a spectroscopy primer

A Spectroscopy Primer

An Introduction to Atomic, Rotational, Vibrational, Raman,

Electronic, Photoelectron and NMR Spectroscopy

by

Robert J. Le Roy

Department of Chemistry, University of Waterloo

Waterloo, Ontario N2L 3G1, Canada

c© Robert J. Le Roy, 2003-2011

i

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

1.1 Light and the Electromagnetic Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Physics in 1900 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.2 Wave Properties of Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.3 The Quantum Theory of Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.4 A Brief Note on Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Quantum Theory of Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.1 The Spectrum of the Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.2 The Bohr Theory of the Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2.3 de Broglie Wavelengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3 Wave Mechanics and the Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3.1 A Particle in a One-Dimensional Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3.2 Orbital or Rotational Motion: A Particle on a Ring . . . . . . . . . . . . . . . . . . . 17

1.4 Electronic Structure of Atoms and Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.4.1 Hydrogenic Atomic Orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.4.2 Multi-Electron Atoms and Atomic Spectroscopy . . . . . . . . . . . . . . . . . . . . . 21

1.4.3 Molecular Energies and the Born-Oppenheimer Approximation . . . . . . . . . . . . . 22

1.5 Spectroscopy at last . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2 Rotational Spectroscopy 27

2.1 Classical Description of Molecular Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.1 Why Does Light Cause Rotational Transitions? . . . . . . . . . . . . . . . . . . . . . . 27

2.1.2 Relative Motion and the Reduced Mass . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.1.3 Motion of a Rotating Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.2 Quantum Mechanics of Molecular Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2.1 The Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2.2 Energy Levels, Selection Rules, and Transition Energies . . . . . . . . . . . . . . . . . 32

2.2.3 Illustrative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.3 Complications ! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.4 Degeneracies and Intensities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.5 Rotational Spectra of Polyatomic Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

v

vi CONTENTS

2.5.1 Linear Molecules are (Relatively) Easy to Treat! . . . . . . . . . . . . . . . . . . . . . 42

2.5.2 Illustrative Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.5.3 Non-Linear Polyatomic Molecules are More Difficult . . . . . . . . . . . . . . . . . . . . . 45

2.6 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3 Vibrational Spectroscopy 51

3.1 Classical Description of Molecular Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.1.1 Why Does Light Cause Vibrational Transitions? . . . . . . . . . . . . . . . . . . . . . 51

3.1.2 The Centre of Mass and Relative Motion . . . . . . . . . . . . . . . . . . . . . . . . . 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 Eigenvalues and Properties of the Morse Potential . . . . . . . . . . . . . . . . . . . . 57

3.3.2 Overtones and Hot Bands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.3.3 Higher-Order Anharmonicity and the Dunham Expansion . . . . . . . . . . . . . . . . 60

3.4 Dissociation Energies and Birge-Sponer Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.5 Vibrations in Polyatomic Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.6 Rotational Structure in Vibrational Spectra of Diatomics . . . . . . . . . . . . . . . . . . . . 67

3.7 Why Are Vibrational Level Spacings so Large? . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

4 Raman Spectroscopy 75

4.1 “Light-As-A-Wave” Description of Raman Scattering . . . . . . . . . . . . . . . . . . . . . . . 75

4.2 “Light-As-A-Particle” Description of Raman Scattering . . . . . . . . . . . . . . . . . . . . . 78

4.3 Rotational Raman Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.4 Vibrational Raman Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

4.5 Raman Spectra of Polyatomic Molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

4.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

5 Electronic Spectroscopy 83

5.1 Why Does Light Cause Electronic Transitions? . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.2 Vibrational-Rotational Structure in Electronic Spectra . . . . . . . . . . . . . . . . . . . . . . 84

5.3 Vibrational Propensity Rules in Electronic Transitions . . . . . . . . . . . . . . . . . . . . . . 88

5.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

6 Photoelectron Spectroscopy 95

6.1 Photoelectron Spectroscopy:

The Photoelectric Effect Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

6.2 Koopmans’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

6.3 Vibrational Fine Structure in Photoelectron Spectra . . . . . . . . . . . . . . . . . . . . . . . 97

6.4 Molecular Orbitals and Photoelectron Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6.5 Some Complications in Photoelectron Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

6.6 X-Ray Photoelectron Spectroscopy (XPS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

6.7 Auger Electron Spectroscopy (AES) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

6.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

7 NMR Spectroscopy 111

7.1 Basics of NMR Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

7.1.1 Angular Momentum and Nuclear Spin . . . . . . . . . . . . . . . . . . . . . . . . . . 111

7.1.2 Magnetic Moments and Nuclei in a Magnetic Field . . . . . . . . . . . . . . . . . . . 112

7.1.3 NMR Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

7.2 Chemical Shifts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

CONTENTS vii

7.2.1 Electronic Shielding of Nuclei and ‘Chemical Shifts’ . . . . . . . . . . . . . . . . . . . 117

7.2.2 What Determines Chemical Shifts, and The Chemical Shift Scale . . . . . . . . . . . 117

7.2.3 Working With Chemical Shifts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

7.3 Spin-Spin Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

7.3.1 Basics: Coupling from a Single Neighbour . . . . . . . . . . . . . . . . . . . . . . . . 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

7.4 Molecular Structures from NMR Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

7.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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. The experiment; B. The observations . . . . . . . . . . . . . . . 5

1.4 The hydrogen atom emission spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.5 Hydrogen atom energy levels and transitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.6 Properties of three square-well particle-in-a-box systems . . . . . . . . . . . . . . . . . . . . . 14

1.7 Level energies and wave functions for four particle-in-a-box systems . . . . . . . . . . . . . . . 16

1.8 Particle-on-a-ring orbits and wave functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.9 Definition of spherical polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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

1.12 Atomic orbital energies in some many-electron atoms . . . . . . . . . . . . . . . . . . . . . . . 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

2.1 Component of the dipole field of a rotating polar diatomic molecule . . . . . . . . . . . . . . 28

2.2 Centre of mass and relative coordinates for a two-body system . . . . . . . . . . . . . . . . . 28

2.3 Rotational energies and level spacings for a linear rigid rotor . . . . . . . . . . . . . . . . . . 32

2.4 Microwave absorption spectrum of CO gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.5 Microwave emission spectrum of gaseous HF . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.6 Microwave absorption spectrum of H–C≡C–C≡C–C≡N . . . . . . . . . . . . . . . . . . . . . 36

2.7 Graphical determination of B0 and D0 for CO . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.8 Angular momentum projections for J = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.9 Boltzmann rotational population distribution for CO at T = 293K . . . . . . . . . . . . . . . 41

2.10 Four types of linear molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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

3.5 Vibrational extrapolations and the Birge-Sponer plot . . . . . . . . . . . . . . . . . . . . . . . 63

3.6 Vibrational normal modes of: A. water H2O and B. acetylene C2H2 . . . . . . . . . . . . 65

3.7 Room temperature absorption spectrum of DCl . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.8 Spectrum and energy level pattern for an infrared band . . . . . . . . . . . . . . . . . . . . . 68

3.9 NaCl emission spectrum showing band heads . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

3.10 High temperature (1800K) 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 light in Raman scattering . . . . . . . . . . . . . . . . . . . . . . 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

5.2 Vibrational bands in the electronic spectrum of SrS . . . . . . . . . . . . . . . . . . . . . . . . 85

5.3 Rotational structure near the (0,0) band head in the A 1Σ+ −X 1Σ+ spectrum of CuD . . . . 87

5.4 Definition of the “stationary point” for a particular (v′, v′′) electronic transition. . . . . . . . 89

5.5 Potential curves and turning points for the Br2 B(3Π0+u)−X(1Σ+

g ) 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-

state potential energy functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 H+2 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

6.6 XPS spectra for C atoms in different molecular environments . . . . . . . . . . . . . . . . . . 106

6.7 Schematic level energy diagram illustrating XPS core-hole decay . . . . . . . . . . . . . . . . 107

6.8 He(I) photoelectron spectrum of CO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

7.1 Space quantization of angular momentum �P for R=32 . . . . . . . . . . . . . . . . . . . . . . . 112

7.2 Nuclear spin energy levels in a magnetic field for I = 12 . . . . . . . . . . . . . . . . . . . . . . 114

7.3 NMR spectra of various atomic nuclei in 250 MHz and 600 MHz spectrometers. . . . . . . . . 115

7.4 1H NMR spectrum of ethanol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

7.5 Chemical shifts of 1H and 13C nuclei in various environments . . . . . . . . . . . . . . . . . . 118

7.6 Energy level diagram for a proton without and with coupling to a neighbouring proton of a

different type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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

7.12 60 MHz NMR spectrum of an unknown organic compound . . . . . . . . . . . . . . . . . . . . 126

List of Tables

1.1 Conversion factors for energy units encountered in spectroscopy . . . . . . . . . . . . . . . . . 7

1.2 Wave functions of hydrogenic orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.1 Predicted and observed microwave spectrum of CO . . . . . . . . . . . . . . . . . . . . . . . . 34

2.2 Experimental microwave transition energies for ground state (v = 0) CO . . . . . . . . . . . . 37

2.3 Interstellar molecules detected by their rotational spectra . . . . . . . . . . . . . . . . . . . . 49

3.1 Characteristic group vibrational energies νi for common chemical function groups . . . . . . . 66

4.1 Labels for various types of rotational transitions . . . . . . . . . . . . . . . . . . . . . . . . . 79

6.1 Molecular parameters for some states of HCl, HCl+, N2 and N+2 . . . . . . . . . . . . . . . . 101

6.2 Molecular parameters for some states of O2 and O+2 . . . . . . . . . . . . . . . . . . . . . . . 103

6.3 Compare calculated orbital binding energies and experimental ionization energies for CO . . . 105

6.4 Ionization energies of the 1s electrons of the first-row elements . . . . . . . . . . . . . . . . . 105

6.5 Nitrogen 1s chemical shifts relative to the core ionization energy for N(1s) in gaseous N2 . . . 106

7.1 Table of the NMR properties of various nuclear isotopes . . . . . . . . . . . . . . . . . . . . . 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 m s−1

Planck Constant h = 6.626 068 96×10−34 joules s

� = h/2π = 1.054 571 628×10−34 joules s

Rydberg Constant R∞ = 2.179 871 97×10−18 joules = 109737.315685 27 cm−1

RH = 2.178 685 42×10−18 joules = 109677.58341 cm−1

Boltzmann Constant kB = 1.380 650 4×10−23 joules K−1 = 0.695 035 6 cm−1 K−1

Molar Gas Constant R = 8.314 472 joules mol−1 K−1

Avogadro Number NA = 6.022 141 79×1023Bohr Radius a0 = 5.291 772 0859×10−11 m = 0.529 177 208 59 A

Atomic mass unit 1 u = 1mu = 1.660 538 782×10−27 kg ≡ 112 m(12C)

Mass of the Proton mp = 1.672 621 637×10−27 kg = 1.007 276 466 77 u

Mass of Electron me = 9.109 382 15×10−31 kg = 5.485 799 0943×10−4 u

Fundamental Charge e = 1.602 176 487×10−19 coulombs

Inertial Constant Cu = 16.857 629 u cm−1 A2

Measures

ν frequency (hertz, Hz)

λ wavelength (metre, m)

ν wavenumber (cm−1)

E energy (J = joule = kg m2 s−2, eV, cm−1)

E energy in cm−1

F rotational energy in cm−1

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 = kgm s−2 or Jm−1); F in spectroscopists’ units (cm−1 A−1)

T temperature (Kelvin, K)

xiii

xiv LIST OF SYMBOLS

Quantum Numbers

n principal

� orbital angular momentum

m� magnetic

J rotational

v vibrational

S total electronic spin angular momentum

mS z-component of S

I total nuclear spin angular momentum

mI z-component of I

Molecular Parameters

V (r) molecular potential energy

De dissociation energy

re equilibrium bond length

k, k bond force constant in J/m2, cm−1/A2

μ reduced mass

β Morse potential exponent

I moment of inertia

Bv rotational constant in cm−1

Dv centrifugal distortion constant in cm−1

νe vibrational frequency in Hz

ωe vibrational frequency in wavenumbers, cm−1

ωexe vibrational anharmonicity, in cm−1

Be inertial rotational constant in wavenumbers, cm−1

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.

How Do We Do It? We measure the discrete frequencies and intensities of the electromagnetic radiation

that is absorbed, emitted or scattered by atoms or molecules.

