energy obtained on putting one quantum into each of the bending modes

vs. putting two quanta in one and none in the other would be 2 x2,2 . Thus, even for a degenerate mode there are signiﬁcant complications.

As discussed earlier, the antisymmetric stretch and bending modes of CO2 will be infrared active, but for the symmetric stretch mode there will be no allowed IR transitions.

(iii) A non-linear triatomic molecule: H2O As in the preceding case, N=3, but since this molecule is non-linear it will have only the 3 × 3 − 6=3 independent vibrational modes shown in Fig. 3.6-A. In this case we note that the symmetric stretch mode is identiﬁed as “ν1”, the bending mode as “ν2” and the antisymmetric stretch as “ν3”. The water molecule has a permanent dipole moment, and it is clear that for all three of these modes that dipole will oscillate in magnitude (for ν1 and ν2) or direction (for ν3), so all three of these modes will be IR active.

(iv) A linear tetra-atomic molecule: acetylene H–C≡C–H

Since N=4 a n d C 2H2 is linear, it will have a total of 3×4−5=7 vibrational modes. Any non-cyclic N– atom molecule will have N−1 bond-stretch modes (No. stretching-type modes = No. bonds), while the remaining 3N−4 (for linear molecules) or 3N−5 (for non-linear molecules) modes will be associated with bending motion. In the present case this means that we will have three stretching-type modes and four bending modes, as seen in Fig. 3.6-B. As is always the case for linear molecules, these bending 66 CHAPTER 3. VIBRATIONAL SPECTROSCOPY

modes come in degenerate pairs associated with atoms moving in vs. perpendicular to the plane of the paper.

One question raised by these examples is, what is the convention for labeling the modes?Toanswerit properly would require the use of group theory for classifying the symmetries of the various types of motion, a topic which is beyond the scope of this course. However, a wide range of cases are covered by the simple rules that: (i) the most symmetric types of motion are counted ﬁrst, (ii) within a given symmetry type the modes are numbered in order of decreasing frequency, and (iii) for linear triatomic molecules the bending mode is always labelled ν2. This is as far as this topic will be pursued here.

Local Mode or Group Vibrations Each normal vibrational mode of a polyatomic molecule involves simultaneous co-ordinated motion of every atom in the molecule. In practice, however, one pair or one sub-group of atoms will often have very large amplitude motion, while other atoms move relatively little. When this occurs, such modes are classified as local mode or group vibrations, and they usually have characteristic frequencies (or frequency ranges) that carry over from one molecule to another. This provides us with a wonderful tool for identifying the makeup of large molecules, since the observation of certain characteristic frequencies in their infrared spectra advertises the presence of that chemical functional group in the molecule. Consider, for example, the linear molecule H–C≡N. We know that it must have 3 × 3 − 5 = 4 vibrational modes, of which 3−1=2 will be bond-stretch modes while the other 2 are bending modes. By analogy with CO2 or H2O, we would expect those stretching modes to consist of an approximately symmetric mode in which both bonds stretch and compress in phase, plus an antisymmetric-type mode in which one bond stretches while the other compresses, and vice versa. This is indeed the case, with the symmetric stretch ss −1 mode having an energy ofν ˜1 =˜ν1 = 3 441 cm and for the antisymmetric stretch mode an energy of as −1 ν˜3 =˜ν3 = 2 119 cm (recall that for linear triatomics, the bending mode is always labelled ν2). However, examination of the precise nature of these normal modes (a sophisticated mathematical treatment reserved for a higher-level course) shows that ν1 involves large-amplitude C–H stretching and low-amplitude C≡N motion, while the reverse is true for ν3. Moreover, looking up the properties of diatomic molecules we see −1 −1 that ωe = 2 862 cm for CH and 2 068 cm for CN, while the infrared spectra of a whole range of other molecules with C–H functional groups have IR transitions with energies in the range 2 800 − 3 400 cm−1. Thus, it seems clear that even in this very simple polyatomic molecule, a local mode picture is appropriate for describing the vibrational stretching modes. Table 3.1 presents a list of the characteristic vibrational transition energies associated with particular chemical functional groups. The first seven entries in the first column are all associated with the stretching

Table 3.1: Characteristic group vibrational energiesν ˜i for common chemical function groups. chemical approximate chemical approximate group frequency [cm−1] group frequency [cm−1] ≡C-H 3300 S-H 2580 (phenyl)-C-H 3060 –C≡N 2250 =CH2 3030 –C≡C– 2220 –CH3 2970 (antisym. stretch) –C=O 1750-1600 –CH3 2870 (sym. stretch) C=C 1650 –CH2– 2930 (antisym. stretch) C=N 1600 –CH2– 2860 (sym. stretch) C-C, C-N, C-O 1200-1000 –CH2– 1470 (deformation) C-Cl 725 –CH3 1460 (antisym. deform.) C-Br 650 –CH3 1375 (sym. deform.) C-I 550 -O-H 3600 C=S 1100 NH2 3400 C-F 1050 3.6. ROTATIONAL STRUCTURE IN VIBRATIONAL SPECTRA OF DIATOMICS 67

Figure 3.7: Room temperature absorption spectrum of D35Cl (the stronger line of each pair) and D37Cl. of C–H bonds in various chemical environments, while the next three are frequencies associated with various H–C–H bending modes; even though the external bonding of the C atom changes from one case to another, within each of these groups there is still a large degree of similarity. [It is also interesting to note that the ratios of the characteristic frequencies for the carbon-to-carbon√ single,√ double and triple bonds, 1200 : 1650 : 2220 ≈ 1:1.38 : 1.85 approximately matches the ratios 1 : 2: 3 ≈ 1:1.41 : 1.73 , which is qualitatively what one would expect, since ωi ∝ k˜ .] These characteristic features are particularly useful when applying IR spectroscopy to solid or liquid materials.

3.6 Rotational Structure in Vibrational Spectra of Diatomics

Up to this point we have discussed vibrational spectroscopy in much the same manner as we described rotational spectra, as if it consisted of a single sharp line for each unique v−v transition. However, 35 2 35 −1 this is not the case. Consider, for example, the case of D Cl (or H Cl), for which ωe=2145.133 cm , −1 −1 −1 ωexe=2 7 .1593 cm , ωeye=0 .079 93 cm and ωeze=− 0.003 03 cm . A harmonic oscillator model would predict that its vibrational spectrum consisted of a single line (or superposition of lines) at a transition energy of −1 ν˜=ωe=2145.133 cm . Allowing for anharmonicity, it would be expected to have a strong fundamental transition at 2 091.059 cm−1, and weaker first and second overtone transitions at 4 128.430 and 6 112.449 cm−1, while if the sample was sufficiently hot to yield measurable populations in excited vibrational levels, there could be weak “hot band” transitions at 2 037.372 and 1 984.019 cm−1. In contrast, Fig. 3.7 shows that its absorption spectrum at room temperature (under which conditions no hot bands would be observed) consists of about 30 sharp, roughly equally spaced lines spread across a 320 cm−1 interval. The reason for this was alluded to earlier when it was pointed out that there is no angular momentum associated with vibrational stretching. As a result, for angular momentum to be conserved when an IR photon is absorbed or emitted, a simultaneous rotational transition must occur. Since rotational levels have relatively small energy level separations, a substantial number of them tend to be populated at most temperatures, and when a collection of molecules in thermal equilibrium undergo a vibrational transition, we obtain a “band” or group of spectral lines such as that seen in Fig. 3.7. It is because vibrational transitions of gas phase molecules always occur with a wealth of rotational fine structure that we normally refer to vibrational spectra as consisting of “bands”. 68 CHAPTER 3. VIBRATIONAL SPECTROSCOPY

P branch R branch

⎩ ⎪ ⎪ ⎪ ⎪ ⎪ J=⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ ⎩ ⎪ ⎪ ⎪ Δ⎪ J=⎨ −⎪ 1 ⎪ ⎪ ⎧ Δ +1 ⎧ ⎪ 7 ⎪ ⎪ ⎪ 6 ⎪ ⎪ υ′ ⎨ 5 ⎪ ⎪ 4 ⎪ ⎪ 3 ⎪ 2 ⎪ ⎩ J ′=0 ↑ energy ⎧ ⎪ 7 ⎪ ⎪ 6 ⎪ ⎪ υ′′ ⎨ 5 ⎪ ⎪ 4 ⎪ 3 ⎪ 2 ⎪ ⎩ J ′′=0 band R(3) gap R(2) R(4) P(4) P(3) R(1) P P(2) ←⎯→ R(5) spectrum (5) R(6) P(6) P R(0) P(7) (1) R(7) P(8) ↓ ∼ −1 ν0 ν∼/cm → Figure 3.8: Spectrum and energy level pattern for an infrared band.

The pattern of level energies and allowed transitions, and the associated spectrum of a model infrared (vibrational) band are presented in Fig. 3.8. As indicated there, each of the vibrational levels of any molecular potential energy curve (such as those seen in Figs. 3.4 or 3.5-A) has a ladder of rotational sub-levels associated with it. The transitions observed within a given vibrational band are those allowed by the usual ΔJ = ±1 selection rule. The individual transitions must, of course, be labelled by the vibrational and rotational quantum numbers of the upper and lower states:

ν˜(v,J; v,J)=E(v,J) − E(v,J)i, (3.32) where we use the standard convention of labeling the quantum numbers for the upper levels with a single prime () and those for the lower level in the transition with double primes (). The values of J and J are linked by the normal angular momentum conservation selection rule, and it is customary to label the allowed transitions according to the sign of the change in J and the rotational quantum number of the lower level. In particular, transitions with ΔJ = J − J = Jupper − Jlower = −1 are called P transitions, and those with ΔJ = +1 are called R transitions.11 For P transitions in vibrational spectra the upper level has less rotational energy than the lower one, and for R transitions the opposite is true. (As a result, all transitions in pure rotational spectroscopy are R transitions.) It is very important that these labeling rules be understood and followed, since there would be considerable confusion if a particular E(v,J) ↔ E(v,J) transition was given one name in an absorption spectrum and another in emission. In view of the above, the energy of a particular P -branch transition in a given (v,v) vibrational band would be ν˜P (J )=E(v ,J ) − E(v ,J )=E(v ,J − 1) − E(v ,J ) = {G(v )+Fv (J − 1)}−{G(v )+Fv (J )} = {ignoring centrifugal distortion} =[G(v ) − G(v )] + [Bv (J − 1)(J ) − Bv J (J +1)] 2 =˜ν0(v ,v ) − [Bv + Bv ](J ) − [Bv − Bv ](J ) (3.33) ≈ ν˜0(v ,v ) − [Bv + Bv ](J )(forsmallJ ) 11 Note that we have introduced another standard convention – that the quantum number change is written as Jupper minus Jlower. 3.6. ROTATIONAL STRUCTURE IN VIBRATIONAL SPECTRA OF DIATOMICS 69 while for an analogous R-branch transition ν˜R(J )=E(v ,J ) − E(v ,J )=E(v ,J +1) − E(v ,J ) = {G(v )+Fv (J +1)}−{G(v )+Fv (J )} = {ignoring centrifugal distortion} =[G(v ) − G(v )] + [Bv (J +1)(J +2) − Bv J (J +1)] 2 =˜ν0(v ,v )+ [Bv + Bv ](J +1) − [Bv − Bv ](J +1) (3.34) ≈ ν˜0(v ,v )+ [Bv + Bv ](J +1) (forsmallJ )

whereν ˜0(v ,v ) ≡ [G(v ) − G(v )] is called the band origin, the energy that the transition would have if rotational effects could be ignored. The band origin lies roughly mid-way between the P -andR-branches in a spectrum such as that for DCl shown in Fig. 3.7. The mathematical form of these expressions is in some regards quite similar to that of Eq. (2.17), which was derived for pure rotational transitions. First of all, the quadratic terms found here would disappear for pure rotational spectra, since there v = v , and hence Bv = Bv . Moreover, Eqs. (3.33) and (3.34) also predict that the lines in the spectrum should be equally spaced with a line separation of twice the inertial rotational constant (here [Bv + Bv ] vs. 2B in Eq. (2.17)). The main remaining difference is simply that the pure rotational spectra discussed in Chapter 2 can only have R-branch transitions, while both P - and R-transitions are possible in an infrared (vibrational) spectrum, because the relatively large vibrational energy quantum more than compensates for the decrease in rotational energy. This pattern of (roughly) equally spaced lines marching off to low (P -branch) and high (R-branch) energies relative to the band origin is clearly displayed in Figs. 3.7 and 3.8. We also see that there is no transition associated with the band origin (at 2 091 cm−1 in Fig. 3.7, and indicated by the dotted line in the lower segment of Fig. 3.8), and that there is a “band gap” of width approximately 2[Bv + Bv ], twice the usual line spacing, centred at the band origin and separating the P and R branches of the band. The intensity pattern seen in Figs. 3.7 and 3.8 is explained by the discussion of §2.4. The initial levels of all the transitions seen here are pure rotational sublevels of vibrational level v, and their thermal population distribution will be given by Eq. (2.31). Since centrifugal distortion can be neglected at low J,wemay expect Eqs. (2.33) to predict the initial-state population distribution accurately, which leads us to predict pop Jmax(293 K) = 3.85 ≈ 4 as the initial-state rotational quantum number for the most intense P -andR- branch lines in Fig. 3.7, as is observed.12 This type of rotational intensity pattern is found in virtually all vibration-rotation (or electronic, see Chapter 5) band spectra of molecular systems in thermal equilibrium. Another interesting feature of Fig. 3.7 is the way the that line spacings slowly change as one moves away from the band centre. In particular, in both the P -andR-branches the line spacings are approximately the same at small J, but at larger J values the P -branch line spacings become systematically larger while the R lines get closer and closer together. Consideration of Eqs. (3.33) and (3.34) shows that this behaviour is due to the quadratic terms in J (or J + 1), and reflects the fact that the inertial rotational constant for the upper vibrational level, Bv , is slightly smaller than that for the lower level, Bv . Indeed, at sufficiently high J the quadratic term will actually take over and cause R-branch lines to turn around and begin to march off “to the red” (i.e., to smaller transition energies, or longer wavelengths). This branch turnaround behaviour is very common in the rotational structure of vibrational bands in electronic spectra, since the upper- and lower-level rotational constants Bv are often quite different, which makes the coefficient of the quadratic terms in Eqs. (3.33) and (3.34) relatively large (see Chapter 5). In infrared spectra, however, the Bv constants vary relatively slowly from one level to another, so this quadratic coefficient is typically quite small. As a result, such rotational branch turnarounds are only observed for systems at relatively high temperatures, where a very wide range of J levels is thermally populated. This is the case for the 1300 K emission spectrum of Na35Cl shown in Fig. 3.9, which shows the high frequency turnaround of the R-branches of the vibrational fundamental and the first hot band. Moving to higher and higher frequencies with increasing J the lines get closer and closer together, and then pile up on one another as they begin to turn around and move off to the red (i.e., to lower frequencies) with further increases in J. The pile-up of intensity located where the branch turns around is called a “band head”, and will be discussed

12 Small rotational intensity correction factors of J/(2J +1)for P lines and (J +1)/(2J +1)for R lines reﬁne these prediction at small J. 70 CHAPTER 3. VIBRATIONAL SPECTROSCOPY

Figure 3.9: NaCl emission spectrum showing band heads for the fundamental and ﬁrst hot bands.

further in Chapter 5. Note that the separation between the heads of the (2 − 1) and (1 − 0) bands of Na35Cl seen in Fig. 3.9 is approximately the same anharmonicity frequency shift of ∼ 2 ωexe predicted by Eqs. (3.18) 35 −1 or (3.24) ( ωexe(Na Cl) = 1.78 cm ) for the shift in the band origin. A final feature of Fig. 3.7 which deserves comment here is the small, but roughly constant shift to lower frequencies of the transitions associated with the minority (24% relative abundance) isotopologue 37 √ D Cl. From Eqs. (3.10) or (3.14) we see that all else being equal, ωe ∝ 1/ μ . This implies that for 37 35 37 −1 −1 D Cl, ωe ≈ 2 145.133 μ(D Cl)/μ(D Cl) = 2 141.975 cm ,a3.2cm shift to lower frequencies, which is roughly what is observed. Small additional J-dependent differences arise because potential energy function anharmonicity makes the Bv value slightly larger for the heavier (larger reduced mass) isotopologue. A room temperature spectrum such as that seen in Fig. 3.7 is clearly very instructive, and fitting the observed transition frequencies to Eqs. (3.33) and (3.34) can yield accurate values ofν ˜0(1, 0) = ΔG1/2 and of the rotational constants for the v = 0 and 1 vibrational levels. However, vibrational level spacings tend to be sufficiently large that in most cases only the ground level will have significant populations at “normal” temperatures. If one wishes to obtain information about higher vibrational levels from IR spectroscopy, it is therefore necessary to use experimental methods that produce significant populations of molecules in high vibrational levels. This can be done in a number of ways. One method of generating IR spectra that contain information about a wide range of vibrational levels is to take spectra of very hot gaseous samples. The GeO emission spectrum shown in the left half of Fig. 3.10 was obtained in this manner. This spectrum displays a rough version of the P/R branch structure seen in Fig. 3.7, but it is clearly somewhat irregular, and for very good reasons! One complication is the fact that there are five relatively abundant naturally occurring isotopes of germanium (35.9% is 74Ge; 27.7% is 72Ge; 21.2% is 70Ge; 7.7% is 73Ge; 7.4% is 76Ge) whose overlapping Δv = 1 emission contribute to this spectrum, and another is that in addition to the vibrational fundamental, hot bands are observed for (v,v)=(2, 1) up to (8, 7). For each hot band and each isotopologue, the band origin is slightly shifted, as predicted by Eqs. (3.14), (3.15) and (3.24), and for each vibrational level of each isotopologue, all of the rotational constants have small isotopic and vibrational shifts. In spite of these complexities, the very high resolution of modern instruments, and our understanding of isotopologue scaling effects allowed this extremely dense spectrum to be decomposed into 36 distinct bands characterized by very sharply determined parameters. To illustrate this point, the right half of Fig. 3.10 shows a small segment of the assigned spectra of the R-branches of the fundamental band; this illustrates the power of modern IR spectroscopy for untangling very messy spectra.

As indicated both by the preceding discussion and by that of Chapter 2, the rotational constants Bv and centrifugal distortion constants Dv, Hv, . . . , etc., will in general all depend upon the vibrational level v. 3.7. WHY ARE VIBRATIONAL LEVEL SPACINGS SO LARGE? 71

Figure 3.10: High temperature (1800 K) emission spectrum of GeO. Left: overview spectrum; Right: dispersed segment of the fundamental band showing isotopologue assignments.

One way of representing this behaviour is based on analogs of Eq. (3.23), namely:

1 1 2 Bv = Be − αe(v + 2 )+γe(v + 2 ) + ... (3.35) 1 Dv = De + βe(v + 2 )+... . (3.36)

However, this approach encounters diﬃculties, as one quickly runs out of distinct letters of the Greek alphabet to use in labeling constants, and remembering which Greek letter goes where and what sign convention to use becomes a problem. A much simpler approach is to use the generalized version of the Dunham vibrational energy expression of Eq. (3.26) by writing:

m 1 l m E(v, J)= Km(v)[J(J +1)] = Yl,m v + 2 [J(J +1)] , (3.37) m=0 m=0 l=0 in which the prime ()onthesumoverl indicates that there is no {m, l} = {0, 0} term in this double sum. It will immediately be clear that K0(v) ≡ G(v), K1(v) ≡ Bv , K2(v) ≡−Dv , . . . , etc. While spectroscopists often still refer affectionately to the leading coefficients of the “traditional” expressions, ωe, Be, ωexe, αe, ... , etc., the general expression of Eq. (3.37) is becoming more commonly employed in the scientific literature.

3.7 Why Are Vibrational Level Spacings So Much Larger Than Rotational Level Spacings?

In Chapter 2 we stated that rotational spectra were observed in the microwave (MW) region of the electromagnetic spectrum with transition energies in the 1 − 100 cm−1 region, and in the present chapter we have stated similarly that vibrational transitions occur in the infrared with transition energies of 100−5 000 cm−1. In other words, without presenting any justification we said that for any given molecule, vibrational level spacings are typically 10−100 times larger than rotational level spacings. Similarly, we shall see in Chapter 5 that electronic transitions typically occur in the visible/ultraviolet region of the spectrum, 104 − 106 cm−1. While a ‘civilian’ might be content to remember such facts blindly, a scientist would tend to seek a rational framework to explain it. It turns out that the simple particle-in-a-square-well model discussed in Chapter 1 provides exactly the type of explanation we would like to have. We saw in the early sections of Chapters 2 and 3 that the internal motion of a diatomic molecule could be described exactly in terms of the motion of a pseudo-particle of mass μ moving in space. For vibration this is motion along the radial coordinate r, while for rotation it is orbital motion about the centre of mass of the system at a radius of re. More particularly, the discussion 72 CHAPTER 3. VIBRATIONAL SPECTROSCOPY of §1.3.2 and 2.2.1 showed that diatomic molecule rotation, the orbital motion of a particle of mass μ at a radius of re, could be thought of as motion of a particle trapped in a box of length L = πre . In contrast, the “box” governing the (radial) amplitude of the vibrational motion has a length of around 0.2−0.4 re (see, e.g. Fig. 1.13). Since the box length associated with rotational motion is an order of magnitude larger than that for vibration, our particle-in-a-square-well energy formula tells us that rotational level spacings will be of order 100 times smaller than those for vibration. Moreover, since the values of μ for common molecules are typically in the lower portion of the range 1 − 100 u, and re values are typically 1 − 3 A,˚ Eq. (1.27) predicts that vibrational level spacing would be of order 50 − 5000 [cm−1], which is indeed the infrared region (see Fig. 1.14). Similar arguments place rotational level spacings in the range 0.5 − 50 [cm−1], which is what we call the microwave region of the spectrum. Similar arguments explain the relative magnitude of the transition energies for electronic spectra. First −4 of all, the electron has a mass of me ≈ 5.486 × 10 u, while a diatomic molecule reduced mass will have magnitudes of 1 − 100 u. Within a particle-in-a-square-well-box picture, for a fixed box size this difference would make the energy level spacings for the electron roughly 104 − 105 times larger. However, the radius of the orbit for one of the outermost electrons in a molecule (the first one to be excited) would be at least twice the equilibrium internuclear separation re , and this difference increases the effective box length relative to that for rotation (which we saw was πre) by a factor of two and hence reduces the level spacing by a factor of four. Thus, electronic excitation energies are expected to be of order 2 500 − 25 000 times larger than rotational level spacings for the same molecule, which would make them 25 − 250 times larger than its vibrational level spacings. Thus, we see that with very simplistic particle-in-a-box reasoning, we can rationalize the relative magnitudes of the photon energies required to drive rotational, vibrational and electronic transitions.

