Download A Spectroscopy Primer - Symposium on Chemical Physics

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
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ii
Preface
Spectroscopy is the scientist’s window on the molecular world. As molecules are too small to be seen directly
by the human eye, we rely on their interaction with light (or electromagnetic radiation), to determine their
properties, how they are formed from their constituent atoms, and how they react. Interaction with light
probes the molecules’ characteristic rotational and vibrational motions, and we then attempt to explain that
behaviour in terms of theoretical models. This allows us to determine what atoms a particular molecule is
composed of, the length and strength of its bonds, and more generally, the patterns in which atoms assemble
to form the diverse and myriad molecules on which we rely for industrial applications, for modern drugs,
and for life itself.
These notes begin by examining the interaction between light and matter as predicted by models derived
from quantum mechanics, and by outlining the principles of spectroscopy. We then study the types of
spectra associated with several different regions of the electromagnetic spectrum, and see that absorption or
emission of light in those distinct regions tends to be associated with different types of molecular motion.
We will see how the structures of molecules found both in interstellar space and closer to home can be
determined with rotational (or microwave) spectroscopy. We will then see how vibrational (or infrared) and
Raman spectroscopy may be used to determine the strengths of bonds and to identify characteristic groups
of atoms within molecules. We will also see that the electronic energy binding atoms together in molecules
and molecular ions can be studied using electronic and photoelectron spectroscopy. Finally, we will see how
nuclear magnetic resonance spectroscopy uses the magnetic properties of atomic nuclei within molecules to
learn about the structures of complex molecules, such as proteins, and to image tissues within human beings.
The credit for these notes rests not only with the author, but also with several colleagues who provided
important input. In particular, Professor John Hepburn developed the first offerings of this material at the
University of Waterloo, while Professor William Power’s renovated version of his course notes were a key
template for the current document. In addition, Professors Fred McCourt and Peter Bernath have freely
offered valuable and abundant suggestions and criticism throughout the development process, while Drs. Iain
McNab and John Ogilvie have provided helpful comments and suggestions on early drafts of the manuscript.
I am also particularly grateful to Professor Fred McCourt for a thorough, critical reading of the current
version of this document. Finally, the curiosity and bafflement of several classes of Chemistry 129 and 209
students stimulated many revisions and improvements that now grace these pages. All surviving errors are,
of course, mine. I invite all readers to help to improve these notes further by suggesting changes that they
feel might be helpful.
Robert Le Roy
Waterloo, August 2011
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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 Light and the Electromagnetic Spectrum . . . . . . . . . . . . . . . . .
1.1.1 Physics in 1900 . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.2 Wave Properties of Light . . . . . . . . . . . . . . . . . . . . .
1.1.3 The Quantum Theory of Light . . . . . . . . . . . . . . . . . .
1.1.4 A Brief Note on Units . . . . . . . . . . . . . . . . . . . . . . .
1.2 Quantum Theory of Matter . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 The Spectrum of the Hydrogen Atom . . . . . . . . . . . . . .
1.2.2 The Bohr Theory of the Atom . . . . . . . . . . . . . . . . . .
1.2.3 de Broglie Wavelengths . . . . . . . . . . . . . . . . . . . . . .
1.3 Wave Mechanics and the Schrödinger Equation . . . . . . . . . . . . .
1.3.1 A Particle in a One-Dimensional Box . . . . . . . . . . . . . . .
1.3.2 Orbital or Rotational Motion: A Particle on a Ring . . . . . .
1.4 Electronic Structure of Atoms and Molecules . . . . . . . . . . . . . .
1.4.1 Hydrogenic Atomic Orbitals . . . . . . . . . . . . . . . . . . . .
1.4.2 Multi-Electron Atoms and Atomic Spectroscopy . . . . . . . .
1.4.3 Molecular Energies and the Born-Oppenheimer Approximation
1.5 Spectroscopy at last . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1
1
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2
3
7
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17
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21
22
24
25
2 Rotational Spectroscopy
2.1 Classical Description of Molecular Rotation . . . . . . . . . . .
2.1.1 Why Does Light Cause Rotational Transitions? . . . . .
2.1.2 Relative Motion and the Reduced Mass . . . . . . . . .
2.1.3 Motion of a Rotating Body . . . . . . . . . . . . . . . .
2.2 Quantum Mechanics of Molecular Rotation . . . . . . . . . . .
2.2.1 The Basics . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.2 Energy Levels, Selection Rules, and Transition Energies
2.2.3 Illustrative Applications . . . . . . . . . . . . . . . . . .
2.3 Complications ! . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Degeneracies and Intensities . . . . . . . . . . . . . . . . . . . .
2.5 Rotational Spectra of Polyatomic Molecules . . . . . . . . . . .
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27
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31
31
32
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35
39
42
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CONTENTS
vi
2.6
2.7
2.5.1 Linear Molecules are (Relatively) Easy to Treat! . . . . . .
2.5.2 Illustrative Applications . . . . . . . . . . . . . . . . . . . .
2.5.3 Non-Linear Polyatomic Molecules are More Difficult . . . . . .
Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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42
44
45
47
49
3 Vibrational Spectroscopy
3.1 Classical Description of Molecular Vibrations . . . . . . . . . . .
3.1.1 Why Does Light Cause Vibrational Transitions? . . . . .
3.1.2 The Centre of Mass and Relative Motion . . . . . . . . .
3.1.3 The Classical Harmonic Oscillator . . . . . . . . . . . . .
3.2 Quantum Mechanics of the Harmonic Oscillator . . . . . . . . . .
3.3 Anharmonic Vibrations and the Morse Oscillator . . . . . . . . .
3.3.1 Eigenvalues and Properties of the Morse Potential . . . .
3.3.2 Overtones and Hot Bands . . . . . . . . . . . . . . . . . .
3.3.3 Higher-Order Anharmonicity and the Dunham Expansion
3.4 Dissociation Energies and Birge-Sponer Plots . . . . . . . . . . .
3.5 Vibrations in Polyatomic Molecules . . . . . . . . . . . . . . . . .
3.6 Rotational Structure in Vibrational Spectra of Diatomics . . . .
3.7 Why Are Vibrational Level Spacings so Large? . . . . . . . . . .
3.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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51
51
51
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60
62
64
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71
72
4 Raman Spectroscopy
4.1 “Light-As-A-Wave” Description of Raman Scattering . .
4.2 “Light-As-A-Particle” Description of Raman Scattering
4.3 Rotational Raman Spectra . . . . . . . . . . . . . . . . .
4.4 Vibrational Raman Spectra . . . . . . . . . . . . . . . .
4.5 Raman Spectra of Polyatomic Molecules . . . . . . . . .
4.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . .
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75
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78
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81
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5 Electronic Spectroscopy
5.1 Why Does Light Cause Electronic Transitions? . . . .
5.2 Vibrational-Rotational Structure in Electronic Spectra
5.3 Vibrational Propensity Rules in Electronic Transitions
5.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . .
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83
83
84
88
93
6 Photoelectron Spectroscopy
6.1 Photoelectron Spectroscopy:
The Photoelectric Effect Revisited . . . . . . . . . .
6.2 Koopmans’ Theorem . . . . . . . . . . . . . . . . . .
6.3 Vibrational Fine Structure in Photoelectron Spectra
6.4 Molecular Orbitals and Photoelectron Spectra . . . .
6.5 Some Complications in Photoelectron Spectra . . . .
6.6 X-Ray Photoelectron Spectroscopy (XPS) . . . . . .
6.7 Auger Electron Spectroscopy (AES) . . . . . . . . .
6.8 Problems . . . . . . . . . . . . . . . . . . . . . . . .
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7 NMR Spectroscopy
7.1 Basics of NMR Spectroscopy . . . . . . . . . . . . . . . .
7.1.1 Angular Momentum and Nuclear Spin . . . . . .
7.1.2 Magnetic Moments and Nuclei in a Magnetic Field
7.1.3 NMR Spectra . . . . . . . . . . . . . . . . . . . . .
7.2 Chemical Shifts . . . . . . . . . . . . . . . . . . . . . . . .
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111
111
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113
117
CONTENTS
7.3
7.4
7.5
7.2.1 Electronic Shielding of Nuclei and ‘Chemical Shifts’ . . . . . . .
7.2.2 What Determines Chemical Shifts, and The Chemical Shift Scale
7.2.3 Working With Chemical Shifts . . . . . . . . . . . . . . . . . . .
Spin-Spin Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3.1 Basics: Coupling from a Single Neighbour . . . . . . . . . . . .
7.3.2 ‘Equivalence’, and Coupling from Multiple Equivalent Nuclei . .
7.3.3 Spin-Spin Coupling to More Than One Type of Neighbour . . .
Molecular Structures from NMR Spectra . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
vii
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117
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120
120
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125
viii
CONTENTS
List of Figures
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
The electric field of light oscillates in space and in time . . . . . . . . . . . . . . . .
Black-body radiation: observed distributions and the Rayleigh-Jeans law prediction
The photoelectric effect: A. The experiment; B. The observations . . . . . . . . . .
The hydrogen atom emission spectrum . . . . . . . . . . . . . . . . . . . . . . . . . .
Hydrogen atom energy levels and transitions . . . . . . . . . . . . . . . . . . . . . .
Properties of three square-well particle-in-a-box systems . . . . . . . . . . . . . . . .
Level energies and wave functions for four particle-in-a-box systems . . . . . . . . . .
Particle-on-a-ring orbits and wave functions . . . . . . . . . . . . . . . . . . . . . . .
Definition of spherical polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . .
Effective radial wave functions of H atom orbitals for n = 1 − 4 . . . . . . . . . . .
Angular behaviour of H atom orbitals for n = 1 − 3 . . . . . . . . . . . . . . . . . .
Atomic orbital energies in some many-electron atoms . . . . . . . . . . . . . . . . . .
Potential energy curves for the ten lowest energy electronic states of Li2 . . . . . . .
Regions of the electromagnetic spectrum and associated types of spectroscopy . . . .
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3
4
5
9
10
14
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19
20
20
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23
24
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
Component of the dipole field of a rotating polar diatomic molecule . . . . . . . .
Centre of mass and relative coordinates for a two-body system . . . . . . . . . . .
Rotational energies and level spacings for a linear rigid rotor . . . . . . . . . . . .
Microwave absorption spectrum of CO gas . . . . . . . . . . . . . . . . . . . . . . .
Microwave emission spectrum of gaseous HF . . . . . . . . . . . . . . . . . . . . . .
Microwave absorption spectrum of H–C≡C–C≡C–C≡N . . . . . . . . . . . . . . .
Graphical determination of B0 and D0 for CO . . . . . . . . . . . . . . . . . . . .
Angular momentum projections for J = 2 . . . . . . . . . . . . . . . . . . . . . . .
Boltzmann rotational population distribution for CO at T = 293 K . . . . . . . . .
Four types of linear molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Structure of H2 O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Rotational moments of inertia for some symmetric non-linear polyatomic molecules
Gas phase molecular structure of azulene determined from rotational spectroscopy
The rotational emission spectrum of cyanodiacetylene in Sagittarius B2 . . . . . .
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28
28
32
33
35
36
38
40
41
42
46
47
48
48
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
Dipole moment of a vibrating polar diatomic molecule . . . . . . . . . . . .
Vibrational modes of CO2 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Harmonic oscillator eigenvalues and wavefunctions for three model systems
Vibrational levels and transitions of a Morse potential energy function . . .
Vibrational extrapolations and the Birge-Sponer plot . . . . . . . . . . . . .
Vibrational normal modes of: A. water H2 O and B. acetylene C2 H2 . .
Room temperature absorption spectrum of DCl . . . . . . . . . . . . . . . .
Spectrum and energy level pattern for an infrared band . . . . . . . . . . .
NaCl emission spectrum showing band heads . . . . . . . . . . . . . . . . .
High temperature (1800 K) emission spectrum of GeO . . . . . . . . . . . .
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52
52
55
58
63
65
67
68
70
71
ix
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LIST OF FIGURES
x
4.1
4.2
4.3
4.4
Oscillating induced dipole moment of a rotating non-polar molecule
Oscillating induced dipole moment of a vibrating non-polar molecule
Incident ν0 and scattered νs light in Raman scattering . . . . . . . .
Schematic illustration of rotational and vibrational Raman spectra .
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76
77
78
81
5.1
5.2
5.3
5.4
5.5
5.6
5.7
Schematic illustration of rotational and vibrational structure in electronic spectroscopy . . . .
Vibrational bands in the electronic spectrum of SrS . . . . . . . . . . . . . . . . . . . . . . . .
Rotational structure near the (0,0) band head in the A 1 Σ+ − X 1 Σ+ spectrum of CuD . . . .
Definition of the “stationary point” for a particular (v , v ) electronic transition. . . . . . . .
) − X(1 Σ+
Potential curves and turning points for the Br2 B(3 Π0+
g ) system . . . . . . . . . .
u
Classical prediction for the time a vibrating molecule spends at a particular radius . . . . . .
Dependence of vibrational band intensities on the relative radial positions of upper- and lowerstate potential energy functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
85
87
89
90
91
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96
98
100
102
104
106
107
109
Space quantization of angular momentum P for R=32 . . . . . . . . . . . . . . . . . . . . . . .
Nuclear spin energy levels in a magnetic field for I = 12 . . . . . . . . . . . . . . . . . . . . . .