Why Do We Do It? 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 ther-

modynamics, 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 = 10−9m)

• 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 = ν λ = 2.997 924 58×108 m s−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],

ν =1

102 λ=

ν

102 c, (1.2)

a factor of 102 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

0.0 0.5 1.0 1.5 2.0 2.5 3.0

field

viewed at

fixed point

in time

distance /μm

+E0

−E0

0

E(x,t)

wavelength λ

0 2 4 6 8 10

field

viewed at

fixed point

in space

time /10-15 s

+E0

−E0

0

E(x,t)

oscillationperiod 1/ν

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

E(x, t) = E0 cos

(2πν

cx − 2πν t + φ0

), (1.3)

in which t represents time and φ0 is a constant phase factor. For light of frequency ν = 3×1014 s−1 , 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 and ν = 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


1013 1014 1015

Intensity

Ι(ν,T) /

10-7 Jm-2

ν /Hz

0

10

20

visibleinfrared ultraviolet

T=300K(×104 )

T=6000K(×10 )

T=15000KRayleigh-Jeanslaw (15000K)

5

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

I(ν, T ) =2πkT

c2ν2 . (1.4)

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) = h ν = h c 102 ν (1.5)

in which h is a tiny scaling factor, that same derivation gave the distribution law

I(ν, T ) =2πhν3

c21

ehν/kT − 1. (1.6)

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 1Hz = 1 s−1, the units of h are more commonly expressed as joules×seconds, or J·s.


frequency ν [s-1]→

↑maximumelectronkineticenergy

0

-W0

B

ν0

e-

e-

e-metalcathode withvoltage=0

positiveanode tocollectfast e-

negativegrid formeasuringe- kineticenergy

A

incidentlight offrequency ν

emittedelectrons

A

current meter

Figure 1.3: The photoelectric effect: A. The experiment; 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 ex-

plained if one assumed that the energy associated with light of frequency ν was “quantized” in tiny bundles

of size ε(ν) = h ν , in which the constant h can be determined from the slope of the type of plot shown in

Fig. 1.3B. 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, by W.H. Cropper, Oxford University

Press, 1970.


that there exists a characteristic minimum energy required to tear an electron free from a particular metal.

This threshold energy

W0 = h ν0 (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. 1.3B

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) is in fact a

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 scat-

tered 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 re-

lationship, 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

ε(ν) (∼ mλ c2) = pλ c = h ν . (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.


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 = h ν = h c/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, fre-

quencies 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 pro-

portionality 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

1 joule (J) = 1 6.241 509 65×1018 5.034 117 47×1022 1.509 190 45×10331 eV = 1.602 176 487×10−19 1 8065.544 65 2.417 989 454×1014

1 cm−1 = 1.986 445 501×10−23 1.239 841 875×10−4 1 2.997 924 58×10101 Hz = 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= 10−9m) is commonly

used to characterize the wavelength of light, molecular dimensions are most commonly reported in units of

A ( 1 A= 10−10m), 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

to use atomic mass units, mu = 1u ≡ m(12C)/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/(2M d2) [J], in which � = 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], wheremu is the mass in kg of one atomic mass unit (see p. xiii). Similarly,


because we wish to quote distances in Angstroms, 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/(2M d2) becomes

(� [J · s])22(M [kg]) (d [m])2

=

{�2

2 (mu [kg])

1020

102 hc

}1

(M [u])(d [A]

)2 =Cu

(M [u])(d [A]

)2 =16.857 629

(M [u])(d [A]

)2in 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

expression. This conversion of �2/(2M d2

)[J] to Cu/ (M [u])

(d [A]

)2[cm−1] 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 10−10m = 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

∼ 10−14m, while the electrons move about it in orbits of diameter ∼ 10−10m 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

λ = A

(n1

2

n12 − 4

)in which n1 = 3, 4, 5, 6, . . . (1.10)

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):

ν = RH

(1

(2)2− 1

(n1)2

), (1.11)

in which (recall Eqs. (1.1) and (1.2)) ν is the wavenumber of the emitted spectral line, and the constant

RH = 109 677.583 41 cm−1 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

ν = RH

(1

(n2)2− 1

(n1)2

), (1.12)

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

⏐⏐⏐⏐⏐⏐⏐⏐⏐⏐⏐⏐⏐⏐⏐⏐⏐⏐

0 20000 40000 60000 80000 100000 120000

10020040010004000wavelength λ / nm

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

values n1 = n2 + 1, n2 + 2, n2 + 3, . . ., etc. Because the quantity 1/n2 decreases rapidly as n increases,{1

1,1

4,1

9,

1

16,

1

25,

1

36, . . .

}= {1, 0.25, 0.111111, 0.0625, 0.040, 0.027777, . . .}

each of these series converges to a characteristic limit ν∞ = RH/(n2)2. Except for the n2=2 series that was

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, where θ 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

expression (in cm−1):

En = −(

e4 μH

32π2 ε02 �2 102 hc

)1

n2= − RH

1

n2(1.13)


93.78nm

410.3nm

102.57

nm97.25nm

94.98nm

121.56

nm

1094.nm

486.2nm

434.2nm

656.5nm

1875.nm

1282.nm

0 1 2 3 4 5 6 7-120000

-100000

-80000

-60000

-40000

-20000

0energy/ cm-1

r / Å

Lyman

BalmerRitz-Paschen

Coulomb potential

n =1

n=2

n=3n=4

n=∞

− e 2V(r) = ⎯⎯⎯⎯⎯⎯

(4πε0 102 hc) r

Figure 1.5: Hydrogen atom energy levels and transitions.

in which e is the electron charge, ε0 is the permittivity of vacuum (a known constant which arises in

electrostatics), and μH = memp/(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 � , where n (= 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

ΔE(n1, n2) = En1 − En2 = RH

[1

(n2)2− 1

(n1)2

]= ν (1.14)

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

and n2 . The theory also predicts that the radius of the orbit associated with quantum number n is n2×a0,where a0 = 0.529 177 208 59 A is the radius of the Bohr orbit in the ground (n = 1 ) level of an 1H atom.

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

consisting of a nucleus of mass mAnuc and charge +Ze, the energy levels are given by the formula

EAn = − RA

1

n2= − Z2

(μA

μH

)RH

1

n2, (1.15)

in which μA = memAnuc/(me + mA

nuc) is the reduced mass of this system.7 This generalization allowed

Bohr to explain the differences between the transition energies of the 1H and 2H 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

μ7Li+2 = m7Li+3

nuc me/(m

7Li+3

nuc +me

)= 5.485 370 093×10−4 u ,

which is only 0.0466% larger than the value of μH=5.482 813 061×10−4 u, and only 0.0013% larger than

μ6Li+2=5.485 298 698×10−4 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”

λ = λp =h

mv≡ h

p(1.16)

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

7Ignoring the very small relativistic mass correction, mAnuc = MA − Zme , in which MA is the standard atomic mass of

atomic isotope A.


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 Schrodinger 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 Schrodinger developed and published in a series of papers in 1926.

Conceptually, mathematically, and in a sense physically, Schrodinger’s mechanics was altogether differ-

ent from Heisenberg’s mechanics. While Heisenberg’s method is algebraic, Schrodinger’s method begins

with a differential equation. Heisenberg’s method builds on discrete and discontinuous quantities, whereas

Schrodinger’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, Schrodinger’s differential

equation is very explicitly a “wave equation”.

As might be expected, Heisenberg and Schrodinger 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 Schrodinger’s theory,

the more disgusting it appears to me.”

Similarly, although he published a formal mathematical proof of the equivalence of the two theories later in

1926, Schrodinger 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 Schrodinger 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.

Schrodinger’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 Schrodinger’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), Schrodinger’s differential equation for the particle/wave’s

displacement y(x) has the form

− �2

2m

d2 y(x)

dx2+ V (x) y(x) = E y(x) . (1.17)

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 for 0 < x < L , but V (x) = ∞ for x ≤ 0 and x ≥ L . This physical system is

schematically illustrated by Fig. 1.6 (see p. 14) which shows three thick-walled boxes, each with a particle

bouncing back and forth from left to right. In this diagram, the vertical axis is energy, and the vertical thick

line segments indicate that at those positions the potential energy goes to infinity, so the moving particle

cannot escape, and is condemned forever to bounce back and forth, rebounding perfectly elastically every

time it hits the wall.

Inside the box where the potential energy V (x) is zero, the Schrodinger differential equation describing

the particle becomes simply

d2 y

dx2= −

(2mE

�2

)y or

d2 y

dx2= − k2 y , (1.18)

in which k2 = 2mE/�2 is a constant. This differential equation is one of the simplest one can encounter;

its solution is the function y(x) whose second derivative is minus a constant (k2) times the function itself.

Two well known functions possessing this property are

y(x) = sin(k x) and y(x) = cos(k x) . (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 Schrodinger’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 Schrodinger’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 Schrodinger’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.


0 0 0L=1.0 L=1.5 L=1.0x→ x→x→

energy

m= 2.25m=1.0 m= 1.0

n= 4

n= 3

n= 2

n=1

n⎯ 7

6

5

4

3

21

↑V(x)

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) and cos(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 < 0 and x > 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) +B cos(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(k x) . Applying the second boundary

condition in Eq. (1.22) then yields the condition

ψ(x = L) = A sin(k L) = 0 . (1.24)

Our knowledge of trigonometric functions tells us that Eq. (1.24) is satisfied whenever k L = nπ where

n is an integer, or if A = 0 . However, if either n = 0 or A = 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 Schrodinger equation for this

system which satisfy our four Rules of wave mechanics are those associated with discrete values of the


constant k corresponding to positive integer values of n :8

k =

√2mE

�2=

nπ

Lfor n = 1, 2, 3, 4, . . . (1.25)

Rearranging this expression shows that these allowed solutions can only occur if the system energy has a

discrete value given by the equation

E = En =

(�2

2m

π2

L2

)n2 (1.26)

In “spectroscopists’ units”, with mass in u, length in Angstroms and energy in cm−1, this yields the expression

that we commonly use for calculations:

En = Cu

(π2

m[u] (L[A])2

)n2 [cm−1] (1.27)

in which Cu = 16.857 629 056 [u cm−1 A2] 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 Schrodinger 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 =(π2 �2/2mL2

), 4

(π2 �2/2mL2

), 9

(π2 �2/2mL2

), 16

(π2 �2/2mL2

), . . . etc., and the associated (normal-

ized) 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 Schrodinger 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.

[For this particular ‘particle-in-a-box’ case of a square well potential: En ∝ 1/L2.]

• 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, as in eigenvalue, 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.


n=1

n=2 n=3

n=4r→

−RH

0

− e2V(r) = ⎯⎯⎯

4πε0 r

r→

De→

re

υ=5

υ=0

υ=1

υ=2

υ=4

υ=3

B

D

0 Lx→

n=3

n=2n=1

n=4

n=5

n=6

V(x)∞ ∞

A

υ=5

V(r) = ½ k (r−re)2

re r→

υ=0

υ=1

υ=2

υ=3

υ=4

C

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 Schrodinger 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,


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. 1.7B), 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 Schrodinger equation

for this model problem has the form

− �2

2mr2d2 ψ(φ)

dφ2= E ψ(φ)

ord2 ψ(φ)

dφ2= −

(2mr2E

�2

)ψ(φ) = − b2 ψ(φ) (1.28)

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) for all 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:

E� =�2

2mr2�2 =

(�2

2m

π2

(π r)2

)�2 (1.29)

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 A and

masses in u and to generate energies in cm−1, in which case Eq. (1.29) becomes

E� =

(Cu

m[u](r[A]

)2)�2 [cm−1] (1.30)

in which Cu = 16.857 629 056 [u cm−1 A2] 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 = π r . One important difference, however,


l=2l=0

l=3

l=1

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 if ψ(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 Schrodinger equation. However, note that the time-independent Schrodinger equation for a

general system is written symbolically as

Hψ = E ψ (1.31)

in which H is the Hamiltonian operator for the system, a generalization to three dimensions (and multiple

particles, if appropriate) of the operator H1D = − �2

2md2

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 Schrodinger 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 Schrodinger 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 = memp/(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

x

y

z

+

r

→

θ

φ

→

→

→

(x,y,z)(r,θ,φ) x= r

y= rz= r

sinθ cosφsinθ sinφcosθ

Figure 1.9: Definition of spherical po-

lar coordinates in terms of rectangular

Cartesian coordinates.

theory”, Eqs. (1.13) and (1.15), and the associated wave functions

define the familiar hydrogenic atomic orbitals presented in intro-

ductory Chemistry courses. As the electron in the atom is free

to move in three dimensions, it should be no surprise that these

solutions are characterized by three quantum numbers, n, � and

m�. For the generalized case of an electron orbiting around a bare

nucleus with charge +Ze, where e is the magnitude of the elec-

tron charge (i.e., for the one-electron atom or ions H, He+1, Li+2,

Be+3, . . . etc.), the eigenfunctions of the lowest few levels are listed

in Table 1.2.