3.8 Problems

1. The hydrogen halides have the following fundamental (v =1← 0) band origins (in cm−1):

1H19F 1H35Cl 1H81Br 1H127I 4143.3 2988.9 2649.7 2309.5

Using the harmonic oscillator model, calculate the force constants k˜ of each of the hydrogen-halogen bonds. Are the bonds getting stronger or weaker through the series HF to HI? What does this imply about the the potential energy curves for these molecules?

2. The origins of the fundamental and the ﬁrst “hot” band for HgH occur at 1203.7 and 965.6 cm−1, respectively. Determine estimates of the harmonic and anharmonic vibrational constants ωe and ωexe,

the bond dissociation energy De , and the classical vibrational frequency νe (in Hz) for this molecule.

3. The following lines (in cm−1) were observed in the vibration-rotation spectrum of 1H35Cl: 2752.01, 2775.77, 2799.00, 2821.59, 2843.63, 2865.14, 2906.25, 2925.92, 2944.99, 2963.35, 2981.05, 2998.05, 3014.50. (a) Assign these lines to the corresponding P –andR–branch transitions. (b) Determine the rotational constants B and B and the average bond lengthsr ¯ for the upper and lower vibrational levels of this band. (c) Determine the value of the force constant k˜ for for HCl.

4. The following lines (in cm−1) were observed in the vibration-rotation spectrum of 2H35Cl: 2023.12, 2034.95, 2046.58, 2058.02, 2069.24, 2080.26, 2101.60, 2111.94, 2122.05, 2131.91. Using these data and your results from Question #3, determine if the bond length in H-Cl changes when 2H is substituted for 1H. [Your answer will allow you to justify or discredit the assumption made in this chapter that isotopic substitution does not change the bond.] 3.8. PROBLEMS 73

7 12 14 5. Molecular orbital theory predicts that Li2, C2 and N2 have diﬀerent bond orders. Given that −1 ωe = 357.43, 1854.71 and 2358.57 cm , respectively, for these molecules, determine whether the actual force constants for these bonds follow the predictions of MO theory.

6. The four central lines in the infrared spectrum of 1H35Cl have energies 2843.63, 2865.14, 2906.25 and 2925.92 cm−1. The spacings between these lines are not equal, so anharmonicity does inﬂuence this spectrum. Given that the R(0) and P (2) transitions both end up in the {v =1,J =1} level, and that the P (1) and R(1) transitions both begin in the {v =0,J =1} level, determine the two rotational

constants, Bv=0 and Bv=1, and determine the diﬀerence between the average bond lengths for these two vibrational levels.

−1 127 −1 7. Given that k = 314 N m , re =1.597 Aand˚ m( I) = 126.904 473 u, calculate the energy (in cm ) of the P (2) and R(4) transitions in the fundamental band of IR spectrum of 1H127I.

12 16 8. The vibrational frequency of C Oisknowntobe νe =65.0550 THz, and the fundamental band origin occurs at 2143 cm−1. Determine the positions of the origins of the ﬁrst “hot” and ﬁrst overtone band for 13C16O.

9. An unknown molecule XY has a vibrational frequency of 2331 cm−1 and a force constant of k = −1 2245 N m . The diatomic oxide of X, XO, has a vibrational frequency of 1876 cm−1 and a force constant of k = 1 550 N m−1. Use this information to identify the molecule XY.

10. The IR spectrum of 23Na127I has a strong band with its origin at 284.50 cm−1, a weaker band at 283.00 cm−1, and the relative intensity of the latter increases with temperature. Determine the bond force constant, anharmonicity, zero-point energy, dissociation energies De and D0 of NaI in “spectroscopists’ units”.

11. In the infrared spectrum of 1H35Cl, the P (4) and R(2) lines of the fundamental band occur at 2799.00 −1 −1 and 2944.99 cm , respectively. Determine the values of ωe (in cm ) and of the HCl bond length in the v = 0 vibrational energy level (in A).˚

12. For 2H35Cl the P (4) and R(2) lines have energies 2046.58 and 2122.05 cm−1, respectively. Does the bond force constant or v = 0 bond length change upon substitution of 2Hfor1H in HCl? Would you expect either of these quantities to be diﬀerent, and if so/not, why?

13. How many normal modes of vibration are there in each of the following molecules: CO, H2O, HCCH, H2CCH2,H3CCH3 and C60 ? For the ﬁrst ﬁve of these species, how many of those modes are associated with bending motion?

1 35 14. Assuming that H Cl is reasonably well described by a Morse potential with parameters: De = −1 −1 1 35 2 35 5.250 eV, ωe = 2990.95 cm ,and ωexe =52.82 cm , determine D0 for each of H Cl and H Cl.

15. Which bond would be easier to break, a C-1HbondoraC-2H bond? Brieﬂy explain your reasoning.

16. Draw and label the level energies, the vibration-rotation transitions, and resulting IR spectrum of a typical diatomic molecule, for all lines up to R(5) and P (5). Describe how you would obtain the bond length and the force constant k˜ from the IR spectrum.

17. The infrared spectrum of N2O shows three fundamental vibrational bands centred at 589, 1285 and 2224 cm−1. What does this tell you about the structure of this molecule; i.e., “Is it linear or bent?” and “Is the O atom between the two N atoms or at one end of the molecule?” Compare this case to the IR spectrum you would expect for CO2 .

18. Calculate the percentage diﬀerence between the fundamental vibrational transition energies of 23Na35Cl and 23Na37Cl under the assumption that their force constants are the same.

−1 19. For a molecule whose vibrational data give ωe = 1387.09 and ωexe =83.01 cm , obtain an estimate for De and for the total number of vibrational levels for that molecule. 74 CHAPTER 3. VIBRATIONAL SPECTROSCOPY

20. Determine extended versions of Eqs. (3.33) and (3.34) that take account of the upper and lower level

centrifugal distortion constants, Dv and Dv . Chapter 4

Raman Spectroscopy

What Is It? Raman spectroscopy determines vibrational and rotational level spacings from the frequency shifts of scattered light.

HowDoWeDoIt? Molecular transition energies are observed by measuring the shifts in frequency of light scattered when a molecule is subjected to an intense beam of monochromatic light.

WhyDoWeDoIt? We saw in preceding chapters that molecules with no permanent dipole moment would have no pure rotational spectra, and that molecular vibrational motion for which there was no oscillating electric dipole would have no infrared absorption or emission spectra. Raman spectroscopy allows us to determine vibrational and rotational level spacings for such systems, and hence to determine bond lengths and force constants for molecules whose symmetry prevents them from having normal allowed MW or IR spectra.

We have seen that a molecule with no permanent dipole moment will have no pure rotational (microwave) spectrum, and if it is a diatomic, it will also have no (infrared) vibrational spectrum. Both are true for the molecules N2 and O2 which comprise most of our atmosphere, and the former is true for the greenhouse gas molecules CO2 and CH4; however, we very much need to know their structure and properties. Fortunately, in 1928 an Indian scientist named C.V. Raman discovered a special method of making spectroscopic measurements on such species, a discovery that earned him both a knighthood and the 1930 Nobel Prize in physics. In one sense this topic introduces nothing new, since it is just another way of measuring the rotational and vibrational level spacings: once these spacings are known, the theory and methods presented in Chapters 2 and 3 can be used for determining molecular bond lengths, stretching force constants and dissociation energies. However, the rules governing such spectra are diﬀerent than those presented previously, so the Raman mechanism deserves some attention in its own right.

4.1 “Light-As-A-Wave” Description of Raman Scattering

We begin by describing two perspectives on the Raman effect. The first combines our classical view of light as a wave phenomenon with the fact that a molecule consists of a distribution of negative electronic charge centred about tiny, heavy positive nuclei. When a molecule is subjected to a strong external electric field E, that field will distort the molecule’s diffuse electronic charge distribution to induce a small dipole moment that is proportional to the strength of the field:1

M ind = α E (4.1) in which α is the polarizability of the molecule, a property that indicates how readily its electron distribution is distorted by an external ﬁeld. If this is a static “laboratory” electric ﬁeld, the induced dipole may interact

1 Note that M ind and E are not necessarily parallel, and that in a more sophisticated treatment α would be represented by a3×3 matrix of values.

75 76 CHAPTER 4. RAMAN SPECTROSCOPY

16 18 16 orientation O O O 16 of 18O 16O 16O 18O 18O O molecule 18 16 18 O O O → E d l e i f dipole c

i induced r

t by field c e l e

l 1 / νrot a

n ↑ r

e dipole t

x induced e by field time → 0

Figure 4.1: Oscillating induced dipole moment of a rotating non-polar molecule in an external electric field. with light in approximately the normal manner, and transitions can occur. However, if that external electric field is due to an intense beam of light, this is what we call a “second-order” process, since the light must first create the dipole, and then interact with it; as a result, transitions excited in this way will be relatively weak. However, they can still be observed. Consider, for example, the clockwise rotation of an isotopically heteronuclear O2 molecule in an external electric field, as illustrated by Fig. 4.1.2 Its electron distribution (shaded ellipsoids) will clearly be cylin- drically symmetric about the centre of the bond. Because of this shape, the polarizability along the axis of the molecule, α, is greater than its polarizability perpendicular to that axis, α⊥, and hence the magnitude of the induced dipole moment will be greatest when the molecule is aligned parallel to the external electric field. As a result, the induced dipole will oscillate with a frequency that depends on the natural rotational period of the molecule, 1/νrot . However, because of the cylindrical symmetry of its electron distribution, one full rotation of the molecule is accompanied by two full oscillations of the induced dipole (see Fig. 4.1). As a result, the induced dipole oscillates with a frequency of 2νrot . If the external electric field distorting the molecule is provided by an intense beam of light of frequency ν0 Hz, Eq. (1.3) shows that the electric field felt by the molecule oscillates as E( t)=E 0 cos(2πν0 t + φ0). At the same time, because of the rotation of the molecule, its polarizability along the direction of the field can be written as 0 rot αE0 (t)=α +Δα cos(4πν t) , (4.2) in which α0 =(2α⊥ + α)/3 is the spherical average of the polarizability, Δα =[α − α⊥] is called the polarizability anisotropy,andE0 is a unit vector pointing along the direction of the electric field. Upon substituting these expressions for E(t)andαE0 (t) into Eq. (4.1), we see that the time-dependent induced dipole may be written as

M ind(t)=E 0 {α0 cos(2πν0 t)+Δα cos(4πνrot t)cos(2πν0 t) } (4.3) 1 1 = E0 α0 cos(2πν0 t)+ 2 E0 Δα cos (2π[ν0 +2νrot]t)+ 2 E0 Δα cos (2π[ν0 − 2νrot]t) , with the second version of Eq. (4.3) being obtained from the ﬁrst by utilizing the familiar trigonometric 1 relation: cos a cos b = 2 [cos(a + b)+cos(a − b)] . The second version of Eq. (4.3) shows us that in a very intense beam of light a rotating molecule will have an oscillating induced dipole moment with three diﬀerent frequency components: one at the frequency of the incident light ν0, one at a frequency that is the sum of ν0 plus twice the rotation frequency of the molecule,

2 Note that isotope diﬀerences have nothing to do with the process, and have been introduced here solely to indicate the direction of rotation. 4.1. “LIGHT-AS-A-WAVE” DESCRIPTION OF RAMAN SCATTERING 77

16 16 16 16 16 O 16 O stretching O O O O of 18 18 18 18 molecule 18 O 18 O O O O O → E

d [equilibrium] [equilibrium] [equilibrium] l e i f dipole c

i induced r

t by field c e l e

l 1 / νvib a

n ↑ r

e dipole t

x induced e by field time → 0 Figure 4.2: Oscillating induced dipole moment of a vibrating non-polar molecule in an external electric ﬁeld.

and one that is ν0 minus twice the rotation frequency of the molecule. Just as the oscillating charges in a radio station antenna cause emission of electromagnetic radiation, so these oscillating induced dipoles allow molecules to scatter light at these three frequencies. The scattering of light at the frequency of the incident beam ν0 is called Rayleigh scattering, that associated with the lower frequency ν0 − 2νrot is called Stokes scattering, and that associated with ν0 +2νrot is called anti-Stokes scattering. Since normally |Δα/α0| 1, we expect that the Stokes and anti-Stokes scattering will be much weaker than Rayleigh scattering. The source of this naming convention will be explained below. Most Raman spectroscopy is performed using high-frequency incident light, since historically it was much easier to find intense, monochromatic short-wavelength light sources,3 and theory (not presented here) 4 shows that the fraction of incident light intensity that is scattered is proportional to (ν0) .Itisimportant to remember that in Raman spectroscopy, the incident light only serves as the source of the strong electric field felt by the molecule, and its frequency has no relationship at all to the period of the molecular motion being excited or to the energy of the molecular excitation or de-excitation. In contrast, as we often see in physics, it is the sum and difference of the frequencies of two fields that really matters. As illustrated by Fig. 4.2, our classical description of vibrational Raman spectroscopy is qualitatively quite similar to that presented for rotation. In this case the vibration of the molecular bond is accompanied by a rhythmic stretching and compression of the electron distribution, and hence also by an oscillation of the component of the molecular polarizability along the direction of the field. Equation (4.1) shows us that this in turn gives rise to an induced dipole that oscillates in phase with the vibrational motion,

vib αE0 (t)=¯α + δα cos(2πν t) , (4.4) with δα representing the amplitude of the change in the polarizability during a full cycle of vibrational motion, andα ¯ its average value over the cycle. As in the discussion of rotational Raman scattering, if the external electric ﬁeld is due to an intense beam of monochromatic light, substitution of Eq. (4.4) and our expression for E( t) into Eq. (4.1) yields the following expression for the time-dependent induced dipole moment: 1 1 M ind(t)=E0 α¯ cos(2πν0 t)+ 2 E0 δα cos (2π[ν0 + νvib]t)+ 2 E0 δα cos (2π[ν0 − νvib]t) (4.5)

As for rotation, this implies that there will be Rayleigh scattering at the frequency of the incident light, Stokes scattering at the frequency ν0 − νvib , and anti-Stokes scattering at frequency ν0 + νvib . However, a

3 Recall that the intensity of a light beam is determined by the photon ﬂux, which in turn determines the net strength of the electric ﬁeld of the light. This is completely unrelated to the quantized energy-per-photon, which the Planck-Einstein relation tells us is proportional to the frequency of the light. 78 CHAPTER 4. RAMAN SPECTROSCOPY

Stokes Rayleigh anti-Stokes scattering scattering scattering

"virtual level"

ν 0 νs = ν0+ δE/h ν0 νs = ν0 νs=ν0 ν0 −δE/h

E ′(υ′,J ′) δE E ′′(υ′′,J ′′)

Figure 4.3: Incident ν0 and scattered νs light in Raman scattering involving upper and lower vibration- rotation levels E(v,J)andE(v,J). key diﬀerence from the case of rotational Raman scattering, is that in the present case the frequency shift is simply ±νvib , rather than ±2νrot . As a result, the Raman vibrational selection rule is based on the same very strong preference for Δv = ±1 that governs normal infrared spectroscopy. A more detailed examination of this theory is beyond the scope of this course. What we really want to take away from the above discussion is simply an understanding that the electric ﬁeld of the light can induce a small dipole moment where there was none before, and that a component of this induced dipole will oscillate with the natural motion of the molecule. In the case of rotation, the fact that this component of the dipole oscillation has twice the frequency of the physical molecular motion gives rise to the rotational selection rule ΔJ = ±2 (or zero) associated with Raman spectroscopy, while for vibration the analogous (not so rigorous) selection rule is Δv = ±1.

4.2 “Light-As-A-Particle” Description of Raman Scattering

The above description of Raman spectroscopy is based on the viewpoint of light as a wave phenomenon, characterized by oscillating electric and magnetic ﬁelds propagating through space. An alternate approach is to treat light as a stream of quantum particles, each with an energy and momentum precisely deﬁned by the associated frequency or wavelength. The scattering process is then described by the energy conservation equation

molecule{E(vbef,Jbef)} + photon{ν0} = molecule{E(vaft,Jaft)} + photon{νs = ν0 − δE/h} (4.6) where (vbef,Jbef)and(vaft,Jaft) are, of course, the vibrational and rotational quantum numbers of the molecule before (“bef”) and after (“aft”) the collision with the photon, and δE = E(vaft,Jaft)−E(vbef,Jbef) may be either positive or negative. If positive, δE is the amount of energy gained (in Stokes scattering) by the molecule, and if negative it is the amount of energy lost (in anti-Stokes) by the molecule. In other words, in anti-Stokes scattering the photon picks up energy from a molecule that was initially in an excited vibration-rotation state. Similarly, if (vbef,Jbef)=(vaft,Jaft), δE = 0 and we have Rayleigh scattering in which the scattered photons have exactly the same frequency as the incident light. This is described as elastic scattering, since although the particles (molecule and photon) bounce off one another, there is no transfer of internal energy. Inelastic processes are those in which there is a change of internal energy in one or both of a pair of colliding particles. If vaft = vbef but Jaft = Jbef , we have pure rotational Raman scattering, and if vaft = vbef we have vibrational Raman scattering. As indicated earlier, cases in which δE > 0 are “Stokes scattering” and those for which δE < 0 are “anti-Stokes”. The reason for these latter names is not intuitively obvious. Rayleigh’s name is used for the unshifted scattered light because it was Lord Rayleigh who showed (in 1871) 4.3. ROTATIONAL RAMAN SPECTRA 79 that the intensity of scattered light was proportional to 1/λ4. In contrast, a rule developed for electronic spectroscopy and called “Stokes’ law” states that the frequency of fluorescent light4 is always smaller than or equal to that of the exciting light. Scattered light with frequency less than that of the incident light (ν0) is therefore consistent with Stokes’ law, and hence is called Stokes’ scattering, while scattered light with frequency greater than ν0 contradicts it, and hence earns the name anti-Stokes scattering. This naming convention was adopted even though Stokes’ law was actually formulated for an entirely different physical process. The schematic picture of Raman spectroscopy presented in Fig. 4.3 illustrates the fundamental difference between it and conventional infrared or microwave spectroscopy. In the latter, the energy of the light quantum absorbed or emitted is the molecular energy level spacing δE, while in Raman the magnitude of δE is determined from the difference between the frequencies ν0 and νs , respectively, of the incident and scattered photons. However, both probe the same patterns of level energy spacings, and are interpreted in terms of the same quantum mechanical models for the molecule. The only real difference, other than that Raman spectroscopy tends to be more challenging experimentally, is the different rotational selection rule discussed in the next section. At the same time, it is important to remember that in Raman spectroscopy the frequency of the incident light is completely unrelated to the properties of the molecule, and that the “virtual levels” shown as dashed lines in Fig. 4.3 are fictitious, and do not correspond to any real allowed quantum level of the system.

4.3 Rotational Raman Spectra

As discussed in the two preceding sections, Raman scattering is a two-photon process. In §4.1 the first photon was the source of the electric field that induced the oscillating dipole in the molecule, while the second was the photon emitted due to the resulting oscillating charge distribution. In §4.2 the incoming photon had frequency ν0 , and the scattered photon frequency was νs . Since a photon has an angular momentum quantum number of 1, conservation of angular momentum for the two-step process means that the overall change in the rotational quantum number of the molecule is ΔJ =±2 or 0, since (see Chapter 2) each photon can cause the molecule to change its angular momentum by ±1. Of course, in pure rotational Raman scattering one can only observe transitions with ΔJ =+2, since ΔJ = 0 would mean that nothing had happened to the molecule, and since we define ΔJ =J − J =Jupper − Jlower as the difference between the quantum numbers of the upper and lower levels of the transition, and not in terms of Jbef and Jaft . However, in vibrational Raman spectra all three types of transitions are possible. As our final bit of nomenclature for this chapter we note that just as the labels “P” and “R”, respectively, are used to identify ΔJ = −1 and +1 transitions in vibrational and rotational spectroscopy, so the names “O”, “Q” and “S” are used to label ΔJ = −2 , 0 and +2 transitions in Raman spectroscopy. Table 4.1 summarizes this rotational transition labeling. From the alphabetic sequence seen there it is now evident where the seemingly arbitrary P and R labels introduced in Chapters 2 and 3 came from. The additional labels “N” and “T” for ΔJ = −3 and +3, respectively, arise in three-photon spectroscopy or in floppy molecules with internal rotational motion, topics that are beyond the scope of these notes. Note too that there is no correlation between the labeling of O, Q and S transitions and Stokes vs. anti-Stokes transitions.