NMR spectra of various atomic nuclei in 250 MHz and 600 MHz spectrometers. . . . . . . . .
1
H NMR spectrum of ethanol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chemical shifts of 1 H and 13 C nuclei in various environments . . . . . . . . . . . . . . . . . .
Energy level diagram for a proton without and with coupling to a neighbouring proton of a
different type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.7 Simulated NMR spectrum for neighbouring single H atoms A and B with different chemical
shifts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.8 Energy level diagram for a nucleus without and with coupling to two identical neighbouring
nuclei . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.9 Simulated NMR spectrum for proton A interacting with two equivalent neighbouring protons
B. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.10 Pascal’s triangle and the calculation of weights for split NMR peaks. . . . . . . . . . . . . . .
7.11 1 H NMR spectra of ethyl chloride, n-propyl iodide, iso-propyl iodide and tert -butyl alcohol .
7.12 60 MHz NMR spectrum of an unknown organic compound . . . . . . . . . . . . . . . . . . . .
112
114
115
116
118
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
7.1
7.2
7.3
7.4
7.5
7.6
Molecular orbital level-energy picture of the photoionization of H2 . . . . . . . . . . . . . .
He(I) photoelectron spectrum of H2 and the associated H2 and H+
2 potential energy curves
Molecular orbital diagram and He(I) photoelectron spectrum for HCl . . . . . . . . . . . . .
Molecular orbital diagram and He(I) photoelectron spectrum for N2 . . . . . . . . . . . . .
He(I) photoelectron spectrum and molecular orbital diagram for O2 . . . . . . . . . . . . .
XPS spectra for C atoms in different molecular environments . . . . . . . . . . . . . . . . .
Schematic level energy diagram illustrating XPS core-hole decay . . . . . . . . . . . . . . .
He(I) photoelectron spectrum of CO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
120
121
122
122
123
124
126
List of Tables
1.1
1.2
Conversion factors for energy units encountered in spectroscopy . . . . . . . . . . . . . . . . .
Wave functions of hydrogenic orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
19
2.1
2.2
2.3
Predicted and observed microwave spectrum of CO . . . . . . . . . . . . . . . . . . . . . . . .
Experimental microwave transition energies for ground state (v = 0) CO . . . . . . . . . . . .
Interstellar molecules detected by their rotational spectra . . . . . . . . . . . . . . . . . . . .
34
37
49
3.1
Characteristic group vibrational energies ν̃i for common chemical function groups . . . . . . .
66
4.1
Labels for various types of rotational transitions . . . . . . . . . . . . . . . . . . . . . . . . .
79
6.1
6.2
6.3
6.4
6.5
Molecular parameters for some states of HCl, HCl+ , N2 and N+
. . . . . . . . . . . . . .
2
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Molecular parameters for some states of O2 and O+
2
Compare calculated orbital binding energies and experimental ionization energies for CO .
Ionization energies of the 1s electrons of the first-row elements . . . . . . . . . . . . . . .
Nitrogen 1s chemical shifts relative to the core ionization energy for N(1s) in gaseous N2 .
7.1
Table of the NMR properties of various nuclear isotopes . . . . . . . . . . . . . . . . . . . . . 113
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101
103
105
105
106
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 = 109 737.315 685 27 cm−1
RH = 2.178 685 42×10−18 joules = 109 677.583 41 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 79×1023
Bohr Radius
a0 = 5.291 772 0859×10−11 m = 0.529 177 208 59 Å
Atomic mass unit
1 u = 1 mu = 1.660 538 782×10−27 kg ≡
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
Inertial Constant
e = 1.602 176 487×10−19 coulombs
2
Cu = 16.857 629 u cm−1 Å
Measures
ν
λ
ν̃
E
E
F
G
p
v
L
F
T
1
12
m(12 C)
frequency (hertz, Hz)
wavelength (metre, m)
wavenumber (cm−1 )
energy (J = joule = kg m2 s−2 , eV, cm−1 )
energy in cm−1
rotational energy in cm−1
vibrational energy in cm−1
momentum (kg m s−1 )
speed (m s−1 )
angular momentum (kg m s−1 )
in spectroscopists’ units (cm−1 Å−1 )
force (newton, N = kg m s−2 or J m−1 ); F
temperature (Kelvin, K)
xiii
LIST OF SYMBOLS
xiv
Quantum Numbers
Molecular Parameters
n
m
J
v
S
mS
I
mI
principal
orbital angular momentum
magnetic
rotational
vibrational
total electronic spin angular momentum
z-component of S
total nuclear spin angular momentum
z-component of I
V (r)
De
re
k, k̃
μ
β
I
Bv
Dv
νe
ωe
ωe xe
Be
α
M
IE
Ke−
εorbital
γ
σ
δ
JAB
molecular potential energy
dissociation energy
equilibrium bond length
bond force constant in J/m2 , cm−1 /Å2
reduced mass
Morse potential exponent
moment of inertia
rotational constant in cm−1
centrifugal distortion constant in cm−1
vibrational frequency in Hz
vibrational frequency in wavenumbers, cm−1
vibrational anharmonicity, in cm−1
inertial rotational constant in wavenumbers, cm−1
molecular polarizability
dipole moment
ionization energy
photoelectron kinetic energy
molecular orbital energy
nuclear magnetogyric ratio
chemical shielding constant
chemical shift
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 thermodynamics, allow us to predict thermodynamic properties (enthalpy, entropy, heat capacity, etc.) of
molecular gases, and equilibrium constants for chemical reactions. The spectrum of a given species also
provides an absolutely unique molecular fingerprint, an essential tool for environmental and analytical
chemistry, and for astrophysics.
The observed spectral transition intensities tell us about the populations of various chemical species
in a given system, and more microscopically, about the relative populations of individual molecular
energy levels. The “selection rules” that govern which levels may be coupled by allowed transitions tell
us about the symmetry of the molecular electronic wave functions, and hence help us identify particular
molecular states.
More generally, spectroscopy is our most exquisite probe of the properties of matter. It stimulated the
development of quantum mechanics, the theory that underlies our current understanding of the nature
of atomic-scale matter, and it provides a powerful means for testing basic theories.
1.1
1.1.1
Light and the Electromagnetic Spectrum
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
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
2
• 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 P V = nRT .
• The Periodic Table of the elements had been developed empirically, although the reason for the periodicity, 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 nanometers nm, where 1 nm = 10−9 m )
• Wavenumber (symbol: ν̃) – 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, ν̃ [cm−1 ],
ν̃ =
1
ν
=
,
2
10 λ
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
oscillation
period 1/ν
+ E0
field
viewed at
fixed point
in space
E(x,t)
0
− E0
0
2
4
6 -15 s
time /10
8
10
wavelength λ
+ E0
field
viewed at
fixed point
in time
E(x,t)
0
− E0
0.0
0.5
1.0
1.5
2.0
distance
/ μm
2.5
3.0
Figure 1.1: The electric field of light oscillates in space and in time.
to the effects of the electric fields. For plane polarized light traveling in the x direction, the the solution of
Maxwell’s equations show that the oscillation of its electric field along the axis of polarization is described
by the expression
2πν
x − 2πν t + φ0
,
(1.3)
E(x, t) = E0 cos
c
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 ν̃ = 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
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
4
10
13
ν / Hz
10
14
10
15
T = 15 000 K
Rayleigh-Jeans
law (15 000 K)
20
Intensity
Ι(ν,T) /
-7
-2
10 J m
10
T = 6 000 K
( × 10 )
5
T = 300 K
( × 104 )
0
infrared
visible
ultraviolet
Figure 1.2: Black-body radiation: observed distributions and the Rayleigh-Jeans law prediction.
(solid curves in Fig. 1.2), but the best available theory, the Rayleigh-Jeans law, predicted that the observed
intensity per unit frequency was given by the expression
I(ν, T ) =
2πkT 2
ν .
c2
(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 ν̃
(1.5)
in which h is a tiny scaling factor, that same derivation gave the distribution law
I(ν, T ) =
1
2πhν 3
.
c2 ehν/kT − 1
(1.6)
This function has the correct qualitative behaviour shown by the solid curves in Fig. 1.2: it increases as
ν 2 at small frequencies, passes through a maximum, and dies off exponentially at high frequencies. Planck
also showed that if his scaling factor was given the value h = 6.626×10−34 J/Hz (now called the Planck
constant),1 Eq. (1.6) yielded essentially exact agreement with experiment!
This result was truly remarkable, but for a number of years many people (initially including Planck
himself) were reluctant to accept the full implications of the quantization postulate of Eq. (1.5). In particular,
although people were compelled to accept the results of his derivation, since the agreement with experiment
was so good, many refused to accept the hypothesis of “quantization” on which it was based, and kept trying
1
Since 1 Hz = 1 s−1 , the units of h are more commonly expressed as joules×seconds, or J·s.
1.1. LIGHT AND THE ELECTROMAGNETIC SPECTRUM
incident
light of
frequency ν
A
e-
positive
anode to
collectfast e
5
↑
maximum
electron
kinetic
energy
emitted
electrons
B
-
e
metal
cathode with
voltage = 0
e-
negative
grid for
measuring
e- kinetic
energy
A
0
ν0
frequency ν [s-1] →
- W0
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 explained 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. 1.3 B. The fact that there is a threshold frequency below which no electrons are emitted merely indicates
2 Quoted from The Quantum Physicists and an Introduction to Their Physics, by W.H. Cropper, Oxford University
Press, 1970.
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
6
that there exists a characteristic minimum energy required to tear an electron free from a particular metal.
This threshold energy
(1.7)
W0 = h ν0
is called the “work function” of the metal. Quanta of light of frequency ν0 have sufficient energy to dislodge
the electrons, but no remaining energy to transfer to the electron as kinetic energy.
In this theory, Einstein extended Planck’s idea of quantized energy packets emitted or absorbed by matter
and predicted that the energy carried about by light was also quantized. When accurate data later became
available, it was also found that the empirical constant h determined as the slope of the plot in Fig. 1.3 B
was exactly the same as the empirical constant determined by fitting Eq. (1.6) to the observed intensity
distributions of black-body radiation. This result showed that Planck’s energy quantum ε(ν) 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 University in Saint Louis Missouri.3 He found that when monochromatic (single-wavelength) light of very short
wavelength (X-rays) passed through thin films of solid material, the scattered light had two components: (i)
intense scattered light with exactly the same wavelength as the incident X-rays, and (ii) low intensity scattered light with slightly longer wavelengths, where the magnitude of the wavelength shift varied with the
angle of deflection from the direction of the incident beam. Compton showed that his observations were
quantitatively explained if both the electrons in the material and the quanta or ‘photons’ of light behaved
like classical billiard balls undergoing collisions subject to the normal energy and momentum conservation
laws of classical mechanics. However, this evidence also required him to devise some definition for the
momentum of a photon. This was done by combining Einstein’s famous special relativity mass–energy relationship, E = m c2 , 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 .
3
(1.8)
Not all of the key work establishing quantum mechanics was done in the great universities of Europe!
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.
4
1.1. LIGHT AND THE ELECTROMAGNETIC SPECTRUM
7
Rearranging this expression and making use of the usual frequency/wavelength relationship of Eq. (1.1) yields
Compton’s expression for the momentum of a photon of light of wavelength λ:
pλ = h/λ .
(1.9)
Thus, while Planck and Einstein showed that the energy associated with light of a given frequency (or
wavelength, or colour) was quantized in minute packets of magnitude ε = 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, frequencies and wavenumbers equivalently, jumping back and forth between J, Hz and cm−1 while talking all
the time about “energy”. The Planck equation is the justification for this, as it demonstrates the direct proportionality between the energy and frequency of light. Moreover, use of particular experimental techniques
leads to the use of seemingly unrelated units such as the electron volt (eV) in certain types of spectroscopy
(see Chapter 6). The table below [taken from Rev. Mod. Phys. 80, 633 (2008)] will facilitate conversions
among these various “energy-like” units.
Table 1.1: Conversion factors for energy units encountered in spectroscopy.
joule (J)
1 joule (J) =
1
1 eV = 1.602 176 487×10−19
cm−1
eV
6.241 509 65×10
18
5.034 117 47×10
Hz
22
1.509 190 45×1033
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 58×1010
6.626 068 86×10−34
4.135 667 33×10−15
3.335 640 951×10−11
1
1 Hz =
Although all of the energy units appearing above are sometimes used in molecular spectroscopy, the most
widely used unit is wavenumbers, with units cm−1 , and in most cases it will be the unit used in this text.
Moreover, although the SI unit of length is the meter, and the nanometer (1 nm = 10−9 m) is commonly
used to characterize the wavelength of light, molecular dimensions are most commonly reported in units of
Å ( 1 Å = 10−10 m), and this is the unit that will be use for molecular bond lengths. Similarly, although the
SI unit for mass is kg, in discussing and performing calculations for molecules it is much more convenient
to use atomic mass units, mu = 1 u ≡ m(12 C)/12 . In spite of the above, formal expressions for molecular
level energies encountered in this course are normally derived and written down in SI units, with energy in
joules, mass in kilograms, and length in meters. It would of course be quite tedious if we had to undertake
detailed unit conversions in every calculation, but if we think ahead, this will not be necessary.