It is easy to see that each of these hydrogenic functions11 may

be written as the product of a function of r times a function of θ

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 hydro-

genic 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 wavefunc-

tions 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

−C1/r, with C1 = Ze2/4πε0 , 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, 2H or D, 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, 2 and 3. Here

Z is the atomic number of the nucleus, and ao = 0.519 177 2083 A is the Bohr radius.

n = 1, � = 0, m� = 0; ψ1s =1√π

(Zao

)3/2

e−Zr/ao

n = 2, � = 0, m� = 0; ψ2s =1

4√2π

(Zao

)3/2 (2− Zr

ao

)e−Zr/2ao

� = 1, m� = 0; ψ2pz = 14√2π

(Zao

)3/2 (Zrao

)e−Zr/2ao cos θ

� = 1, m� = ±1; ψ2px = 14√2π

(Zao

)3/2 (Zrao

)e−Zr/2ao sin θ cosφ

ψ2py = 14√2π

(Zao

)3/2 (Zrao

)e−Zr/2ao sin θ sinφ

n = 3, � = 0, m� = 0; ψ3s =1

81√3π

(Zao

)3/2 (27− 18Zr

ao+ 2Z2r2

a2o

)e−Zr/3ao

� = 1, m� = 0; ψ3pz =√2

81√π

(Zao

)3/2 (6Zrao− Z2r2

a2o

)e−Zr/3ao cos θ

� = 1, m� = ±1; ψ3px =√2

81√π

(Zao

)3/2 (6Zrao− Z2r2

a2o

)e−Zr/3ao sin θ cosφ

ψ3py =√2

81√π

(Zao

)3/2 (6Zrao− Z2r2

a2o

)e−Zr/3ao sin θ sinφ

� = 2, m� = 0; ψ3dz2= 1

81√6π

(Zao

)3/2 (Z2r2

a2o

)e−Zr/3ao

(3 cos2 θ − 1

)� = 2, m� = ±1; ψ3dxz =

√2

81√π

(Zao

)3/2 (Z2r2

a2o

)e−Zr/3ao sin θ cos θ cosφ

ψ3dyz =√2

81√π

(Zao

)3/2 (Z2r2

a2o

)e−Zr/3ao sin θ cos θ sinφ

� = 2, m� = ±2; ψ3dx2−y2= 1

81√2π

(Zao

)3/2 (Z2r2

a2o

)e−Zr/3ao sin2 θ cos 2φ

ψ3dxy = 181

√2π

(Zao

)3/2 (Z2r2

a2o

)e−Zr/3ao sin2 θ sin 2φ


r R3d(r)

r R2p(r)

r R3p(r)

r R1s(r)2

10

3

r R2s(r)

2

1

0

-1

r R3s(r)1

0

-1

0 5 10 15 20

r R4s(r)1

0

-1

r /Å0 5 10 15 20

r R4p(r)

r /Å0 5 10 15 20

r R4f (r)

r R4d(r)

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 r Rn�(r) . Thus, the radial part

of the wavefunction for an electron in a hydrogenic atom is effectively a solution of the one-dimensional

Schrodinger 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 r Rn�(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

Figure 1.11: Angular behaviour of H atom

orbitals for n = 1− 3 .

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 os-

cillatory behaviour of the type commonly associated with

classical wave motion. Because it is an ordinary mathemati-

cal function, it can have positive and negative values, as seen


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 molecu-

lar 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 Schrodinger 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 Schrodinger 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 Schrodinger 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


(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 nu-

cleus, 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 approx-

imation 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 ≈ 1800me ), 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 Schrodinger 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 Schrodinger 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 Schrodinger 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


2 4 6 80

10000

20000

30000

r / Å

energy/ cm-1

Li2

Li(2S)+Li(2S)

Li(2S) +Li(2P)

D

re

e

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 Schrodinger equation vary with the

internuclear distance r for all molecular states that correlate with the two lowest Li2 dissociation limits.12 In

particular, at each selected internuclear distance r, the electronic Schrodinger equation was solved numerically

to determine the electronic energy levels shown by the stacks of square points at that value of r. One can

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 Schrodinger 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 Schrodinger equation for vibrational motion that for a diatomic molecule has

the familiar particle-in-a-box form

− �2

2μ

d2 ψ(r)

dr2+ V (r)ψ(r) = E ψ(r) (1.35)

in which μ = mAmB/ (mA +mB) is the effective or vibrational “reduced mass” for a diatomic molecule

formed from atoms A and B with massesmA andmB, respectively. This equation clearly has exactly the same

12 I. Schmidt-Mink, W. Muller and W. Meyer, Chemical Physics 92, 263 (1985).


Figure 1.14: Regions of the electromagnetic spectrum and the type of molecular motion and spectroscopy

associated with each.

form as our generic Schrodinger 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 Schrodinger 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 Schrodinger equation,

Htotal ψtotal = Etotal ψtotal (1.36)

in which Htotal incorporates all interactions within the system, Etotal is one of the observable energy eigen-

values 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 Schrodinger equation si-

multaneously 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 char-

acteristic 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 4.0×10−10m.

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 200nm? 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

kJ·mol−1 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 m s−1, what was

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)


10. The Brackett series of H-atom emission lines consists of a series of lines corresponding to transitions

from the n1 > 4 to the n2 = 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 transi-

tions 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 as νΔn(nlower) ,

where Δn = nupper − nlower .

(b) If the 7Li atom were replaced by a 6Li atom, what would be the “isotope shift” Δν(n) = ν(6Li)−ν(7Li) for each of the transitions considered in part (a)?

(c) For the 6Li atom in this hard-wall box, what would be the wavenumbers of the ν2(3) and ν3(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 elec-

tronic 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, at http://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.

How Do We Do It? Transitions are observed by measuring the frequency and amount of microwave

radiation that is absorbed or emitted by rotating molecules in the gas phase.

Why Do We Do It? 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

↑vertical

componentof

moleculardipole

time→

+

+

+

−+

+ +

− −

−− −

directionof

dipole

orientationof

molecule

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

electromagnetic spectrum.1 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

x

y

z+ m2

m1

r1

r2

R→

→

→

→r

cm

Figure 2.2: Position �Rcm of the cen-

tre of mass “+”, and relative coordi-

nates r for a two-body system.

Consider a system consisting of two particles of mass m1 and m2

located at positions specified by vectors �r1 and �r2, respectively, as il-

lustrated in Fig. 2.2. In studying molecules we are normally interested

in the motion of the atoms relative to one another, and not in their

absolute positions in space. It is therefore convenient to replace these

individual-particle coordinates �r1 and �r2 with coordinates specifying

the relative position of the two nuclei �r = �r2 − �r1 and the position�Rcm of the centre of mass. The centre of mass of a system (indicated

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

simply m1 +m2) located at �Rcm :

(m1 +m2) �Rcm = m1 �r1 +m2 �r2 (2.1)

Combining our definitions of �r and �Rcm, we may then express the

one-particle positions �r1 and �r2 in terms of the centre of mass and relative coordinates �Rcm and �r :

�r1 = �Rcm −(

m2

m1 +m2

)�r and �r2 = �Rcm +

(m1

m1 +m2

)�r (2.2)

Using the usual equations of classical mechanics, we know that the total kinetic energy of our two-particle

system may be written as

KEtot =1

2m1|�v1|2 + 1

2m2|�v2|2 =

1

2m1|�r1|2 + 1

2m2|�r2|2 (2.3)

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 = �ri = d�ri/dt is the velocity of particle–i. Substitution of Eq. (2.2) into Eq. (2.3) yields the result

that

KEtot =1

2m1

∣∣∣∣�Rcm − m2

m1 +m2

�r

∣∣∣∣2 + 1

2m2

∣∣∣∣�Rcm +m1

m1 +m2

�r

∣∣∣∣2=

1

2m1

{∣∣∣�Rcm

∣∣∣2 − 2m2

m1 +m2

�Rcm · �r +

(m2

m1 +m2

)2 ∣∣∣�r∣∣∣2}

+1

2m2

{∣∣∣�Rcm

∣∣∣2 + 2m1

m1 +m2

�Rcm · �r +

(m1

m1 +m2

)2 ∣∣∣�r∣∣∣2}

=1

2(m1 +m2)

∣∣∣�Rcm

∣∣∣2 + 1

2

(m1m2

m1 +m2

) ∣∣∣�r∣∣∣2= 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 �Rcm, 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 m1m2/(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

reduced mass:μ =

m1m2

m1 +m2=

(1

m1+

1

m2

)−1

. (2.5)

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 ; it may be used only 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 cosφ− φ sin θ sinφ

)y = re

(θ cos θ sinφ+ φ sin θ cosφ

)z = −re θ sin θ .

The internal or rotational kinetic energy may then be written as

KErot = 12 μ

∣∣∣�r∣∣∣2 = 12μ

(x2 + y2 + z2

)= . . . . . .

= 12 μ (re)

2(θ2 + φ2 sin2 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.


If we assume, for convenience, that the rotation occurs in the x-y plane, then θ is fixed at θ = 90◦ , so θ = 0

and sin θ = 1 . Moreover, since the constant factor μ (re)2 is a property characteristic of this rigid rotating

system, it is convenient to give it a special name and symbol: the symbol is I = μ (re)2 and the name is the

moment of inertia. In this notation we can therefore write

KErot = 12 I φ

2 =(I φ)2

2 I=|�L|22 I

. (2.7)

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 × (m�r)| = |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

analogous to the familiar classical expression 12mv2 = p2/2m for the translational kinetic energy of a particle

with mass m, speed v, and linear momentum p = mv . The moment of inertia I 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 analogof 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

Erot = KErot =L2

2 I, (2.9)

in which L ≡ |�L| is the magnitude of the orbital angular momentum, also holds for anymulti-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∑i=1

mi

)�Rcm =

N∑i=1

mi �ri (2.10)

and the moment of inertia about an axis through that centre of mass is given by the equation

I =N∑i=1

mi (d⊥i )

2 (2.11)

in which d⊥i is the perpendicular distance from particle–i to that specific axis of rotation. For a linear

molecule, d⊥i 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

moment of inertia reduces to the simple expression I = Id = μ (re)2 , where μ = m1m2/(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 Schrodinger

equation gives the allowed energies as

Erot =(� �)2

2μ(re)2=

(� �)2

2 Id(2.12)

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 dis-

cussion 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

two types of motion are very similar. Note too that since√J(J + 1) = J

√(1 + 1/J)

large 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 me-

chanical expression for rotational energy of a rigid molecular system:

Erot(J) =�2

2I[J(J + 1)] [J] . (2.14)

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:

F (J) ≡ Erot(J) = Erot(J)/(102 hc

)= B [J(J + 1)] [cm−1] , (2.15)

where as before, J = 0, 1, 2, 3, . . . , etc. Since I has units (mass)×(distance)2 , we can write

B =�2

2I

1020

102 hc=

Cu

I [u A2]

[cm−1] , (2.16)

where as usual Cu=16.857 629 [u cm−1 A2] , and I = μ(re)

2 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. Its value (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”.


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.

J= 0

J= 2

J= 3

J= 4

J= 5

J= 6

J= 7

J= 102B

6B

12B

20B

30B

42B

56B

ν=14B

ν=12B

ν=10B

ν=8B

ν=6B

ν=4B∼

∼

∼

∼

∼

∼rotationalenergy→

Figure 2.3: Rotational energies and

level spacings for a linear rigid rotor.

As discussed in Chapter 1, a photon is a particle with zero rest

mass and a momentum of pλ = h/λ = 102 hν . However, an impor-

tant additional property is that it has an intrinsic angular momentum

(the property we call spin4) of exactly 1, in the usual angular mo-

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

two cars colliding while sliding and spinning on a frictionless sheet of

ice). Since the total angular momentum of a system (e.g., molecule

plus photon) must always be conserved, whenever a molecule ab-

sorbs or emits a photon of light, the rotational quantum number J

must change by ΔJ = Jupper− Jlower = ±1 , where Jupper and Jlower

are the values of the rotational quantum number for the upper and

lower levels of the transition. The ± sign appears here because angu-

lar momentum is a vector property, and the result of vector addition

depends on the relative orientations of the two vectors: in this case,

quantum mechanics allows only these two choices for the relative

alignment.

The essential implication of the above selection rule is that only

rotational transitions between adjacent J levels are allowed. For our

rigid-rotor system this means that the allowed transition energies are

νrotJ = ΔF (J) = F (J + 1)− F (J)= B [(J + 1)(J + 2)− J(J + 1)] = 2B (J + 1) (2.17)

It is interesting to note that the frequency of the light absorbed or

emitted in such a transition, νrotJ ≈ J�/(2πI) ≈ L/(2πI) [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).

As illustrated in Fig. 2.3, the allowed transition energies for a linear rigid rotor are νrotJ = 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

ΔνrotJ ≡ νrotJ − νrotJ−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

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

It is then straightforward to calculate

μ =m1 m2

m1 +m2=

12.000 000 × 15.994 915

12.000 000 + 15.994 915u = 6.856 209 u ,

and hence to obtain B as

B =Cu

μ [u](re [A]

)2 =16.857 629

6.856 209× (1.128 322)2cm−1 = 1.931 285 cm−1 .

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 frequencies νJ = 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, at http://www.hbcpnetbase.com.7Recall that 1 u = 1 atomic mass unit = 1.660 538 782× 10−27 kg.


Table 2.1: Predicted and observed microwave spectrum of CO.

J J(J + 1) F (J) = B[J(J + 1)] νrotJ = 2B(J + 1) νrot

J (obs)

0 0 0.03.845 034 3.845 033

1 2 3.845 0347.690 068 7.689 919

2 6 11.535 10211.535 102 11.534 510

3 12 23.070 20415.380 136 15.378 662

4 20 38.450 34019.225 170 19.222 223

5 30 57.675 510...

......

......

The above calculation seems very straightforward, and the predicted J = 1 ← 0 transition energy seen in

Table 2.1 is in very good agreement with the best experimental value νrotJ (pbs) = 3.845 0335 cm

−1. However,

we see in Table 2.1 that the agreement becomes systematically worse with increasing J , and for the J = 5← 4

transition the discrepancy νcalc − νobs = 0.002 947 cm−1 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 of ν = 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, so we obtain

B = 12[163.936 16− 123.129 67] = 20.403 245 cm−1 .

To determine the HF bond length, we then simply re-arrange Eq. (2.16) while utilizing the special expression

for the moment of inertia of a diatomic molecule (I = μre2),

I = Cu/B = 0.826 222 94 [u A2] = μ (re)2 .

The known atomic masses6 yield a reduced mass of

μ(HF) =

(1

1.007 825 032+

1

18.998 403 20

)−1

= 0.957 055 278 u ,

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

J + 1 = νrotJ /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 and 1.