Table 4.1: Labels for various types of rotational transitions.

ΔJ = J − J : −3 −2 −1 0 +1 +2 +3 Label : N O P Q R S T

If we ignore the eﬀects of centrifugal distortion, rotational Raman transitions yield Stokes scattering at energies

4Fluorescence is spontaneous light emission by molecules in an excited state that was populated by an initial absorption transition. 80 CHAPTER 4. RAMAN SPECTROSCOPY

Stokes ν˜S (J)=˜ν0 − ΔFv(J +2← J)=˜ν0 − [Fv(J +2)− Fv(J)]

=˜ν0 − [Bv(J +2)(J +3)− Bv J(J +1)]

=˜ν0 − Bv(4J + 6) (4.7) and anti-Stokes scattering at

anti−S ν˜S (J)=˜ν0 +ΔF (J +2← J)=˜ν0 +[F (J +2)− F (J)]

=˜ν0 + Bv(4J +6) , (4.8) where the subscript “S” reminds us that both cases involve ΔJ =+2 molecular transitions. Thus, we see that the spacing between adjacent Stokes or anti-Stokes lines is 4Bv, twice the line spacing encountered in pure rotational spectroscopy. Moreover, the first Stokes or anti-Stokes line lies a distance 6Bv fromν ˜0. We also see that in Raman spectroscopy there exists an analog of the “band gap” found in the rotational structure of IR spectra, in that the first Stokes and first anti-Stokes lines have a separation of 12Bv . However, this separation is difficult to observe in practice, since the much more intense Rayleigh scattering at frequency ν˜0 often obscures the low–J pure rotational Raman lines. Figure 4.4 presents a schematic picture of pure rotational and fundamental-band vibrational Raman spectra. “All else being equal”, the intensity maximum in the pure rotational anti-Stokes branch will be shifted to slightly lower J =J than in the Stokes branch, since the former correspond to molecules initially in the level Jbef = J = J + 2 and the latter to molecules initially in Jbef = J . For the same reason, the relative intensities of Stokes and anti-Stokes vibrational spectra will be dramatically different, since for a system in thermal equilibrium the population of the v =1 molecules which give rise to the latter will always be very much smaller than that for v =0 molecules. It is for this reason, as well as the very different scales of rotational and vibrational level spacings, that a picture such as Fig. 4.4 must necessarily be “schematic”.

4.4 Vibrational Raman Spectra

Vibrational Raman spectroscopy is the same as rotational Raman spectroscopy, except that the energy difference δE includes a vibrational as well as a rotational level spacing. Alternatively, it can be thought of as being the same as infrared spectra, apart from the fact that the level spacing ΔE =E(v,J) − E(v,J) is observed as a difference rather than directly and the different rotational selection rules. The latter means that we have three rotational branches in both the Stokes and anti-Stokes regions. If we ignore centrifugal distortion, the vibrational Stokes lines occur at energies Stokes νÕ (J)=˜ν0 − E(v ,J − 2) − E(v ,J) =˜ν0 −{[G(v ) − G(v )] + [Fv (J − 2) − Fv (J)]} =˜ν0 −{[G(v ) − G(v )] + [B (J − 2)(J − 1) − B J(J +1)]} (4.9) v v 2 =˜ν0 − [G(v ) − G(v )] − (Bv + Bv )(2J − 1) − (Bv − Bv )(J − J +1) Stokes ν˜Q (J)=˜ν0 − E(v ,J) − E(v ,J) =˜ν0 −{[G(v ) − G(v )] + [Fv (J) − Fv (J)]} =˜ν0 −{[G(v ) − G(v )] − [Bv − Bv ] J(J +1)} (4.10) Stokes ν˜S (J)=˜ν0 − E(v ,J +2)− E(v ,J) =˜ν0 −{[G(v ) − G(v )] + [Fv (J +2)− Fv (J)]} =˜ν0 −{[G(v ) − G(v )] + [B (J +2)(J +3) − B J(J +1)]} (4.11) v v 2 =˜ν0 − [G(v ) − G(v )] + (Bv + Bv )(2J +3) − (Bv − Bv )(J +3J +3) .

The analogous expressions for anti-Stokes transitions are identical, except that the mathematical “−” sign following the symbolν ˜0 in each of these equations is replaced by a “+” sign. Note, however, that while S(J) 4.5. RAMAN SPECTRA OF POLYATOMIC MOLECULES 81

fundamental vibrational band Rayleigh line (Stokes) ⎫ ⎬ ⎭

fundamental Q branch vibrational band (anti-Stokes) S branch O branch pure rotation pure rotation ⎫ ⎬ ⎭ ⎫ ⎬ ⎭ S branch S branch ⎫ ⎬ ⎭ (Stokes) (anti-Stokes) Q branch O branch S branch ⎫ ⎬ ⎭ ⎫ ⎬ ⎭

↑ νo ν → Figure 4.4: Schematic illustration of rotational and vibrational Raman spectra. and Q(J) transitions occur for all non-negative values of J =0,1,2,3,...,etc.,O(J) transitions are only possible for J ≥ 2. The vibrational energy spacings are of course deﬁned by exactly the same harmonic or anharmonic oscillator models discussed in Chapter 3, and the same vibrational selection rules apply: Δv = ±1is strongly allowed, Δv = ±2isweak,Δv = ±3 very weak, . . . , and so on. Overtones and hot bands can also occur, but in practice the fact that all Stokes and anti-Stokes Raman scattering is weak means that most attention is focussed on the fundamental band. Moreover, because they contain precisely the same information and are very much weaker, little practical eﬀort is directed to measuring anti-Stokes vibrational spectra.

4.5 Raman Spectra of Polyatomic Molecules

While the images of Figs. 4.1 and 4.2 and the algebraic expressions of Eqs. (4.7) – (4.11) are framed in terms of a description of diatomic molecules, all of the same arguments apply to polyatomic molecules. However, a few additional points deserve note. In particular, the shapes of their charge distributions means that all diatomic molecules can have pure rotational Raman spectra. On the other hand, for polar diatomics high resolution MW spectra are much easier to obtain, so that is the preferred method for studying such species. The same is true for polar polyatomic molecules. However, while rotational Raman spectra can be obtained for all non-polar diatomics, there is one class of non-polar polyatomic molecules for which this is not the case. In particular, spherical rotors such as CH4 or SF6 (see Fig. 2.12) have essentially isotropic charge distributions, so their polarizability along the direction of an external electric ﬁeld does not oscillate as the molecule rotates; hence they will have no pure rotational Raman spectra as well as no pure rotational absorption spectra. Fortunately, the rotational structure in their vibrational spectra still allows us to determine moments of inertia, and hence their structure and bond lengths. With respect to vibrational spectra, the discussion of §3.5 pointed out that an N–atom polyatomic molecule will in general have (3N − 5) modes of vibration if it is linear and (3N − 6) if it is non-linear. For a polar molecule all of those modes will generally be both infrared and Raman active, in which case they would normally be studied using IR methods. However, as illustrated by Figs. 3.2 and 3.6B,insomemodes non-polar molecules remain non-polar throughout the vibrational cycle, and hence they will be infrared inactive. However, if we think of replacing the O2 molecule in Fig. 4.2 by a CO2 molecule (simply stuﬀ a C atom between the two O’s), we can readily see that the IR inactive symmetric stretch mode would be Raman active. Indeed, while we will not attempt to justify it here, some further theory yields a “Rule of Mutual Exclusion”, which states that 82 CHAPTER 4. RAMAN SPECTROSCOPY

For molecules with a centre of symmetry (e.g., N2 ,CO2 ,SF6 or C2H2), • IR active vibrational modes are Raman forbidden, and • Raman active vibrational modes have no infrared spectra.

For example, for the linear molecule C2H2 (see Fig. 3.6-B), the ν1 , ν2 and ν4 modes are all Raman active and infrared forbidden, and the opposite is true for the ν3 and ν5 modes. Because Raman spectroscopy requires a very intense incident light beam with a near monochromatic frequency distribution, this ﬁeld of study underwent a massive revival with the development of lasers in the 1960’s. One particularly important application is identifying and studying the properties of large molecules in solution. This is greatly facilitated by the fact that although H2O absorbs very strongly in the IR, it is a very weak Raman scatterer. This makes Raman spectroscopy the preferred technique for studying molecules in aqueous solution, as solvent signals will not provide much interference. It has therefore become particularly important for studying biological molecules and for environmental analysis.

4.6 Problems

1. Determine the generalized versions of Eqs. (4.7) and (4.8) for cases in which the leading centrifugal distortion constant Dv cannot be ignored.

2. When excited by a laser of wavelength 180.000 nm, the P2 molecule is observed to have a vibrational Raman spectrum with Q-branch Stokes bands centred at wavelengths 182.538 and 185.123 nm, respectively. Determine the vibrational frequency (in Hz), the harmonic and anharmonic vibrational constants in cm−1, and estimate the bond dissociation energy for this molecule (in cm−1).

3. Draw and label the level energies, the vibration-rotation transitions, and the resulting anti-Stokes Raman spectrum of the fundamental band of a diatomic molecule, for all lines up to J =3.

4. Determine the generalized versions of Eqs. (4.9) – (4.11) for cases in which the leading centrifugal distortion constant Dv cannot be ignored. Chapter 5

Electronic Spectroscopy

What Is It? Electronic spectroscopy uses photons with energies ranging from the visible (eV’s) to ultraviolet (tens of eV) to excite molecules into excited electronic states.

HowDoWeDoIt?Electronic spectroscopy is analogous to absorption or emission spectroscopy in the infrared, except that more energetic photons are involved. In emission, the light is dispersed to allow us to measure the line frequencies, while in absorption the incident light frequency is varied, and the frequencies where intensity loss occurs are measured.

WhyDoWeDoIt?The ﬁne structure in electronic spectroscopy gives us the vibrational and (usually also) rotational level spacings for both the ground-state and electronically excited molecule, which in turn allow us to determine their bond lengths, dissociation energies, potential energy functions, and other properties. It is particularly useful because it gives information about high vibrational levels that cannot be accessed by infrared spectroscopy, and allows us to observe the vibrational and rotational levels of homonuclear molecules such as H2 and N2 for which normal infrared transitions are forbidden.

5.1 Why Does Light Cause Electronic Transitions?

Since electrons are charged particles, in a classical description their orbiting motion within an atom or molecule would give rise to an oscillating electric field that would give the oscillating electric field of an incident light beam something to “push” against. As in our classical wave descriptions of rotational and vibrational excitation, transitions may occur when the frequency of motion and the frequency of the incident light are the same. However, in contrast to the rotational and vibrational cases, an oscillating electronic dipole will always be present within a molecule, no matter what its size, shape or symmetry, so every atom and molecule will have allowed electronic transitions. In spite of this universal existence of electronic transitions, only a fraction of the infinite number of possible electronic transitions1 are actually allowed. This restriction reflects the fact that the integer angular momentum of the photon only allows transitions in which the overall electronic orbital angular momentum changes by one unit. This is the reason that 2s → 3p optical transitions of a Li atom are allowed, but 2s → 3d or 2s → 3s transitions are forbidden. Electronic selection rules for molecules are much more complicated than this, and are intimately concerned with details of the orbital symmetries and electronic spin degeneracies of the initial and final states. However, those details are beyond the scope of these notes, and we shall concern ourselves here only with allowed electronic transitions.

1 Recall that for the hydrogen atom, the principal quantum number n enumerating the discrete energy levels ranges up to inﬁnity. The same inﬁnity of electronic states occurs for molecules.

83 84 CHAPTER 5. ELECTRONIC SPECTROSCOPY

Figure 5.1: Schematic illustration of rotational and vibrational structure in electronic spectroscopy.

5.2 Vibrational-Rotational Structure in Electronic Spectra

Figure 5.1 provides a schematic overview of rotational, vibrational and electronic transitions in a diatomic molecule. As indicated by the discussion of Fig. 1.13 in §1.4.3, each upper and lower electronic state has its own potential energy curve whose distinct shape and radial position gives rise to a unique set of vibrational and rotational constants, and hence also to different patterns of vibrational and rotational energy-level spacings. Pure rotational transitions occur between adjacent rotational sublevels within a ‘stack’ associated with a single vibrational level, as seen on the right hand side of the figure. Vibration-rotation (or infrared) transitions occur between rotational sublevels of different vibrational levels of a single potential energy curve, as illustrated by the R(7) line of the first overtone (2, 0) band shown in this figure. Electronic transitions are then transitions between vibration-rotation sublevels in different electronic states, as illustrated by the P (10) line of the (3, 1) band of the electronic transition shown in this figure. Note that as in infrared spectra, the set of all rotational transitions associated with a given upper (v)andlower(v) level is called a “band” and is labelled by the two vibrational quantum numbers, with the label for the level at higher energy being written first, as in (v,v)or v −v. The fact that vibrational level spacings are much larger than rotational level spacings means that vibrational bands associated with a single lower-state v value and a series of different upper-state v values, or with a common v value and a series of different v values, will have distinctly different transition energies. This is illustrated by Fig. 5.2, which shows a number of vibrational bands in a low resolution spectrum of the A 1Σ+ − X 1Σ+ electronic transition of SrS. The separation between the positions of adjacent bands with a common v value and different v values will be roughly equal to the fundamental vibrational constant ωe for the upper state, while adjacent bands with a common v value will be separated by the analogous lower-state constant, ωe . In particular, in Fig. 5.2 the separations between the peaks associated with the (v,v)=( 0 , 0) and (1, 0) bands, the (1, 0) and (2, 0) bands, and between the (1, 1) and (2, 1) bands, are all roughly equal to ωe. This spectrum also shows that bands in a particular “vibrational sequence”,asetof bands with the same Δv = v − v value, tend to lie relatively close together, since from one band to the next the vibrational energies in the upper and lower electronic state both increase (or both decrease) by one level spacing. This is the reason that the 1–0, 2–1, 3–2 and 4–3 bands in the middle of Fig. 5.2 lie relatively 5.2. VIBRATIONAL-ROTATIONAL STRUCTURE IN ELECTRONIC SPECTRA 85

Figure 5.2: Vibrational bands in the electronic spectrum of SrS. close together. Within a given vibrational sequence, the spacing between adjacent vibrational bands will roughly equal |ωe − ωe |. One question raised by the spectrum in Fig. 5.2 is why each band has a sharp intensity maximum at its high energy (short wavelength) edge, rather than the roughly symmetric double-hump shape seen in the vibration-rotation spectra of Figs. 3.7 or the left hand side of Fig. 3.10, or the broad smoothly changing pattern seen in Fig. 2.5. This is readily understood if we examine the expressions for the energies of the individual transitions observed in electronic spectra. Transition energies in electronic spectroscopy are described in essentially the same manner as were transition energies in vibrational spectroscopy, except that (i) the energies of the upper and lower states each also contains an additive electronic energy contribution, Te, the energy at the minimum of the potential energy curve for that state (see Fig. 5.1), and (ii) the vibration-rotation energies of the upper and lower levels are governed by diﬀerent sets of vibration-rotation constants. As a result, the energies of individual transitions are given by the general expression

− − ν˜ = E (v ,J ) E (v ,J )=[Te + G (v )+Fv (J )] [Te + G (v )+Fv (J )] { − } − = [Te + G (v )] [Te + G (v )] +[Fv (J ) Fv (J )] − =˜ν0(v ,v )+[Fv (J ) Fv (J )] , (5.1) in which the energy diﬀerence deﬁning the band origin

− − ν˜0(v ,v )=[Te + G (v )] [Te + G (v )] = ΔTe + G (v ) G (v ) (5.2) − now contains the difference in electronic energies of the two states, ΔTe=Te Te . Note that in contrast to Eqs. (3.33) and (3.34) in vibrational spectroscopy and Eqs. (4.9) - (4.11) in Raman spectroscopy, prime ()or double prime () labels are now associated not only with the vibrational and rotational quantum numbers of the upper and lower level of each transition, but also with the energies (e.g., Fv (J ) vs. Fv (J )) and with (see below) molecular constants such as Bv or ωe, since both the molecular constants and the level energies patterns that they define differ from one electronic state to another. As in vibrational and rotational spectroscopy, conservation of total angular momentum in an absorption or emission process means that the usual rotational selection rule ΔJ=± 1 also applies to transitions in electronic spectra. However, if the molecule has non-zero electronic angular momentum in one or both states, the electronic degrees of freedom may take up the photon’s angular momentum, in which case the rotational selection rule must be extended to allow also ΔJ=0 transitions. This means that in addition to having P and R branches, each vibrational band of such electronic transitions will also have a Q branch consisting of transitions for which ΔJ=0 (see Table 4.1). Use of the same manipulations applied in §3.6 then yields the 86 CHAPTER 5. ELECTRONIC SPECTROSCOPY following expressions for the allowed transition energies:

ν˜P (J )= E (v ,J ) − E (v ,J )=E(v ,J − 1) − E (v ,J ) − − − 2 =˜ν0(v ,v ) [Bv + Bv ](J ) [Bv Bv ](J ) (5.3) ν˜R(J )= E (v ,J ) − E (v ,J )=E(v ,J +1)− E (v ,J ) − − 2 =˜ν0(v ,v )+ [Bv + Bv ](J +1) [Bv Bv ](J +1) (5.4) ν˜Q(J )= E (v ,J ) − E (v ,J )=E(v ,J ) − E (v ,J ) − − =˜ν0(v ,v ) [Bv Bv ] J (J +1) , (5.5) in which the final expression for each case is obtained by ignoring centrifugal distortion terms. These expressions for P –andR–branch transitions are precisely equivalent to Eqs. (3.33) and (3.34) for vibration-rotation spectra, except that the electronic energy difference ΔTe is now included in the definition ofν ˜0(v ,v ), and that the values of Bv for the two electronic states are based on different sets of molecular constants (i.e., different sets of {Yl,1} values in Eq. (3.37)). The latter fact is of critical importance for understanding the intensity patterns within vibrational bands in electronic spectra. The reason for the difference between the profiles of the vibrational bands seen in Fig. 5.2 and those seen in Fig. 3.7 or 3.10 has, in fact, already been discussed in §3.6. It was noted there that for vibrational transitions within a single potential energy well, Bv is always greater than Bv , so the quadratic-in-J terms in Eqs. (3.33) and (3.34) are both negative. This causes the P –branch lines to become progressively farther apart with increasing J, while the R–branch lines become progressively closer together. If the rotational series can be followed to sufficiently high J,theR branch will eventually turn around and begin marching off “to the red” (i.e., to lower frequencies). This behaviour can be seen in the high-temperature infrared spectrum of NaCl shown in Fig. 3.9 where the fact that the high-JR-branch lines get ever closer together and pile up on one another as the branch turns around is the reason for the sharp intensity maximum at the turnaround point, since multiple transitions occur at essentially the same frequency. This sharp intensity peak due to the turnaround of the progression of rotational lines in a vibrational band is called a “band head”. We note, however, that the infrared spectrum shown in Fig. 3.9 was obtained at the relatively high temperature of 1000◦ C. In contrast, in infrared spectra taken at more modest temperatures the intensities of the R–branch lines usually drop off and become negligible before such a turnaround point is reached. This was the case for the DCl spectrum shown in Fig. 3.7. This occurs because most infrared spectra involve transitions between adjacent vibrational levels whose Bv values are very similar, differing only by ∼ αe (see Eq. (3.35)). As a result, the coefficients of the quadratic-in-J terms in Eqs. (3.33) and (3.34) are relatively small, and the R–branch turnaround only occurs at quite high J values. For vibrational bands in electronic spectra the same considerations apply, but we end up with quite a different result. In this case the Bv values in the upper and lower electronic states are normally quite different, which means that the coefficients of the quadratic-in-J terms in Eqs. (5.3) – (5.5) may be relatively large. As a result, the band-head branch turnaround usually occurs at relatively small J.Moreover,in electronic spectra the inertial constant for the lower level will sometimes be smaller than that for the upper level ( Bv Bv are called “red shaded”. The bands shown in Fig. 5.2 are clearly red-shaded (note that since wavelength is increasing to the right in this figure, the wavenumber or transition energy is increasing to the left). In contrast with the results shown in Fig. 5.2, Fig. 5.3 shows the detailed structure of a band head for a case in which the lines are all fully resolved up to, at, and beyond the band head. However, it is not always possible to obtain fully resolved spectra such as this, so in order to determine the pure vibrational contribution to the level energies, we need to be able to estimate the displacement between the observed band head and the band origin. To do this, we need to determine the value of J associated with the point 5.2. VIBRATIONAL-ROTATIONAL STRUCTURE IN ELECTRONIC SPECTRA 87

Figure 5.3: Rotational structure near the (0,0) band head in the A 1Σ+ − X 1Σ+ spectrum of CuD. where the R–orP –branch turns around, and hence the location of that turnaround point relative to the band origin. This requires the use of a little calculus. For a red-shaded band, for which it is the R branch that turns around, the turnaround point occurs when the R–line transition frequencies stop increasing with J. Thus, taking the ﬁrst derivative with respect to J of the transition energy of Eq. (5.4) and setting it equal to zero

d ν˜R(J) =[B + B ] − 2[B − B ](J + 1) = 0 (5.6) dJ v v v v R yields an expression for Jh , the value of J at the band head, the point where the R–branch turns around: & R − (Jh +1) = [Bv + Bv ] 2[Bv Bv ] . (5.7)

Substituting this result into Eq. (5.4) then yields an expression for the position of the band head: 2 R Bv + Bv − − Bv + Bv ν˜R(Jh )=˜ν0(v ,v )+ [Bv + Bv ] − [Bv Bv ] − 2(Bv Bv ) 2(Bv Bv ) 2 (Bv + Bv ) =˜ν0(v ,v )+ − . (5.8) 4(Bv Bv ) Alternately, for cases in which Bv

Eqs. (5.7) and (5.8) for those two cases. In infrared spectra, Bv and Bv are usually associated with adjacent vibrational levels, so their values are very similar, and the denominators are therefore small. As a result, the band heads are displaced to relatively high J where the level population is suﬃciently low that the transition intensities are no longer observable. This was the case for the DCl spectrum seen in Fig. 3.7. However, Fig. 3.9 shows that these arguments do not apply to infrared spectra obtained at very high temperatures. In particu- 35 −1 −1 lar, for the 1–0 infrared band of Na Cl shown there, Bv=0 =0.217 251 cm and Bv=1 =0.215 637 cm , R −1 so Eqs. (5.7) and (5.8) predict that Jh = 133.10 and thatν ˜R(Jh) − ν˜0(v ,v )=29.03 cm . Inspection of Fig. 3.9 shows that these values are slightly larger than what is found experimentally; this discrepancy merely reﬂects the neglect of centrifugal distortion terms in our derivations (see Problem #2attheendof this chapter). Having the band heads lie at relatively large values of J and the band-head displacements [˜νR(Jh) − ν˜0(v ,v )] be relatively large are characteristic properties of infrared vibrational-rotational spectra. However, whether or not those band heads can be observed in a particular case will depend on the temperature of the system being studied.