The theoretical formulae for the energy associated with many phenomena considered in molecular physics
contain a factor of the form 2 /(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], where mu is the mass in kg of one atomic mass unit (see p. xiii). Similarly,
8
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
because we wish to quote distances in Ångströms, we substitute d [m] = 10−10 d [Å]. Combining these terms
with the factor 1/(102 hc) required to convert from [J] to [cm−1 ], the ubiquitous factor 2 /(2M d2 ) becomes
2
( [J · s])2
1020
1
Cu
16.857 629
=
2 =
2 =
2
2
2 hc
2
(m
[kg])
10
u
2(M [kg]) (d [m])
(M [u]) d [Å]
(M [u]) d [Å]
(M [u]) d [Å]
in which the numerical value of the constant Cu = 16.857 629 056, which we call the “inertial constant”,
is obtained on substituting values of the various physical constants into the initial versions of the above
2
expression. This conversion of 2 / 2M d2 [J] to Cu / (M [u]) d [Å] [cm−1 ] appears repeatedly in the
following chapters.
1.2
1.2.1
Quantum Theory of Matter
The Spectrum of the Hydrogen Atom
In 1900 it was known that atoms were roughly 10−10 m = 0.1 nm in diameter, but it was not clear what
their structure was or where the electrons were located. Then in 1911 Ernest Rutherford proposed the
“nuclear” model of the atom, according to which all positive charge is located in a tiny nucleus of diameter
∼ 10−14 m, while the electrons move about it in orbits of diameter ∼ 10−10 m that define the effective
atomic size.5 However, particles moving in a circular orbit are constantly accelerating towards the center,
and classical electromagnetic theory predicts that charged particles that are accelerating spontaneously
emit light. This prediction is indeed obeyed by atomic-scale particles, and it is the basis for very intense
tunable light source machines known as “synchrotrons”, such as the “Canadian Light Source” facility in
Saskatoon, Saskatchewan (see http://www.lightsource.ca). However, within an atom this would spell
disaster: if the orbiting electrons emitted light they would lose energy, slow down, and eventually spiral into
the nucleus.6 This “collapsing atom” problem appeared to raise serious questions about the validity of the
Rutherford model.
Black-bodies were well known to emit light over a continuous range of frequencies, and the distribution
of their intensities was explained by Planck, as discussed above. However, by the early 1900’s experimental
spectroscopy had also shown that individual types of atoms and molecules absorbed or emitted light at
certain discrete frequencies (or ‘colours’). The simplest atom, hydrogen, was the most intensively studied,
since it should be the easiest to understand. Indeed, around 1885 the Swiss schoolteacher Johannes Balmer
had shown that the lines of the emission spectrum of gaseous H atoms in the visible region could be exactly
explained by the formula
n1 2
λ = A
(1.10)
in which
n1 = 3, 4, 5, 6, . . .
n1 2 − 4
and A is a constant. Today it is more common describe this series of lines using the expression obtained on
inverting the left- and right-hand sides of Eq. (1.10):
1
1
ν̃ = RH
−
,
(1.11)
(2)2
(n1 )2
in which (recall Eqs. (1.1) and (1.2)) ν̃ 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
1
1
ν̃ = RH
−
,
(1.12)
(n2 )2
(n1 )2
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
4000 1000
⏐ ⏐
0
⏐
9
400
200
⏐
⏐
20000
40000
wavelength λ / nm
100
⏐
60000
80000
100000
120000
− / cm-1
wavenumber ν
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 1 1 1
, , ,
,
,
, ...
= {1, 0.25, 0.111111, 0.0625, 0.040, 0.027777, . . .}
1 4 9 16 25 36
each of these series converges to a characteristic limit ν̃∞ = 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
1
2
3
4
5
series
Lyman
Balmer
Ritz-Paschen
Brackett
Pfund
region
far ultraviolet
visible
near-infrared
mid-infrared
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 θ̇/2π, where θ̇ 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 θ̇, 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 θ̇/2π, and then led to the level energy
expression (in cm−1 ):
e 4 μH
1
1
n = −
E
= − RH 2
(1.13)
2
2
2
2
2
32π 0 10 hc n
n
434. 2 nm
410. 3 nm
486. 2 nm
656. 5 nm
-20000
n=4
n=3
1094. nm
n=∞
0
energy
/ cm-1
1282. nm
1875. nm
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
10
Ritz-Paschen
Balmer
n=2
-40000
Lyman
Coulomb potential
121.56 nm
102.57 nm
97.25 nm
94.98 nm
93.78 nm
-60000
2
−e
V(r) = ⎯⎯⎯⎯⎯⎯
(4π ε0 102 hc) r
-80000
-100000
n=1
-120000
0
1
2
3
r/Å
4
5
6
7
Figure 1.5: Hydrogen atom energy levels and transitions.
in which e is the electron charge, 0 is the permittivity of vacuum (a known constant which arises in
electrostatics), and μH = me mp /(me + mp ) is what we call the “reduced mass” of this two-particle system
(see § 2.1.2). Note that the “tilde” () over a symbol for energy indicates that its units are those of the
wavenumber, cm−1 , while the factor of 102 in the denominator of Eq. (1.13) converts from the SI unit of
inverse length (m−1 ) to the common “spectroscopists’ unit” of cm−1 . Substituting the known values of the
various physical constants into Bohr’s expression for RH yields exactly the same value of this constant that
Rydberg had determined empirically by fitting the observed positions of the lines in the H-atom spectrum
to Eq. (1.12)! Another result yielded by this derivation is that the magnitude of the angular momentum
of the electron (L) is an integer multiple of , L = n , 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
1
1
ΔE(n1 , n2 ) = En1 − En2 = RH
−
= ν̃
(1.14)
(n2 )2
(n1 )2
agrees exactly with the empirical Rydberg expression of Eq. (1.12). Thus, radiation with wavenumber
1 , n2 ) is emitted or absorbed when the electron undergoes a transition between quantum states n1
ΔE(n
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 Å is the radius of the Bohr orbit in the ground ( n = 1 ) level of an 1 H 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 mA
nuc and charge +Ze, the energy levels are given by the formula
μA
1
1
A
2
En = − RA 2 = − Z
RH 2 ,
n
μH
n
(1.15)
A
7
in which μA = me mA
This generalization allowed
nuc /(me + mnuc ) is the reduced mass of this system.
1
Bohr to explain the differences between the transition energies of the H and 2 H 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 7 Li+2 and 6 Li+2 . For example, consider the one-electron 7 Li+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
7 +3
7 +3
Li
Li
μ7Li+2 = mnuc
me / mnuc
+ 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 7 Li+2 than for 6 Li+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
h
≡
mv
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
7 Ignoring the very small relativistic mass correction, mA
nuc = MA − Z me , in which MA is the standard atomic mass of
atomic isotope A.
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
12
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 Schrödinger Equation
Bohr’s successful combination of a classical description of the motion of the electron and nucleus with
his seemingly arbitrary postulate regarding quantization had yielded a successful ‘quantum-plus-mechanics’
description of one-electron atoms. However, this model did not work for more complicated systems, and after
a decade of efforts to extend or generalize it, it became clear that it was a dead end, and that an entirely
new approach was required. The first successful more general theory to appear was the “matrix mechanics”
proposed by Werner Heisenberg in 1925. However, because of its mathematical complexity and sophistication
it is not normally considered to be a good starting point for a beginner’s approach to quantum mechanics.
A somewhat more accessible approach was the differential equation formulation that the Austrian physicist
Erwin Schrödinger developed and published in a series of papers in 1926.
Conceptually, mathematically, and in a sense physically, Schrödinger’s mechanics was altogether different from Heisenberg’s mechanics. While Heisenberg’s method is algebraic, Schrödinger’s method begins
with a differential equation. Heisenberg’s method builds on discrete and discontinuous quantities, whereas
Schrödinger’s mechanics is based on a quantity that is continuous. On the wave-particle duality question,
the Heisenberg procedure seems to side with the particle viewpoint; in contrast, Schrödinger’s differential
equation is very explicitly a “wave equation”.
As might be expected, Heisenberg and Schrödinger initially had difficulty accepting the validity of each
other’s theories. In a letter to theoretical physicist Wolfgang Pauli, Heisenberg wrote:2
“The more I ponder about the physical part of Schrödinger’s theory,
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, Schrödinger had doubts about the physical content of Heisenberg’s matrix mechanics, and wrote:2
“I was discouraged, if not repelled, by what appears to me a rather difficult method of
transcendental algebra, defying any (intuitive) visualization.”
Erwin Schrödinger took de Broglie’s proposed matter waves to mean that a proper description of particle
behaviour would be given by a ‘wave function’ – a mathematical description of its wave nature – one of
whose properties was that the particle could exist only in certain quantized states, as in the Bohr model.
Schrödinger’s description also provided a rationale for these discrete states: they were states in which
the electron waves would not destructively interfere. No rigorous derivation is possible for Schrödinger’s
differential equation (nor for Heisenberg’s method); he simply wrote it down, based on his deep understanding
of the mathematics of classical wave theory and a profound intuitive acceptance of the wave nature of
particles. However, what he dreamed up provides the basis for all of our current understanding of atomic
matter and of chemistry. We next describe the application of his method to two key illustrative cases: (i)
the “particle-in-a-box” problem, a simple model system whose description illustrates how “quantization”
arises, and (ii) the hydrogenic atom, which provides the basis for our description of atomic orbitals and of
the chemist’s periodic table.
1.3.1
A Particle in a One-Dimensional Box
For the special case of a particle of mass m with total energy E moving in one dimension along the x
coordinate subject to a potential energy field V (x), Schrödinger’s differential equation for the particle/wave’s
displacement y(x) has the form
−
2 d2 y(x)
+ V (x) y(x) = E y(x) .
2m dx2
(1.17)
1.3. WAVE MECHANICS AND THE SCHRÖDINGER 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 Schrödinger differential equation describing
the particle becomes simply
2m E
d2 y
d2 y
=
−
= − k2 y ,
(1.18)
y
or
dx2
2
dx2
in which k 2 = 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 (k 2 ) 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 Schrödinger’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 Schrödinger’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 Schrödinger’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 onedimensional problems, such as the particle-in-a-box problem that we have just discussed.
3. The wave function ψ is a continuous mathematical function of all coordinates of the system.
4. The probability of finding the system with a particular configuration (i.e., with a particular set of spatial
coordinates) is proportional to |ψ|2 .
The wave equation solution ψ is a mathematical function that can have positive or negative
(or complex number) values, but the square of its absolute value is always non-negative:
|ψ|2 ≥ 0 . Since the particle must be somewhere, the probabilities must all add up to
unity. For our one-dimensional system this requirement is expressed mathematically by the
“normalization” condition
+∞
−∞
|ψ|2 dx = 1
(1.20)
This shows that |ψ|2 is actually a “probability density”, or in our one-dimensional case, the
probability per unit distance.
14
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
V(x)
n
⎯ 7
↑
energy
m =1.0
m = 1.0
m = 2.25
6
n=4
5
n=3
4
n=2
3
n =1
2
1
0
x → L =1.0
0
x→
L =1.5
0 x → L =1.0
Figure 1.6: Potential energy function V (x) (solid curves), discrete level energies (dashed lines) and wave
functions (dotted curves) for three “square-well” particle-in-a-box systems.
For the case of our particle confined to a rigid one-dimensional box, we can see that the general function
ψ(x) = A sin(kx) + B cos(kx)
(1.21)
in which A and B are arbitrary numerical constants, satisfies the first three rules. It is a continuous function
of all of the coordinates of the system (here, x), and it is easy to verify that if sin(kx) 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 = n π 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 Schrödinger equation for this
system which satisfy our four Rules of wave mechanics are those associated with discrete values of the
1.3. WAVE MECHANICS AND THE SCHRÖDINGER EQUATION
15
constant k corresponding to positive integer values of n :8
k =
nπ
2mE
=
2
L
for
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 π 2
2 m L2
n2
(1.26)
In “spectroscopists’ units”, with mass in u, length in Ångströms and energy in cm−1 , this yields the expression
that we commonly use for calculations:
n = Cu
E
π2
m[u] (L[Å])2
n2 [cm−1 ]
(1.27)
2
in which Cu = 16.857 629 056 [u cm−1 Å ] 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 Schrödinger equation
and must be continuous with the interpretation of |ψ(x)|2 as a probability density that gives rise to the
phenomenon of quantization of energy in molecular systems.
Equation (1.26) shows that the energy of our particle is only allowed to have the discrete values En =
2 2
π /2mL2 , 4 π 2 2 /2mL2 , 9 π 2 2 /2mL2 , 16 π 2
2 /2mL2 , . . . etc., and the associated (normalized) wave functions (or “eigenfunctions”) are ψn (x) = 2/L sin(nπx/L) . For three related “square-well”
particle-in-a-box models, Fig. 1.6 shows the allowed energy levels (horizontal dashed lines) and the associated
wave functions (dotted curves).9 They illustrate properties that apply to virtually all quantum systems.
• For every allowed “eigenstate” (or distinct solution of the Schrödinger 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.]