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 18O or 12C 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


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

Bv =Cu

μ(rv)2(2.20)

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 692.481cm−1 for v = 0 to 665.937cm−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

rotational lines is approximately 2.54GHz or 0.0847cm−1, which implies that the B value is 0.04236 cm−1,

and hence that the moment of inertia is I(HC5N) = Cu/0.04236 = 390 u A2.8 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

to increase. This in turn leads to an increase in the moment of inertia I = I(J) = μ (rv(J))2, and since the

spacing between rotational lines is 2Bv(J) = 2(Cu/I) , at higher values of J the rotational line spacings will

become progressively smaller. This behaviour is evident in Fig. 2.5, where the νrotJ=14 − νrotJ=13 line spacing

of 36.0 cm−1 is visibly larger than the νrot25 − νrot24 line separation of 26.4 cm−1, and it explains why these

rotational transition energies (e.g., νrot24 = 904.385 cm−1) are not simple 2(J+1) multiples of the rotational

constant B = B0 = 20.403 cm−1 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

Fv(J) = Bv[J(J + 1)]−Dv [J(J + 1)]2, (2.21)

where as usual Bv ≡ Cu/Iv(J=0) , and 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

νrotJ = ΔFv(J) ≡ F (J + 1)− F (J)=

{Bv[(J + 1)(J + 2)]−Dv[(J + 1)(J + 2)]2

}− {Bv[J(J + 1)]−Dv[J(J + 1)]2

}= 2Bv(J + 1)− 4Dv(J + 1)3 (2.22)

and the rotational line spacings

ΔνrotJ = νrotJ − νrotJ−1 = 2Bv − 4Dv(3J2 + 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

listed in Table 2.2. The fact that the line spacings ΔνrotJ 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.

J transition νrotJ ΔνrotJ = νrotJ − νJ−1 (3J2 + 3J + 1)

0 1← 0 3.845 033 — —

1 2← 1 7.689 919 3.844 886 7

2 3← 2 11.534 510 3.844 591 19

3 4← 3 15.378 662 3.844 152 37

4 5← 4 19.222 223 3.843 561 61

5 6← 5 23.065 043 3.842 820 91


intercept =2B0 = 3.845059

slope = −4D0 = −2.459×10−5

0 25 50 75 100

3.843

3.844

3.845

(J2+3J+1)

ΔνJ/ cm−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

νrot0 = 2B0(J + 1)− 4D0(J + 1)3

= 2B0 − 4D0 = 3.845 033

or D0 = [2B0 − 3.845 033] /4 (2.24)

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

νrot5 = 12B0 − 864D0 = 23.065 043

Using Eq. (2.24) for D0, we obtain

νrot5 = 23.065043 = 12B0 − 864 [2B0 − 3.845 033] /4 (2.25)

Solving Eq. (2.25) for B0 and substituting the result into Eq. (2.24) then yields B0 = 1.922 529 cm−1 and

D0 = 6.138 × 10−6 cm−1.

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

of ΔνrotJ vs. (3J2 + 3J + 1) should be linear, with intercept 2Bv and slope −4Dv . As shown in Fig. 2.7, this

approach yields the same B0 value, but a D0 value of 6.147×10−6 cm−1 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

Beffv (J) ≡ dE(v, J)

d[J(J + 1)]= Bv − 2Dv[J(J + 1)] ≡ Cu

μ [rv(J)]2(2.26)

Rearranging this result to solve for rv(J), and then multiplying the numerator and denominator by rv(J=0)

yields the expression

rv(J) = rv(J=0)

(Cu/μ [rv(J=0)]

2

Bv − 2Dv[J(J + 1)]

)1/2

(2.27)

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

rv(J) = rv(J = 0)

/(1− 2[J(J + 1)]

Dv

Bv

)1/2

(2.28)

For the ground vibrational level of CO, the results in Fig. 2.7 show us that D0/B0 = 3.30×10−6 , so it will

require a relatively large value of J (≈ 55) to cause even a 1% increase in the CO bond length. In contrast,

for the ground vibrational level of HF the ratio D0/B0 = 1.0×10−4, so 1% stretching occurs for J=9 and

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

Fv(J) = Bv[J(J + 1)]−Dv [J(J + 1)]2 +Hv [J(J + 1)]3 + Lv [J(J + 1)]4 + . . . (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.


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 − 1 and J . 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, in that for a given J , 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-

0h

1h

Lz= 2h

-1h

-2h

−

−

−

−

−

z

L→

Figure 2.8: Angular momen-

tum projections for J=2 .

entum quantum number J that defines the magnitude of |�L|, and the quan-

tum numberMJ 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

(or space-fixed) z axis. For present purposes, we are only concerned with

knowing that the number of degenerate sublevels associated with any given

value of J is gJ=2J+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.

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

only difference is that in discussing the electron orbits, the total angular

momentum quantum number is given the label � (instead of J), and the

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), and the seven f (for �=3) degenerate

substates.

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

particular quantum state i with energy Ei is proportional to e−Ei/kB T ,

in which kB = 1.380 650 4×10−23 [JK−1] = 0.695 035 6 [cm−1 K−1

] is known as the Boltzmann constant.10 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

particular energy level Ei is proportional to gi e−Ei/kB T ,

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

fi(T ) = gi e−Ei/kB T

/Q(T ) , (2.30)

in which the quantity Q(T ) =∑

i gi e−Ei/kB T , 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”

R = 8.314 472 Jmol−1 K−1 = NA kB , where NA = 6.022 141 79× 1023 is the Avogadro number.

2.4. DEGENERACIES AND INTENSITIES 41

B

Bυ

υ

gJ e−B J(J+1)/k T

0 10 200

5

10

J

popmax

gJ= (2J+1)J (T)

10× e−B J(J+1)/k T

Figure 2.9: Boltzmann rotational population distribution for CO at T = 293K.

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=2J+1 . The fraction of molecules in the level

with energy Fv(J) is therefore given by

fJ(T ) = (2J + 1) e−Fv(J)/kBT/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 , and the term

e−BvJ(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 ?

Calculus tells us that the location of this maximum will be the value of J , denoted Jpopmax(T ), for which the

derivative of fJ(T ) with respect to J is zero. Hence, for a rigid rotor:

d

dJ

{(2J + 1) e−Bv J(J+1)/kBT

}= 0 = e−Bv(J

2+J)/kBT

{2 + (2J + 1)

[−Bv

kBT(2J + 1)

]}(2.32)

Removing the exponential common factor and solving for J = Jpopmax(T ) yields the expression

J = Jpopmax(T ) =

√kBT

2Bv− 1

2(2.33)

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 energy ν, while the


full emission intensity expression includes a factor of ν4. 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

J absmax(T ) ≈

√3 Jpop

max(T ) and J emmax(T ) ≈

√5 Jpop

max(T ) (2.34)

2.5 Rotational Spectra of Polyatomic Molecules

2.5.1 Linear Molecules are (Relatively) Easy to Treat!

m1 m2

m2 m1 m2

m1 m2 m2 m1

m1 m2 m3

d

d12 d12

d12 d22 d12

d12 d23

A

B

C

D

Figure 2.10: Four types of linear

molecules.

For a general polyatomic molecule consisting of N atoms distributed

in space, introducing centre-of-mass and relative coordinates in the

same manner as in our diatomic molecule derivation in §2.1.2 yields

the same separation of the total energy into the sum of the kinetic

energy of the centre of mass plus the internal energy. That internal

energy in turn separates into a sum of the internal vibrational energy

plus the overall rotational energy of the molecule. For any linear

molecule, that rotational energy is also given by Eq. (2.9), where

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

HC5N of Fig. 2.6) is identical to that for a diatomic molecule. In

fact, the only difference between the treatment of a rigid diatomic

molecule and a rigid linear polyatomic is in the way we describe the

moment of inertia I.

As mentioned in §2.1.3, for an arbitrary multi-particle system

the position of the centre of mass is given by Eq. (2.10) and the

moment of inertia by Eq. (2.11). For the particular case of a diatomic

molecule, the latter happens to simplify to the expression Id=μ(re)2

used in the early parts of this Chapter, but that expression cannot

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 �Rcm = (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: A Diatomic Molecule.

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):

I = m1 (−xcm)2 +m2 (d− xcm)2

= m1m2

2 d2

(m1 +m2)2+m2

m12 d2

(m1 +m2)2

=m1m2

m1 +m2d2 = μ d2 = Id . (2.35)

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

I = m2(−d12)2 +m1(0)2 +m2(d12)

2 = 2m2 (d12)2 . (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)

2(m1 +m2)= d12 +

12 d22 , (2.38)

which confirms our intuitive prediction. Combining this result with our formal definition of the moment

of inertia then yields

I = m1(−d12 − 12 d22)

2 +m2(− 12 d22)

2 +m2(12 d22)

2 +m1(d12 +12 d22)

2

= 2m1 (d12 +12 d22)

2 + 2m2(12 d22)

2 . (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 , atom m2 at x2 = (d12 − xcm) , andatom m3 at x3 = (d12 + d23 − xcm) . Our definition of the moment of inertia then yields

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

2 +m3(d23)2 −

[(m1 d12 −m3 d23)

2 / (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


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 sim-

pler 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 A and rC≡N=1.159 A. Assuming that these bond lengths are essentially unchanged in cyanodi-

acetylene, if we choose the origin of our coordinate system to lie at the H atom, the position of the centre of

mass is (using the masses of 1H, 12C and 14N for mH, 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 + 2 rC≡C + rC−C) +mC (rC−H + 2 rC≡C + 2 rC−C)

+ mN (rC−H + 2 rC≡C + 2 rC−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

d⊥i to the centre of mass, we can readily calculate

I(HC5N) = 1.007 825(−4.2630)2 + 12.000 000[(−3.2050)2 + (−2.0000)2 + (−0.6220)2

+(0.5830)2 + (1.9610)2]+ 14.003 074(3.1200)2

= 380.759 [u A2] .

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

B(HC5N) ≈ Cu/I = 0.04427 [cm−1]

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 14N by 15N (0.37%

abundance) or on replacing one of the 12C atoms by 13C (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:

B0(16O12C32S) = 0.202 864 [cm−1]

B0(16O12C34S) = 0.197 910 [cm−1] .

2.5. ROTATIONAL SPECTRA OF POLYATOMIC MOLECULES 45

We know that for any molecule, Bv = Cu/I , so the two moments of inertia are

Ia ≡ I(16O12C32S) = Cu/0.202 864 = 83.098 18 [u A2]

Ib ≡ I(16O12C34S) = Cu/0.197 910 = 85.178 26 [u A2] .

To simplify the following expressions, we define mO = m(16O), mC = m(12C), mS = m(32S) and

m∗S = m(34S), all in units u. If we explicitly write out Eq. (2.41) for our two isotopologues, we obtain

Ia = mO(rCO)2 +mS(rCS)

2 − (mO rCO −mS rCS)2

mO +mC +mS

Ib = mO(rCO)2 +m∗

S(rCS)2 − (mO rCO −m∗

S rCS)2

mO +mC +m∗S

.

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

Ia (mO,mC,mS) = (mO mC +mO mS) (rCO)2 (2.42)

+ (mO mS +mC mS) (rCS)2 + 2mO mS rCO rCS

Ib (mO,mC,m∗S) = (mO mC +mO m∗

S) (rCO)2 (2.43)

+ (mO m∗S +mC m∗

S) (rCS)2 + 2mO m∗

S rCO rCS .

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

Ia (mO,mC,mS)m∗

S

mS= (mO mC +mO mS)

m∗S

mS(rCO)

2 (2.44)

+ (mO m∗S +mC m∗

S) (rCS)2 + 2mO m∗

S 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

Ib (mO,mC,m∗S) − Ia (mO,mC,mS)

m∗S

mS= mO mC

(1− m∗

S

mS

)(rCO)

2

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

895.055 3899 (rCS)2 + 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-Linear Polyatomic Molecules are More Difficult . . . . . .

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.


×

y

x

θ=104.5°

0.958Å

Figure 2.11: Structure of H2O.

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

plane of the molecule, and that the centre of mass must lie on the y

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 the16O atom, the two 1H atoms are located at the coordinate positions

(xH, yH) = (±rOH sin(θ/2),−rOH cos(θ/2))

= (±0.7575,−0.5865) .

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

xcm =mH×(−0.7575)+mO×0 +mH×(+0.7575)

2mH +mO= 0

ycm =mH×(−0.5865)+mO×0 +mH×(−0.5865)

2mH +mO= − 0.06564 ,

while symmetry tells us that zcm = 0 . The centre of mass is shown as an “×” in Fig. 2.11; the fact that it is

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:

Ix = mH(−0.5865 + 0.0656)2 +mO(0.0656)2 +mH(−0.5865 + 0.0656)2 = 0.6158 [u A

2] ≡ IA

Iy = mH(−0.7575)2 +mO(0)2 +mH(0.7575)

2 = 1.1566 [u A2] ≡ IB

Iz = Ix + Iy = 1.7724 [u A2] ≡ IC .