In contrast with the infrared case, in electronic spectra the upper- and lower-level Bv values are usually quite diﬀerent, so that the denominators of the fractions in Eqs. (5.7) – (5.10) are relatively large. As a result, those fractions will be relatively small and the band heads will lie close to the band origins. For example, for the 0 – 0 band of the A 1Σ+ −X 1Σ+ spectrum of 63CuD shown in Fig. 5.3, the published rotational constants A −1 X −1 are B0 =3.475 22 cm and B0 =3.992 52 cm . Utilizing these values in Eq. (5.8) yields the prediction R −1 that Jh =6.2and˜νR(Jh) − ν˜0(v ,v )=26.95 cm , results that are in good agreement with what we see in Fig. 5.3. Similarly, higher resolution spectra for the 1– 0 band of the A 1Σ+ − X 1Σ+ electronic transition of SrS −1 −1 seen in Fig. 5.2 allow us to determine that Bv=0 =0.120 565 cm and Bv=0 =0.113 422 cm .Wemay R −1 then predict that the band head will occur at J = Jh =15.38 and that [˜νR(Jh) − ν˜0(v ,v )] = 1.916 cm . Thus, the (1, 0) vibrational band head lies at a relatively modest J value, and the band head lies very close to the band origin. This case is typical of many electronic spectra.2 Consequently, even when one cannot fully resolve rotational structure and determine the precise positions of the band origins (as in Fig. 5.2), the positions and relative spacings of the band heads still yield quite good estimates of the vibrational level spacings in the upper and lower electronic states.

5.3 Vibrational Propensity Rules in Electronic Spectra: The Classical Franck-Condon Principle

In Chapters 3 and 4 we saw that vibrational transition intensities in infrared and Raman spectra were governed by the following selection rules: (1) Δv = ±1 transitions are strongly allowed, and (2) |Δv| > 1 transitions are much weaker, and their intensity decreases very rapidly as |Δv| increases. Those rules arose from the special mathematical property of the Schr¨odinger equation that requires wavefunctions for diﬀerent vibrational levels of a given potential energy curve to be precisely “orthogonal” to one another. As a full discussion of this property is beyond the scope of the present work, we simply state here that these restrictions do not apply to vibrational wavefunctions associated with potential energy curves that do not have exactly the same radial position and shape. As a result, all else being equal, all v ↔ v transitions are allowed. However restrictions that are not related to the numerical values of v and v values are imposed by the Franck-Condon principle which states that:

“Nuclear positions and momenta do not change during an optical transition.”

Some justiﬁcation for this statement is provided by consideration of the magnitudes of the quantities involved. The time associated with an optical absorption or emission process is ∼ 10−15 s, while the period of vibrational motion is roughly 10−13 s and rotational motion is 1–2 orders of magnitude slower than that. Thus, during the absorption or emission process the nuclei have no time to move, so the transition occurs

2 B ≈ B There are, of course, bands in electronic spectra for which by accident v v . When this occurs, the pattern of line intensities in the vibrational band will be similar to those seen in ordinary infrared vibrational spectroscopy (e.g., see Fig. 3.7). However, the sharp band-head structure seen in Figs. 5.2 and 5.3 is much more common. 5.3. VIBRATIONAL PROPENSITY RULES IN ELECTRONIC TRANSITIONS 89

Figure 5.4: Deﬁnition of the “stationary point” for a particular (v,v) electronic transition.

‘vertically’, at a fixed radial distance on a potential energy diagram. Similarly, the Compton relationship of Eq. (1.9) shows that photons of visible light have momenta of order 10−27 kg·m/s, while the average radial momentum for a vibrating diatomic molecule will be of order 10−21 kg·m/s. Thus absorption or emission of a photon cannot change the radial momentum of a vibrating molecule significantly. This means that a given (v,v) transition can only occur when the system has an instantaneous radial configuration for which the radial momentum, and hence also the radial kinetic energy [Ev − V (r)], is the same immediately before and after the transition. As is illustrated by Fig. 5.4, for a given (v,v) transition there will usually exist only one radial distance, called a “stationary point”, for which this condition is satisfied. In classical mechanics, the total energy of a system is the sum of its potential energy plus its kinetic energy. What we call “classical turning points” are the radial distances at which the total energy equals the potential energy, Ev = V (r) , and hence the radial kinetic energy is precisely zero. For example, Fig. 5.5 shows the potential curves and the energies of a number of upper- and lower-state vibrational levels 3 1 + + − 2 involved in the B( Π0u ) X( Σg ) electronic band system of Br . The inner and outer turning points for each of the vibrational levels shown there are the distances at which the horizontal line representing the vibrational energy meets the potential energy curve. Because negative kinetic energies are not allowed in classical mechanics, the system cannot escape from the “classically allowed” region between its inner and outer turning points. Thus, since the Franck-Condon principle tells us that the internuclear distance does not change during a transition, unless the classically allowed region for the upper vibrational level overlaps that for the lower one, transitions are forbidden. In terms of Fig. 5.5, this means, for example, that ground-state vibrational levels with v ≤ 2 are not allowed to have transitions into B–state levels with v 13 . These considerations are the basis for our first selection rule for vibrational transitions in electronic spectroscopy. Selection Rule 1: Transitions cannot occur unless the radial intervals between the inner and outer classical turning points of the two levels have a significant degree of overlap. A quantum mechanical statement of this rule would be: the radial wavefunctions in the upper and lower levels must have significant spatial overlap. The above arguments determine the region of internuclear distance the molecule must find itself in for agivenaparticular(v,v) electronic transition to occur (at/near a stationary point). However, it says nothing about the relative probability or intensity of such a transition. Recall that molecular vibration can be described mathematically as the motion of a particle of mass μ along the one-dimensional radial coordinate r. Within a classical picture, at any instant the radial speed vr of a vibrating molecule is related to the radial kinetic energy by the expression

rad 1 2 KE = μ (vr) =[E − V (r)] . (5.11) v 2 v 90 CHAPTER 5. ELECTRONIC SPECTROSCOPY

25000

3 B + Π 0u 20000 30 15 20 10 5 v’= 0 15000 60 energy 52 -1 /cm 44

10000 36 X 1Σ+ 28 g

20 5000 16 12 8 Br2 4 0 v"= 0 2.0 2.5 3.0 3.5 4.0 4.5 r /Å 3 1 + 2 + − Figure 5.5: Potential curves and turning points for the Br B( Π0u ) X( Σg )system

As a result, the time that the vibrating molecule spends with its internuclear distance in the tiny interval between r and r + dr is 1 μ δtr ≡ fv(r) dr = dr = dr , (5.12) vr 2[Ev − V (r)] in which fv(r) defines the probability of finding the system at a particular value of r. For a representative vibrational level with energy Ev in a potential energy curve V (r), Fig. 5.6 shows the characteristic behaviour of the distribution function fv(r). As with Figs. 1.6, 1.7 and 3.3, this figure is a superposition of two separate types of plot: the first is an energy vs. distance plot showing the vibrational level energy Ev and the variation of the potential energy with distance, while the second (here) is a plot of fv(r) vs. r, with the zero of its vertical axis placed at the energy Ev of the vibrational level in question. It is clear that fv(r) →∞ at the inner and outer turning point of every level. Elementary calculus tells us that these are “integrable singularities”, in that the area under the curve is finite. Nonetheless, it is clear that (within this classical picture) the molecule spends most of its time with its bond length close to one of those turning points. As a result, although the stationary point for a given (v,v) transition can in principle occur at any radial distance, the transition will be most intense if it lies near one of the classical turning points where the vibrating molecule spends most of its time. This leads to our second selection rule for vibrational transitions in electronic spectroscopy.

Selection Rule 2: Vibrational transitions in electronic spectra will be most intense when the upper and lower vibrational levels have turning points that are nearly coincident.

In practice, quantum mechanics blurs the above rules. In particular, the assumption that the probability distribution deﬁned by our function fv(r) is only qualitatively correct, especially for small v values. Moreover, the wavefunctions seen in Figs. 1.6, 1.7 and 3.3, show that in its lowest ( v = 0 ) level, the most probable place to ﬁnd a molecule is near its equilibrium distance, half-way between the two turning points. As may be seen in Fig. 5.5, the fact that the v = 0 level lies very near the potential minimum means that those turning 5.3. VIBRATIONAL PROPENSITY RULES IN ELECTRONIC TRANSITIONS 91

∞ ∞

fυ(r)

↑ energy

Eυ

inner outer turning KE turning point υ point V(r)

r/Å→ Figure 5.6: Classical prediction for the amount of time fv(r) δr that a vibrating molecule spends within incremental distance δr of a particular radius r. points will lie relatively close to their midpoint. As a result, when applying Selection Rule 2 to transitions involving a v = 0 vibrational level, we usually think of its turning points as eﬀectively lying at the potential minimum, re. 3 1 + + − 2 Application of the above considerations to the B( Π0u ) X( Σg )bandsystemofBr illustrated by Fig. 5.5 leads to the following illustrative predictions. • B–state level v(B) = 0 will emit most strongly into ground state levels near v(X) = 17 , and will have negligible intensity for transitions into levels with v(X) 12 .

• Ground-state level v(X) = 4 will absorb most strongly into B–state levels near v(B)=8.

• B–state level v(B) = 5 will have very strong emission both into ground-state level v(X) = 6 , due to emission from its inner turning point, and into high vibrational levels near v(X) = 38 , due to emission associated with distances near its outer turning point.

Finally, Fig. 5.7 shows a diagram modeled on figures from the famous diatomic spectroscopy monograph by (Canadian!) Nobel Prize winner Gerhard Herzberg, that illustrates the implications of the Franck-Condon principle for the case of absorption from the v = 0 level on a lower potential energy curve. The four upper- state potentials considered there are identical to one another, as are the four lower-state potentials, but the former are shallower and broader than the latter. The only difference between these cases is the radial displacement of the upper-state potential relative to the lower-state one. Parts A–D of the upper half of Fig. 5.7 show the quantum-mechanically calculated vibrational intensity patterns for these four cases. In particular, in Case B the two potential minima lie at exactly the same internuclear distance, and the absorption intensity is by-far the greatest (by a factor of ∼ 100) for Δv =0, drops very sharply for Δv = 1 , recovers slightly for Δv = 2 , and then dies off monotonically with further increases in Δv. While the trend for v = 0, 2, 3 and 4 is qualitatively what the Franck-Condon arguments presented above would lead us to expect, the anomalous weakness of the (1, 0) transition reflects the fact that quantum mechanical wavefunction orthogonality considerations cannot be totally discounted when the two potential minima are aligned. In contrast with Case B, for the three other cases considered in Fig. 5.7 the maximum in the quantum 2 mechanical probability density |ψv=0(r)| for the lower-state v = 0 level lies below the inner or outer wall of the upper-state potential, so the maximum vibrational transition intensity is associated with higher v values. When the lower-state potential minimum lies below the inner wall of the upper-state potential, as in Cases C and D, transitions into a wide range of upper-state levels are observed. However, the anharmonicity of the 92 CHAPTER 5. ELECTRONIC SPECTROSCOPY

(2,0) (1,0) A (3,0) (0,0) (5,0)

(0,0) I/20 B (2,0) (1,0) (3,0) (4,0)

(2,0) (3,0) (1,0) (4,0) (0,0) (5,0) (6,0) (8,0) C (10,0) (12,0)

(4,0) (5,0) (6,0) (8,0) (3,0) (10,0) (2,0) D (1,0) (12,0) (0,0)

40000 45000 1 50000 55000 ν /cm−

A B C D

Figure 5.7: Dependence of vibrational band intensities on the relative radial positions of upper- and lower- state potential energy functions. upper-state potential energy function means that the outer turning points for different vibrational levels are located relatively much farther apart than are the corresponding inner turning points. As a result, when the lower-state minimum lies below the outer wall of the upper-state potential, transitions are observable only for a relatively sparse set of upper-state levels. Note that the sharp intensity drop-off near dissociation for Case D reflects the fact that the extreme potential function anharmonicity for levels lying near dissociation means that the area under the fv(r) integrand near the inner turning point is very much smaller than that associated with the outer turning point. A quantum mechanical illustration of this point is provided by Panel D of Fig. 1.7 (on p. 16), where we see that for the highest vibrational level shown, the amplitude of 5.4. PROBLEMS 93 the outermost loop of the wavefunction is much greater than that of the innermost loop. In the following chapter we will see that the types of patterns shown in Fig. 5.7 play an important role in interpreting the vibrational structure observed in photoelectron spectroscopy.

5.4 Problems

1. The positions of the band heads seen in Fig. 5.2 are summarized by the following (Deslandres) table:

v(X 1Σ+)=0 1 2 3 v(A 1Σ+) = 0 13910 13525 1 14244 13863 2 14594 14200 3 14541 14152 4 14487 14103 5 14446

Use these results to determine estimates of ωe and ωe for this band system.

2. (a) Determine the extended versions of Eqs. (5.3) – (5.5) which take account of the leading centrifugal distortion constant Dv and Dv . (b) For a case in which Bv >Bv , use your results from part (a) to determine an expression for the R band head quantum number Jh which takes account of Dv and Dv contributions. (c) The leading centrifugal distortion constant for the two lowest levels of ground-state NaCl are −7 −7 −1 D0 =3.11524 × 10 and D1 =3.10850 × 10 cm , while the associated values of Bv and Bv R have been given on p. 88. Using your solution to part (b), determine an improved estimate of Jh for this band, and compare your result to that obtained on p. 88.

3. The leading molecular constants for the X 1Σ+ and A 1Σ+ states of CuH are listed below. Use these constants to answer the following questions. (a) Determine the band origins for the (0,3), (3,0) and (0,0) bands in the A − X electronic spectrum, and show whether these bands will be “red-shaded” or “blue-shaded”.

(b) For the three bands of part (a), what is the value of the rotational quantum number Jh associated with the band head, and what is the displacement of the band head from the band origin? (c) What are the positions of the R(5) and P (9) lines of the (3, 0) band (in units cm−1)?

Molecular constants for the X 1Σ+ and A 1Σ+ electronic states of CuH, all in units cm−1.

state Te ωe ωexe Be αe X 1Σ+ 0.0 1941.610 37.887 7.9448 0.2557 A 1Σ+ 23 407.976 1717.543 53.06.9294 0.2576 94 CHAPTER 5. ELECTRONIC SPECTROSCOPY Chapter 6

Photoelectron Spectroscopy

What Is It? Photoelectron spectroscopy (PES) uses photons of high energy, from the ultraviolet (tens of eV) to X-ray (thousands of eV), to dislodge electrons from molecules. It is precisely analogous to normal electronic spectroscopy, except that the upper state of a transition is one of the electronic states of the molecular ion formed upon removing an electron.

HowDoWeDoIt? In ‘conventional’ photoelectron spectroscopy, the quantity that is measured is the kinetic energy of the ejected electrons, rather than the frequency of the absorbed or emitted light. The diﬀerence between the energy of the incident photon and the kinetic energy of the ejected photoelectron gives the energy of the resulting molecular ion. Recent years have seen increasing use of a related technique called ‘threshold photoelectron spectroscopy’, in which the frequency of the incident light is tuned to determine the minimum energy required for an electron to be released. However, the present discussion considers only conventional PES.

WhyDoWeDoIt? Molecular ions are important reactive species in the atmosphere and many other environments, but they are diﬃcult to produce and study by conventional techniques. Photoelectron spectroscopy allows us to test the predictions of molecular orbital theory, and the observed ﬁne structure tells us about the vibrational level spacings in the various electronic states of the molecular ion. It is also an important tool for identifying particular atoms and molecules in solids and on surfaces.

6.1 Photoelectron Spectroscopy: The Photoelectric Eﬀect Revisited

Photoelectron spectroscopy differs qualitatively from the types of spectroscopy discussed up to this point, since rather than directly observing the emission or absorption of light, it measures the appearance and kinetic energy of ejected electrons. We saw in §1.1.3 that when light is focused on a metal surface, electrons are emitted if the energies of the light quanta are larger than the work function of that metal, W0,the minimum energy required to remove an electron from the surface. Any extra energy contained in the incident light appeared as kinetic energy of the emitted electrons, which are called photoelectrons. When the same experiment is performed on individual molecules rather than on bulk metal, it is called Photoelectron Spectroscopy or PES. The methodology for PES is analogous to that used for studying the photoelectric effect. If radiation of sufficient energy is absorbed by a molecule, it can dislodge an electron to yield a molecular ion plus a free photoelectron. The kinetic energy of the emitted electron is the key quantity that provides information about the molecular ion, so conventional PES experiments must be performed using a light source that emits photons of a single known frequency, which we label ν0. An electrostatic detector, which measures how much the trajectory of a moving charge is deflected on passing through a magnetic field, is used to determine the 1 2 velocity ve , and hence also the kinetic energy KEe−=2 me(ve) of the photoelectrons. The difference between the energy of the incident photons and the measured kinetic energy of the ejected electron is the ionization energy (IE) of the molecule, the minimum energy required to remove an electron:

1 2 IE = hν0 − KEe− = hν0 − 2 me(ve) . (6.1)

95 96 CHAPTER 6. PHOTOELECTRON SPECTROSCOPY

ionization threshold 1 m v2 2− e e

2sA 2sB hν0 Hatom Coulomb potential σu* 1s A 1sB

σg

HatomA H2 HatomB

Figure 6.1: Molecular orbital level-energy picture of the photoionization of H2.