8 While 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.
10 Our 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.
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
16
∞
∞
V(x)
r→
0
n=4
n=6
n=3
n=2
n=5
2
−e
V(r) = ⎯⎯⎯
4π ε0 r
n=4
n=3
A
n=2
n=1
0
− RH
L
x→
B
n=1
V(r) = ½ k (r−re)2
υ=5
De →
υ=5
υ=4
υ=4
υ=3
υ=2
υ=3
C
υ=2
υ=1
D
υ=1
υ=0
υ=0
re
r→
re
r→
Figure 1.7: Level energies and wave functions for particle-in-a-box problems associated with differently
shaped potential energy functions. A – square-well potential with infinitely rigid walls; B – Coulomb
potential and effective radial wave functions for an electron in an H atom; C – harmonic-oscillator potential
for molecular vibration; D – Morse potential for molecular vibration.
The above properties also apply to the wave functions for the orbital motion of an electron in an atom
and the rotation of a molecule (see §1.3.2), for the radial motion of an electron in an atom, and for vibrational
motions in a molecule. For the two latter cases the differential equation governing the radial or stretching
motion can be written in precisely the same form as Eq. (1.17), the only difference being that the potential
energy function differs from one case to another. In the various cases, the precise mathematical dependence
of the level energy on the trap size (L in Eq. (1.26) or Fig. 1.6) or particle mass m differ, but qualitative
features remain the same. For example, Fig. 1.7 shows the energies (horizontal dashed lines) and wave
functions (dotted curves) for the lowest few levels of a particle of mass m trapped in potential energy wells
(generalized boxes) with four representative shapes. Case A is the conventional “square-well” problem
discussed above, with V (x) = 0 inside the box; case B is for the radial motion of an electron in an H atom
(c.f. Fig. 1.5); case C is the harmonic-oscillator model for molecular vibration, which is discussed in Chapter
3, and case D shows the solutions of the radial Schrödinger equation for molecular vibration obtained using
the more realistic Morse function model for molecular vibration, also discussed in Chapter 3. Following the
spectroscopic convention for labeling eigenstates of vibrational motion, the levels in segments C and D of
Fig. 1.7 are labeled v = 0, 1, 2, 3, . . . , rather than n = 1, 2, 3, 4, . . . etc. However, this naming convention
has no effect on the properties listed above.
It is immediately clear that although the four properties listed above apply to all cases illustrated in
Fig. 1.7, the pattern of level energy spacings depends very strongly on the shape of the box. In particular,
1.3. WAVE MECHANICS AND THE SCHRÖDINGER EQUATION
17
in cases B–D the width of the well (effectively, the width of the box) increases with energy, and the rate
at which it does so increases from C to D to B. As a result, whereas the spacings between adjacent levels
increase with energy for the square-well problem of Case A, this trend is halted for C and reversed for
cases D and B, with the most extreme reversal being for the Coulomb potential (Case B), which is the
case for which the width of the potential energy well (the effective box length) increases most rapidly with
increasing energy. We will see in Chapter 3 that this dependence of level spacings on well width can be used
to determine the properties of the potential-energy function governing molecular vibration.
A final observation about the results shown in Figs. 1.6 and 1.7 is the fact that the lowest allowed energy
level never lies at the bottom or energy zero of the potential energy function. Except for the Coulomb case,
for which the potential function goes to minus infinity (Fig. 1.7 B), the gap between the energy of the lowest
allowed level and the potential minimum is called the “zero-point energy” of the system. This existence
of a finite zero-point energy is a special property of some quantum systems that has no analog in classical
mechanics. However, as we shall see in the next section, it does not appear in the quantum description of
rotational or orbital motion.
1.3.2
Orbital or Rotational Motion: A Particle on a Ring
Orbital motion of an electron or rotational motion of a molecule occur in three-dimensional space and require
more sophisticated mathematical treatments than are appropriate here. However, the simple one-dimensional
problem of a particle of mass m moving in a flat circular orbit with radius r provides a realistic illustrative
model for these motions. Your first course in quantum mechanics will teach you that the Schrödinger equation
for this model problem has the form
−
or
2 d2 ψ(φ)
2m r2 dφ2
d2 ψ(φ)
dφ2
=
=
E ψ(φ)
2m r2 E
−
ψ(φ) = − b2 ψ(φ)
2
(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 = 2m r2 E/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:
2
π2
2
2
=
(1.29)
2
E =
2m r2
2m (π r)2
in which = 0, 1, 2, 3, . . . , and the associated wave functions can be written as ψ(φ) = 1/2π sin( φ + δ) ,
where δ is an arbitrary phase constant. For practical work we wish, of course, to have distances in Å and
masses in u and to generate energies in cm−1 , in which case Eq. (1.29) becomes
Cu
2
−1
E =
(1.30)
2 [cm ]
m[u] r[Å]
2
in which Cu = 16.857 629 056 [u cm−1 Å ] 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,
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
18
l=0
l =1
l=2
l=3
l=4
l=5
Figure 1.8: Particle-on-a-ring orbits (simple circles) and the wave functions (oscillating curves) associated
with the eight lowest-energy levels.
is that in the present case the = 0 state with a total energy of zero is allowed; i.e., there is no zero-point
energy for rotation. This reflects the fact that in the square-well problem, if the energy lay at the potential
minimum the wavefunction inside the box could only join continuously to that outside if its amplitude was
everywhere zero, a result that would contradict the normalization condition of Eq. (1.20). In contrast, for
continuity to be satisfied in the present case, all the wavefunction must do is join to itself smoothly
whenever
the angle φ increases by 2π, and this is possible for E = 0 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 Schrödinger equation. However, note that the time-independent Schrödinger equation for a
general system is written symbolically as
ψ = Eψ
H
(1.31)
is the Hamiltonian operator for the system, a generalization to three dimensions (and multiple
in which H
1D = − 2 d2 2 + V (x) appearing in Eq. (1.17). As with the
particles, if appropriate) of the operator H
2m dx
one-dimensional problems discussed in §1.3, Eq. (1.31) always has many different solutions corresponding
to discrete allowed energy eigenvalues. This section reviews the properties of some familiar atomic and
molecular orbital solutions of the Schrödinger equation that you encountered in your introductory chemistry
course(s) in the light of the quantum-mechanical language introduced in the preceding section.
1.4.1
Hydrogenic Atomic Orbitals
The Schrödinger equation for the hydrogen atom may be written as a three-dimensional version of Eq. (1.17)
for a pseudo-particle with effective mass μH = me mp /(me + mp ) , where me and mp are the masses of the
electron and proton, respectively (note that since mp me , μH ≈ me ). Because an atom is inherently
spherically symmetric, it is most convenient to describe this system using the spherical polar coordinates r, θ
and φ in place of the conventional Cartesian coordinates x, y and z (see Fig. 1.9). Although mathematically
somewhat more complicated than Eq. (1.17), this differential equation can also be solved exactly in closed
form. This solution yields precisely the same energy level expression obtained from Bohr’s “old quantum
1.4. ELECTRONIC STRUCTURE OF ATOMS AND MOLECULES
19
→
theory”, Eqs. (1.13) and (1.15), and the associated wave functions
z
define the familiar hydrogenic atomic orbitals presented in introductory Chemistry courses. As the electron in the atom is free
+ (x,y,z)
to move in three dimensions, it should be no surprise that these
x = r sinθ cosφ
(r,θ,φ)
θ
solutions are characterized by three quantum numbers, n, and
→
y = r sinθ sinφ
r
z = r cosθ
m . For the generalized case of an electron orbiting around a bare
→
y
nucleus with charge +Ze, where e is the magnitude of the elecφ
tron charge (i.e., for the one-electron atom or ions H, He+1 , Li+2 ,
+3
Be , . . . etc.), the eigenfunctions of the lowest few levels are listed
→
x
in Table 1.2.
Figure 1.9: Definition of spherical poIt is easy to see that each of these hydrogenic functions11 may
lar coordinates in terms of rectangular
be written as the product of a function of r times a function of θ
Cartesian coordinates.
times a function of φ times a constant factor which imposes the
three-dimensional analog of the normalization condition of Eq. (1.20). Moreover, if we associate the hydrogenic quantum number |m | with the one-dimensional orbital quantum number of §1.3.2, we see that the
φ–dependent parts of these hydrogenic wavefunctions have the same form as the simple flat-orbit wavefunctions of §1.3.2.
The radial part of each of these eigenfunctions, Rn,l (r) , is itself the product of an exponential term times
a member of the family of “Laguerre polynomials”, a class of functions whose properties have been thoroughly
studied by mathematicians. However, a little mathematical manipulation of the differential equation that
yields these radial functions converts it to the familiar form of Eq. (1.17) with x replaced by r and V (x) by
−C1 /r, with C1 = Ze2 /4π0 , the Coulomb coefficient for an electron interacting with a nucleus of charge
11 The word “hydrogenic” labels atomic systems consisting of a single electron interacting with a nucleus of charge +Ze (e.g.,
H, 2 H 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 Å is the Bohr radius.
3/2
n = 1, = 0, m = 0;
ψ1s = √1π aZo
e−Zr/ao
3/2 −Zr/2ao
2 − Zr
n = 2, = 0, m = 0;
ψ2s = 4√12π aZo
ao e
3/2 Zr
−Zr/2ao
= 1, m = 0;
ψ2pz = 4√12π aZo
cos θ
ao e
3/2 Zr
−Zr/2ao
= 1, m = ±1;
ψ2px = 4√12π aZo
sin θ cos φ
ao e
3/2 Zr
−Zr/2ao
sin θ sin φ
ψ2py = 4√12π aZo
ao e
3/2 Z 2 r2
n = 3, = 0, m = 0;
27 − 18 Zr
e−Zr/3ao
ψ3s = 81√1 3π aZo
ao + 2 a2o
3/2 √
Z 2 r2
6 Zr
e−Zr/3ao cos θ
= 1, m = 0;
ψ3pz = 81√2π aZo
ao − a2o
3/2 √
Z 2 r2
= 1, m = ±1;
6 Zr
e−Zr/3ao sin θ cos φ
ψ3px = 81√2π aZo
ao − a2o
√
3/2
Z 2 r2
6 Zr
e−Zr/3ao sin θ sin φ
ψ3py = 81√2π aZo
ao − a2o
3/2 2 2 Z r
= 2, m = 0;
e−Zr/3ao 3 cos2 θ − 1
ψ3dz2 = 81√1 6π aZo
a2o
3/2 2 2 √
Z r
e−Zr/3ao sin θ cos θ cos φ
ψ3dxz = 81√2π aZo
= 2, m = ±1;
a2o
3/2 2 2 √
Z r
ψ3dyz = 81√2π aZo
e−Zr/3ao sin θ cos θ sin φ
a2
3/2 o 2 2 Z r
e−Zr/3ao sin2 θ cos 2φ
ψ3dx2 −y2 = 81√1 2π aZo
= 2, m = ±2;
a2
3/2 2 2 o
Z r
ψ3dxy = 81√1 2π aZo
e−Zr/3ao sin2 θ sin 2φ
a2
o
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
20
3
2
1
0
r R1s(r)
2
r R2p(r)
r R2s(r)
1
0
-1
r R3d(r)
r R3p(r)
r R3s(r)
1
0
-1
1
r R4s(r)
r R4p(r)
r R4f (r)
0
-1
0
r R4d(r)
5
10
15
r /Å
20 0
5
10
15
r /Å
20 0
5
10
15
r /Å
20
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
Schrödinger equation for a particle of mass μH trapped in a “box” defined by a Coulomb potential energy
function. Because of this relationship to our familiar particle-in-a-box problem (see also Fig. 1.7), we have
chosen to consider the product 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
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 threedimensional hydrogen atomic orbitals may also be considered
to be solutions of a particle-in-a-box problem.
At this point it is important to recall what the atomic
wave function ψ really is: it is a mathematical description
of the properties of the electron, and it generally has oscillatory behaviour of the type commonly associated with
classical wave motion. Because it is an ordinary mathemati- Figure 1.11: Angular behaviour of H atom
cal function, it can have positive and negative values, as seen
orbitals for n = 1 − 3 .
1.4. ELECTRONIC STRUCTURE OF ATOMS AND MOLECULES
21
Figure 1.12: Atomic orbital energies in some many-electron atoms.
in Figs. 1.6 − 1.8, 1.10 and 1.11. The algebraic sign of the wave function is important when we consider how
orbitals overlap to hybridize (e.g., to yield sp or sp3 orbitals) or to yield bonding or anti-bonding molecular orbitals. However, insofar as the wavefunction represents a physical property, it is its square |ψ|2 , the
“probability density” of finding a system in a particular configuration, that matters.
1.4.2
Multi-Electron Atoms and Atomic Spectroscopy
The orbital wave functions that are commonly used to describe electronic structure in atoms or molecules
are solutions to the simple one-electron hydrogenic atom Schrödinger equation described above. However,
although orbitals in multi-electron atoms have the same types of symmetry seen there, and many similar
properties, they are different for two reasons. The first is simply the fact that the larger nuclear charge +Ze
gives rise to an electron–nucleus attraction which is Z times stronger than that in a simple H-atom; this
observation also applies to our general hydrogenic (one-electron) atoms, as seen in the exact wave function
and eigenvalue expressions of Table 1.2 and Eq. (1.15). More serious complications arise from the effects of
electron–electron repulsion and the fact that inner-shell electrons partially shield the nucleus from the outer
ones, so that they effectively feel only a portion of its actual charge. On the one hand, these considerations
greatly increase the complexity of the Schrödinger equation so that it cannot be solved exactly in closed
form, not even for the two-electron He atom. On the other hand, numerical quantum-chemistry computer
programs can solve the resulting Schrödinger equation to very high precision for a wide range of cases, and
for simple atoms those computational results can be made almost as exact as one could desire.