This definition of Iz reflects the simple Pythagoras theorem result that the ⊥ distance (in the plane) from

atom i to the z axis (which is ⊥ to the plane of the paper) is simply√(xi − xcm)2 + (yi − ycm)2 , and the

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, or C 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,

F sph(J) = Bsph[J(J + 1)] =

(Cu

I

)[J(J + 1)] , (2.45)

and that the degeneracy of the rotational energy levels is now gJ = (2J + 1)2 . Similarly, for a symmetric

top

F sym = F sym(J,K) = B⊥[J(J + 1)] +(B‖ −B⊥

)K2

=

(Cu

I⊥

)[J(J + 1)] +

(Cu

I‖− Cu

I⊥

)K2 (2.46)

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. (as usual), and K = 0, 1, 2, . . . , J (i.e., 0 ≤ K ≤ J ). In this case the energy

level degeneracy is 2(2J + 1) for K > 0 and simply (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


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., 1H vs. 2H and 12C vs. 13C in various locations,

and 14N vs. 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-H8-Atomics HCOOCH3, CH3–C≡C–C≡N9-Atomics CH3OCH3, CH3CH2OH, N≡C–C≡C–C≡C–C≡C-H11-Atomics N≡C–C≡C–C≡C–C≡C–C≡C-H

2.7 Problems

1. The rotational constant B for 12C2 has been determined to be 1.8198 cm−1. 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) of the J = 3← 2 transition for this species.

4. The first rotational transition J = 1← 0 of 12C16O 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.]

6. Given that 12C16O, 14N16O and 16O2 have approximately the same bond lengths (1.16±0.04 A), predictthe 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 12C16O and 13C16O 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 = 1000K. 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 .


11. The equilibrium bond length of LiH is 1.595 A. Determine the inertial rotational constants for 7Li1H,6Li1H and 7Li2H, 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)?

14. The rotational constant Bv of 127I35Cl is 0.1142 cm−1. Determine the bond length in this molecule.

Determine the rotational constant Bv for 127I37Cl, 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

Bv = 10.5933− 0.3070(v+ 12 ) + 0.00163(v+ 1

2 )2 cm−1. What are the average bond lengths rv in each

of the three lowest vibrational levels, v = 0, 1 and 2 ?

20. Show that for a two-particle system the general definition of Eq. (2.11) reduces to our simple two-

particle or diatomic molecule result Id = μ (re)2 .

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 temper-

ature 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 6000K.]

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 A and 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.

How Do We Do It? Transitions are observed by measuring the amount of infrared radiation that is ab-

sorbed or emitted by vibrating molecules in solid, liquid, or gas phases.

Why Do We Do It? 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

↑vertical

componentof

moleculardipole

time→

equilibriumdipole moment

+

− − − − − − − −

++

++

++

+

strengthof

dipole

lengthof

molecule

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.

symmetric stretch

antisymmetric stretch

bending mode

Figure 3.2:

Vibrational modes of CO2.

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

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

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-

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-

trast, for both the antisymmetric stretch and bending modes, the molecular

distortion that occurs during the vibrational cycle will give rise to a tempo-

rary 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

the internal energy. For a rigid molecule that internal energy is simply the rotational energy Erot = L2/2I 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 = m1m2/(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 Schrodinger equation for the system into a Schrodinger 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 eigen-

values of the electronic Schrodinger equation yielded the potential energy curves V (r) = Eel(r) governing

the vibrational motion of the nuclei, which appears in the effective radial Schrodinger 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

F = − k(r − re) = − dV (r)

dr, (3.1)

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:

VHO(r) = 12 k(r − re)2 . (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)):

F = μd2r

dt2= − k (r − re) (3.3)

that may be rewritten as

d2(r − re)dt2

= −(k

μ

)(r − re) = − κ2(r − re) , (3.4)

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(κ t+ 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, and y is replaced by (r − re) as the

name of the dependent variable; however, the underlying mathematics is exactly the same.


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 is r(t) ≡ dr(t)/dt = 2πνeA cos(2πνet + δ) , and

since the instantaneous radial kinetic energy is 12μ(r)

2 , the sum of the potential energy and kinetic energy

is constant:

Etot = KEradial + VHO(r(t)) = 12μ(r)

2 + 12k (r(t) − re)2

= 12μ [2πνeA cos(2πνe t+ δ]2 + 1

2k [A sin(2πνe t+ δ)]2 = 12k A

2 (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 Schrodinger equation

Hvib ψ(r) = − �2

2μ

d2 ψ(r)

dr2+ V (r)ψ(r) = Evib ψ(r) (3.7)

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

Evib = EHOvib (v) =

(�√k/μ

)(v + 1/2) = h νe (v + 1/2) (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:

GHO(v) ≡ EHOvib (v)/

(102 hc

)= ωe(v + 1/2) [cm−1] (3.9)

in which

ωe =

√2k

�2

2μ

1

102 hc=

√2Cu k [cm−1 A−2]

μ [u][cm−1] (3.10)

and Cu = 16.857 629 [u cm−1 A2] is the ubiquitous numerical factor introduced in §1.1.4. Note that we

choose to write the force constant as k when it is in “spectroscopists’ units”. In SI units k would normally have

the units newtons/meter (or Nm−1). However, it is equally valid to multiply numerator and denominator

by 1m and write it with units Jm−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., rather than 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

υ=0

υ=1

υ=2

υ=3

υ=4

υ=5

V(r)= k (r-re )2 ; μ=μ0

(r - re )0.0- 0.2 0.2

A

1−2

υ=0

υ=1

υ=2

υ=3

0.0- 0.2 0.2

ωe

ωe

ωe

ωe / 2

B

V(r)= k ′(r-re )2 ; μ=μ0

1−2{k ′= 2 k}

(r - re )

υ=0

υ=1

υ=2

υ=3

0.0- 0.2 0.2

C

V(r)= k ′(r-re )2 ; μ= 0.9μ0

{k ′= 2 k}

(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

energy, but rather has a zero point energy of GHO(0) = 12 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

ΔGHOv+1/2 ≡ GHO(v + 1)−GHO(v)

= 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

ΔEHOv+1/2(v) = h νe [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 §3.1.1For 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

states of HgI and IBr, ωe = 125.6 and 268.7 cm−1, respectively, while the ground electronic states of HgH

and HF have ωe=1 387.1 and 4 138.5 cm−1, 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.


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

from one state to another (e.g., see Fig. 1.13). For example, ωe = 1 405.498 cm−1 for the X 1Σ+ ground

electronic state of 7LiH, while for the electronically excited A 1Σ+ state of this same species, ωe = 234.4 cm−1;

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)

in the X 1Σ+ ground electronic state, ωe=1 420.048, 1 054.940 and 1 074.309cm−1, respectively.4 Note that

the zero point energy of 12 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

νvib = 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 har-

monic oscillator those terms allow |Δv| > 1 transitions at energies νvib = 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 mo-

mentum 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=1902Nm−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:

μ =

(1

12.0+

1

15.994 915

)−1

[u] =6.856 392 × 10−3 [kg/mole]

6.022 136 7× 1023 [molec/mole]= 1.138 531× 10−26 [kg/molec] .

4 A careful student will note that these ωe values are not precisely those predicted by multiplying the value for 7LiH 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

h νe = �

√k

μ= 1.054 572 66× 10−34

√1 902 [Nm−1]

1.138 531× 10−26 [kg]= 4.310 317× 10−20 [J] ,

or νe = 4.310 317×10−20/6.626 075 5×10−34 = 6.505 083×1013 [Hz] = 65.060 83 [THz]. Our familiar frequency

to wavenumber conversion then yields

ωe = νe/102 c =

(6.505 083× 1013/102 × 2.997 924 58× 108

)= 2169.9 [cm−1]

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 [Nm−1]=1 902 [Jm−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:

k = 1902 [Jm−2]10−20

102 hc= 9.574 892× 105 [cm−1 A−2] .

Substituting this result into Eq. (3.10) then yields the same result obtained before:

ωe =√

2×16.857 629× 9.574 892×105/6.856 392 = 2 169.9 [cm−1] .

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=1−0vibrational spacing of 7LiH was ΔG1/2 = 1359.708 cm−1. 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

μ =

(1

7.016 003 0+

1

1.007 825 035

)−1

= 0.881 238 16 [u] .

Since we are given only one vibrational spacing, we must treat the system as a harmonic oscillator and assume

that ωe = 1359.708 cm−1. Substituting this value and the reduced mass into Eq. (3.10) and solving for k then

yields

k = (1 359.708)2×0.881 238 16/(2×16.857 629) = 48 323.47 [cm−1 A−2] .

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

V (r) = De

[1− e−β(r−re)

]2= De

[e−2β(r−re) − 2 e−β(r−re) + 1

], (3.12)


υ=0

υ=1

υ=2

υ=3

υ=4

υ=5

υ=6υ=7

fundamental

first overtone

second overtone

third overtone

hot bands}

DD

rer→

V (r)Morse

0

e

↑

zero point energy

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

Schrodinger 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 Schrodinger equation may be solved analytically to give an exact closed-form expression for its

vibrational level energies:

GMorse(v) = ωe(v + 1/2)− ωexe(v + 1/2)2 , (3.13)

in which6

ωe = �

√2De β2

μ

1

102 hc=

√√√√4CuDe [cm−1](β [A

−1])2

μ [u][cm−1] (3.14)

ωexe = β2 �2

2μ

1

102 hc=

Cu (β [A−1])2

μ [u][cm−1] . (3.15)

Squaring both sides of Eq. (3.14) and dividing by Eq. (3.15) cancels out the factor Cu β2 and yields the

relation

DMorsee = (ωe)

2/4ωexe . (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 In particular, real intermolecular potentials have attractive inverse-power long-range tails such as V (r) ∼ D−C6/r6 , and

not the rapidly-dying exponential tail of a Morse function V (r) ∼ D− 2De e−β(r−re) .6 Note that the constant “ωexe” should always be treated as a single symbol, and not as the product ωe×xe .


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),

e−x = 1 +

∞∑n=1

(−x)nn!

= 1− x+1

2x2 − 1

6x3 +

1

24x4 + . . . ,

our Morse potential energy function may be expanded as

V (r) ≈ De

{[β(r − re)]2 − [β(r − re)]3 + 7

12[β(r − re)]4 − 1

4[β(r − re)]5 + . . .

}. (3.17)

For very small |r − re|, the cubic and higher-power terms will be much smaller than the leading quadratic

term, and this potential collapses to the harmonic oscillator function of Eq. (3.2), with k = kMorse = 2De β2 .

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

ΔGMorsev+1/2 = GMorse(v + 1)−GMorse(v)

={ωe(v + 1 + 1/2)− ωexe(v + 1 + 1/2)2

}− {ωe(v + 1/2)− ωexe(v + 1/2)2

}= ωe − 2ωexe(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)

dv= ωe − 2ωexe(v + 1/2) = 0 . (3.19)

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


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) (which equals [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

transition was known. However, given both that quantity (namely, ΔG1/2 = 1 359.708 cm−1) and the energy

of the first overtone transition ΔG(2 ← 0) = 2 674.560 cm−1, 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)

ΔG(2← 0) = 2 674.560 = GMorse(v = 2)−GMorse(v = 0)

={ωe(2 + 1/2)− ωexe(2 + 1/2)2

}− {ωe(0 + 1/2)− ωexe(0 + 1/2)2

}= 2ωe − 6ωexe . (3.22)

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

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

and substituting this back into Eq. (3.21) yields

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

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

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

vD = 1404.564/(2×22.428) − 1/2 = 30.813

for De and vD. The total number of bound vibrational levels of 7LiH (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


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):

G(v) = ωe(v + 1/2)− ωexe(v + 1/2)2 + ωeye(v + 1/2)3 + ωeze(v + 1/2)4 + . . . . (3.23)

This in turn yields the vibrational spacing equation7

ΔGv+1/2 = ωe − 2ωexe(v + 1) + ωeye(3v2 + 6v + 13/4

)+ ωeze

(4v3 + 12v2 + 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 attempt-

ing 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

Np < Nd such a fitting procedure also effectively smooths over experimental uncertainties that give rise to

small irregularities in the data, which effectively ensures that those irregularities do not yield implausible

parameter values or “wiggly” expressions.

One very widely-used alternative to the harmonic or Morse oscillator models is the generalized polynomial

potential energy function introduced by J.L. Dunham in 1932:

VDun(r) = a0

(r − rere

)2{1 + a1

(r − rere

)+ a2

(r − rere

)2

+ a3

(r − rere

)3

+ . . .

}. (3.25)

This expression may be thought of as a generalization of the Taylor series expansion for the Morse potential

(see Eq. (3.17)) to a case in which all of the higher-order polynomial coefficients are independent of one

another, rather than being defined by (known) numerical factors and powers of β. Dunham showed that the

vibrational level energies for this potential could be expressed in the form

GDun(v) = Y1,0 (v + 1/2) + Y2,0 (v + 1/2)2+ Y3,0 (v + 1/2)

3+ Y4,0 (v + 1/2)

4+ . . .

=∑l=1

Yl,0 (v + 1/2)l, (3.26)

and he determined explicit analytic expressions for the various Yl,0 coefficients in terms of the potential

expansion parameters {ai}:Y1,0 = 2

√a0Be (3.27)

Y2,0 =3

2Be

{a2 − 5(a1)

2/4}

(3.28)

Y3,0 =(Be)

3/2

4√a0

{10a4 − 35 a1 a3 − 17(a2)

2/2 + 225(a1)2 a2/4− 705(a1)

4/9}

(3.29)

. . .

in which Be=Cu/(re)2 (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 aipotential 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 Note that as for ωexe,6 the variable names ωeye 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 .