This quantity is the molecular analog of the work function W0 that was discussed in § 1.1.3. It defines the energy of the molecular ion produced by this process, and is a measure of the binding energy of the electron in the molecular orbital in which it originally resided. The fact that those binding energies are quantized and can be semi-quantitatively explained in terms of simple molecular orbital theory arguments attests to the utility of that theory. Very high-energy photons are required to drive this type of process. One of the light sources most widely used for this purpose is the He(I) lamp, whose dominant emission is due to the 1s12p1 → 1s2 transition of atomic helium. This transition occurs at a wavelength of 58.43 nm, and yields photons with an energy of 21.2182 eV. Note that since the observable in this kind of spectroscopy is the kinetic energy of the ejected electrons, electron volts (or eV) are the most convenient units of energy to work with, where 1 eV = 8065.544 65 cm−1 =1.602 176 487×10−19 joules is the change in the kinetic energy of an electron when it passes through an electric potential of 1 Volt. Since the He(I) emission line falls in the ultraviolet region of the electromagnetic spectrum, we refer to PES experiments using this type of source as Ultraviolet Photoelectron Spectroscopy (UPS). Shining light from a He(I) source onto a sample of H2 molecules yields + H2 molecular ions plus photoelectrons with kinetic energies around 5.794 eV. The 15.426 eV difference between the photon energy and this kinetic energy is the ionization energy of H2. Photoelectron spectroscopy provides evidence that supports the elementary ‘molecular orbital’ (MO) theory which is discussed in most Introductory Chemistry courses. Consider, for example, an H2 molecule, whose level energies are are schematically illustrated by Fig. 6.1. The left- and right-hand portions of this figure show the lowest energy levels and the attractive Coulomb potential that binds the electrons in the two component H atoms. As usual, each electron is represented by an arrow that points up or down to 1 1 indicate whether its spin quantum number is ms=+ 2 or − 2 . The middle segment of the figure then shows ∗ how the energies of the associated bonding σg and anti-bonding σu molecular orbitals that are formed when the atoms come together to form a molecule, split apart to lie below and above the energies of the parent isolated-atom orbitals. In the ground state of the neutral H2 molecule, both electrons occupy the σg bonding orbital, and the lowering of the total system energy associated with this bonding is the source of the binding energy of the H2 molecule. It is important to avoid confusing ionization energies with dissociation energies. Ionization corresponds to + − the removal of an electron from a molecule to create a molecular ion, a process such as H2 +hν0 → H2 +e , which is illustrated in the middle panel of Fig. 6.1. In contrast, dissociation is the process of breaking a bond to yield two separate atomic or molecular species, a process such as H2 → H + H . Since the ground-state H2 molecule has two electrons in the σg bonding orbital, in the context of Fig. 6.1, the dissociation energy of 6.2. KOOPMANS’ THEOREM 97

the neutral molecule D0(H2) is twice the difference between the energy of the σg molecular orbital and the energy of the parent 1s atomic orbitals (dotted lines in Fig. 6.1). When one of these electrons is removed, + we expect the binding energy of the resulting H2 molecular ion to be roughly half that of its neutral parent, since it has only one bonding electron. The hydrogen molecule is a particularly simple example, as there is only one type of orbital from which an electron can be removed. The photoelectron spectra of heavier molecules are more complex, as they have more than one different type of occupied orbital. For example, when N2 molecules are subjected to light from a He(I) lamp, photoelectrons with kinetic energies of approximately 5.64, 4.52 and 2.47 eV are emitted, which implies that this molecule had three different ionization energies: 15.58, 16.70 and 18.75 eV. As is discussed later, these three ionization processes are the result of removing an electron from the highest ∗ occupied molecular orbital, σ2p , or from one of the lower-energy (more strongly bound) π2p or σ2s molecular + orbitals. In each case, only one electron is removed, and an N2 ion is produced. However, since the three + cases involve the removal of an electron from a different molecular orbital, each case produces an N2 ionina different electronic state. Fortunately, the three different ionization processes can all be observed in a single laboratory experiment, because there is a large number of N2 molecules in the gas sample, and a fraction of the ionizations may involve excitation from each orbital. There is no electronic selection rule for photoelectron spectroscopy; all possible ionizations of a molecule + − hν0 + M → M + e are allowed.

6.2 Koopmans’ Theorem

As was mentioned above, the measured ionization energies tell us about the energies of the molecular orbitals in which the photoelectrons originally resided. This is the basis of what is known as Koopmans’ Theorem, which states: For a closed-shell molecule, the ionization energy of an electron in a particular orbital is approximately equal to the orbital binding energy. In other words, IE = εorbital , (6.2) where εorbital is the binding energy of the electron in the initial molecular state. This statement assumes that the orbital energies of the other electrons in the molecule are not affected when the photoelectron is removed. In other words, it assumes that the orbital energies are exactly the same in the product molecular ion and the neutral parent molecule. While a good first approximation, this is not precisely true, since removal of the photoelectron leaves the system with less electron–electron repulsion energy and with less shielding of the nuclear charge. Corrections due to these considerations are especially important if the electron being removed does not come from the highest occupied molecular orbital. As a result, the observed ionization energies may be somewhat smaller than the actual orbital binding energy in the parent molecule, with the difference being due to the slight lowering of the orbital energies which occurs on forming the ion.

6.3 Vibrational Fine Structure in Photoelectron Spectra

A deeper understanding of photoelectron spectroscopy is obtained on comparing the properties of the initial neutral molecule with those of the molecular ion formed by the PES process. The right hand side of Fig. 6.2 + illustrates the photoionization of a ground state H2(v = 0) molecule to produce an H2 molecular ion in + vibrational level v : + + − 1 2 H2(v =0)+ hν0 → H2 (v )+e 2 me (ve) . (6.3) The left-hand side of this ﬁgure then shows the resulting photoelectron spectrum. For v+ ≥ 10 , and again for v+ ≥ 13 , a separate scan with enhanced amplitude was obtained by increasing the data collection time in the experiment. 98 CHAPTER 6. PHOTOELECTRON SPECTROSCOPY

+ KEe− H(1s) + H

υ+=6 υ+=3 υ+=0

IE(υ+=2)

hν0 IEadi

H(1s) + H(1s)

υ″=0

Figure 6.2: Left: He(I) photoelectron spectrum of H2 with small peaks due to N2 impurity marked by arrows. + Right: Potential energy curves for H2 and H2 , and an illustrative photoionization transition.

The potential energy function and level energy picture in Fig. 6.2 is precisely analogous to those for vibrational structure in ordinary electronic spectra, which were illustrated by Fig. 5.7. In the present case, the system must end up in one of the discrete vibrational levels of the molecular ion, and the different vibrational energies associated with different values of its vibrational quantum number v+ give rise to the vibrational fine structure observed in photoelectron spectra. Since photoelectron spectroscopy normally deals with relatively cold samples, the observed transitions all originate in the ground vibrational level of the ground electronic state of the neutral molecule, so the observed fine structure is due to excitation from there into different vibrational levels of the molecular ion. Moreover, in contrast to “normal” (absorption or emission) electronic spectroscopy, rotational substructure is usually not resolved in photoelectron spectra, so the interpretation need only concern itself with the positions and relative intensities of peaks due to different final-state vibrational levels. The left-hand portion of Fig. 6.2 shows that when neutral H2 molecules are ionized, a number of peaks + appear in the spectrum. This series of lines corresponds to the formation of H2 molecules with a range of different vibrational energies. Conservation of energy therefore allows us to rewrite Eq. (6.1) as ' ( + + + hν0 =KEe− +IEadi + G (v ) − G (0) , (6.4) in which the “adiabatic ionization energy” IEadi is the lowest-energy ionization process possible – the one that leaves the product molecular ion in its ground vibrational level. This expression shows that if the ionization process leaves the molecular ion in an excited vibrational level, the ejected electron will have less kinetic energy. In the context of Fig. 6.2, this explains why the ionization energy and assigned vibrational quantum numbers both increase from right to left, while the electron kinetic energy increases from left to right. Note that when using Eq. (6.4), it is important to remember that all of the different types of energies must be expressed in the same units. In particular, while it is natural to measure KEe− and to report IE in eV, we have seen that vibration-rotation energies are normally expressed in wavenumbers, cm−1,and that the energy of light may be expressed in joules, or by specifying its frequency or wavenumber. Thus, propitious use of the conversion factors listed in Table 1.1 on p. 7 will often be required. 6.3. VIBRATIONAL FINE STRUCTURE IN PHOTOELECTRON SPECTRA 99

The vibrational level energies of a molecular ion are described in exactly the same manner as those for a neutral molecule (see Chapter 3). Thus, within a Morse potential approximation, the vibrational level energies of a molecular ion are given by the expression + + + + 1 + + 1 2 G (v )=ωe (v + 2 ) − ωexe (v + 2 ) . (6.5)

Thus, Eqs. (6.4) and (6.5) allow a set of measured KEe− peak positions to be employed to determine the + + adiabatic ionization energy IEadi and the vibrational constants ωe and ωexe of a molecular ion, from which Eq. (3.10) can give us the equilibrium force constant for the vibrational stretching motion in the ion. Note that as illustrated here, the symbols for energy and other properties of a molecular ion are sometimes (although not always) given a superscript label “ + ” in order to remind us that they are referring to the properties of the molecular cation.

Exercise (i): The three highest-energy peaks in the photoelectron spectrum of H2 generated by a He(I) lamp are

found to correspond to KEe− =5.772, 5.502 and 5.246 eV. From this information, determine the value of IEadi + + −1 + in eV, the values of ωe and ωexe in cm , and the value of the vibrational force constant k˜ for H2 . Answer: Assuming that small Franck-Condon factors do not prevent its observation, the largest value of

KEe− is associated with the adiabatic ionization into the lowest vibrational level of the ion. Using the known energy of photons generated by a He(I) lamp, we obtain + IE0 = hν0 − KEe− (v =0)= 21.2182 − 5.772 = 15.446 eV . Since the observed peak spacings are the vibrational level spacings of the molecular ion, use of Eq. (3.18) yields + + + ΔG1/2 =(5.772 − 5.502) × 8065.54465 = 2177 = ωe − 2ωexe (0 + 1) + + + ΔG3/2 =(5.502 − 5.246) × 8065.54465 = 2065 = ωe − 2ωexe (1 + 1) . + −1 + −1 Solving these equations then yields ωe = 2289 cm and ωexe =56cm . Finally, substituting this value of + + ωe and the reduced mass μ(H2 )=0.503 775 338 [u] into Eq. (3.10) yields + 2 μ (ωe ) 4 −1 −2 k˜ = =7.83×10 [cm A˚ ] . 2 Cu

The intensity pattern of the vibrational ﬁne structure peaks in a photoelectron spectrum are readily explained in terms of the classical Franck-Condon principle discussed in § 5.3. In photoelectron spectroscopy, ionization almost always occurs from the v=0 level of the neutral parent molecule, so we have the type of simple cases illustrated by Fig. 5.7. As is shown there, the most intense absorption occurs into vibrational levels of the upper state whose inner (or outer) turning point lies directly above the equilibrium distance

(re) of the initial neutral-molecule potential. This intensity maximum defines what is called the “vertical ionization energy”. + For the case of H2, Fig. 6.2 shows that the most intense ionization occurs for transitions into the v =2 + + vibrational level of H2 . This tells us that the minimum of the H2 potential energy curve is displaced relative to that for H2. From the spectrum alone, we might not know whether this radial displacement was inward or outward. However, the fact that the observed transitions involve a fairly large number of vibrational levels suggests that it is probably outward, since the steepness of the inner wall of a potential means that the inner turning points of the various levels will lie relatively close to one another, and hence would all be accessible for vertical transitions at distances near the ground-state re value. A more definitive conclusion is provided by the molecular orbital theory description illustrated by Fig. 6.1. We know that H2 has a bond + 1 order of 1, since it has two bonding electrons, while the bond order in H2 is only 2 , since it has only one + bonding electron. As a result, we expect that H2 will have a longer bond length and smaller dissociation + + −1 energy than does neutral H2. The fact that the value of ωe (H2 ) = 2230 cm is roughly half as large −1 + as ωe(H2) = 4161 cm confirms that the bond in H2 is much weaker than that in H2. This in turn + tells us that the potential energy curve for H2 is indeed displaced outward from that for H2, as is seen in Fig. 6.2. Thus, we see that the intensity patterns of the fine structure in PES do give us information about differences between the potential curves of the neutral molecule and the molecular ion, and hence also about the molecular orbital energies in the parent molecule. In closing this section, we note that reports of PES ionization energies sometime speak of two different ionization processes: 100 CHAPTER 6. PHOTOELECTRON SPECTROSCOPY

σ3p*

H(1s) Cl(3p5)

σ3p ⇑bonding

σ3p*

H(1s) Cl(3p5)

σ3p

Figure 6.3: Left: Molecular orbital diagram for HCl. Right: He(I) photoelectron spectrum of HCl.

• Adiabatic ionization is the ionization process associated with the symbol IEadi in Eq. (6.4) and Fig. 6.2, which leaves the molecular ion in its ground vibrational level v+=0. This is the quantity referred to by Koopman’s theorem. As is illustrated by Case D in Fig. 5.7, in some systems the intensity of this peak may be very low (or unobservable!), so it may be diﬃcult to know whether or not the weak highest- + energy KEe− peak observed actually corresponds to v = 0 . This is the reason for the conditional statement at the beginning of the answer given for Exercise (i) on p. 99.

• Vertical ionization is the ionization process that gives the most intense vibrational peak in the spectrum; + + for the case of H2 we see that this corresponds to ionization into v =2.

6.4 Molecular Orbitals and Photoelectron Spectra

Let us now examine further the relationship between the molecular orbitals of a molecule and the vibrational structure of its photoelectron spectrum. The left-hand portion of Fig. 6.3 shows the molecular orbital diagram for HCl and the right half its He(I) photoelectron spectrum. Within this molecular orbital diagram, the lower segment shows the relative energies and occupancy of the highest occupied orbitals in the separated atoms, ∗ while the upper segment shows the occupancy of the bonding (σ3p) and anti-bonding (σ3p) molecular orbitals formed when the atoms come together to form the molecule. In the molecule, the bonding σ3p orbital is occupied by two electrons, one contributed by each atom. The valence electrons in the ground state of the neutral HCl molecule therefore consist of the one pair of electrons in the σ3p bonding molecular orbital, plus the two pairs of non-bonding “lone-pair” 3p electrons remaining on the Cl atom. Photoionization of HCl involves removal of an electron from one or the other of these orbitals. Note that within each band in the PES spectra of Figs. 6.2 and 6.3, the vibrational quantum number associated with the diﬀerent peaks increases from right to left, since more energy is required if the ion left behind by the departing electron is to be found in a higher vibrational level.

Since the 3p lone-pair orbitals on the Cl atom lie at a higher energy than does the σ3p bonding orbital, it will be easier to remove an electron from the former. This process will therefore deﬁne the lowest ionization energy, and will yield the highest-energy photoelectrons. However, those 3p lone-pair electrons are not expected to participate much in the bonding, so the properties of the resulting HCl+ molecule, which has 6.4. MOLECULAR ORBITALS AND PHOTOELECTRON SPECTRA 101

2 3 the electronic configuration {(σ2p) , 3p }, should be fairly similar to those of the neutral parent molecule, 2 4 whose electron configuration was {(σ3p) , 3p }. This is confirmed by the results shown in the first two rows of Table 6.1. This situation corresponds approximately to Case B in Fig. 5.7, which explains why the intensity pattern for the low-energy (or high KEe− ) band seen in Fig. 6.3 is totally dominated by the Δv=0 peaks. (The splittings of the peaks associated with transitions into the v+ = 0 and 1 peaks of the X 2Π state of HCl+ will be explained later.)

Similar arguments indicate that removal of an electron from the σ3p bonding molecular orbital of HCl will 1 occur at higher energy (yielding slower photoelectrons), and will reduce the bond order to 2 . This predicted substantial weakening of the bond is conﬁrmed by the dramatic reductions in the vibrational level spacing constant ωe and bond dissociation energy D0, relative to the values for the neutral molecule, as shown in the third row of Table 6.1. This in turn suggests that the minimum in the potential curve for the resulting + 1 4 molecular ion HCl {σ3p , 3p } will be displaced to larger internuclear distance. This relatively large increase in re from the neutral molecule to the ion qualitatively corresponds to Case C in Fig. 5.7, and it explains why the transitions into the higher-energy A 2Σ+ state of the molecular ion show substantial intensity for a fairly wide range of vibrational levels. A ﬁnal comment about the HCl spectrum concerns the fact that the vibrational peaks associated with the lower-energy band are split into doublets. This is due to the fact that in the molecular ion state produced by this process, the electrons have both orbital angular momentum and spin angular momentum for precession about the molecular axis. These two types of angular momentum may be aligned either parallel or anti- parallel to one another, and the energy splitting between the two cases gives rise to a ‘spin-orbit splitting’ of the v+=0 and 1 peaks in the X 2Π spectrum. For the higher-energy band centred around 16.5 eV, the molecular ion state has no net electronic orbital angular momentum, so such splittings do not occur there.

+ + Table 6.1: Molecular parameters for some states of HCl, HCl ,N2 and N2 .

−1 state ωe/cm re/A˚ D0/eV T0/eV

1 + a HCl (X Σg ) 2990.91.2746 4.4336 0.0 + 2 HCl (X Πi) 2673.71.3147 4.653 12.768 HCl+(A 2Σ+) 1606.51.5142 1.169 16.252

1 + a N2 (X Σg ) 2358.61.0977 9.7537 0.0 + 2 + N2 (X Σg ) 2207.41.1164 8.7127 15.5808 + 2 + + N2 (A Πui) 1903.5 1.1749 7.5949 16.6986 + 2 + b N2 (B Σu ) 2420.81.0747 5.5429 18.7506 a The reference energy is the zero point level of the neutral parent molecule. b See the discussion in §6.5.

As a second example, let us consider the case of N2, for which a schematic molecular orbital diagram and the photoelectron spectrum generated with a He(I) lamp are shown in Fig. 6.4. The presence of three groups of transitions in this spectrum indicates that three different electronic states of the ion lie within 21.2182 eV (the He(I) photon energy) of the ground state of neutral N2. These states are produced by removing an electron from one or another of the three highest occupied molecular orbitals of the parent molecule; the absence of a fourth band indicates that the binding energy of an electron in the σ2s valence electron orbital is greater than 21.2182 eV. The energies of the molecular ion states associated with these electronic bands increase from right to left in Fig. 6.4, and as in Figs. 6.2 and 6.3, within each band the vibrational quantum number associated with the different peaks also increases from right to left. For all three bands, the sharp intensity breakoff at the low-energy edge of the band indicates that the lowest-energy peak is indeed associated with ionization into the ground v+=0 level of that electronic state of the ion. This assignment means that those peak positions define the threshold ionization energy IEadi, and that the spacings to their neighbours can be used + + to determine values for the vibrational constants ωe and (and for the middle spectrum, also ωexe )forthese three electronic states. 102 CHAPTER 6. PHOTOELECTRON SPECTROSCOPY

Figure 6.4: Left: Molecular orbital diagram for N2. Right: He(I) photoelectron spectrum of N2.

+ Two other prominent features of these N2 electronic bands are the vibrational intensity patterns and the characteristic peak spacings. Table 6.1 compares the leading vibrational constant, the bond length, the binding energy (D0), and the energy of the zero point level (T0) of the three observed molecular ion states + 2 + with those of the parent N2 molecule. Removal of a σ2p bonding electron to produce N2 (X Σg )leadsto modest reductions in the values of ωe and D0, and to a small increase in re; this is what we expect when 1 the overall bond order is reduced from 3 to 2 2 . However, the fact that these changes are relatively small indicates that the σ2p orbital makes only a modest contribution to the binding energy. The very sharp drop in intensity from the v+=0 to 1 peaks for this band is what we expect for cases in which there is little difference between the equilibrium bond lengths of the initial and final states. The photoelectron band system of N2 centred near 17 eV is associated with removal of an electron from the π2p bonding orbital. The relatively large increase in re and decreases in ωe and D0 compared to the values for the neutral molecule (see Table 6.1), reflect the fact that the π2p electrons contribute much more to the 2 + binding of N2 than do the σ2p electrons (see the MO diagram in Fig. 6.4). As a result, the product A Πui 2 + molecular ion is distinctly less strongly bound than is the X Σg ground-state ion. The resulting larger 2 + outward displacement of the A Πui state potential then gives rise to the more extended type of vibrational spectrum illustrated by Fig. 5.7 B. ∗ Finally, the molecular parameters in Table 6.1 show that removal of a σ2s anti-bonding electron from neutral N2 increases the strength of the bond in the ion, making the vibrational motion a little stiffer, and the bond length a little shorter than in either the ground state of the molecular ion or the parent neutral molecule. The fact that the bond length is only slightly shorter than that in the neutral parent molecule 2 + explains why we again have the same type of intensity pattern found for transitions into the X Σg state. 1 Note, however, that although the bond order formally increases from 3 to 3 2 on going from the neutral 2 + D molecule to the B Σu state of the molecular ion, the bond strength 0 actually becomes substantially smaller. The reason for this counter-intuitive behaviour is discussed in the next section. In summary, therefore, we can characterize three types of ionization processes: those involving removal of an electron from a bonding (σ or π) orbital, from a non-bonding (lone-pair) orbital, or from an anti-bonding (σ ∗ or π ∗) orbital. • Loss of a bonding electron decreases the bond order, thereby reducing the vibrational spacings and increasing the bond length in the resulting cation, relative to those properties of the parent molecule. 6.5. SOME COMPLICATIONS IN PHOTOELECTRON SPECTRA 103

These changes will be largest for the orbital that contributes most to the bonding. The results for N2

suggest that its π2p electrons contribute more to the bonding than do its σ2p electrons. • Loss of a non-bonding electron has little effect on bond order, vibrational spacings, or bond length. 1 + + 2 However, the modest changes between the properties of HCl(X Σ )andthoseofHCl (X Πi)(see Table 6.1) indicate that the lone-pair electrons do contribute somewhat to the bonding in this system. • Loss of an anti-bonding electron increases the bond order, thereby increasing the vibrational spacings and decreasing the bond length of the cation, by comparison with the properties of the parent molecule. + 2 + The changes for N2 (B Σu ) seen in Table 6.1 are relatively modest, since the relevant anti-bonding orbital is not in the outermost valence electron subshell. These effects are much more pronounced for + 2 + the case of O2 (X Σ ), which is discussed below. Thus, the peak spacings and intensity pattern of the vibrational fine structure in an ionization band provides information on the contribution to the bonding in the neutral molecule of the orbital from which the electron was removed.