For elements in the first rows of the periodic table, Fig. 1.12 shows the accurate calculated energies of the
valence (outer shell) electrons. On the extreme left side of this figure we see that the He+ (1s) binding energy
is four times larger than that for H(1s), a straightforward manifestation of the effect of the Z 2 factor in
Eq. (1.15). However, the analogous binding energy of a 1s electron in a neutral He atom is only approximately
twice (instead of four times) as large as that for an H atom, partly because of the electron–electron repulsion
energy, and partly because each of those two 1s electrons partially shields the nucleus from the other. For the
two 1s electrons of a neutral Li atom the same considerations apply, making their binding energy far smaller
than the factor of nine times stronger than that for H that is predicted by Eq. (1.15) for a one-electron Li+2
system. Moreover, the fact that those two 1s electrons are so tightly bound to the nucleus means that they
shield it very efficiently from the outermost 2s electron whose binding energy, as a result, is only about twice
22
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
(instead of nine times!) as strong as that for the 2s orbital of an H atom. These same considerations explain
most of the trends among the level energies as one continues across this diagram. The most significant
additional effect, seen for the 2p electrons for elements from C to F, is that when the p subshell has between
2 and 5 electrons, the p orbitals in a given atom are no longer all energetically equivalent to one another.
The trends in behaviour seen in Fig. 1.12 are certainly interesting in their own right. However, the main
objective of this discussion is to point out that as in the simple H atom, complex many-electron atoms also
have ladders of energy levels, and that the associated wave functions also oscillate increasingly rapidly in
space and have more nodal surfaces as the energy increases. For a multi-electron atom the total electronic
energy is the sum of the energies of all its electrons, and as with the H atom, there always exist an infinite
number of empty allowed eigenstates lying above the highest occupied ones.
In atomic spectroscopy, light whose photon energy matches one of the level spacings in the atom promotes
one or more of the electrons from one of the occupied lower-energy orbitals into one of the unfilled higherenergy orbitals. For hydrogenic atoms those orbital energies are accurately given by Eq. (1.15), but for
multi-electron atoms there are no analogous explicit expressions for the orbital energies. On the other hand,
accurate measurements have been made of the allowed spectroscopic transition energies of all of the atoms
in the periodic table, and those results pose a persistent challenge to theoreticians as the latter pursue
the development of ever better and more accurate computational methods. However, the most important
practical application of atomic spectroscopy is the fact that the unique spectroscopic fingerprints of allowed
transitions for each species provides a marvelously general technique for identifying the atomic composition
of unknown molecules and materials.
1.4.3
Molecular Energies and the Born-Oppenheimer Approximation
A molecule differs from a multi-electron atom in that instead of having a single positively charged nucleus, it has two or more positively charged nuclei about which the electrons are distributed. This makes
the description of molecules much more complicated than for atoms because one must take account of the
simultaneous motion of both the nuclei and the several electrons. Fortunately, a remarkably effective approximation method, the Born-Oppenheimer approximation, allows the nuclear and the electronic motion to be
treated separately, and permits a relatively straightforward treatment of the nuclear motion – the molecular
vibrations and overall rotation of the molecule.
In 1927, Max Born and Robert Oppenheimer recognized that since nuclei are much more massive than
electrons ( mp ≈ 1800 me ), if they had comparable energies the nuclei would move much more slowly than
electrons. This situation suggested that for each instantaneous configuration of those slowly-moving nuclei
one might ignore their motion and solve the Schrödinger equation for the electrons alone. The detailed
properties of those solutions would, of course, vary as the nuclei slowly moved about, but taking account
of such changes would be much simpler than trying to solve the Schrödinger equation for all electrons and
nuclei at the same time.
In the mathematical language of quantum mechanics, the Born-Oppenheimer approximation assumes
that the total Hamiltonian for the system may be written as the sum of a Hamiltonian describing the
behaviour of the electrons and one describing the nuclear motion
total = H
electrons + H
nuclei
H
(1.32)
and that the total wave function can be written as a product of functions characterizing the electronic and
the nuclear motion
ψtotal = ψelectrons × ψnuclei
(1.33)
The electronic wave functions ψelectrons are the solutions of the electronic Schrödinger equation
electrons ψelectrons = Eel ψelectrons
H
(1.34)
for the nuclei fixed in one particular configuration. However, both these wave functions and the electronic
energy eigenvalues Eel will vary as the nuclei move.
Just as an atom has many electronic energy levels, so does a molecule. For a molecule, however, those
energies depend on the relative positions of the various nuclei. For the diatomic molecule Li2 , Fig. 1.13
1.4. ELECTRONIC STRUCTURE OF ATOMS AND MOLECULES
23
Li2
30000
energy
/ cm-1
2
2
Li( S) + Li( P)
20000
Li(2S) + Li(2S)
10000
De
0
2
re
4
r/Å
6
8
Figure 1.13: Potential energy curves for the ten lowest energy electronic states of Li2 .
shows how the energies Eel = Eel (r) calculated by solving the electronic Schrödinger equation vary with the
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 Schrödinger 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 Schrödinger equation, Eq. (1.34), and on the dependence of its energy eigenvalues
on internuclear separation. However, an amazing feature of this approximation is the extremely simple form
of the resulting differential equation governing the nuclear motion. As you will learn in your first quantum
mechanics course, on substituting Eqs. (1.32) and (1.33) into Eq. (1.31) and then neglecting certain minor
terms, one obtains an effective Schrödinger equation for vibrational motion that for a diatomic molecule has
the familiar particle-in-a-box form
−
2 d2 ψ(r)
+ V (r) ψ(r) = E ψ(r)
2μ dr2
(1.35)
in which μ = mA mB / (mA + mB ) is the effective or vibrational “reduced mass” for a diatomic molecule
formed from atoms A and B with masses mA and mB , respectively. This equation clearly has exactly the same
12
I. Schmidt-Mink, W. Müller and W. Meyer, Chemical Physics 92, 263 (1985).
24
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
Figure 1.14: Regions of the electromagnetic spectrum and the type of molecular motion and spectroscopy
associated with each.
form as our generic Schrödinger equation Eq. (1.17) for the one-dimensional motion of a particle subject to
the potential energy function V (x)! Hence, solution of the electronic Schrödinger equation Eq. (1.34) yields
the potential-energy function V (r) that governs the dynamical behaviour of the nuclei, and the differential
equation for the latter is essentially the same as that for a particle trapped in a one-dimensional box. Chapter
3 exploits this result further.
1.5
Spectroscopy at last . . .
By now you must realize that all the information that we need concerning a molecule and its quantized states
can be obtained from the Schrödinger equation,
total ψtotal = Etotal ψtotal
H
(1.36)
total incorporates all interactions within the system, Etotal is one of the observable energy eigenin which H
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 Schrödinger equation simultaneously for all of the types of motion that a molecule can have. In particular, a molecule can have
electronic, vibrational, rotational and nuclear-spin-orientation degrees of freedom, each with their own characteristic sets of level energies. Fortunately, within an extended Born-Oppenheimer type of approximation,
each type of motion can be treated independently, which allows us to treat molecular energies for the different
types of motion separately. Moreover, as illustrated by Fig. 1.14, transitions between levels associated with
the different types of motion tend to lie in distinctly different regions of the electromagnetic spectrum. (A
fascinating survey of the properties of the different regions of the electromagnetic spectrum may be found
on the NASA www site at http://missionscience.nasa.gov/ems/.)
nuclei can be written as
As a straightforward extension of Eq. (1.32), the Hamiltonian for nuclear motion H
total to be written
a sum of terms representing four different types of nuclear behaviour, thereby allowing H
as:
electrons + H
nuclei = H
electrons + H
vibration + H
rotation + H
nuclear−spin
total = H
(1.37)
H
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)
rotation , Chapters 3 and 4 the spectra
In the following, Chapter 2 will discuss the spectra associated with H
associated with Hvibration and its coupling with Hrotation , Chapters 5 and 6 the types of spectra arising
electrons + H
vibration + H
rotation , while Chapter 7 discusses the
from the combination of the effects of H
nuclear spin , which are completely uncoupled from the other types of motion.
manifestations of H
1.6
Problems
1. Calculate the frequency and wavenumber of electromagnetic radiation with a wavelength of 4.0×10−10 m.
What region of the electromagnetic spectrum does this fall into? What is the energy of the photons
from this radiation in joules/photon and kilo-joules/mole?
2. Given that it takes a minimum energy of 4.40 eV to dislodge electrons from the surface of chromium
metal, what is the maximum kinetic energy of electrons emitted from a chromium surface when it is
irradiated with ultraviolet radiation of wavelength 200 nm? What is the de Broglie wavelength of the
emitted electron?
3. Two energy levels in a molecule are separated by 7.50×10−22 J. What is the energy separation in
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 87 Rb, 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)
(b)
(c)
(d)
(e)
(f)
1.00
600.
550.
400.
200.
150.
cm (microwave)
nm (red)
nm (yellow)
nm (blue)
nm (ultraviolet)
pm (X-ray)
26
CHAPTER 1. LIGHT, QUANTIZATION, ATOMS AND SPECTROSCOPY
10. The Brackett series of H-atom emission lines consists of a series of lines corresponding to transitions
from the n1 > 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 9 Be+3
(b) for the “Pfund series” of 56 Fe+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 Å, assuming all possible transitions are ‘allowed’,
what are the energies in cm−1 of the four lowest-energy transitions ?
[Note: this is an approximate model for the energies of π electrons in linear conjugated hydrocarbons.]
14. Because a photon carries one unit of angular momentum, allowed rotational or angular motion transitions must correspond to changes of ±1 in the relevant angular momentum quantum number.
Consider a system consisting of a particle moving on a ring (as discussed in § 1.3.2), for which three
adjacent transitions are observed to occur with wavenumbers 349.8, 489.6 and 629.7 cm−1 ,
(a) What are the values of the quantum number for the upper and lower levels of these transitions?
(b) If the particle is an H atom, what is the radius of the ring around which that H atom is moving?
(c) If the H atom of part (b) were replaced by a 133 Cs 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 Å. For this type of system, the most
strongly allowed spectroscopic transitions are those for which Δn = ±1 .
(a) If the particle is a 7 Li 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 ν̃Δn (nlower ) ,
where Δn = nupper − nlower .
(b) If the 7 Li atom were replaced by a 6 Li atom, what would be the “isotope shift” Δν̃(n) = ν̃(6 Li) −
ν̃(7 Li) for each of the transitions considered in part (a)?
(c) For the 6 Li atom in this hard-wall box, what would be the wavenumbers of the ν̃2 (3) and ν̃3 (2)
transitions?
16. The visible emission of Li, Na and K atoms occurs at 670.7 nm, 589.2 nm and 405.0 nm, respectively.
Assuming that these emissions are all due to electronic transitions from the valence p orbitals to the
valence s orbitals, determine the valence s − p energy spacings (in joules) for each of these atoms.
17. For long-chain conjugated hydrocarbon molecules, a surprisingly good approximation to the first electronic excitation energy of the extended π bond is to treat it as a particle-in-a-box problem, with the
electron being trapped in a rigid box whose length is the sum of the lengths of the bonds sharing the
π electrons. Using appropriate bond lengths based on those of cyanoacetylene (see e.g., Chapter 9 of
Handbook of Chemistry and Physics, 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
2.1.1
Classical Description of Molecular Rotation
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
CHAPTER 2. ROTATIONAL SPECTROSCOPY
28
orientation
of
molecule
−
+
−
+
+
−
−
−
+
+
−
+
direction
of
dipole
↑
vertical
component
of
molecular
dipole
time →
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, H2 O, C6 H5 OH, 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 , C6 H6 or C60 , will be “rotationally inactive”, in that while they do still rotate, that rotational motion
cannot gain energy by absorbing photons from an incident light field, or lose it by emitting radiation.
2.1.2
Relative Motion and the Reduced Mass
m1
→
Consider a system consisting of two particles of mass m1 and m2
z
r
located at positions specified by vectors r1 and r2 , respectively, as il→
r1 →
m2
lustrated in Fig. 2.2. In studying molecules we are normally interested
R
cm
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
→
r2
individual-particle coordinates r1 and r2 with coordinates specifying
the relative position of the two nuclei r = r2 − r1 and the position
cm of the centre of mass. The centre of mass of a system (indicated
R
y
by the symbol “+” in Fig. 2.2) is the “balance point”, defined by the
fact that relative to an arbitrary origin, the total leverage about that
origin of all the particles comprising the system is equal to that for
x
a single “effective” particle with the mass of the whole system (here
cm :
cm of the censimply m1 + m2 ) located at R
Figure 2.2: Position R
tre of mass “+”, and relative coordi cm = m1 r1 + m2 r2
(m1 + m2 ) R
(2.1)
nates r for a two-body system.
cm , we may then express the
Combining our definitions of r and R
cm and r :
one-particle positions r1 and r2 in terms of the centre of mass and relative coordinates R
m2
m1
cm −
cm +
r1 = R
r
and
r2 = R
r
(2.2)
m1 + m2
m1 + m2
+
Using the usual equations of classical mechanics, we know that the total kinetic energy of our two-particle
system may be written as
KEtot =
1
1
1
1
1
m1 |v1 |2 + m2 |v2 |2 = m1 |ṙ 1 |2 + m2 |ṙ 2 |2
2
2
2
2
The reason that rotational transitions fall in the microwave region of the spectrum is discussed in §3.7.