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

Schrodinger 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

v=0− 6 . However, although we know that the curve in Fig. 3.5-B must extrapolate to an asymptote at D0,9

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,

the six known experimental ΔGv+1/2 values are 476.00, 450.24, 423.19, 394.74, 364.90 and 333.74 cm−1. 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

G(6)−G(0) =

5∑v=0

ΔGv+1/2 = 2442.81 [cm−1] .

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

find that the value of our linear function there is 333.74− 12 (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

region is given by the formula for the area of a right-angle triangle, 12×318.16×10.211 = 1 624.29 cm−1. 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

D0 = 2442.81+ 1624.29 = 4067.10 cm−1 .

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

υ=6

5

4

3

2

1

0

V(r)

D

?⇓⇑

A

?D0

0 5 10 150

1000

2000

3000

4000

υ

E (υ)vib

∼

B

Birge-Sponer plot

5 10 15

100

200

300

400

500

ΔEυ+½

υ

C

∼

00

Figure 3.5: Vibrational extrapolations and the Birge-Sponer plot.

spreadsheet or small computer program yields the polynomial expression

ΔGv+1/2 = 488.339− 24.3609 v− 0.68152 v2 ,

and from the roots of this quadratic we find that the intercept of this curve is vD=14.314 . From elementary

calculus we may then determine the area under the curve for this quadratic extrapolation to be:

D0 −G(v=6) =

∫ 14.314

6

(488.339− 24.3609 v− 0.68152 v2

)dv = . . . = 1439.98 [cm−1] ,

which in turn yields

D0 = 2442.81 + 1439.98 = 3882.79 cm−1

as our quadratic-extrapolation estimate of D0 .

The 184 cm−1 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 LeRoy 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

ΔGv+1/2 � X0 (vD − v)(n+2)/(n−2), (3.30)

in which n is the integer power defining the limiting long-range inverse-power behaviour of an intermolecular

potential V (r) ∼ D − Cn/rn , and has a value between 1 and 6.10 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

the functional form ΔGv+1/2 ∝ (vD − v)7/3 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.


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, �Rcm = (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 leadsus 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 vibra-

tional modes. As a result, for a general polyatomic molecule the total vibrational energy is conventionally

written as

G(v1, v2, v3, . . . ) =∑i

ωi(vi + 1/2) +∑i

∑j≥i

xi,j(vi + 1/2)(vj + 1/2) + . . . (3.31)

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 par-

ticular 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 indepen-

dent 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 as νi, 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 spacing νi.

Let us now consider a few particular cases.

(i) A diatomic molecule

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

BA

ν1 C−H symmetric stretchν1 =3374cm

-1

ν2 C≡C stretchν2 =1974cm

-1

ν3 C−H asymmetric stretchν3 =3287cm

-1

ν4 trans bendν4 =612cm

-1

ν5 cis bendν5 =729cm

-1

+ − +−

+ + −−

∼

∼

∼

∼

∼ν3ν3 =3756cm

-1∼

ν2ν2 =1595cm

-1∼

ν1ν1 =3657cm

-1∼

Figure 3.6: Vibrational normal modes of: A. water H2O and 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

Eq. (3.31) associated with independent bending quantum numbers va2 and vb2. 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–HSince N=4 and C2H2 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


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? To answer it

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

mode having an energy of ν1 = νss1 = 3 441 cm−1 and for the antisymmetric stretch mode an energy of

ν3 = νas3 = 2 119 cm−1 (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≡Nmotion, while the reverse is true for ν3. Moreover, looking up the properties of diatomic molecules we see

that ωe = 2 862 cm−1 for CH and 2 068 cm−1 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 energies νi for common chemical function groups.

chemical approximate chemical approximategroup 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,

this is not the case. Consider, for example, the case of D35Cl (or 2H35Cl), for which ωe=2 145.133 cm−1,

ωexe=27.1593 cm−1, ωeye=0.079 93 cm

−1 and ωeze=− 0.003 03 cm−1. A harmonic oscillator model would pre-

dict that its vibrational spectrum consisted of a single line (or superposition of lines) at a transition energy of

ν=ωe=2 145.133 cm−1. Allowing for anharmonicity, it would be expected to have a strong fundamental tran-

sition at 2 091.059cm−1, and weaker first and second overtone transitions at 4 128.430 and 6 112.449cm−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 1 984.019cm−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”.


P(1)P(2)P(3)P(4)

P(5)P(6)

P(7) R(0)

R(1)R(2) R(3) R(4)

R(5)R(6)

R(7)

∼ν / cm−1 →

P(8)

spectrum

∼ν0↓

bandgap

←⎯→

J ′=0234

5

6

7

J ′′=0234

5

6

7⎪

⎩

⎧

⎪

⎪⎪⎪⎪⎪⎨⎪⎪⎪

⎩

⎧

υ′

υ′′

↑energy

⎧

P branchΔ J = −1

R branchΔ J = +1

⎪

⎪⎪

⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪

⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪⎩ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧⎪ ⎪

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 = ±1selection 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 ′′ arelinked 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 bandwould be

νP (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)]

= ν0(v′, v′′) − [Bv′′ +Bv′ ] (J ′′) − [Bv′′ −Bv′ ] (J ′′)2 (3.33)

≈ ν0(v′, v′′) − [Bv′′ +Bv′ ] (J ′′) (for small J ′′)

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

νR(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)]

= ν0(v′, v′′) + [Bv′′ +Bv′ ] (J ′′ + 1) − [Bv′′ −Bv′ ] (J ′′ + 1)2 (3.34)

≈ ν0(v′, v′′) + [Bv′′ +Bv′ ] (J ′′ + 1) (for small J ′′)

where ν0(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 - and R-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 allthe 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 , we may

expect Eqs. (2.33) to predict the initial-state population distribution accurately, which leads us to predict

Jpopmax(293K) = 3.85 ≈ 4 as the initial-state rotational quantum number for the most intense P - and R-

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 - and R-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 1300K 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 .


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 ∼ 2ωexe predicted by Eqs. (3.18)

or (3.24) (ωexe(Na35Cl) = 1.78 cm

−1) 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

D37Cl. From Eqs. (3.10) or (3.14) we see that all else being equal, ωe ∝ 1/√μ . This implies that for

D37Cl, ωe ≈ 2 145.133√μ(D35Cl)/μ(D37Cl) = 2 141.975 cm−1, a 3.2 cm−1 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 of ν0(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 (1800K) emission spectrum of GeO. Left: overview spectrum; Right: dis-

persed segment of the fundamental band showing isotopologue assignments.

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

Bv = Be − αe(v +12 ) + γe(v +

12 )

2 + . . . (3.35)

Dv = De + βe(v +12 ) + . . . . (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:

E(v, J) =∑m=0

Km(v)[J(J + 1)]m =∑m=0

∑l=0

′Yl,m

(v + 1

2

)l[J(J + 1)]m , (3.37)

in which the prime (′) on the sum over l 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 electro-

magnetic 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 000cm−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


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

of all, the electron has a mass of me ≈ 5.486 × 10−4 u, while a diatomic molecule reduced mass will have

magnitudes of 1 − 100u. 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– and R–branch transitions.

(b) Determine the rotational constants B′ and B′′ and the average bond lengths r 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

5. Molecular orbital theory predicts that 7Li2,12C2 and 14N2 have different bond orders. Given that

ωe = 357.43, 1854.71 and 2358.57 cm−1, 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.

7. Given that k = 314 Nm−1, re = 1.597 A and m(127I) = 126.904 473 u, calculate the energy (in cm−1)

of the P (2) and R(4) transitions in the fundamental band of IR spectrum of 1H127I.

8. The vibrational frequency of 12C16O is known to be 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 =

2245Nm−1. The diatomic oxide of X, XO, has a vibrational frequency of 1876 cm−1 and a force

constant of k = 1 550 Nm−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 “spectro-

scopists’ units”.

11. In the infrared spectrum of 1H35Cl, the P (4) and R(2) lines of the fundamental band occur at 2799.00

and 2944.99 cm−1, respectively. Determine the values of ωe (in cm−1) 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 2H for 1H 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?

14. Assuming that 1H35Cl is reasonably well described by a Morse potential with parameters: De =

5.250 eV, ωe = 2990.95 cm−1, and ωexe = 52.82 cm−1, determine D0 for each of 1H35Cl and 2H35Cl.

15. Which bond would be easier to break, a C-1H bond or a C-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.

19. For a molecule whose vibrational data give ωe = 1387.09 and ωexe = 83.01 cm−1, obtain an estimate

for De and for the total number of vibrational levels for that molecule.


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.

How Do We Do It? 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.

Why Do We Do It? 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 measure-

ments 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

a 3×3 matrix of values.

75

76 CHAPTER 4. RAMAN SPECTROSCOPY

externalelectricfield

E→

16O

18O

16O

16O

18O

16O

18O

18O

18O

16O

16O

18O

dipoleinducedby field

orientationof

molecule

↑dipoleinducedby field

time→0

1 /νrot

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) = �E0 cos(2π ν0 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

αE0(t) = α0 + Δα cos(4π νrot t) , (4.2)

in which α0 = (2α⊥ + α‖)/3 is the spherical average of the polarizability, Δα = [α‖ − α⊥] is called the

polarizability anisotropy, and E0 is a unit vector pointing along the direction of the electric field. Upon

substituting these expressions for �E(t) and αE0(t) into Eq. (4.1), we see that the time-dependent induced

dipole may be written as

�M ind(t) = �E0 {α0 cos(2π ν0 t) + Δα cos(4π νrot t) cos(2π ν0 t) } (4.3)

= �E0 α0 cos(2π ν0 t) + 12�E0 Δα cos (2π[ν0 + 2νrot]t) + 1

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

relation: cos a cos b = 12 [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

externalelectricfield

E→

16O

18O18

O18O

16O 16

O

18O

16O

18O 18

O

16O 16

O

[equilibrium][equilibrium] [equilibrium]

dipoleinducedby field

stretchingof

molecule

↑dipoleinducedby field

time→0

1 /νvib

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)

shows that the fraction of incident light intensity that is scattered is proportional to (ν0)4 . It is important

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,

αE0(t) = α + δα cos(2π νvib t) , (4.4)

with δα representing the amplitude of the change in the polarizability during a full cycle of vibrational

motion, and α 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:

�M ind(t) = �E0 α cos(2π ν0 t) + 12�E0 δα cos (2π[ν0 + νvib]t) + 1

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.


ν0 νs=ν0ν0 νs =

ν0−δE/h

ν0

δE

Stokesscattering

Rayleighscattering

anti-Stokesscattering

νs = ν0+ δE/h

E ′′(υ′′,J ′′)

E ′(υ′,J ′)

"virtuallevel"

Figure 4.3: Incident ν0 and scattered νs light in Raman scattering involving upper and lower vibration-

rotation levels E′(v′, J ′) and E′′(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.


νStokesS (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

νanti−SS (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 from ν0. 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

ν0 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

νStokesO (J) = ν0 −{E(v′, J − 2)− E(v′′, J)

}= ν0 − {[G(v′)−G(v′′)] + [Fv′(J − 2)− Fv′′ (J)]}= ν0 − {[G(v′)−G(v′′)] + [Bv′(J − 2)(J − 1) − Bv′′ J(J + 1)]} (4.9)

= ν0 −{[G(v′)−G(v′′)] − (Bv′′ +Bv′) (2J − 1) − (Bv′′ − Bv′) (J2 − J + 1)

}νStokesQ (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)

νStokesS (J) = ν0 −{E(v′, J + 2)− E(v′′, J)

}= ν0 − {[G(v′)−G(v′′)] + [Fv′(J + 2)− Fv′′ (J)]}= ν0 − {[G(v′)−G(v′′)] + [Bv′(J + 2)(J + 3) − Bv′′ J(J + 1)]} (4.11)

= ν0 −{[G(v′)−G(v′′)] + (Bv′′ +Bv′) (2J + 3) − (Bv′′ − Bv′) (J2 + 3J + 3)

}.

The analogous expressions for anti-Stokes transitions are identical, except that the mathematical “−” sign

following the symbol ν0 in each of these equations is replaced by a “+” sign. Note, however, that while S(J)

4.5. RAMAN SPECTRA OF POLYATOMIC MOLECULES 81

ν →νo

Rayleigh line

Q branch

S branchO branch

S branch O branch

Q branch

pure rotationS branch(Stokes)

pure rotationS branch(anti-Stokes)

⎫ ⎬ ⎭⎫ ⎬ ⎭

fundamentalvibrationalband

(anti-Stokes)

⎫ ⎬ ⎭

⎫ ⎬ ⎭

⎫ ⎬ ⎭

⎫ ⎬ ⎭

fundamentalvibrationalband

(Stokes)

↑

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 = ±1 is

strongly allowed, Δv =±2 is weak, 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, in some modes

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


For molecules with a centre of symmetry (e.g., N2 , CO2 , SF6or 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 180.000nm, the P2 molecule is observed to have a vibrational

Raman spectrum with Q-branch Stokes bands centred at wavelengths 182.538 and 185.123nm, re-

spectively. 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 ultravi-

olet (tens of eV) to excite molecules into excited electronic states.

How Do We Do It? 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.