6.5 Some Complications in Photoelectron Spectra

The preceding discussion of the photoelectron spectrum of H2,HClandN2 illustrates a good degree of consistency between the nature of observed photoelectron spectra and predictions of molecular orbital theory. However, additional complications can substantially blur the satisfying generalities implied by that zeroth- order picture. One illustration of this point is the spin-orbit splitting of the vibrational peaks associated with the ground state of the HCl+ ion that was seen in Fig. 6.3. Such splittings can arise whenever the molecular ion state has both non-zero electronic orbital angular momentum (i.e., it is not in a Σ state) and non-zero total electron spin angular momentum (i.e., it is not a spin-singlet state), and their magnitudes increase sharply for species formed from atoms in the lower rows of the periodic table. For example, in the analogous PES spectra of HBr and HI, the magnitude of this X 2Π spin-orbit splitting increases dramatically, 2 + + and becomes substantially larger than the vibrational peak spacing, while for the A Πui state of N2 (middle band in Fig. 6.4), these splittings are too small to be experimentally resolved. Complications of another type arise for O2. In this case, the only orbitals that are energetically accessible to He(I) excitation are those formed from the 2p electrons of the parent O atoms. However, while the MO diagram in Fig. 6.5 shows that only three such orbitals are occupied, the PES spectrum shows (at least) four distinct bands. The lowest ionization energy for this species is readily assigned to removal of an electron ∗ D from the π2p anti-bonding MO; the substantial increases in both the bond strength 0 and the vibrational 2 + + constant ωe shown in Table 6.2 are clearly consistent with this assignment. As for the A Πu state of N2 discussed above, the spin-orbit splittings in this species are too small to be resolved. However, it is not immediately clear why there should be (at least) three higher-energy bands, since this species has only two other types of occupied molecular orbitals.

This conundrum is resolved when we realize that when a π2p or σ2p electron is removed, the energy of the resulting molecular ion state will depend on whether the spin of the remaining unpaired electron in ∗ that orbital is parallel or anti-parallel to the spins of the two parallel-spin electrons in the anti-bonding π2p 3 orbitals. The former case (parallel) would yield a quartet state with total spin quantum number S=2 ,and

+ Table 6.2: Molecular parameters for some states of O2 and O2 .

−1 state ωe/cm re/A˚ D0/eV T0/eV 3 − O2(X Σg ) 1580.41.2077 5.1156 0.0 + 2 O2 (X Πg) 1906.11.1169 6.662 12.0717 + 4 O2 (a Πui) 1035.11.3814 2.629 16.1042 + 2 O2 (A Πu) 899.01.4090 1.694 17.0395 + 4 − O2 (b Σg ) 1197.01.2794 2.530 18.1706 + 2 − O2 (B Σg ) 1156. 1.298 1.760 20.297 104 CHAPTER 6. PHOTOELECTRON SPECTROSCOPY

Figure 6.5: He(I) photoelectron spectrum of O2. Insert: Molecular orbital diagram for O2.

1 the latter a doublet state with S=2 , and Hund’s rule suggests that in each case, the quartet state should have the lower energy. It is this spin-spin splitting that gives rise to the extra bands found in the PES spectrum of O2. At the same time, the nature of the bonding in the resulting species depends primarily on the orbital from which the electron was removed, so we would expect the properties of the doublet and quartet states produced on removing an electron from a given parent orbital to be fairly similar to one another.

Removing an electron from the σ2p orbital leaves an unpaired electron with no net orbital angular momentum about the molecular ion axis, so the resulting molecular ion is in a Σ state. As this is the lowest of the three occupied MO s, it will be responsible for the two bands with highest ionization energy, and the similarity of their peak spacings and intensity patterns suggests that the resulting molecular ions do { 1 2 ∗ 2} indeed have the same overall electronic structure, which would be (π2p) (σ2p) (π2p) . These intensity patterns also suggest that their bond lengths only differ modestly from that for the ground state of the neutral molecule, as is confirmed in Table 6.2. Since Hund’s rule suggests that the higher-spin state should 2 − be relatively more stable, it is reasonable to find that the highest-energy band is assigned as the B Σg state 4 − and the one centred near 18.5 eV as the b Σg state. Similarly, removing an electron from the middle occupied orbital (π2p) leaves an unpaired electron with one unit of orbital angular momentum precessing about the molecular axis, so the resulting species is in a Π state. The smaller peak spacings and extended vibrational profile of the bands centred near 16.8 eV indicate that the associated electronic states have ‘softer’ bonds, and that their potential minima are displaced from that of the neutral molecule by substantially more than is the case for the two Σ states. This is consistent with the results for N2,whichshowedthatremovalofaπ2p orbital had a larger effect on the character of the bond than the removal of a σ2p electron. For this case, the spin-spin splitting is smaller than was the case for the 2Σand3Σ excited states, so that the 2Πand4Π bands overlap; however, at the resolution of the spectrum shown here they cannot readily be distinguished from one another. In summary, therefore, we see that the four sets of peaks in Fig. 6.5 are due to transitions into five different molecular-ion states associated with removal of an electron from one or another of the three highest occupied orbitals of the neutral O2 molecule. We are able to rationalize qualitatively the similarity in peak 2 − 4 − 2 4 spacing and band profile for the Σg and Σg states, and for the Πu and Πu states, and their differences from the properties of the ground-state molecular ion and the neutral parent molecule. However, while the + bond energies D0 of the three lowest O2 states vary qualitatively in a way that may be justified by simple MO arguments, that is not the case for the two highest-energy states considered here. 2 − 4 − In order to account for the fact that the B Σg and b Σg states have larger vibrational spacings, but 2 4 similar or smaller dissociation energies than the corresponding Πu and Πu states, we must take account of two other concepts. One is that the three lowest molecular ion states dissociate to yield atomic fragments 6.6. X-RAY PHOTOELECTRON SPECTROSCOPY (XPS) 105

Table 6.3: Comparison of calculated orbital binding energies εorbital and experimental adiabatic ionization energies IEadi for CO. ∗ orbital σ2p π2p σ2s 2 + 2 2 + ion state X Σ A Πu B Σu εorbital / eV 15.09 17.40 21.87 IEadi / eV 14.02 16.59 19.71

O(3P )andO+(4S) in their ground electronic states, but the two highest molecular ion states yield one 2 − 3 + 2 fragment or the other in an excited state. In particular, the B Σu state dissociates to yield O( P2)+O ( D), 4 − 1 + 4 while the b Σu state dissociates to O( D)+O ( S) . The second consideration is the fact that the electronic character of a given molecular state, that is, the ordering and energies of its orbitals, may change as a bond stretches from the equilibrium value towards dissociation. These two effects also explain why the relative + well depth D0 of the highest-energy N2 state considered in Table 6.1 does not correlate with the difference between its other properties and those for the other states. The final complication considered here is simply the fact that Koopman’s theorem, Eq. (6.2), the assumption that ionization energies are a direct measure of the orbital binding energy in the parent molecule, is only an approximation. As mentioned in §6.2, changes in electron-electron interactions and electron shielding of the nuclei mean that the molecular orbitals of the cation in general are not the same as the molecular orbitals of the parent neutral molecule. Table 6.3 provides a simple illustration of this point for the case of CO, whose molecular orbital structure is similar to that for N2. In this case the “orbital relaxation” which occurs when an electron is removed from the parent molecule reduces the energies of all three ion states, but leaves them with the same energy ordering. That is, however, not always the case. Our discussion of the PES spectrum of N2 was based on the assumption that the orbital ordering shown on the left-hand side of Fig. 6.4 is correct; that assumption is indeed consistent with a Koopman’s theorem interpretation of the photoelectron spectrum. However, when molecular orbital calculations are made more and more accurate and pushed to the Hartree-Fock limit, one finds that the π2p orbital of neutral N2 actually lies slightly above (rather than below, as suggested by Fig. 6.4) the σ2p orbital. In order to reverse this situation and achieve a level ordering consistent with experiment, the theory has to include ‘configuration interaction’, which effectively means that we have to give up the simple molecular orbital picture we have been using. Another way of thinking about this situation is to say that the relaxation of the σ2p orbital on forming the ion is significantly greater than that of the π2p orbital. Fortunately, however, this re-ordering is an unusual case, and although Koopman’s theorem definitely has its limitations, in most cases it provides a useful first-order explanation of the ionization energies observed in photoelectron spectroscopy.

6.6 X-Ray Photoelectron Spectroscopy (XPS)

Up to now we have considered only photo-ionization of valence electrons, whose binding energies are sufficiently small that they can be liberated by ordinary ultraviolet radiation with photon energies of 5 − 100 eV. However, if a radiation source that produces X-rays (100−10 000 eV) is employed, then core electrons, whose binding energies range from 50−10 000 eV, will also be dislodged. These core electrons are not affected much by chemical bonding, so that their binding energies and other properties are mainly defined by the nature of the particular atom. As is shown by Table 6.4, those binding energies vary quite dramatically from one atom to the next. However, most of this variation is explained simply by the increase of the nuclear charge with atomic number. In particular, the 1s core electrons lie much closer to the nucleus than do any others,

Table 6.4: Ionization energies of the 1s electrons of the ﬁrst-row elements.

atom He Li Be B C N O F atomic number Z 23456789 IE(1s) / eV 25 55 111 188 285 399 532 686 {IE(1s)/Z 2} / eV 6.14 6.11 6.94 7.52 7.92 8.14 8.32 8.47 106 CHAPTER 6. PHOTOELECTRON SPECTROSCOPY

Figure 6.6: XPS spectra for C atoms in different molecular environments. and hence the effective nuclear charge they see will be only very slightly screened by the other electrons. If we could ignore that screening, their binding energies would be defined by the extended Bohr formula of Eq. (1.15) in §1.2.2, which predicts that those binding energies are proportional to Z2, with Z being the nuclear charge. While this screening cannot be totally neglected, the fact that the ratio shown in the last row of Table 6.4 varies relatively slowly from one atom to the next illustrates the dominant effect of the nuclear charge on the magnitude of the 1s electron ionization energies. The linewidths of common X-ray sources are relatively large, of order 0.5 − 1.0 eV. This is due to the fact that the core-hole states produced by the XPS process are very short-lived, so the associated level energies are broadened by uncertainty principle considerations. As a result, vibrational fine structure cannot be resolved in XPS spectroscopy. However, as the very large differences in the core-level ionization energies for different atoms means that their XPS spectra can readily be distinguished from one another, this technique may be utilized for chemical element analysis. The degree to which the nucleus of a given type of atom is screened by its outer electrons depends upon the nature of the atoms to which it is bonded. If those neighbours are highly electronegative electron- withdrawing species, some of the electron density shielding the nucleus will be drawn away, and the 1s XPS peak will be shifted to higher energy. These “chemical shifts” can often be resolved, and provide remarkably effective signatures of the bonding environment for the atom in question. At the same time, comparison of Tables 6.4 and 6.5 shows that these chemical shifts are far smaller than the differences between the core ionization energies of the different atoms, which means that these two types of shifts cannot be confused with one another. Table 6.5 shows that a given atom will have slightly different IE(1s) values when found in different compounds, and Fig. 6.6 shows that, depending on their bonding and neighbours, the shifts for a given type of atom can vary within a single molecule. Once again, we see that bonding to the most electronegative elements, F and O, gives rise to the largest (most positive) shifts, and we also see that the peaks associated with the methyl ( –CH3 )andmethylene(–CH2– ) carbons can be distinguished from one another. The results for acetone and sodium azide seen in Fig. 6.6 also illustrate the fact that the area of a given peak is proportional to the number of atoms having a given type of environment. In view of its importance for chemical and structural analysis, XPS is often called “Electron Spectroscopy for Chemical Analysis” (ESCA). → + The difference between the ionization energy of an element A within a molecule, Am Am ,andthat → + of the free atom, Af Af , is called the chemical shift,ΔIEn,, and is given by

Table 6.5: Nitrogen 1s chemical shifts relative to the core ionization energy for N(1s) in gaseous N2.

Δ{IE(1s)} /eV NF3 NO2 NNOONClNON2 NNO HCN NH3 CH3NH (CH3)3N compound 4.3 3.0 2.6 1.5 0.8 0.0 −1.3 −3.1 −4.3 −4.8 −5.2 6.7. AUGER ELECTRON SPECTROSCOPY (AES) 107

valence electron vacancies

X-ray fluorescence Auger process 2nd electron ejected

ionization threshold 2p

primary core electron vacancy 1s

Figure 6.7: Schematic level energy diagram illustrating XPS core-hole decay.

' ( + − − + − ΔIEn, = IEn,(Am) IEn,(Am) IEn,(Af ) IEn,(Af ) , (6.6) in which n and are the quantum numbers for that atomic orbital. Assuming that Koopmans’ Theorem is valid, this expression reduces to

ΔIEn,(Am) ≈−εn,(Am)+εn,(Af ) , (6.7) where ε are the core orbital bonding energies.

6.7 Auger Electron Spectroscopy (AES)

When an electron is removed from a core orbital of an atom, either by X-ray photoionization or by some other mechanism, the system will have a spontaneous tendency to relax by having one of the higher-energy electrons drop down to fill the core vacancy. The short lifetime of this metastable state is one reason for the limited resolution of XPS. Energy conservation in this process may be achieved in the two ways illustrated by Fig. 6.7, either by emission of an X-ray photon whose energy matches the core/valence level spacing, or by ejection of a second electron whose kinetic energy accounts for the difference between that level spacing and the binding energy of the second electron. This radiationless second mechanism is called Auger electron emission. The very large difference between the binding energies of the core and valence electrons – hundreds vs. tens of eV (Fig. 6.7 is not drawn to scale!) – means that most of the energy released by the Auger process is carried by the electron kinetic energy. The large differences between the core level binding energies of different atoms (see Table 6.4) then means that the measured electron kinetic energies provide a clear signature for the atomic composition of the species of interest. X-ray fluorescence and Auger electron emission are bases for important methods of materials analysis which complement ordinary XPS or ESCA studies. They are particularly important for studying solids and the surfaces of materials. For this type of experiment, the primary step of removing the core electron is usually performed using an intense beam of incident electrons (rather than X-rays) for two reasons. One is the fact that the incident electrons lose energy quickly when they enter the solid, so the method most directly probes the elemental composition of the material at the surface. The second is that the incident electron beam may be focused to a very small spot size, of order 10 − 100 nm, and this high degree of 108 CHAPTER 6. PHOTOELECTRON SPECTROSCOPY spatial resolution allows the elemental composition of the surface of a material to be mapped in great detail. Auger electron spectra also exhibit chemical shifts, similar to XPS, and thus can be used to characterize the electronic environments of identical atoms in different regions of a sample. Because electrons do not travel far through solids, generally 20 A˚ at most, AES is used almost exclusively to study species on surfaces.

6.8 Problems

1. Argon, Ar, ionizes to Ar+ at 15.759 eV. If radiation of 21.2182 eV (from a He(I) lamp) 1s directed at the Ar atoms, what will be the kinetic energy in Joules of an emitted electron?

2. What is the molecular orbital electronic conﬁguration for CO? What is the electron conﬁguration for CO+ when an electron is removed from the highest occupied MO of CO? From the second highest MO? From the third highest MO? Determine the bond orders for the parent molecule and for each of the positively-charged ions described above, and indicate any changes in bond length that you expect when these ions are formed.

3. HBr undergoes two ionizations when irradiated with a He(I) lamp ( hν =15.759 eV), one emitting electrons with a kinetic energy of approximately 9.2 eV, and another emitting electrons with a kinetic energy of approximately 6.4 eV. What are the two ionization energies? Using the MO diagram for HCl as a guide, which orbitals of HBr are these emitted electrons leaving? Which of these ionizations would you expect to exhibit vibrational ﬁne structure, and why?

−1 + 4. Using the photoelectron spectrum of H2 in Fig. 6.2, determine ωe and ωexe (in cm )forH2 . 5. The photoelectron spectrum of NO can be described as follows: using He(I) radiation, there is a strong peak at kinetic energy 4.69 eV, and a long series of lines starting at kinetic energy 5.56 eV and ending at 2.2 eV. A shorter series of six lines begins at kinetic energy 12.0 eV and ends at 10.7 eV. Account for this spectrum, using the MO diagram for NO (assume that the MO diagram for N2 given in Fig. (6.4) on p. 102 is a good model for the NO orbital energies).

6. Xenon, Xe, ionizes to Xe+ and emits electrons when irradiated with a He(I) lamp (21.2182 eV), the fastest of which travel with a velocity of 1.788 × 106 ms−1. What is the ionization energy of Xe?

7. The He(I) ultraviolet photoelectron spectrum of CO seen in Fig. 6.8 exhibits three distinct ionization bands, for which the ionizations are given below: First ionization band: lines at 14.018 and 14.289 eV. Second ionization band: lines at 16.536, 16.725, 16.913, 17.096, 17.277 and 17.454 eV. Third ionization band: lines at 19.688 and 19.896 eV. Assign each of these ionizations to the corresponding molecular orbitals. Using the harmonic oscillator model, determine the vibrational frequencies (in cm−1) for each of the ions formed. For reference, the vibrational frequency of CO is 0.269 eV. Do the changes in vibrational frequencies parallel the changes you anticipate based on the bond orders of the ions?

8. Interpret the photoelectron spectrum of CO, given in Fig. 6.8, in terms of the ionization energies, the ionizations to which they correspond, and changes in structure (if any) that result from each of the ionizations. Assume the MO diagram of N2 provides a good model of the MO diagram of CO.

9. Using the information from the previous question, sketch the approximate potential energy curves of CO and each of the ions formed in the UPS experiment.

10. To monitor the ozone layer, it has been proposed that a satellite containing an X-ray photoelectron spectrometer be placed into orbit approximately 30 km above the earth. This equipment will be set up to detect only XPS signals from O3. Recalling that ozone is a bent molecule, predict the appearance of the O 1s ionization spectrum of ozone, including the approximate energy of the ionizations, the number of peaks, and their relative positions and intensities in the spectrum. 6.8. PROBLEMS 109

Figure 6.8: He(I) photoelectron spectrum of CO.

+ 11. The irradiation of H2 with ultraviolet light produces H2 molecular ions with a variety of vibrational energies. Explain why the intensity of the v =0→ v+ = 2 transition (the most intense) is stronger than that of the v =0→ v+ = 0 transition. Each of these two ionizations has a particular name; what are they called?

12. Krypton, Kr, ionizes to Kr+ when excited by a He(I) lamp. Given that the velocity of the electrons emitted from the Kr atoms is 1.12688×106 ms−1, what is the ionization energy of Kr?

13. Hydrogen iodide, HI, shows two sets of ionizations: one, with little vibrational fine structure, beginning at IE = 10.4 eV, and another, with substantial vibrational fine structure, beginning at 13.8 eV. Using the MO diagram of HCl in Fig. 6.3 as a model for HI, assign the ionizations, and comment on the fine structure observed in the spectrum.

14. The photoelectron spectra of N2 and CO are given in Figs. 6.4 and 6.8. Using the MO diagram of N2 as a model for both molecules, contrast the bonding character of each of the orbitals ionized, with particular attention to: diﬀerences in the energies of ionization; changes in the vibrational frequencies −1 upon ionization. For reference, the vibrational frequencies of N2 and CO are 2359 and 2170 cm , respectively.

15. By using the ionization energies for N2, CO, NO and O2 determined from their UPS spectra, plot the relative energies of the molecular orbitals for these four molecules.

1 16. By using the data for the photoelectron spectrum of H2 given in Figure 6.2 as well as any other data you consider necessary (and there are other data you will need!), predict the energies of the ionizations 2 2 + of H2 to the ﬁrst ﬁve vibrational energy levels of H2 .

+ 17. Analyze the vibrational ﬁne structure of HCl (Figure 6.3) to determine ωe and ωexe for HCl .

− 18. Predict the N(1s) XPS spectrum that you anticipate would be obtained for the azide ion, N3 ,interms of approximate energies and intensities. 110 CHAPTER 6. PHOTOELECTRON SPECTROSCOPY Chapter 7

NMR Spectroscopy

What Is It? Nuclear magnetic resonance (or NMR) is based on the interaction between the magnetic dipoles of nuclei and an external magnetic ﬁeld.

HowDoWeDoIt?When radio waves of the correct frequency are applied, transitions occur between the energy levels associated with the diﬀerent possible orientations of the nuclear magnetic dipoles within the magnetic ﬁeld.

WhyDoWeDoIt?The energies of the NMR transitions are very sensitive to changes in the chemical bonding environment around the nuclei, and to the presence of other neighboring nuclei. NMR is thus an important tool for characterizing the structural connectivity of the atoms in a molecule.