(2.3)
2.1. CLASSICAL DESCRIPTION OF MOLECULAR ROTATION
29
where vi = ṙi = dri /dt is the velocity of particle–i. Substitution of Eq. (2.2) into Eq. (2.3) yields the result
that
2
2
1
m2
1
m1
m1 Ṙcm −
KEtot =
ṙ + m2 Ṙcm +
ṙ
2
m1 + m2
2
m1 + m2
2 2
m
1
2 m2 2
2
Ṙcm · ṙ +
m1 Ṙcm −
=
ṙ 2
m1 + m2
m1 + m2
2 m
1
2 m1 2
2
1
Ṙcm · ṙ +
+ m2 Ṙcm +
ṙ 2
m1 + m2
m1 + m2
1
1
m1 m2
2
2
(m1 + m2 ) Ṙcm +
=
ṙ 2
2 m1 + m2
= KEcm + KErel
(2.4)
where the subscript “cm” stands for ‘centre of mass’, and the subscript “rel” stands for ‘relative motion’. In
other words, the total kinetic energy for our two-particle system is simply the sum of
cm , whose mass is the sum of the masses
• the kinetic energy of a particle located at the centre of mass R
of all of the particles in their system, plus
• the kinetic energy of relative motion, which for our two-particle system is the kinetic energy of a
particle having the effective mass m1 m2 /(m1 + m2 ) that is located at position r2s relative to a fixed
origin.
This effective mass associated with the relative motion in a two-particle system occurs so ubiquitously that
we give it a special name and symbol; the symbol is the Greek letter “mu” (written μ) and the name is the
−1
reduced mass:
1
m1 m2
1
μ =
=
+
.
(2.5)
m1 + m2
m1
m2
Note that while our definition of the centre of mass readily generalizes to many-particle systems, there is
no multiple-particle analog of the quantity μ ; 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:
ẋ ≡ dx/dt) while recalling that re is constant, we obtain
ẋ = re θ̇ cos θ cos φ − φ̇ sin θ sin φ
ẏ = re θ̇ cos θ sin φ + φ̇ sin θ cos φ
ż
= −re θ̇ sin θ .
The internal or rotational kinetic energy may then be written as
2
KErot = 12 μ ṙ = 12 μ ẋ2 + ẏ 2 + ż 2 = . . . . . .
= 12 μ (re )2 θ̇2 + φ̇2 sin2 θ
(2.6)
2 Note that this derivation is also the origin of the reduced masses μ and μ that appeared in the discussion of the Bohr
H
A
model for a hydrogenic atom in §1.2.2.
CHAPTER 2. ROTATIONAL SPECTROSCOPY
30
If we assume, for convenience, that the rotation occurs in the x-y plane, then θ is fixed at θ = 90◦ , so θ̇ = 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 =
1
2
I φ̇2 =
2
|L|
(I φ̇)2
=
.
2I
2I
(2.7)
to represent the angular momentum of the system. In a
In Eq. (2.7) we have introduced the symbol L
more systematic three-dimensional derivation, we would find that
≡ |r × |L|
p = |r × (m ṙ)| = |I ω | = Iφ̇ ,
(2.8)
is perpendicular to the plane of rotations. Here, ω is a vector generalization of our φ̇
which shows that L
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 12 mv2 = 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
the analog
mass m, the angular velocity φ̇ the analog of the linear velocity v, and the angular momentum |L|
of the linear momentum p.
For a rigid molecular system rotating freely in space, the potential energy is identically zero, so that
Erot = KErot . While we will not attempt to derive it here, the final result obtained above for the case of
two particles, the fact that
L2
,
(2.9)
Erot = KErot =
2I
is the magnitude of the orbital angular momentum, also holds for any multi-particle system.
in which L ≡ |L|
In particular, for a system of N particles, the position of the centre of mass is defined by a straightforward
generalization of Eq. (2.1), namely,
N
N
cm =
mi R
mi ri
(2.10)
i=1
i=1
and the moment of inertia about an axis through that centre of mass is given by the equation
I =
N
2
mi (d⊥
i )
(2.11)
i=1
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 = m1 m2 /(m1 + m2 ) .
While we will not attempt to derive it here, these same points hold in the exact quantum mechanical
description of an N –particle system, and they provide the basis for our description of rotational spectra.
2.2. QUANTUM MECHANICS OF MOLECULAR ROTATION
2.2
2.2.1
31
Quantum Mechanics of Molecular Rotation
The Basics
In Bohr’s theory of the H atom, the central postulate required to make it work (unjustifiable at the time!)
was that the orbital angular momentum of the electron was only allowed to have discrete values which were
integer multiples of = h/2π = 1.054 571 628 × 10−34 J s. However, the discussion of §1.3.2 shows that for
the case of a particle of mass μ rotating in a plane at a fixed radius r = re , solution of the Schrödinger
equation gives the allowed energies as
Erot =
( )2
( )2
=
2
2μ(re )
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 (nonnegative) integer value of . More generally, quantum mechanics tells us that all forms of angular momentum
are quantized such that the component of the angular momentum along any space-fixed axis has values that
differ by increments of .
Of course, molecular rotation actually occurs in 3 dimensions. A straightforward extension of the discussion of §1.3.2 shows that for orbital or rotational motion in three-dimensional space, the allowed values
of the angular momentum are
= L = L3D = J(J + 1)
for
J = 0, 1, 2, 3, . . . etc.
(2.13)
|L|
Note that while the symbol is commonly used to represent the total orbital angular momentum quantum
number of an electron in an atom, it is a near-universal convention in spectroscopy to use the different symbol
J to represent the total angular momentum of a molecule, even though the mathematical descriptions of the
large J
two types of motion are very similar. Note too that since J(J + 1) = J (1 + 1/J) −−−→ J , this angular
momentum quantization is only slightly different from our result for orbital motion in two dimensions.
Coupling this three-dimensional angular momentum result with Eq. (2.9) then yields the quantum mechanical expression for rotational energy of a rigid molecular system:
Erot (J) =
2
[J(J + 1)] [J] .
2I
(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) ≡
=
rot (J) = Erot (J)/ 102 hc
E
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
Cu
2 1020
=
[cm−1 ] ,
2
2I 102 hc
I [u Å ]
B =
(2.16)
2
where as usual Cu=16.857 629 [u cm−1 Å ] , 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 C would be the numerical value of the inertial rotational constant B for a diatomic molecule with a reduced
u
mass of 1 u (approximately true for D2 ) and a bond length of 1 Å; this is the reason for calling it the “inertial constant”.
CHAPTER 2. ROTATIONAL SPECTROSCOPY
32
2.2.2
Energy Levels, Selection Rules, and Transition Energies
rotational energy →
From Eq. (2.15) we see that the allowed rotational levels of a molecule have energies of 0, 2B, 6B, 12B,
20B, . . . , etc., as shown in Fig. 2.3. The systematic increase in the level spacings would appear to make it
absolutely trivial to make rotational assignments, i.e., to identify the upper and lower level quantum numbers
associated with observed transitions. However, this is not quite as simple as it might seem.
As discussed in Chapter 1, a photon is a particle with zero rest
J=7
56B
mass and a momentum of pλ = h/λ = 102 hν̃ . However, an important additional property is that it has an intrinsic angular momentum
∼
(the property we call spin4 ) of exactly 1, in the usual angular moν = 14B
mentum unit of . We also know from classical mechanics and from
our everyday experience that in any multi-particle collision, the total
J=6
42B
amounts of linear and angular momenta are conserved (e.g., imagine
two cars colliding while sliding and spinning on a frictionless sheet of
ν∼ = 12B
ice). Since the total angular momentum of a system (e.g., molecule
plus photon) must always be conserved, whenever a molecule absorbs or emits a photon of light, the rotational quantum number J
J=5
30B
must change by ΔJ = Jupper − Jlower = ±1 , where Jupper and Jlower
∼
ν
= 10B
are the values of the rotational quantum number for the upper and
lower levels of the transition. The ± sign appears here because anguJ=4
20B
lar momentum is a vector property, and the result of vector addition
∼ = 8B
depends on the relative orientations of the two vectors: in this case,
ν
quantum mechanics allows only these two choices for the relative
J=3
12B
alignment.
∼
ν = 6B
The essential implication of the above selection rule is that only
J=2
6B
∼
rotational transitions between adjacent J levels are allowed. For our
ν = 4B
2B
J=1
rigid-rotor system this means that the allowed transition energies are
J=0
0
ν̃Jrot = ΔF (J) = F (J + 1) − F (J)
Figure 2.3: Rotational energies and
= B [(J + 1)(J + 2) − J(J + 1)] = 2B (J + 1)
(2.17) level spacings for a linear rigid rotor.
It is interesting to note that the frequency of the light absorbed or
emitted in such a transition, νJrot ≈ J/(2πI) ≈ L/(2πI) [Hz], is equal to the classical rotation frequency
θ̇/(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 ν̃Jrot = 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
Δν̃Jrot
≡
rot
ν̃Jrot − ν̃J−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 12 C16 O, 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 Å, 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(12 C) = 12.000 000 u
and
m(16 O) = 15.994 915 u .
It is then straightforward to calculate
μ =
m1 m2
12.000 000 × 15.994 915
u = 6.856 209 u ,
=
m1 + m2
12.000 000 + 15.994 915
and hence to obtain B as
B =
Cu
16.857 629
cm−1 = 1.931 285 cm−1 .
2 =
6.856 209 × (1.128 322)2
μ [u] re [Å]
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 ν̃J = 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.
7 Recall that 1 u = 1 atomic mass unit = 1.660 538 782 × 10−27 kg.
CHAPTER 2. ROTATIONAL SPECTROSCOPY
34
Table 2.1: Predicted and observed microwave spectrum of CO.
J
J(J + 1)
F (J) = B[J(J + 1)]
0
0
0.0
1
2
3.845 034
2
6
11.535 102
3
12
23.070 204
4
20
38.450 340
5
..
.
30
..
.
57.675 510
..
.
ν̃Jrot = 2B(J + 1)
ν̃Jrot (obs)
3.845 034
3.845 033
7.690 068
7.689 919
11.535 102
11.534 510
15.380 136
15.378 662
19.225 170
..
.
19.222 223
..
.
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 ν̃Jrot (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 ν̃calc − ν̃obs = 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 ν̃ = 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. Equation (2.18) indicates that the separation between the two spectroscopic lines is equal to 2B, so we obtain
B =
1
[163.936
2
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 = μre 2 ),
I = Cu /B = 0.826 222 94 [u Å2 ] = μ (re )2 .
The known atomic masses6 yield a reduced mass of
−1
1
1
+
μ(HF) =
= 0.957 055 278 u ,
1.007 825 032
18.998 403 20
so we then obtain
re =
I/μ =
0.826 222 94 /0.957 055 278 Å = 0.929 137 8 Å .
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 = ν̃Jrot /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 13 C 16 O, 12 C 18 O or 12 C 17 O. 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% 13 C), 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
CHAPTER 2. ROTATIONAL SPECTROSCOPY
36
Figure 2.6: Microwave absorption spectrum of H–C≡C–C≡C–C≡N, showing vibrational satellites.
for identification of a particular chemical species.
Complication
#
2: Vibrational Stretching and Vibrational Satellites
A second type of complication is illustrated by the presence of two sets of (roughly) equally spaced lines
in the HF emission spectrum of Fig. 2.5. It is to be expected that when a molecule is excited into higher
vibrational energy levels, the amplitude of its vibrational motion will increase. Because of the asymmetry of
typical intermolecular potential energy curves, this means that the average bond length will increase with
the degree of vibrational excitation. For example, in Fig. 1.7-D on p. 16 it is clear that the average bond
length in the v = 5 vibrational level is much greater than that in the v = 0 “ground” level. This in turn
means that the B value in this v = 5 level will be much smaller, and hence that the spacings (of roughly
2B, see Eq. (2.18)) between the lines in its pure rotational spectrum will be distinctly smaller than those for
the lower vibrational levels.
Because of this dependence of the average bond length on vibrational level, it is customary to introduce
the subscript “v” and use the symbol Bv to represent the inertial rotational constant. For a diatomic
molecule, this means that Eq. (2.16) may be re-written as
Cu
Bv =
(2.20)
μ(r v )2
in which rv is the effective average bond length for a moleule in vibrational level v. For HF, this vibrational
stretching causes the J = 18 ← 17 transition to shift from 692.481 cm−1 for v = 0 to 665.937 cm−1 for
v = 1 . From Eq. (2.20) we can see that this corresponds to a 3.8% decrease in Bv from v = 0 to 1, which
in turn corresponds to an increase of 1.9% in the effective bond length r v .
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 r v ) 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 HC5 N, 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 2.54 GHz or 0.0847 cm−1 , which implies that the B value is 0.04236 cm−1 ,
and hence that the moment of inertia is I(HC5 N) = Cu /0.04236 = 390 u Å2 .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 !