Why Do We Do It? 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′) and lower (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 vibra-

tional bands associated with a single lower-state v′′ value and a series of different upper-state v′ values, orwith 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

ω′e for the upper state, while adjacent bands with a common v′ value will be separated by the analogous

lower-state constant, ω′′e . 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 ω′e. This spectrum also shows that bands in a particular “vibrational sequence”, a set of

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 |ω′e − ω′′

e |.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 tran-

sition 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 ′′) = [T ′e +G′(v′) + F ′

v′ (J ′)] − [T ′′e +G′′(v′′) + F ′′

v′′ (J ′′)]

= {[T ′e +G′(v′)] − [T ′′

e +G′′(v′′)]} + [F ′v′(J ′)− F ′′

v′′ (J ′′)]

= ν0(v′, v′′) + [F ′

v′ (J ′)− F ′′v′′ (J ′′)] , (5.1)

in which the energy difference defining the band origin

ν0(v′, v′′) = [T ′

e +G′(v′)] − [T ′′e +G′′(v′′)] = ΔTe + G′(v′) − G′′(v′′) (5.2)

now contains the difference in electronic energies of the two states, ΔTe=T′e − T ′′

e . Note that in contrast to

Eqs. (3.33) and (3.34) in vibrational spectroscopy and Eqs. (4.9) - (4.11) in Raman spectroscopy, prime (′) ordouble 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., F ′v′(J ′) vs. F ′′

v′′(J ′′)) and with

(see below) molecular constants such as B′′v′′ or ω′

e, 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


following expressions for the allowed transition energies:

νP (J′′) = E′(v′, J ′)− E′′(v′′, J ′′) = E(v′, J ′′ − 1)− E′′(v′′, J ′′)

= ν0(v′, v′′) − [B′′

v′′ +B′v′ ] (J ′′) − [B′′

v′′ −B′v′ ] (J ′′)2 (5.3)

νR(J′′) = E′(v′, J ′)− E′′(v′′, J ′′) = E(v′, J ′′ + 1)− E′′(v′′, J ′′)

= ν0(v′, v′′) + [B′′

v′′ +B′v′ ] (J ′′ + 1) − [B′′

v′′ −B′v′ ] (J ′′ + 1)2 (5.4)

νQ(J′′) = E′(v′, J ′)− E′′(v′′, J ′′) = E(v′, J ′′)− E′′(v′′, J ′′)

= ν0(v′, v′′) − [B′′

v′′ −B′v′ ] J ′′(J ′′ + 1) , (5.5)

in which the final expression for each case is obtained by ignoring centrifugal distortion terms. These expres-

sions for P– and R–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 of ν0(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 , the R 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-J R-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 (B′′v′′ < B′

v′ ), which means that the quadratic terms in Eqs. (5.3)-(5.5) would both be positive, rather

than negative. In that case it would be the P–branch whose lines become ever closer together with increasing

J until they turn around and march off “to the blue” (to higher frequencies), while it is the R–branch lines

that become progressively farther apart with increasing J . Bands of this latter type are called “blue shaded”

bands, while those associated with the more common case of B′′v′′ > B′

v′ 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– or P–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 νR(J)

d J= [B′′

v′′ +B′v′ ] − 2[B′′

v′′ −B′v′ ](J + 1) = 0 (5.6)

yields an expression for JRh , the value of J at the band head, the point where the R–branch turns around:

(JRh + 1) = [B′′

v′′ +B′v′ ]

/2[B′′

v′′ − B′v′ ] . (5.7)

Substituting this result into Eq. (5.4) then yields an expression for the position of the band head:

νR(JRh ) = ν0(v

′, v′′) + [B′′v′′ +B′

v′ ]

(B′′

v′′ +B′v′

2 (B′′v′′ −B′

v′)

)− [B′′

v′′ −B′v′ ]

(B′′

v′′ +B′v′

2 (B′′v′′ −B′

v′)

)2

= ν0(v′, v′′) +

(B′′v′′ +B′

v′)2

4 (B′′v′′ −B′

v′). (5.8)

Alternately, for cases in which B′′v′′ < B′

v′ it is the P–branch that turns around, and we obtain

JPh = − [B′′

v′′ +B′v′ ] /2[B′′

v′′ −B′v′ ] = [B′

v′ +B′′v′′ ] /2[B′

v′ −B′′v′′ ] (5.9)

and

νP (JPh ) = ν0(v

′, v′′) − (B′′v′′ +B′

v′)2

4 (B′v′ −B′′

v′′). (5.10)

Equations (5.7) and (5.8) also apply to the R–branch turnaround that is sometimes observed in infrared

spectra (e.g., see Fig. 3.9). The reason for the differences between the appearance of vibrational bands in in-

frared and electronic spectra is merely the different magnitudes of the denominators on the right hand sides of


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-

lar, for the 1–0 infrared band of Na35Cl shown there, Bv′′=0 = 0.217 251 cm−1 andBv′=1 = 0.215 637 cm−1,

so Eqs. (5.7) and (5.8) predict that JRh = 133.10 and that νR(Jh) − ν0(v′, v′′) = 29.03 cm−1. 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 #2 at the end of

this chapter). Having the band heads lie at relatively large values of J and the band-head displacements

[νR(Jh) − ν0(v′, v′′)] be relatively large are characteristic properties of infrared vibrational-rotational spec-

tra. 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

are BA0 = 3.475 22 cm−1 and BX

0 = 3.992 52 cm−1. Utilizing these values in Eq. (5.8) yields the prediction

that JRh = 6.2 and νR(Jh)− ν0(v′, v′′) = 26.95 cm−1, 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

seen in Fig. 5.2 allow us to determine that B′′v′′=0 = 0.120 565 cm−1 and B′

v′=0 = 0.113 422 cm−1. We may

then predict that the band head will occur at J = JRh = 15.38 and that [νR(Jh)− ν0(v′, v′′)] = 1.916 cm−1.

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 Schrodinger 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 There are, of course, bands in electronic spectra for which by accident B′′v′′ ≈ B′

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

involved in the B(3Π0+u) −X(1Σ+

g ) electronic band system of Br2. 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

a given a particular (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

KEradv =

1

2μ (vr)

2 = [Ev − V (r)] . (5.11)


2.0 2.5 3.0 3.5 4.0 4.5

0

5000

10000

15000

20000

25000

v"=04

8121620

v’=05

1530

1020

X 1Σg+

B 3Π 0u+

Br2

r /Å

energy/ cm-1

28

36

44

5260

Figure 5.5: Potential curves and turning points for the Br2 B(3Π0+u)−X(1Σ+

g ) 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

δtr ≡ fv(r) dr =

(1

vr

)dr =

√μ

2[Ev − V (r)]dr , (5.12)

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

V(r)

Eυ

fυ(r)

KEυ

outerturningpoint

innerturningpoint

∞∞

r / Å→

↑energy

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.

Application of the above considerations to the B(3Π0+u) − X(1Σ+

g ) band system of Br2 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

mechanical probability density |ψv′′=0(r)|2 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


40000 45000 50000 55000

D(0,0)

(1,0)(2,0)

(3,0) (4,0) (5,0) (6,0) (8,0)(10,0)

(12,0)

C(0,0)(1,0) (2,0) (3,0)

(4,0)(5,0)

(6,0) (8,0)(10,0) (12,0)

(0,0)I /20 B

(2,0)(1,0) (3,0) (4,0)

A(0,0)

(1,0)

(2,0)

(3,0)(5,0)

ν / cm−1

BA 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 ω′e and ω′′

e 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 D′v′ and D′′

v′′ .

(b) For a case in which B′′v′′ > B′

v′ , use your results from part (a) to determine an expression for the

band head quantum number JRh which takes account of D′

v′ and D′′v′′ contributions.

(c) The leading centrifugal distortion constant for the two lowest levels of ground-state NaCl are

D0 = 3.11524× 10−7 and D1 = 3.10850× 10−7 cm−1, while the associated values of Bv′′ and Bv′

have been given on p. 88. Using your solution to part (b), determine an improved estimate of JRh

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 53.0 6.9294 0.2576

Chapter 6

Photoelectron SpectroscopyWhat 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.

How Do We Do It? 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.

Why Do We Do It? 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

velocity ve , and hence also the kinetic energy KEe−=12 me(ve)

2 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:

IE = hν0 − KEe− = hν0 − 12 me(ve)

2 . (6.1)

95

96 CHAPTER 6. PHOTOELECTRON SPECTROSCOPY

1sB

H atom B

2sB

σu

σg

hν0

meve21−2

H2

*

1sA

H atom A

2sA

ionizationthreshold

H atomCoulombpotential

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

H+2 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

indicate whether its spin quantum number is ms=+ 12 or − 1

2 . The middle segment of the figure then shows

how the energies of the associated bonding σg and anti-bonding σ∗u 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 → H+2 +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 H+2 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 σ∗2s molecular

orbitals. In each case, only one electron is removed, and an N+2 ion is produced. However, since the three

cases involve the removal of an electron from a different molecular orbital, each case produces an N+2 ion in a

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 H+2 molecular ion in

vibrational level v+ :

H2(v = 0) + hν0 → H+2 (v

+) + e−{

12me (ve)

2}

. (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.


H(1s) +H(1s)

H(1s) +H

υ+=0

IE(υ+=2)

hν0

KE

υ″=0

IE

+e−

adi

υ+=6υ+=3

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 H+2 , 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 H+2 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

G+(v+) = ω+e (v

+ + 12 )− ωex

+e (v

+ + 12 )

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 ω+e and ωex

+e 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

in eV, the values of ω+e and ωex

+e in cm−1, and the value of the vibrational force constant k for H+

2 .

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

ΔG+1/2 = (5.772 − 5.502) × 8065.54465 = 2177 = ω+

e − 2ωex+e (0 + 1)

ΔG+3/2 = (5.502 − 5.246) × 8065.54465 = 2065 = ω+

e − 2ωex+e (1 + 1) .

Solving these equations then yields ω+e = 2289 cm−1 and ωex

+e = 56 cm−1. Finally, substituting this value of

ω+e and the reduced mass μ(H+

2 ) = 0.503 775 338 [u] into Eq. (3.10) yields

k =μ (ω+

e )2

2Cu= 7.83×104 [cm−1 A

−2] .

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 H+2 . This tells us that the minimum of the H+

2 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

order of 1, since it has two bonding electrons, while the bond order in H+2 is only 1

2 , since it has only one

bonding electron. As a result, we expect that H+2 will have a longer bond length and smaller dissociation

energy than does neutral H2. The fact that the value of ω+e (H

+2 ) = 2230 cm−1 is roughly half as large

as ωe(H2) = 4161 cm−1 confirms that the bond in H+2 is much weaker than that in H2. This in turn

tells us that the potential energy curve for H+2 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:


⇑bonding

H(1s)Cl(3p5)

σ3p

σ3p

*

H(1s)Cl(3p5)

σ3p

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 H+2 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 (σ∗3p) 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

the electronic configuration {(σ2p)2, 3p3}, should be fairly similar to those of the neutral parent molecule,

whose electron configuration was {(σ3p)2, 3p4}. 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

occur at higher energy (yielding slower photoelectrons), and will reduce the bond order to 12 . 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

molecular ion HCl+ {σ3p1, 3p4} 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 N+2 .

state ωe/cm−1 re/A D0/eV T0/eV

HCl (X 1Σ+g ) 2990.9 1.2746 4.4336 0.0 a

HCl+(X 2Πi) 2673.7 1.3147 4.653 12.768

HCl+(A 2Σ+) 1606.5 1.5142 1.169 16.252

N2 (X 1Σ+g ) 2358.6 1.0977 9.7537 0.0 a

N+2 (X 2Σ+

g ) 2207.4 1.1164 8.7127 15.5808

N+2 (A 2Π+

ui) 1903.5+ 1.1749 7.5949 16.6986

N+2 (B 2Σ+

u ) 2420.8 1.0747 5.5429 b 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 ω+e and (and for the middle spectrum, also ωex

+e ) for these

three electronic states.


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

Two other prominent features of these N+2 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

with those of the parent N2 molecule. Removal of a σ2p bonding electron to produce N+2 (X

2Σ+g ) leads to

modest reductions in the values of ωe and D0, and to a small increase in re; this is what we expect when

the overall bond order is reduced from 3 to 2 12 . 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 andD0 compared to the values

for the neutral molecule (see Table 6.1), reflect the fact that the π2p electrons contribute much more to the

binding of N2 than do the σ2p electrons (see the MO diagram in Fig. 6.4). As a result, the product A 2Π+ui

molecular ion is distinctly less strongly bound than is the X 2Σ+g ground-state ion. The resulting larger

outward displacement of the A 2Π+ui state potential then gives rise to the more extended type of vibrational

spectrum illustrated by Fig. 5.7B.

Finally, the molecular parameters in Table 6.1 show that removal of a σ ∗2s 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

explains why we again have the same type of intensity pattern found for transitions into the X 2Σ+g state.

Note, however, that although the bond order formally increases from 3 to 3 12 on going from the neutral

molecule to the B 2Σ+u state of the molecular ion, the bond strength D0 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.

However, the modest changes between the properties of HCl(X 1Σ+) and those of HCl+(X 2Πi) (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.