7.1 Basics of NMR Spectroscopy

7.1.1 Angular Momentum and Nuclear Spin

NMR spectroscopy is made possible by the fact that quantum mechanics requires angular momentum to be quantized. These notes have previously discussed three diﬀerent types of angular momentum. In Chapter 2 we discussed the mechanical rotation of molecules, and saw that quantization of that angular momentum led to quantization of rotational energy, and hence to characteristic patterns of lines in rotational and (in Chapters 3–5) vibrational-rotational spectra. In § 1.4.1 we reviewed the properties of electrons in a hydrogen(ic) atom and the two types of angular momenta that need to be considered there. The ﬁrst type, which is characterized by the quantum number , was the angular momentum associated with the orbital motion of the electron; we can think of it as being the mechanical orbiting of an electron about the nucleus. The second type of angular momentum was the ‘spin’ angular momentum of the electron, an intrinsic property of an electron that is not associated with any physical coordinates or motion. The present chapter focuses on the properties of nuclei which, like electrons, posses an intrinsic ‘spin’ angular momentum. We begin by reviewing some of the properties associated with any type of angular momentum. Angular momentum is a vector property. However, the Heisenberg Uncertainty Principle of quantum mechanics tells us that for any given type of angular momentum P, all we can ever know about it are its magnitude and one of its vector components Px , Py or Pz . Quantum mechanics also tells us that the magnitude of the angular momentum P may be written as |P| = R(R +1) , (7.1)

1 3 in which R must be either a non-negative integer (i.e., R=0,1,2,3,...)orhalfofanoddinteger(R=2 , 2 , 5 2 , . . . ) number. By convention, we always select the z component of angular momentum as the one which we know, and its allowed values are Pz = mR ,where

mR = − R, − R+1 , − R+2 , ... , R−2,R−1,R . (7.2)

111 112 CHAPTER 7. NMR SPECTROSCOPY

Thus, for any given value of R (i.e., of |P| ), there are (2 R+1) allowed values of mR (i.e., of Pz ). Figure 7.1 illustrates the four possible angular momentum alignment sub-states of a system with total angular 3 momentum quantum number R = 2 . For mechanical physical motion such as the rotation of a molecule or the orbital motion of an electron, only integer values of R are allowed. This explains the quantum labeling of rotational level energies and of hydrogenic orbitals ( =0 f o r s, =1 f o r p, =2 f o r d, . . . etc.). The associated 3 − (2J+1 ) va l u e s o f mJ are the source of the rotational degeneracy factor 2 § discussed in 2.4, while the (2 +1)valuesof m are the source of the ↑ → spatial degeneracy factors of 1, 3, 5, . . . , respectively, for hydrogenic s, P/ h− mR p, d,...,orbitals. 1 The type of angular momentum important for NMR spectroscopy − 2 is nuclear spin, and transitions between these quantum states yield the spectra observed in NMR. You have already encountered spin in the context of electron spin, mS, the last quantum number used when filling atomic orbitals. This quantum number is actually the z–component or 1 −− projection of the electron spin angular momentum, whose magnitude is 2 1 defined by the total spin quantum number S = 2 . For an electron, therefore, the degeneracy rule of Eq. (7.2) tells us that an electron can only have the 2S+1=2 va l u e s : m =+ 1 or − 1 . We commonly refer to S 2 2 3 −− these two states as ‘spin-up’ or α, and ‘spin-down’ or β, respectively. Spin 2 angular momentum has no classical analogue in terms of the mechanical ⎯35 − × − → motion of a particle. While it is tempting to think of it as corresponding √ 23 to the spinning motion of the particle, it really isn’t that simple. For an atomic nucleus the total angular momentum quantum number Figure 7.1: Space quantization of angular momentum P for R=3 . is denoted I and its z–component mI , and as discussed above, there are 2 (2I+1) different mI sub states associated with each value of I: mI=− I, −I+1 , −I+2,... , I−1, and I. The value of I may be either an integer or half of an odd integer, depending upon the structure of the nucleus. There are general rules for predicting some nuclear spin properties, but in most cases we need to refer to a table of isotope properties (such as Table 7.1) to learn their values. General Rule 1. If a nucleus has both an even atomic number and an even mass number, its nuclear spin is zero: i.e., I = 0. This is the case for 12Cand16O. 1 3 5 General Rule 2. If a nucleus has an odd mass number, it has half-integer spin: i.e., I=2 , 2 , 2 , ... , 1 1 3 13 15 19 31 3 11 23 etc. For example, I=2 for H, H, C, N, F, and P; similarly I = 2 for Band Na, while 5 17 27 I = 2 for Oand Al. General Rule 3. If a nucleus has an odd atomic number and an even mass number, it has integer spin I>0. For example, I=1 f o r 2Hand14N, while I=3 f o r 10Band50V.

1 Nuclei with spin I=2 are among the most common and easiest to study, and will be the focus of most of 1 the following discussion. Nuclei with I>2 are called quadrupolar nuclei, and are more complicated to deal with, and will not be considered in any detail here.

7.1.2 Magnetic Moments and Nuclei in a Magnetic Field Every nucleus has a magnetic moment μ whose strength μ=|μ| is proportional to the magnitude of the total spin angular momentum I : μ = γ |I| = γ I(I +1) . (7.3)

The proportionality constant γ is known as the ‘magnetogyric ratio’, and it has SI units of [radians T−1 s−1] where T refers to Tesla, which is the SI unit for magnetic ﬁeld strength. Values of γ for a number of diﬀerent nuclei are listed in Table 7.1. The mathematical sign of γ indicates whether the magnetic moment of that nucleus points in the same (for positive values) or opposite (for negative values) direction as the spin angular 7.1. BASICS OF NMR SPECTROSCOPY 113

Table 7.1: Table of the NMR properties of various nuclear isotopes.

Nuclear Atomic Nuclear Natural Magnetogyric Isotope Number Spin, I Abundance Ratio, γ (%) (rad T−1 s−1) 1 7 electron – 2 – −17608.4×10 1 7 neutron – 2 – −18.3257×10 1 1 7 H1 2 99.985 26.7519×10 2H 1 1 0.015 4.1066×107 3 1 7 H1 2 (radioactive)28.535×10 7 3 7 Li 3 2 92.58 10.3975×10 11 3 7 B5 2 80.42 8.5843×10 13 1 7 C6 2 1.108 6.7283×10 14N 7 1 99.63 1.9338×107 15 1 7 N7 2 0.37 −2.712×10 17 5 7 O8 2 0.037 −3.6279×10 19 1 7 F9 2 100 25.181×10 27 5 7 Al 13 2 100 6.9760×10 29 1 7 Si 14 2 4.70 −5.3188×10 31 1 7 P15 2 100 10.841×10 59 7 7 Co 27 2 100 6.317×10 momentum vector. However, for the purpose of all discussions in this chapter, we can treat all values of γ as being positive. When nuclei are placed in an external magnetic ﬁeld B , those with non-zero spin will interact with the ﬁeld, just as any two magnets interact. The resulting interaction energy is

E = − μ · B = − (μx Bx + μy By + μz Bz) . (7.4)

It is a universal convention to define the coordinate system such that the z axis points in the direction of the external magnetic field. This means that Bx=By=0, and it is convenient to write Bz=|B |≡B0 . As a result, the interaction energy between a given nucleus and the magnetic field may be written as

E = E(mI )= − μz B0 = − (γIz) B0 = − γmI B0 . (7.5)

This means that when a nucleus with nuclear spin quantum number I is placed in an external magnetic ﬁeld of strength B0 , it has a ladder of equally spaced energy levels

E(mI )= − γ (I)B0 , − γ (I − 1)B0 , − γ (I − 2)B0 , ... , + γ (I − 1)B0 , + γ (I)B0 . (7.6)

1 13 19 1 For a H(or Cor F or . . . ) nucleus, I=2 , and the dependence of the two level energies on the magnetic 1 ﬁeld strength is illustrated by Fig. 7.2. Note that for all cases in which I=2 , it is customary to label the 1 1 state with mI=+ 2 as α and that with mI=− 2 as β .

7.1.3 NMR Spectra

Application of the same angular momentum conservation arguments introduced in § 2.2.2 gives rise to a selection rule for transitions between nuclear spin energy levels:

ΔmI = ± 1 . (7.7) 114 CHAPTER 7. NMR SPECTROSCOPY

β(m = − ½) 5 I ↑ E 0 ΔE (B/ α)

-5 α(mI =+½)

-10 B/ ↑ 0 → / Bα 1 Figure 7.2: Nuclear spin energy levels in a magnetic ﬁeld for I = 2 .

Since the energy levels for a given nucleus are equally spaced, only one possible transition energy is allowed, namely,

ΔE = E(mI ) − E(mI +1) = γ B0 = hν0 . (7.8) The frequency of light that would cause such transitions is known as the Larmor frequency of that particular type of nucleus in that particular magnetic ﬁeld:

ΔE γ B0 γ B0 −1 ν0 = = = [s ] . (7.9) h h 2 π

Values of B0 used in NMR typically range from 1 to 20 T. Combining the Larmor frequency equation (7.9) with values of γ from Table 7.1 shows that ν0 ranges from tens to hundreds of MHz, depending on the particular nucleus and ﬁeld strength. This corresponds to transition energies of order 10−25 Joules or 0.01 cm−1, much smaller energies than we have encountered in the types of spectroscopy discussed in previous chapters of this text.

Exercise (i): 1 13 19 What are the Larmor frequencies for H, Cand F nuclei in an 8.0000 T magnetic ﬁeld ? 7 −1 −1 1 γ B0 26.7519×10 [rad T s ] (8.0000 [T])) For H nuclei: ν0 = = 2 π 2 π [rad] 8 =3.40616×10 [Hz] = 340.62 [MHz] 7 −1 −1 13 γ B0 6.7283×10 [rad T s ] (8.0000 [T]) For C nuclei: ν0 = = 2 π 2 π [rad] 7 =8.56674×10 [Hz] = 85.667 [MHz]

An NMR spectrum is clearly a function of two variables, the magnetic field strength B0 and the radiation frequency ν. Consequently, spectra may in principle be collected as a function of either quantity. (This differs from all of the other types of spectroscopy we have discussed up to now, in which signal was always collected as a function of radiation frequency.) Consider a case in which two nuclei, a proton (a 1H nucleus) and a 13C nucleus are present in the same sample at the same field strength. If the spectrometer is set up such that the proton resonates at a given frequency and field value, then what must be done to bring the 13C nucleus into resonance? There are clearly two options. First, the radiation frequency could be left the same and the field strength adjusted until the Larmor frequency for 13C is achieved. Since the γ for 13Cismuch smaller than that for protons, the same will be true for its Larmor frequency; hence the field will need to be increased if the radiation frequency is left unchanged. Alternatively, if magnetic field is left constant, the fact that the Larmor frequency for 13C is smaller than that for protons means that the radiation frequency will have to be decreased. 7.1. BASICS OF NMR SPECTROSCOPY 115

Exercise (ii): Given a spectrometer for which 1H nuclei resonate in a 1.409 T field. (a) What must the field be increased to in order for 13C nuclei to resonate? Firstly, we must determine the Larmor frequency of the proton in this spectrometer. 7 −1 −1 1 26.7519×10 [rad T s ] (1.409 [T]) ν0( H) = =60.00 [MHz] 2π If 13C were to achieve resonance at that frequency, then necessarily 13 13 13 γ( C) B0( C) ν0( C) = 60.00 [MHz] = 2π Hence, the required field strength would be 6 13 2 πν0 2π(60.00×10 [Hz]) B0( C) = = =5.603 [T] γ(13C) 6.7283×107 [rad T−1 s−1]

(b) Alternatively, what must the radiation frequency be changed to in order for the 13C nuclei to resonate ?

The Larmor equation (7.9) tells us that for a given ﬁeld strength ν0 ∝ γ .Hence 13 7 13 1 γ( C) 6.7283×10 ν0( C) = ν0( H) × =60.000 [MHz] × =15.09 [MHz] γ(1H) 26.7519×107

The effects of increasing the magnetic field and decreasing the radiation frequency are equivalent, in terms of the NMR spectrum. Because NMR spectra were first measured as a function of magnetic field at fixed resonance frequency, NMR spectra were plotted with magnetic field increasing to the right, which for a given pattern of peak positions corresponds to frequency increasing to the left. Even today, this convention, with frequencies increasing to the left, is used in NMR spectroscopy. Modern NMR spectrometers use superconducting magnets which are best held at constant magnetic field strengths. Thus, in practice it is the radiation frequency that is varied, and the NMR spectrum is collected as a function of frequency, as in the other forms of spectroscopy we have discussed. However, NMR spectrometers are rarely classified in terms of their magnetic field strengths, but rather in terms of the Larmor frequencies of the nuclei observed. Since the proton 1H is the most commonly studied nucleus in NMR spectroscopy, NMR spectrometers are usually classified according to the frequency at which protons absorb radiation, or resonate, in the given (fixed!) magnetic field. Figure 7.3 presents schematic NMR spectra of several different nuclei in 250 MHz and 600 MHz spectrometers. Note that as predicted by Eq. (7.9), the relative peak positions for the different nuclei are identical, but their absolute frequency values scale with the magnitude of the magnetic field. Note too that in both cases, the Larmor frequency for 1Histhesame as the ‘name frequency’ for that spectrometer. What are the magnetic field strengths for the two spectrometers considered in Fig. 7.3 ?

3H 1H 19F 11B 31P 13C 2H

250 MHz spectrometer

400 300 200 100 0

←ν0 /MHz

3H 1H 19F 11B 31P 13C 2H

600 MHz spectrometer

600 500 400 300 200 100 0

←ν0 /MHz Figure 7.3: NMR spectra of various atomic nuclei in 250 MHz and 600 MHz spectrometers. 116 CHAPTER 7. NMR SPECTROSCOPY

Figure 7.4: 1H NMR spectrum of ethanol.

If all there was to NMR spectroscopy was first to create energy level spacings by applying a known magnetic field to the nuclei, and then to measure those spacings, the technique would be of no use at all to chemists, since these properties are functions of nuclear structure, and not of electronic or molecular structure. The reason that NMR is such a powerful spectroscopic technique lies in two other phenomena or interactions which are evident in the proton NMR spectrum of ethanol shown in Fig. 7.4. Although the nuclei giving rise to the peaks in Fig. 7.4 are all protons (1H), we see that there are three separate groups of peaks. All else being equal, NMR signal intensities are directly proportional to the number of nuclei within the sample that absorb radiation at that frequency. Thus NMR spectra are useful for determining the relative abundances of different types of nuclei in a sample. The total area associated with each group as obtained by integrating across that group of peaks is indicated by the height of the ‘step function’ superimposed on the peaks for each group. This suggests that the single peak at highest frequency (since it is farthest to the left) is associated with the one –O–H proton in ethanol, that the quartet of peaks in the middle is associated with the two –CH2– protons, and that the triplet of peaks at lowest frequency is associated with the three –CH3 protons. Two different physical phenomena are responsible for the patterns seen in Fig. 7.4. Chemical Shifts δ . Identical nuclei situated at distinct chemical sites in a molecule (i.e., with different neighbours and bonding environments) have slightly different transition energies. The resulting peak shifts reflect the different electron distributions about the nuclei at the different types of sites. This effect is responsible for breaking the proton NMR spectrum of ethanol into three different groups of peaks.

Spin-Spin Couplings, JAB : Since each nucleus with non-zero spin behaves as a tiny bar magnet, its magnetic field can cause small additional shifts of the level energies of neighbouring nuclei. This transmission of magnetic information occurs via the electrons in the bonds linking the coupled nuclei and establishes which types of nuclei are close to one another, since the strength of the interaction drops off rapidly with distance. This effect is responsible for the singlet vs. quartet vs. triplet splitting patterns of the three groups of peaks for ethanol seen in Fig. 7.4. These phenomena and their implications with regard to chemical structure determination are described in the next two subsections. 7.2. CHEMICAL SHIFTS 117

7.2 Chemical Shifts

7.2.1 Electronic Shielding of Nuclei and ‘Chemical Shifts’ In a molecule we do not have bare nuclei, but rather nuclei surrounded by electrons. The immensely strong magnetic ﬁeld of the spectrometer B0 will drive those electrons to circulate, and the resulting current will giverisetoasmallinduced magnetic ﬁeld which we may express as

Bind = σ B0 , (7.10) whose magnitude is directly proportional to B0 , but which points in the opposite direction. The proportionality constant σ is called the chemical shielding constant for that particular type of nucleus in that specific molecular environment, because it is a measure of the tendency of the local electronic environment to shield the atomic nucleus from the applied external field. Because of the presence of this induced field, the strength of the net effective magnetic field Beff seen by the nucleus becomes Beff = B0 − Bind =(1− σ) B0 . (7.11)

Since Beﬀ < B0 , the actual resonance frequency for a given type of nucleus (labelled A) in a given molecular environment is therefore shifted from the value for the ‘free nucleus’ to the value B B A γA eﬀ γA 0 ν0 = = (1 − σ ) . (7.12) 2 π 2 π A

Thus, σA represents the fractional decrease in ν0 from its ‘free-nucleus’ value, due to the speciﬁc molecular environment of nucleus A; i.e., to the identity of its atomic neighbours, local bond orders and bond angles. It has a diﬀerent value for each chemically distinct site within a molecule.

From the above discussion it is clear that σA has no units; i.e., it is a different dimensionless constant for each chemically distinct nucleus. Its values are typically of order ∼ 10−5, so it is often reported in units −6 ‘parts per million’. For example, if σA=6 2 .1×10 , then we say that the chemical shift of nucleus A is “62.1 parts per million”. While the Larmor frequency of nucleus A in the actual molecular environment depends on the strength of the applied field B0, σA does not, and Eq. (7.12) may be rearranged and written in the form A A ν0 (bare nucleus) − ν0 (in molecule) σA = A . (7.13) ν0 (bare nucleus) It is possible to measure transition frequencies with very high precision, especially in the “radio wave” region of the electromagnetic spectrum associated NMR spectroscopy. However, it is very difficult to determine accurate experimental values of σA because it is very difficult to get a stable population of bare nuclei A to sit still to allow us to measure the value of ν0 (bare nucleus) in the particular magnetic field of a given spectrometer. As a result, instead of using the bare nucleus as the reference species it is customary to define a chemical shift parameter δA by comparison with the Larmor frequency of that nucleus in some chosen standard chemical environment. Moreover, to avoid having constantly to write down large powers of ten, this quantity is defined in units of ‘parts-per-million’ as A − A ν0 (in molecule) ν0 (in reference species) × 6 δA(in molecule) = A 10 [ppm] (7.14) ν0 (in reference species) The NMR signal of the chosen reference species is, by convention, taken as the zero of the ‘chemical shift scale’ for the nucleus of interest.

7.2.2 What Determines Chemical Shifts, and The Chemical Shift Scale The greater the shielding of the nucleus, the smaller the Larmor frequency relative to that for the bare nucleus. For convenience, reference compounds are chosen to be stable, relatively inert chemical species in which there is a high degree of shielding. The latter property is a matter of convenience which means that 1 13 29 most measured δA values will be positive. For Hand Cand Si, the conventional reference compound is tetramethylsilane (TMS), Si(CH3)4 , and its NMR signal is, by convention, taken as the zero of the ‘chemical shift scale’ for these nuclei. 118 CHAPTER 7. NMR SPECTROSCOPY

Figure 7.5: Chemical shifts of 1Hand13C nuclei in various environments.

In general, chemical shift values depend on two things: 1. The nature of the neighbouring atoms More strongly electronegative neighbours will pull electrons away from the nucleus of interest. With less shielding electron density to contribute current, Bind will be smaller and Beﬀ larger, so the resulting

transition frequencies (and δA values) will be larger. In particular, we recall that (ignoring the inert gases) electronegativities increase towards the upper right-hand corner of the periodic table. This explains the pattern of proton chemical shifts seen in the two following lists.

CH4 CH3Cl CH2Cl2 CHCl3 CH3FCH3Br CH3I δ =0.23 δ =3.05 δ =5.33 δ =7.26 δ =4.26 δ =2.68 δ =2.16

2. The nature of nearby bonds Double bonds, triple bonds, and aromatic rings (denoted Ar– ) are increasingly strong electron withdrawing groups. Their presence also reduces the electron density around the nucleus of interest and causes the resulting transition frequency (and δA value) to be larger. In larger molecules there are no ‘hard-and-fast’ rules about the size of chemical shifts for a particular type of atom in a specific local environment. However, Fig. 7.5 illustrates the fact that there are well-defined ranges of δ values for particular types of bonded atoms. For protons, δ values typically range from 0 − 12.Ina60MHzspectrometerthisgivesrisetoaLarmor frequency shift range (relative to the reference species, TMS) of ∼ 600 Hz. In contrast, in a 600 MHz spectrometer this frequency range becomes ∼ 6000 Hz. This spreading out of the spectrum (see Fig. 7.3) allows for higher-resolution measurements and for the separation of fine structure which could be difficult to resolve with a lower-field spectrometer. This is also important if the patterns of intensities of split peaks to be discussed in § 7.3 are to be observed. 7.2. CHEMICAL SHIFTS 119

Heavier atomic nuclei (such as 13C, 19F, 31P, . . . , etc.) are surrounded by many more electrons than is an 1H nucleus, so there is much more electron density available to provide the ‘counter current’ that gives rise to the induced magnetic ﬁeld, Bind . As a result, the associated ranges of shielding parameters and δ values are much larger than those for protons. In particular, for 13C the range of δ values is ∼ 300 (see Fig. 7.5), while for 31Pitis∼ 1000. However, 1H atoms are the most abundant species in many chemical environments, and hence the remainder of this chapter will focus on proton NMR spectra.

7.2.3 Working With Chemical Shifts

The δ scale readily allows us to relate NMR spectra obtained at diﬀerent magnetic ﬁeld strengths. However, we must always work in Hz and ppm interchangeably, so it is necessary to become familiar with switching between the two sets of units.

Exercise (iii): Consider a molecule containing two types of protons A and B that have chemical shifts of 3.00 ppm and 6.55 ppm, respectively. What are the frequency shifts (from the TMS reference signal) of their NMR transitions in magnetic ﬁelds of 2.35 T and 9.40 T ?

(a) First, we must use the Larmor formula Eq. (7.9) to determine the reference frequencies for these two spectrometers. 26.7519×107 [rad T−1 s−1] (2.35 T ) For the 2.35 T spectrometer: ν0(reference) = = 100.0[MHz] 2π 26.7519×107 [rad T−1 s−1] (9.40 T ) For the 9.40 T spectrometer: ν0(reference) = = 400.0[MHz] 2π (b) Now we rearrange Eq. (7.14) and use it to deﬁne the desired frequency shifts.