Complication
#
37
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 r v = r v (J) will tend
2
to increase. This in turn leads to an increase in the moment of inertia I = I(J) = μ (rv (J)) , and since the
spacing between rotational lines is 2Bv (J) = 2(Cu /I) , at higher values of J the rotational line spacings will
rot
rot
− ν̃J=13
line spacing
become progressively smaller. This behaviour is evident in Fig. 2.5, where the ν̃J=14
−1
rot
rot
−1
of 36.0 cm is visibly larger than the ν̃25 − ν̃24 line separation of 26.4 cm , and it explains why these
rot
rotational transition energies (e.g., ν̃24
= 904.385 cm−1 ) are not simple 2(J+1) multiples of the rotational
−1
constant B = B0 = 20.403 cm determined in example (ii) on p. 34. It also explains the discrepancies
referred to at the end of each of the illustrative examples of §2.2.3.
This rotational stretching, or centrifugal distortion, is normally accounted for by including an additional
term in the rotational energy expression of Eq. (2.15), to yield
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
ν̃Jrot
= 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
rot
Δν̃Jrot = ν̃Jrot − ν̃J−1
= 2Bv − 4Dv (3J 2 + 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 12 C16 O
listed in Table 2.2. The fact that the line spacings Δν̃Jrot 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
ν̃Jrot
0
1
2
3
4
5
1←0
2←1
3←2
4←3
5←4
6←5
3.845 033
7.689 919
11.534 510
15.378 662
19.222 223
23.065 043
Δν̃Jrot = ν̃Jrot − ν̃J−1
—
3.844 886
3.844 591
3.844 152
3.843 561
3.842 820
(3J 2 + 3J + 1)
—
7
19
37
61
91
CHAPTER 2. ROTATIONAL SPECTROSCOPY
38
intercept = 2B0 = 3.845 059
3.845
∼
Δν
J
/ cm−1
slope = − 4 D0 = − 2.459×10−5
3.844
3.843
0
25
50
(J2+3J+1)
75
100
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
ν̃0rot
or
D0
=
2B0 (J + 1) − 4D0 (J + 1)3
=
2B0 − 4D0 = 3.845 033
=
[2B0 − 3.845 033] /4
(2.24)
Similarly, for the 6 ← 5 transition, J = 5 , and hence
ν̃5rot = 12B0 − 864D0 = 23.065 043
Using Eq. (2.24) for D0 , we obtain
ν̃5rot = 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 Δν̃Jrot vs. (3J 2 + 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, r v (J). In particular, let us define
Bveff (J)
≡
J)
dE(v,
Cu
= Bv − 2Dv [J(J + 1)] ≡
d[J(J + 1)]
μ [r v (J)]2
(2.26)
Rearranging this result to solve for rv (J), and then multiplying the numerator and denominator by r v (J=0)
yields the expression
1/2
Cu /μ [r v (J=0)]2
r v (J) = r v (J = 0)
(2.27)
Bv − 2Dv [J(J + 1)]
2.4. DEGENERACIES AND INTENSITIES
39
However, r v (J=0) is just the bond length that the molecule would have if there was no rotational stretching,
so the numerator on the right hand side is our definition of Bv , and this equation yields
1/2
Dv
r v (J) = r v (J = 0)
1 − 2[J(J + 1)]
(2.28)
Bv
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
= J(J + 1) , for integer values
by quantum mechanics only to have one of the discrete values L = |L|
is a vector quantity,
of the total angular momentum quantum number J (= 0, 1, 2, 3, . . . etc.). Since L
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.
CHAPTER 2. ROTATIONAL SPECTROSCOPY
40
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
is pointing, and not about its magnitude (i.e., not about the magnitude of the
the direction in which L
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.
could be the one that is “known”, for mathematical
While any one of the three Cartesian components of L
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 and the quanentum quantum number J that defines the magnitude of |L|,
z
tum number MJ that defines the value of Lz . For a rotating molecule with
onto the laboratory
J = 2 , Fig. 2.8 illustrates the allowed projections of L
Lz= 2 −
h
→
(or space-fixed) z axis. For present purposes, we are only concerned with
L
knowing that the number of degenerate sublevels associated with any given
1−
h
value of J is gJ=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.
0−
h
This concept of degenerate MJ levels should be quite familiar, as it is
the same property encountered in the discussion of the quantum numbers
-1 −
h
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
-2 −
h
projection quantum number was m (rather than MJ ). In that case the
(2+1)-fold degeneracy associated with a given value of gave rise to the
three p (for =1), the five d (for =2), and the seven f (for =3) degenerate Figure 2.8: Angular momentum projections for J=2 .
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 ,
−1
in which kB = 1.380 650 4×10−23 [J K−1 ] = 0.695 035 6 [cm−1 K ] 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
(2.30)
fi (T ) = gi e−Ei /kB T Q(T ) ,
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 J mol−1 K−1 = NA kB , where NA = 6.022 141 79 × 1023 is the Avogadro number.
2.4. DEGENERACIES AND INTENSITIES
pop
Jmax
(T)
10
41
gJ = (2J+1)
−Bυ J(J+1)/kBT
gJ e
5
−Bυ J(J+1)/kBT
10 × e
0
0
10
J
20
Figure 2.9: Boltzmann rotational population distribution for CO at T = 293 K.
The molecular partition function is a rough measure of the number of quantum states of that molecule
having significant equilibrium populations at the given temperature. It is a very important property, since it
determines the macroscopic thermodynamic behaviour of a system. We won’t discuss the partition function
further in this course, but you will encounter it in upper-year chemistry courses, where it is employed to
connect microscopic and macroscopic properties of matter. You will see there that our ability to make
reliable predictions of the equilibrium thermodynamic properties of many systems depends critically on the
knowledge of the patterns of energy levels {Ei } determined from spectroscopic measurements.
For a rotating linear molecule the energy levels are specified by the total angular momentum quantum
number J, and their energies are given by Eq. (2.15), (2.21) or (2.29) (depending on how precisely the pattern
of level energies is known), with the level degeneracies being gJ=2J+1 . The fraction of molecules in the level
with energy Fv (J) is therefore given by
(2.31)
fJ (T ) = (2J + 1) e−Fv (J)/kB T Q(T ) .
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−Bv J(J+1)/kB T , which decreases exponentially as J increases. The competition between these two terms
gives the overall behaviour shown by the solid curve in Fig. 2.9, while the influence of the initial-state
population on experimental line intensities may be seen in Figs. 2.4 and 2.5 (a cleaner example will be seen
in the vibration-rotation spectra discussion of Chapter 3).
Since Eq. (2.31) provides us with a simple analytic expression for the (fractional) equilibrium population
of any given rotational level, it is a straightforward matter to address the following question. For a system in
thermal equilibrium at temperature T , what is the value of J for the most highly populated rotational level ?
pop
(T ), for which the
Calculus tells us that the location of this maximum will be the value of J, denoted Jmax
derivative of fJ (T ) with respect to J is zero. Hence, for a rigid rotor:
−Bv
d −Bv J(J+1)/kB T
−Bv (J 2 +J)/kB T
= 0 = e
(2J + 1) e
(2J + 1)
(2.32)
2 + (2J + 1)
dJ
kB T
pop
Removing the exponential common factor and solving for J = Jmax
(T ) yields the expression
kB T
1
pop
(T ) =
−
J = Jmax
2 Bv
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 ν̃, while the
CHAPTER 2. ROTATIONAL SPECTROSCOPY
42
full emission intensity expression includes a factor of ν̃ 4 . These factors have little effect on the rotational line
intensities in vibrational or electronic spectra (see Chapters 3 and 5), but they can be quite important for
pure rotational spectra. In particular, using the rigid rotor expression of Eq. (2.17), a slight generalization
of the above discussion shows that the value of J for the most intense line in a pure rotational absorption
or emission spectrum, respectively, is given by11
√
√
abs
em
J max
(T ) ≈ 3 J pop
and
J max
(T ) ≈ 5 J pop
(2.34)
max (T )
max (T )
2.5
Rotational Spectra of Polyatomic Molecules
2.5.1
Linear Molecules are (Relatively) Easy to Treat!
For a general polyatomic molecule consisting of N atoms distributed
d
in space, introducing centre-of-mass and relative coordinates in the
A
same manner as in our diatomic molecule derivation in §2.1.2 yields
m1
m2
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
d12
d12
energy in turn separates into a sum of the internal vibrational energy
plus the overall rotational energy of the molecule. For any linear
m2
m1
m2
molecule, that rotational energy is also given by Eq. (2.9), where
are given by Eq. (2.13). Thus, up
again the allowed values of |L|
d12
d12
d22
to this point the treatment of an arbitrary linear molecule (such as
HC5 N of Fig. 2.6) is identical to that for a diatomic molecule. In
m1
m2
m2
m1
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.
d12
d23
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
m1
m2
m3
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 Figure 2.10: Four types of linear
used in the early parts of this Chapter, but that expression cannot molecules.
be used for a molecule consisting of more than two particles. We
therefore begin by demonstrating the use of Eqs. (2.10) and (2.11) for the four types of linear molecule
illustrated in Fig. 2.10. Since these molecules are all linear, the three-dimensional centre-of-mass position
cm = (xcm , ycm , zcm ) may be replaced by a one-dimensional coordinate, and for convenience, we
vector R
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.
B
C
D
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
m2 2 d2
m1 2 d2
= m1
+ m2
2
(m1 + m2 )
(m1 + m2 )2
m1 m2
=
d2 = μ d2 = Id .
m1 + m2
(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 + 2 m2 )/(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 = 2 m2 (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 )
= d12 + 12 d22 ,
(2.38)
=
2(m1 + m2 )
which confirms our intuitive prediction. Combining this result with our formal definition of the moment
of inertia then yields
I =
m1 (−d12 −
=
2 m1 (d12 +
1
2
1
2
d22 )2 + m2 (− 12 d22 )2 + m2 ( 12 d22 )2 + m1 (d12 +
1
2
d22 )2
d22 )2 + 2 m2 ( 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 ) , and
atom 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 definitions 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
CHAPTER 2. ROTATIONAL SPECTROSCOPY
44
first glance this would seem to be rather uninteresting, since such symmetric molecules would have no
permanent dipole moment, and hence no normal (i.e., dipole allowed) rotational spectra. However, as
we shall see in Chapters 3 and 4, the rotational level spacings of such species may be observed as fine
structure in vibrational or electronic spectra, or more directly by Raman spectroscopy. Thus, it is still
quite relevant to understand how their moments of inertia, that are obtained from the experimentally
determined Bv values, are related to the molecular bond lengths.
2.5.2
Illustrative Applications
Exercise (iv): Predict the microwave spectrum of cyanodiacetylene H–C≡C–C≡C–C≡N.
The microwave spectrum of cyanodiacetylene was shown in Fig. 2.6. If we had first observed that spectrum
in a gaseous mixture in chemical effluent from an industrial plant, how would we know what it was due to?
One way to identify unknown species in spectra is to use predictions based on general chemical knowledge to
predict the structure, and hence the moment(s) of inertia and the spectra of various possible culprits. We will
now do that for cyanodiacetylene, H–C≡C–C≡C–C≡N.
From the data table in Chapter 9 of the Handbook of Chemistry and Physics, we find that for the simpler molecule cyanoacetylene (H–C≡C–C≡N), the bond lengths are given as rC−H=1.058 Å, rC≡C=1.205 Å,
rC−C=1.378 Å and rC≡N=1.159 Å. Assuming that these bond lengths are essentially unchanged in cyanodiacetylene, if we choose the origin of our coordinate system to lie at the H atom, the position of the centre of
mass is (using the masses of 1 H, 12 C and 14 N 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 [Å] .
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 Å, respectively. From this knowledge of each atom’s distance
d⊥
i to the centre of mass, we can readily calculate
I(HC5 N) = 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 Å2 ] .
This result then gives us a prediction for the rotational constant of cyanodiacetylene:
B(HC5 N) ≈ 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 HC5 N. Further confirmation could be obtained by predicting the isotopic shifts
of the moment of inertia (and hence of the transition frequencies) obtained on replacing 14 N by 15 N (0.37%
abundance) or on replacing one of the 12 C atoms by 13 C (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 (16 O12 C32 S)
=
0.202 864 [cm−1 ]
B0 (16 O12 C34 S)
=
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
2
Ia ≡ I(16 O12 C32 S)
=
Cu /0.202 864 = 83.098 18 [u Å ]
Ib ≡ I(16 O12 C34 S)
=
Cu /0.197 910 = 85.178 26 [u Å ] .
2
To simplify the following expressions, we define mO = m(16 O), mC = m(12 C), mS = m(32 S) and
m∗S = m(34 S), all in units u. If we explicitly write out Eq. (2.41) for our two isotopologues, we obtain
(mO rCO − mS rCS )2
mO + mC + mS
(mO rCO − m∗S rCS )2
−
.
mO + mC + m∗S
Ia
=
mO (rCO )2 + mS (rCS )2 −
Ib
=
mO (rCO )2 + m∗S (rCS )2
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)
2
+ (mO mS + mC mS ) (rCS ) + 2 mO mS rCO rCS
Ib (mO , mC , m∗S )
=
(mO mC + mO m∗S ) (rCO )2
+
(mO m∗S
+
mC m∗S ) (rCS )2
(2.43)
+
2 mO m∗S
rCO rCS .
Multiplication of Eq. (2.42) by the ratio m∗S /mS then yields
Ia (mO , mC , mS )
m∗S
mS
=
m∗S
(rCO )2
mS
+ (mO m∗S + mC m∗S ) (rCS )2 + 2 mO m∗S rCO rCS .