The changes for N+2 (B

2Σ+u ) 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

the case of O+2 (X

2Σ+), 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, HCl and N2 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,

and becomes substantially larger than the vibrational peak spacing, while for the A 2Π+ui state of N

+2 (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

from the π ∗2p anti-bonding MO; the substantial increases in both the bond strength D0 and the vibrational

constant ωe shown in Table 6.2 are clearly consistent with this assignment. As for the A 2Π+u state of N+

2

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 π ∗2p

orbitals. The former case (parallel) would yield a quartet state with total spin quantum number S=32 , and

Table 6.2: Molecular parameters for some states of O2 and O+2 .

state ωe/cm−1 re/A D0/eV T0/eV

O2(X3Σ−

g ) 1580.4 1.2077 5.1156 0.0

O+2 (X 2Πg) 1906.1 1.1169 6.662 12.0717

O+2 (a 4Πui) 1035.1 1.3814 2.629 16.1042

O+2 (A 2Πu) 899.0 1.4090 1.694 17.0395

O+2 (b 4Σ−

g ) 1197.0 1.2794 2.530 18.1706

O+2 (B 2Σ−

g ) 1156. 1.298 1.760 20.297


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

the latter a doublet state with S=12 , 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

indeed have the same overall electronic structure, which would be {(π2p)1 (σ2p)2 (π ∗2p)

2}. 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

be relatively more stable, it is reasonable to find that the highest-energy band is assigned as the B 2Σ−g state

and the one centred near 18.5 eV as the b 4Σ−g 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, which showed that removal of a π2p 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 2Σ and 3Σ excited states, so that the 2Π and 4Π 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

spacing and band profile for the 2Σ−g and 4Σ−

g states, and for the 2Πu and 4Πu 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 O+2 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.

In order to account for the fact that the B 2Σ−g and b 4Σ−

g states have larger vibrational spacings, but

similar or smaller dissociation energies than the corresponding 2Πu and 4Πu 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 σ ∗

2s

ion state X 2Σ+ A 2Πu B 2Σ+u

εorbital / eV 15.09 17.40 21.87

IEadi / eV 14.02 16.59 19.71

O(3P ) and O+(4S) in their ground electronic states, but the two highest molecular ion states yield one

fragment or the other in an excited state. In particular, the B 2Σ−u state dissociates to yield O(3P2)+O+(2D) ,

while the b 4Σ−u state dissociates to O(1D)+O+(4S) . 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 N+2 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 assump-

tion 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 suffi-

ciently 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 2 3 4 5 6 7 8 9

IE(1s) / eV 25 55 111 188 285 399 532 686

{IE(1s)/Z2} / eV 6.14 6.11 6.94 7.52 7.92 8.14 8.32 8.47


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 ) and methylene ( –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 → A+m , and that

of the free atom, Af → A+f , 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 NNO ONCl NO N2 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

primary coreelectron vacancy

1s

2s

2p

ionization threshold

Auger process2nd electron ejected

X-ray fluorescence

valence electronvacancies

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

ΔIEn,� =[IEn,�(A

+m)− IEn,�(Am)

]− [IEn,�(A

+f )− 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


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?

4. Using the photoelectron spectrum of H2 in Fig. 6.2, determine ωe and ωexe (in cm−1) for H+2 .

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 m s−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 H+2 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

upon ionization. For reference, the vibrational frequencies of N2 and CO are 2359 and 2170 cm−1,

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.

16. By using the data for the photoelectron spectrum of 1H2 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

of 2H2 to the first five vibrational energy levels of 2H+2 .

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, N−3 , in terms

of approximate energies and intensities.

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.

How Do We Do It? 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.

Why Do We Do It? 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)

in which R must be either a non-negative integer (i.e., R=0, 1, 2, 3, . . . ) or half of an odd integer (R=12 ,

32 ,

52 , . . . ) 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 (2R+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

momentum quantum number R = 32 .

1− −2

3 5− × −2 3

↑mR

P/ h−

1−2

√⎯

→

→

3− −2

3−2

Figure 7.1: Space quantization of

angular momentum �P for R=32 .

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 hy-

drogenic orbitals ( �=0 for s, �=1 for p, �=2 for d, . . . etc.). The associated

(2J+1) values of mJ are the source of the rotational degeneracy factor

discussed in § 2.4, while the (2� + 1) values of m� are the source of the

spatial degeneracy factors of 1, 3, 5, . . . , respectively, for hydrogenic s,

p, d, . . . , orbitals.

The type of angular momentum important for NMR spectroscopy

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

projection of the electron spin angular momentum, whose magnitude is

defined by the total spin quantum number S = 12 . For an electron,

therefore, the degeneracy rule of Eq. (7.2) tells us that an electron can

only have the 2S+1=2 values: mS=+ 12 or − 1

2 . We commonly refer to

these two states as ‘spin-up’ or α, and ‘spin-down’ or β, respectively. Spin

angular momentum has no classical analogue in terms of the mechanical

motion of a particle. While it is tempting to think of it as corresponding

to the spinning motion of the particle, it really isn’t that simple.

For an atomic nucleus the total angular momentum quantum number

is denoted I and its z–componentmI , and as discussed above, there are

(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 12C and 16O.

General Rule 2. If a nucleus has an odd mass number, it has half-integer spin: i.e., I=12 ,

32 ,

52 , . . . ,

etc. For example, I=12 for 1H, 3H, 13C, 15N, 19F, and 31P; similarly I = 3

2 for 11B and 23Na, while

I = 52 for 17O and 27Al.

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 for 2H and 14N, while I=3 for 10B and 50V.

Nuclei with spin I=12 are among the most common and easiest to study, and will be the focus of most of

the following discussion. Nuclei with I > 12 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)

electron – 12 – −17608.4×107

neutron – 12 – −18.3257×107

1H 1 12 99.985 26.7519×107

2H 1 1 0.015 4.1066×1073H 1 1

2 (radioactive) 28.535×1077Li 3 3

2 92.58 10.3975×10711B 5 3

2 80.42 8.5843×10713C 6 1

2 1.108 6.7283×10714N 7 1 99.63 1.9338×10715N 7 1

2 0.37 −2.712×10717O 8 5

2 0.037 −3.6279×10719F 9 1

2 100 25.181×10727Al 13 5

2 100 6.9760×10729Si 14 1

2 4.70 −5.3188×10731P 15 1

2 100 10.841×10759Co 27 7

2 100 6.317×107

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 = − (γ Iz)B0 = − γ mI �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)

For a 1H (or 13C or 19F or . . . ) nucleus, I=12 , and the dependence of the two level energies on the magnetic

field strength is illustrated by Fig. 7.2. Note that for all cases in which I=12 , it is customary to label the

state with mI=+ 12 as α and that with mI=− 1

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)


-10

-5

0

5

10

↑E

ΔE

B0→/

β(mI=−½)

α(mI=+½)

↑Bα

( Bα)

/

/

Figure 7.2: Nuclear spin energy levels in a magnetic field for I = 12 .

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 = h ν0 . (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:

ν0 =ΔE

h=

γ �B0

h=

γ B0

2 π[s−1] . (7.9)

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): What are the Larmor frequencies for 1H, 13C and 19F nuclei in an 8.0000 T magnetic field ?

For 1H nuclei: ν0 =γ B0

2 π=

(26.7519×107 [rad T−1 s−1]

)(8.0000 [T]))

2π [rad]

= 3.40616×108 [Hz] = 340.62 [MHz]

For 13C nuclei: ν0 =γ B0

2π=

(6.7283×107 [rad T−1 s−1]

)(8.0000 [T])

2π [rad]

= 8.56674×107 [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 13C is much

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.

ν0(1H) =

(26.7519×107 [rad T−1 s−1]

)(1.409 [T])

2π= 60.00 [MHz]

If 13C were to achieve resonance at that frequency, then necessarily

ν0(13C) = 60.00 [MHz] =

γ(13C) B0(13C)

2π

Hence, the required field strength would be

B0(13C) =

2π ν0γ(13C)

=2π(60.00×106 [Hz])

6.7283×107 [rad T−1 s−1]= 5.603 [T]

(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

ν0(13C) = ν0(

1H)× γ(13C)

γ(1H)= 60.000 [MHz]× 6.7283×107

26.7519×107 = 15.09 [MHz]

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 1H is the same

as the ‘name frequency’ for that spectrometer.

What are the magnetic field strengths for the two spectrometers considered in Fig. 7.3 ?

0100200300400500600← ν0 / MHz

600 MHzspectrometer

1H 19F3H 11B 13C 2H31P

0100200300400← ν0 / MHz

250 MHzspectrometer

1H3H 19F 11B 13C 2H31P

Figure 7.3: NMR spectra of various atomic nuclei in 250 MHz and 600 MHz spectrometers.


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

give rise to a small induced 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 propor-

tionality 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

νA0 =γA Beff

2 π=

γA B0

2 π(1− σA) . (7.12)

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

‘parts per million’. For example, if σA=62.1×10−6, 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

σA =νA0 (bare nucleus) − νA0 (in molecule)

νA0 (bare nucleus). (7.13)

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 deter-

mine accurate experimental values of σA because it is very difficult to get a stable population of bare nuclei

to sit still to allow us to measure the value of νA0 (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

δA(in molecule) =

(νA0 (in molecule) − νA0 (in reference species)

νA0 (in reference species)

)× 106 [ppm] (7.14)

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

most measured δA values will be positive. For 1H and 13C and 29Si, 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.


Figure 7.5: Chemical shifts of 1H and 13C 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 CH3F CH3Br 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 with-

drawing 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. In a 60 MHz spectrometer this gives rise to a Larmor

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 31P it is ∼ 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.

For the 2.35 T spectrometer: ν0(reference) =

(26.7519×107 [radT−1 s−1]

)(2.35 T )

2π= 100.0 [MHz]

For the 9.40 T spectrometer: ν0(reference) =

(26.7519×107 [radT−1 s−1]

)(9.40 T )

2π= 400.0 [MHz]

(b) Now we rearrange Eq. (7.14) and use it to define the desired frequency shifts.

ΔνA0 = ν0(in molecule)− ν0(TMS) = δ×10−6 × ν0(TMS)

Hence, in the 100MHz spectrometer:

For the 3.00 ppm proton : ΔνA0 = 3.00×10−6 × 100.0×106 = 300 [Hz]


Similarly, in the 400MHz spectrometer:



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.


½ JAB

½ JAB

no fields externalfield B0

nearbyproton β(B)(mB=−½)

nearbyproton α(B)(mB=+½)

β(mA=−½)

α(mA=+½)

α(A), β(A)

(mA= ±½)

/

ν0A

ν0A+ ½ JABν0

A− ½ JAB

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

induced by light of frequency νA0 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 , by an amount

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

associated with the α→ β transition of nucleus A becomes either νA0−JAB/2 or νA0+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

with frequency νA0 associated with the transition α(A) → β(A) into two lines with frequencies differing by

JAB [s−1].1 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

JAB JAB

ν0A − ν0

B = (δA−δB) × ν0spectrometer

ν0Bν0

A ← νFigure 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 :

• α1(B)α2(B) with mtotB = +1

• α1(B)β2(B) with mtotB = 0

• β1(B)α2(B) with mtotB = 0

• β1(B)β2(B) with mtotB = −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

frequencies: νA0−JAB , νA0 and νA0+JAB (see Fig. 7.8). Moreover, we expect the intensity of the middle peak

in this spectrum to be twice as big as the others, since there are two equally probable ways of obtaining the

perturbing field associated with mtotB =0 . At the same time, as there is only one nucleus of type A and hence

only two possible values of mtotA =mA=± 1

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


no fields externalfield

nearbyprotons BmBtot= 0

nearbyprotons BmBtot=−1

β(mA=−½)

α(mA=+½)

α(A), β(A)

(mA= ±½)

½ JAB

½ JAB

ν0A

ν0A+ JABν0

A− JAB

nearbyprotons BmBtot=+1

½ JAB

½ JAB

ν0A

B0/

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

transitions all occur at the same frequency, νA0 . Thus, although the levels are split, the transitions are not,

JAB

JAB

ν0A − ν0

B = (δA−δB) × ν0spectrometer

JAB

ν0Bν0

A ← νFigure 7.9: Simulated NMR spectrum for proton A interacting with two equivalent neighbouring protons B.

7.3. SPIN-SPIN COUPLING 123

1

1 1

21 1

331 1

14641

1 5 1010 5 1N= 5

N= 4

N= 3

N= 2

N= 1

number ofequivalentatoms

sum ofweights

weightsnumberof peaks

2

3

4

5

6

2

4

8

16

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

there are eight possible unique arrangements and four possible mtotB values:

• mtotB = + 3

2 is obtained one way: α1(B)α2(B)α3(B)

• mtotB = + 1

2 is obtained three ways: α1(B)α2(B)β3(B), α1(B)β2(B)α3(B), β1(B)α2(B)α3(B)

• mtotB = − 1

2 is obtained three ways: α1(B)β2(B)β3(B), β1(B)α2(B)α3(B), β1(B)β2(B)α3(B)

• mtotB = − 3

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

will split the absorption associated with proton A into four peaks centred about νA0 , 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

result can be achieved – in this case, the number of distinct ways a given value of mtotB 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 different peaks.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

the five peaks in Fig. 7.9 are 1 : 2 : 1 : 4 : 4 , and for the same reason, the five peaks in the proton NMR

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 .


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. If that same proton A 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 spec-

troscopic 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 100.000000 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

4. What is the energy difference in Joules and cm−1 of the two nuclear spin energy levels of 1H at B0 =

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. In a 40 MHz spectrometer, a compound with two different 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 1H Larmor

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.


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. Which is

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.

a spectroscopy primer

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