A −6 Δν0 = ν0(in molecule) − ν0(TMS) = δ×10 × ν0(TMS)

Hence, in the 100 MHz spectrometer:

A −6 6 For the 3.00 ppm proton : Δν0 =3.00×10 × 100.0×10 = 300 [Hz]

A −6 6 For the 6.55 ppm proton : Δν0 =6.55×10 × 100.0×10 = 655 [Hz] Similarly, in the 400 MHz spectrometer:

A −6 6 For the 3.00 ppm proton : Δν0 =3.00×10 × 400.0×10 = 1200 [Hz]

A −6 6 For the 6.55 ppm proton : Δν0 =6.55×10 × 400.0×10 = 2620 [Hz]

Comparing the magnitudes of these shifts with the scale of the Larmor frequency range seen in Fig. 7.3 makes it clear that the NMR signals from diﬀerent types of atoms will never overlap and interfere with one another. Also, as indicated by the ‘ppm’ units used for δ values, it is clear that the displacements of molecular proton peaks from the reference peak are many orders of magnitude smaller than the absolute ν0(TMS]) transition frequency itself.

If the proton NMR spectrum of ethanol shown in Fig. 7.4 is now re-examined in light of our discussion of chemical shifts, we have ample reasons to conﬁrm the assignments of the three groups of peaks there. In particular, because it is attached to a strongly electronegative O atom, we expect the –O–H proton to have the largest (positive – to the left) frequency shift; this is consistent with the fact that the peak with relative area ≈ 1 lies farthest to the left. Similarly, all else being equal we expect –CH3 protons to be more strongly shielded than –CH2– protons. Moreover, the –CH2– protons in ethanol are only one bond away from the highly electronegative O atom. Hence, we expect the group of peaks with area ≈ 3 to lie farthest to the right, as is the case. Thus, from these considerations alone, it is clear that NMR spectra can be very useful for chemical identiﬁcation. However, that capability is greatly enhanced when we also take account of the spin-spin interactions described in the next subsection. 120 CHAPTER 7. NMR SPECTROSCOPY

β(mA = − ½) ½JAB

A ν0 α(A), β(A) A A ν0 − ½JAB ν0 + ½JAB (mA = ± ½)

½JAB α(mA = + ½) no fields external nearby nearby field /B0 proton α(B) proton β(B) (mB = + ½) (mB = − ½)

Figure 7.6: Energy level diagram for proton A without (left half) and with (right half) coupling to a neighbouring proton B of a diﬀerent type

7.3 Spin-Spin Coupling

7.3.1 Basics: Coupling from a Single Neighbour

The final phenomenon to be discussed here is concerned with the splittings and relative intensity patterns within the groups of peaks for –CH2–and–CH3 protons seen in Fig. 7.4. When two bar magnets are brought together, there are two extreme relative orientations: a low-energy (attractive) one in which the magnets are aligned anti-parallel such that the ‘north pole’ of one is closest to the ‘south pole’ of the other, and a high-energy (repulsive) one in which the magnets are parallel, with the north and south poles side-by-side. The same situation occurs for magnetic nuclei (those with non-zero nuclear spin), except that instead of the magnetic fields acting directly through space, the influence of the fields of the neighbouring nuclei is transmitted through the electrons in the intervening bond(s). In this fashion, the energy of proton A can be either increased or decreased depending on the relative orientation of a nearby proton B that may be in either its “spin-up” or “spin-down” state. This situation is schematically illustrated by Fig. 7.6. In the absence of an external magnetic field, the spin-up and spin-down states of proton A have the same energy (left hand side). Turning on the magnetic field (second segment from the left, see also Fig. 7.2) splits these levels, and the transition between them A induced by light of frequency ν0 may be observed (second segment from left). The effect of the magnetic field of proton B is then to shift these levels up or down, depending on the values of mB ,byanamount which conventionally is written as ΔEAB = h(JAB /4) , the interaction energy of the spins of protons A and B. As illustrated in the two right-hand segments of Fig. 7.6, this means that the frequency of radiation A A associated with the α → β transition of nucleus A becomes either ν0 −JAB /2orν0 +JAB/2 , depending on the orientation of the spin of proton B. Thus, the presence of proton B has the effect of splitting the line A with frequency ν0 associated with the transition α(A) → β(A) into two lines with frequencies differing by −1 1 JAB [s ]. This quantity JAB (in units Hz) is called the spin-spin coupling constant for this particular pair of nuclei. If proton B is in a different local chemical environment than proton A, its NMR spectrum will be affected similarly by the nuclear spin orientation of proton A, resulting in its α(B) → β(B) transition being split into two peaks with the same separation, JAB. The net result is that the presence of spin-spin coupling

1The fact that we wish to have a simple expression for this line splitting (since it is the observable) is the reason the factor of 1/4 was included in the expression for the level shift. 7.3. SPIN-SPIN COUPLING 121

A B spectrometer ν0 −ν0 = (δA−δB)×ν0

JAB JAB

A B ν0 ←ν ν0 Figure 7.7: Simulated NMR spectrum for neighbouring single H atoms A and B with diﬀerent chemical shifts. between A and B changes the NMR spectrum from two lines, one for each type of nucleus, to four lines, two for each type of proton, due to the two possible relative orientations of the neighboring nucleus. Since there are equal numbers of the two types of nuclei, the intensities of the original A and B peaks would be the same, and since each nucleus has an equal probability of being in state α or β, the two components of each pair have equal intensity. Figure 7.7 shows the nature of the resulting spectrum for this type of case, which is called an “AX spectrum”. There are three main features to note. 1. Each type of proton involved in a given type of spin-spin coupling is aﬀected equally; i.e., the splitting

JAB = JBA . 2. Each splitting pattern is symmetric about the chemical shift for that type of nucleus.

3. The coupling constant JAB does not depend on the magnetic ﬁeld, but only on the natures of the two types of nuclei and their physical separation. In particular, its magnitude does not depend on the strength of the external magnetic ﬁeld B0 . Hence values of JAB are always expressed in absolute frequency units, Hz, rather than in ppm.

In addition, we add the general point that perturbations due to nuclei that are more than three bonds away are too small to be seen. Thus, all the splittings we discuss herein are associated with protons attached to directly bonded atoms.

7.3.2 ‘Equivalence’, and Coupling from Multiple Equivalent Nuclei If a proton of type A has two equivalent nearby protons in a local environment of type B, the energy levels for proton A take on the pattern shown in Fig. 7.8. Since there are two equivalent protons of type B,they have four possible relative alignments which give rise to three possible values for the total magnetic moment which perturbs the energies of proton A : • tot α1(B) α2(B) with mB =+1 • tot α1(B) β2(B) with mB =0 • tot β1(B) α2(B) with mB =0 • tot − β1(B) β2(B) with mB = 1

Since the nuclei of type B are ‘equivalent’, they have the same values of the splitting constant JAB,so the energy level shifts will all be the same. The net eﬀect is that we end up with three possible transition A A A frequencies: ν0 −JAB , ν0 and ν0 +JAB (see Fig. 7.8). Moreover, we expect theintensityofthemiddlepeak in this spectrum to be twice as big as the others, since there are two equally probable ways of obtaining the tot perturbing ﬁeld associated with mB =0. At the same time, as there is only one nucleus of type A and hence tot ± 1 only two possible values of mA =mA= 2 , the NMR signal for protons of type B will be split into two peaks of equal area, with a peak separation of JAB. A simulated spectrum for this type of case, known as an “AX2 spectrum”, is shown in Fig. 7.9. Note that since there are two atoms of type B and only one of type A, the sum of the areas of the peaks of type 122 CHAPTER 7. NMR SPECTROSCOPY

½J β(mA = − ½) AB

½JAB

A ν0 α(A), β(A) A A A ν0 − JAB ν0 ν0 + JAB (mA = ± ½)

½JAB

α(mA = + ½) ½JAB no fields external nearby nearby nearby field /B protons B protons B protons B 0 tot tot tot mB = +1 mB = 0 mB = −1 Figure 7.8: Energy level diagram for a nucleus without and with coupling to two identical neighbouring nuclei

B is twice that for the group of peaks associated with protons of type A. An example of a molecule which would give this type of proton NMR spectrum is O=(CH)–CH2I. Since the preceding discussion has introduced the concept of ‘equivalent’ nuclei, it is perhaps about time that we deﬁned what it means. We say that nuclei of a given type of atom are ‘equivalent’ if they are related by symmetry and reside at sites with identical electronic and bonding environments. Thus, we would say that all of the protons in a given –CH3 group are equivalent, and that the two protons in any –CH2–group are equivalent. We would also say that all six of the –CH3 protons in n-butane CH3–CH2–CH2–CH3 are equivalent, and that the same is true for its four –CH2– protons. However, the two groups of –CH3 protons at opposite ends of 2-bromo-n-butane CH3–(CHBr)–CH2–CH3 are not equivalent, since one group is much closer to the electronegative Br atom than is the other. Before we go any further, it is important to state one fundamental point. Spin-spin coupling between chemically equivalent nuclei does not cause splitting in NMR spectra. As a result, the NMR spectra of methane CH4 and of ethane C2H6 each consists of a single peak. The reason for this is not that the associated JAA constants are zero. Rather, it is that the symmetry properties of the wavefunctions for a set of identical particles only allow transitions which happen to have the same frequency. For the case of two equivalent atoms we can explain this result in terms of the level energy diagram of Fig. 7.6. If the two nuclei are equivalent, the two cases on the right-hand side of the diagram are eﬀectively a single case in which both upper and lower levels exist. However, quantum mechanical symmetry selection rules only allow transitions between the lower levels of each pair and between the upper levels of each pair, and these A transitions all occur at the same frequency, ν0 . Thus, although the levels are split, the transitions are not,

A B spectrometer ν0 −ν0 = (δA−δB)×ν0

JAB

JAB JAB

A B ν0 ←ν ν0 Figure 7.9: Simulated NMR spectrum for proton A interacting with two equivalent neighbouring protons B. 7.3. SPIN-SPIN COUPLING 123

number of weights equivalent 1 number sum of atoms of peaks weights N=1 11 2 2 N=2 112 3 4 N=3 113 3 4 8 N=4 1 4 6 4 1 5 16 N=5 1510 10 5 1 6 32

Figure 7.10: Pascal’s triangle and the calculation of weights for split NMR peaks. and there is only a single NMR line associated with this type of proton (assuming there are not also other types of proton present). Consider now the case of the molecule O=(CH)–CH3 . It clearly has two types of proton, the methyl protons (B) forming one group and the O=(CH)– proton (A) being the other, and its proximity to the double-bonded and electronegative O atom means that the latter will have a much larger chemical shift parameter: δA δB . If we now consider all possible alignments of the three methyl protons, we see that tot there are eight possible unique arrangements and four possible mB values: • tot 3 mB =+2 is obtained one way: α1(B) α2(B) α3(B) • tot 1 mB =+2 is obtained three ways: α1(B) α2(B) β3(B), α1(B) β2(B) α3(B), β1(B) α2(B) α3(B) • tot − 1 mB = 2 is obtained three ways: α1(B) β2(B) β3(B), β1(B) α2(B) α3(B), β1(B) β2(B) α3(B) • tot − 3 mB = 2 is obtained one way: β1(B) β2(B) β(B) Generalizing from the discussion of Figs. 7.6 and 7.8, we see that the magnetic field of the protons of type B A will split the absorption associated with proton A into four peaks centred about ν0 , with peak separations of JAB and relative intensities of 1 : 3 : 3 : 1 . In cases with more and more equivalent nuclei, there clearly will be more and more split peaks. The number of such peaks is given by the ‘N+1 rule’, which simply states that N equivalent atoms will split the NMR signal of neighbouring protons into N+1 peaks, while their relative intensities may be readily calculated using “Pascal’s triangle”. Pascal’s triangle provides a way of counting the number of ways in which a given tot result can be achieved – in this case, the number of distinct ways a given value of mB can be generated from N equivalent protons. This device is illustrated below in Fig. 7.10. In Pascal’s triangle, each row contains one more element than the row above it; the number appearing at both ends of each row is 1, and each internal element is the sum of the two closest elements in the row above it. For a given number of equivalent protons, the numbers in that row of Pascal’s triangle give the relative intensities of the N+1 d i ff e r e n t p e a k s . 2 At the same time, it is important to remember that the total intensity of the NMR transitions for a given type of proton is proportional to the relative number of that type of proton in the molecule. This requires one to normalize the relative intensities associated with the group of peaks for a given type of proton to reflect the proper total intensity for that type of nucleus. This is the reason that the relative heights of thefivepeaksinFig.7.9are1:2:1:4:4,andforthesamereason,thefivepeaksintheprotonNMR spectrum of O=(CH)–CH3 will have the relative intensities (from left to right) 1 : 3 : 3 : 1 : 12 : 12 . The same principles explain the pattens and relative intensities for other peaks seen in the NMR spectrum of ethanol in Fig. 7.3 and of the four sample compounds considered in Fig. 7.11.

7.3.3 Spin-Spin Coupling to More Than One Type of Neighbour Within the simple ﬁrst-order treatment presented herein, if a given type of proton is coupled to more than one type of neighbouring proton, we simply treat the eﬀects as being additive. In particular, this means

2These are also the ‘binomial coeﬃcients’ associated with the diﬀerent terms xm yN−m when one expands the algebraic expression (x + y)N . 124 CHAPTER 7. NMR SPECTROSCOPY

Figure 7.11: 1H NMR spectra of ethyl chloride, n-propyl iodide, iso-propyl iodide and tert-butyl alcohol. that one ﬁrst predicts the splitting pattern due to one type of neighbour in the manner discussed above. Then, for each peak in the resulting spectrum, one predicts the splitting due to the second type of neighbour completely independently. This can give rise to a relatively large number of peaks. For example, if proton A is perturbed by a nearby group of –CH3 protons, its spectrum will be split into four peaks with relative intensities 1 : 3 : 3 : 1 and peak separations JAB.IfthatsameprotonA is also perturbed by a nearby group of –CH2– protons with coupling constant JAC , each one of those four peaks will split into three sub-peaks with relative intensities 1 : 2 : 1 and peak spacings JAC. Thus, the spectrum of proton A will consist of a total of 12 peaks with relative intensities ranging from 1 to 2 to 3 to 6, and the ordering will depend on the relative magnitudes of JAB vs. JAC . Note, however, that when multiple sources of splittings are encountered, the order in which they are taken into account is immaterial – the same ﬁnal result is always obtained.

7.4 Molecular Structures from NMR Spectra

In the 60 years since it was first demonstrated, NMR has developed into one of the most versatile spectroscopic tools for structure determination available to chemists. Much of its utility has been due to the establishment of empirical rules relating the values of chemical shifts and spin-spin coupling constants to the local environments of the nuclei. For example, a proton near a highly electronegative atom, such as oxygen or fluorine, will in general have a larger chemical shift. Other empirical rules regarding the strengths of bonds and local electron density have been used, and you will learn much more about these as you go on in Chemistry. Perhaps the most unique aspect of NMR spectra, however, is the phenomenon of J coupling. That nuclei can communicate with their neighbors, signalling what parts of the molecule they are close to, is a tremendous tool for determining structures of unknown molecules. NMR spectroscopists have exploited this feature by developing new experiments that rapidly and readily provide a “fingerprint” for molecular structure. These experiments have been used to great advantage in biochemistry and molecular biology, 7.5. PROBLEMS 125 where structures of proteins and DNA have been determined solely from their NMR spectra. Several characteristic examples are given below. Common pieces of organic molecules, composed solely of carbon and hydrogen and called alkyl groups, give 1H NMR spectra that act as signatures of their presence in a molecule. These groups, such as CH3CH2 (ethyl), CH3CH2CH2 (n-propyl), (CH3)2CH (iso-propyl) and (CH3)C (tert-butyl), occur often in organic chemistry, and NMR has proven to be a useful tool in determining the presence of these groups, as well as many others. NMR is applied not only to molecules, however. Over the last twenty years, a new technique based on NMR spectroscopy, called Magnetic Resonance Imaging (MRI) has been developed, permitting scientists to use the response of magnetic nuclei, such as the protons in H2O, to probe the soft tissue of organisms, including patients in hospitals. Consider the differences with conventional X-rays, which are only deflected by hard matter such as bones. MRI can focus on the soft tissue in an organism, without removing the tissue or harming it, and has been particularly well-developed in the area of brain research, including multiple sclerosis. By providing pictures of the soft brain tissue, while the brain is still within the patient, physicians have learned a tremendous amount about the functioning of the brain and the action of various diseases.

7.5 Problems

1. Calculate the magnetic ﬁeld needed to induce the following nuclei to undergo NMR transitions at a frequency of 100.000 000 MHz. (a) 1H(b)15N(c)29Si

2. Calculate the Larmor frequencies of the following nuclei in a magnetic ﬁeld B0 = 11.7 T. (a) 13C(b)19F(c)31P

3. Two 1H nuclei have chemical shifts of 1.5 and 7.2 ppm. Calculate the frequency diﬀerence in Hz of these two signals in magnetic ﬁelds of: (a) 1.409 T (b) 4.697 T (c) 7.046 T

−1 1 4. What is the energy difference in Joules and cm of the two nuclear spin energy levels of HatB0 = 4.70 T? The highest magnetic fields available today for NMR are around 20 T. How much higher in field must one go before 1H NMR transitions occur at the same order of energy as rotational transitions, that is, transitions that have energies of approximately 1 cm−1?

5. Sketch the NMR spectrum you would expect at B0 = 11.7 T for three protons with the following NMR characteristics: chemical shifts of 1.0, 2.5 and 7.0 ppm. Would you expect these lines to come closer together or move further apart in frequency (Hz) if the magnetic ﬁeld was dropped to 7.05 T?

6. Using the data from the previous two questions, predict the proton NMR spectrum of the AMX2 system, for which there are two equivalent protons at the X site rather than one.

7. The proton NMR signal of nitromethane (CH3NO2) in a 60 MHz spectrometer lies 259.8 Hz to higher frequency than the TMS proton signal. What is the δ value for nitromethane protons, and what would its frequency shift be in a 100 MHz spectrometer?

8.Ina40MHzspectrometer,acompoundwithtwodiﬀerent types of protons gives rise to two sets of proton NMR signals: one doublet (two lines) at Δν =29.6 and 34.3 Hz relative to TMS, and another doublet at Δν = 391.1 and 395.8 Hz. What is the magnitude of the coupling constant J ?Where would the signals for these protons lie in a 60 MHz spectrometer?

9. Our Chemistry Department has a variety of NMR spectrometers, deﬁned in terms of their 1HLarmor frequencies as 60 MHz, 90MHz, 200 MHz, 250 MHz, 300 MHz and 500 MHz instruments. Determine the magnetic ﬁeld strength for each of these instruments, and the Larmor frequency of 13C nuclei on each. 126 CHAPTER 7. NMR SPECTROSCOPY

Figure 7.12: 60 MHz NMR spectrum of an unknown organic compound.

10. The proton NMR spectrum at 60 MHz of an unknown compound (shown above) has been obtained, and is described as follows. It consists of a triplet (group of three lines) centred at 4.40 ppm, a sextet (six lines) centred at 2.05 ppm, and another triplet centred at 1.02 ppm. Three structures have been proposed for this compound: CH3CH2CH=CHNO2,CH3CH2CH2NO2,and(CH3)2CHNO2.Whichis the correct structure (justify your conclusion)? For this case, what should the relative heights of the 12 peaks be?

11. A compound with the chemical formula C4H8O2 has a proton NMR spectra that consists of a single peak at δ =3.56. Deduce its structure, and explain your reasoning.

12. A second compound with the chemical formula C4H8O2 has proton NMR spectra that consists of a quartet at 4.93 ppm, a singlet at 1.93 ppm and a triplet at 1.21 ppm, while the overall areas of the singlet and the triplet group are the same. Deduce its structure, and explain your reasoning.

13. Predict the 1H NMR spectrum at 200 MHz of a compound containing three diﬀerent protons (called an “AMX” spin system) with the following chemical shifts: δA = 4.95 ppm, δM = 6.0 ppm and δX = 7.0 ppm. Draw the spectrum to scale, including the relative intensities, indicating the positions of the peaks in both ppm and Hz from TMS (whose peak is placed at 0 ppm).

14. The previous question neglected any J coupling between the protons. Predict the AMX spectrum under

the same conditions as those given above, with the added information that JAM = 3.5 Hz, JAX =6 Hz and JMX = 1.5 Hz. Draw the spectrum to scale, including the relative intensities, and indicating the positions of the peaks in both ppm and Hz from TMS.

15. Using the data from the previous two questions, predict the proton NMR spectrum of the A2MX spin system, where there are two equivalent nuclei at site A instead of one. Draw the spectrum to scale, including the relative intensities, and indicating the positions of the peaks in both ppm and Hz from TMS.

16. Describe the splitting patterns of the proton NMR spectra for each of the following: CH3-CHO; CH3-CHCl-CH3;CH3-CH2-O-CH2-CH3.

17. Benzene and acetone have proton chemical shifts of 7.37 and 2.17 ppm, respectively. What is the frequency difference between these two proton signals at 2.35 T and 11.7 T? What is the actual difference in magnetic field strength experienced by these two types of protons at each of these magnetic fields? 7.5. PROBLEMS 127

18. Use the Boltzmann factors described in Chapter 2 (Rotational Spectroscopy) to determine the ratio of

the populations of the two nuclear spin energy levels (mI = + 1/2 and - 1/2) at 300 K and magnetic ﬁeld strengths equivalent to Larmor frequencies of 100 MHz and 500 MHz.

19. In the context of your answers to the previous two questions, identify why it is an advantage to perform NMR experiments at the highest magnetic ﬁeld strengths possible.