(mO mC + mO mS )
(2.44)
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∗
m∗S
= mO mC 1 − S (rCO )2
mS
mS
Since IA and IB and all of the atomic masses are known, we can readily solve this equation to obtain
rCO = 1.167 415 Å. 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 Å (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.
CHAPTER 2. ROTATIONAL SPECTROSCOPY
46
y
Consider the case of the water molecule, whose structure is shown
in Fig. 2.11. Because of its symmetry, it is intuitively obvious that the
centre of mass and two of our principal axes of rotation will lie in the
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 the
16
O atom, the two 1 H atoms are located at the coordinate positions
(xH , yH ) =
=
( ±rOH sin(θ/2), −rOH cos(θ/2))
0.
95
×
x
8Å
θ = 104.5°
Figure 2.11: Structure of H2 O.
(±0.7575, −0.5865) .
Applying Eq. (2.10) for the x and y coordinates of the centre of mass, we obtain
xcm
=
ycm
=
mH ×(−0.7575) + mO ×0 + mH ×(+0.7575)
= 0
2 mH + mO
mH ×(−0.5865) + mO ×0 + mH ×(−0.5865)
= − 0.06564 ,
2 mH + mO
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 16 O atom is due to the large differences between the masses of the two
types of atoms.
With the centre of mass located, a little additional arithmetic yields the values of the three moments of
inertia as:
2
Ix
=
mH (−0.5865 + 0.0656)2 + mO (0.0656)2 + mH (−0.5865 + 0.0656)2 = 0.6158 [u Å ] ≡ IA
Iy
=
mH (−0.7575)2 + mO (0)2 + mH (0.7575)2 = 1.1566 [u Å ] ≡ IB
Iz
=
Ix + Iy = 1.7724 [u Å ] ≡ IC .
2
2
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,
Cu
sph
F (J) = Bsph [J(J + 1)] =
[J(J + 1)] ,
(2.45)
I
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⊥ K 2
Cu
Cu
Cu
=
−
(2.46)
[J(J + 1)] +
K2
I⊥
I
I⊥
2.6. CONCLUDING REMARKS
47
Figure 2.12: Rotational moments of inertia for some symmetric non-linear polyatomic molecules; mtot is the
total mass of the molecule.
in which J = 0, 1, 2, 3, . . . etc. (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
48
CHAPTER 2. ROTATIONAL SPECTROSCOPY
Figure 2.13: Gas phase molecular structure of azulene determined from rotational spectroscopy.
independent isotopologue measurements required will clearly grow rapidly. In particular, one needs one
independent inertial constant for each independent geometric parameter (bond length or bond angle) of the
system. This clearly must have been quite challenging for a molecule such as azulene C6 H5 NH2 , 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., 1 H vs. 2 H and 12 C vs. 13 C in various locations,
and 14 N vs. 15 N), 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 planetary 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
Triatomics
Tetratomics
5-Atomics
6-Atomics
7-Atomics
8-Atomics
9-Atomics
11-Atomics
2.7
OH, CO, CN, CS, SiO, SO, SiS, NO, NS, CH, CH+
H2 O, HCN, HNC, OCS, H2 S, N2 H+ , SO2 , HNO, C2 H, HCO, HCO+ , HCS+
NH3 , H2 CO, HNCO, H2 CS, HNCS, N≡C–C≡C, H3 O+ , C3 H(linear), C3 H(cyclic)
N≡C–C≡C-H, HCOOH, CH2 =NH, H-C≡C–C≡C, NH2 CN
CH3 OH, CH3 CN, NH2 CHO, CH3 SH
CH3 –C≡C-H, CH3 CHO, CH3 NH2 , CH2 =CHCN, N≡C–C≡C–C≡C-H
HCOOCH3 , CH3 –C≡C–C≡N
CH3 OCH3 , CH3 CH2 OH, N≡C–C≡C–C≡C–C≡C-H
N≡C–C≡C–C≡C–C≡C–C≡C-H
Problems
1. The rotational constant B for 12 C2 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, 35 Cl19 F, 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 14 N1 H radical has a bond length of 1.037 Å. 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 12 C16 O 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 13 C17 O. [Shortcuts are permitted if you can justify them.]
6. Given that 12 C16 O, 14 N16 O and 16 O2 have approximately the same bond lengths (1.16±0.04 Å), predict
the relative magnitudes of their rotational constants B.
7. Three adjacent rotational transition lines for the ground vibrational level of 1 H79 Br 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 Å, at what frequencies (in Hz) will the first four transitions of 12 C16 O lie? What precision is
needed in order to distinguish the first transition of 12 C16 O and 13 C16 O in order to determine the
relative abundances of the two carbon isotopes?
9. Predict the relative intensities of the first 6 rotational transitions of 12 C16 O for a spectrum obtained
at a temperature of T = 1000 K. Use the value for the inertial rotational constant obtained in Exercise
(i) (on p. 33), and assume the rigid rotor model is valid.
10. The microwave spectrum of
the C–N bond length.
in this case I(14 N) = 1 .
12
C14 N exhibits a series of lines with spacings of 3.7992 cm−1 . Determine
CHAPTER 2. ROTATIONAL SPECTROSCOPY
50
11. The equilibrium bond length of LiH is 1.595 Å. Determine the inertial rotational constants for 7 Li1 H,
6 1
Li H and 7 Li2 H, 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, 1 H2 H, has a bond length of 0.741 42 Å, 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 Å, 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 127 I35 Cl is 0.1142 cm−1 . Determine the bond length in this molecule.
Determine the rotational constant Bv for 127 I37 Cl, assuming the bond length does not change from
one isotopologue to another.
15. Using the data from question # 14, calculate the energies of the first five rotational energy levels and
the first four rotational transitions of 127 I35 Cl, Will the rotational transitions for 127 I37 Cl occur at
higher or lower energies?
16. The rotational spectrum of 1 H127 I 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 1 H19 F 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 Åand the H–N bond length is 0.976 Å, at what frequency (in MHz)
would the lowest energy rotational transition for 1 H14 N12 C occur?
19. The vibrational level dependence of inertial rotational constant of H35 Cl is given by the expression
Bv = 10.5933 − 0.3070(v + 12 ) + 0.00163(v + 12 )2 cm−1 . What are the average bond lengths r v 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 twoparticle or diatomic molecule result Id = μ (re )2 .
21. If the
12
C16 O molecule has a bond length of 1.131 Å, where does its centre of mass lie?
22. Show how we get from the first line of Eq. (2.6) to the second.
23. From Eq. (2.34) and the HF rotational constant determined in Exercise (ii) (see p. 34), what temperature would be associated with the emission spectrum of Fig. 2.5? What would your result have been
if we had considered only the thermal population factor?
[Note that the surface temperature of the sun is around 6000 K.]
24. Compared to 12 C16 O, what is the relative abundance of 13 C18 O? 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 1 H− 12 C ≡ 12 C− 12 C ≡ 14 N ,
r(H−C) = 1.058 Å, r(C ≡ C) = 1.205 Å, r(C−C) = 1.378 Å and r(C ≡ N) = 1.159 Å.
(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 absorbed 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
3.1.1
Classical Description of Molecular Vibrations
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
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
52
length
of
molecule
+
−
+
−
+
−
+
−
+
+
−
−
+
−
+
−
strength
of
dipole
↑
vertical
component
of
molecular
dipole
equilibrium
dipole moment
time →
Figure 3.1: Dipole moment of a vibrating polar diatomic molecule which is fixed and aligned in space.
bond(s) will cause that dipole moment to oscillate. In particular, every chemically heteronuclear1 diatomic
molecule (such as HCl, NaBr or OH) will be vibrationally or “infrared” active, since the differences in atomic
polarizabilities will ensure that there is always some non-zero permanent dipole moment whose magnitude
will oscillate when the bond stretches. This same argument tells us that (chemically) homonuclear diatomics
such as H2 , N2 or Rb2 will be vibrationally or infrared inactive.
For polyatomic molecules the situation is more complicated, since they
have more than one type of vibrational motion (or vibrational “mode”)
that must be considered. As an illustrative example, consider the molecule
symmetric stretch
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
antisymmetric stretch
each other, the “antisymmetric stretch” mode in which one bond compresses
while the other stretches, and vice versa, and the bending mode. It is immebending mode
diately clear that the symmetry of the molecule is maintained throughout the
course of the symmetric stretching vibration, so the vibrational motion does
not give rise to any temporary or “instantaneous” dipole moments. In conFigure 3.2:
trast, for both the antisymmetric stretch and bending modes, the molecular
Vibrational modes of CO2 .
distortion that occurs during the vibrational cycle will give rise to a temporary dipole moment which oscillates in magnitude and direction (from left to
right for the antisymmetric stretch, and up-and-down or in-and-out for bending) as the motion proceeds.
Thus, these two latter modes will be vibrationally (or infrared) active, while the symmetric stretch mode is
not. In the context of Fig. 3.1, these IR active modes correspond to a case in which the oscillating dipole
moment allows transitions to occur even though the average or equilibrium dipole moment is zero.
3.1.2
The Centre of Mass and Relative Motion
As illustrated by our discussion of the classical mechanics of a two-particle system in §2.1, the total energy of
a multi-particle system may always be separated into the kinetic energy of motion of the overall system, plus
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 7 Li6 Li or
HD, since the difference in nuclear masses does not give rise to any asymmetry of the overall charge distribution.
3.1. CLASSICAL DESCRIPTION OF MOLECULAR VIBRATIONS
53
a diatomic molecule, the internal energy (rotation plus vibration) is described in terms of the motion of a
single pseudo-particle of mass μ = m1 m2 /(m1+m2 ) located at the relative coordinate vector r. Rotation
was concerned with the changing orientation of r in space, while our discussion of vibration will focus on
the one-dimensional stretching of r. In other words, the vibration of a diatomic molecule can be described
as the motion of particle of mass μ in one dimension, along the coordinate r.
Near the end of Chapter 1 we discussed the Born-Oppenheimer approximation, which allowed us to
separate the overall Schrödinger equation for the system into a Schrödinger equation for the motion of the
electrons plus one for the motion of the nuclei. We also saw there that the r-dependence of the energy eigenvalues of the electronic Schrödinger equation yielded the potential energy curves V (r) = Eel (r) governing
the vibrational motion of the nuclei, which appears in the effective radial Schrödinger equation of Eq. (1.35).
Thus, the description of vibrational motion for a diatomic molecule, in either classical or quantum mechanics,
is concerned with a particle of mass μ moving in one dimension along the radial coordinate r, subject to a
potential energy function V (r). For a polyatomic molecule the situation is similar, except that we need to
take account of multiple internal coordinates for relative motion, and we need special techniques (beyond
the scope of this course) to determine the particular effective mass associated with each of those different
modes of motion.
3.1.3
The Classical Harmonic Oscillator
The basic starting point for any description of vibrational motion in classical or quantum mechanics is the
harmonic oscillator model. It is based on Hooke’s Law, which states that when a system is displaced from
equilibrium, the restoring force increases linearly with the displacement. For a diatomic molecule stretching
(or compressing) away from its equilibrium distance r = re , this restoring force F can be written as
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) =
1
2
k(r − re )2 .
(3.2)
Combining Eq. (3.1) with Newton’s famous second law, force = mass×acceleration , we obtain a differential equation with a very familiar form (see Eq. (1.18)):
F = μ
d2 r
= − k (r − re )
dt2
(3.3)
that may be rewritten as
d2 (r − re )
= −
dt2
k
(r − re ) = − κ2 (r − re ) ,
μ
(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(2πνe t + 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.
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
54
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 ṙ(t) ≡ dr(t)/dt = 2πνe A cos(2πνe t + δ) , and
since the instantaneous radial kinetic energy is 12 μ(ṙ)2 , the sum of the potential energy and kinetic energy
is constant:
Etot
= KEradial + VHO (r(t)) =
=
3.2
1
2 μ [2πνe
A cos(2πνe t +
2
2
1
1
2 μ(ṙ) + 2 k (r(t) − re )
δ]2 + 12 k [A sin(2πνe t +
δ)]2 =
2
1
2k A
(3.6)
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 Schrödinger equation
Ĥvib ψ(r) = −
2 d2 ψ(r)
+ V (r) ψ(r) = Evib ψ(r)
2μ dr2
(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
HO
(v) = k/μ (v + 1/2) = h νe (v + 1/2)
(3.8)
Evib = Evib
in which νe is exactly the same frequency (in Hz) appearing in the classical mechanical harmonic oscillator
solution of Eq. (3.5), and the vibrational quantum number v has allowed values of v = 0, 1, 2, 3, . . . .
As is the case for rotation, spectroscopists usually express vibrational energies in units cm−1 , and it is
customary to use the letter ‘G’ to represent vibrational energies in those units:
HO
GHO (v) ≡ Evib
(v)/ 102 hc = ωe (v + 1/2) [cm−1 ]
in which
!
1
2
=
2k
2μ 102 hc
ωe =
2
!
2 Cu k̃ [cm−1 Å−2 ]
[cm−1 ]
μ [u]
(3.9)
(3.10)
and Cu = 16.857 629 [u cm−1 Å ] is the ubiquitous numerical factor introduced in §1.1.4. Note that we
choose to write the force constant as k̃ when it is in “spectroscopists’ units”. In SI units k would normally have
the units newtons/meter (or N m−1 ). However, it is equally valid to multiply numerator and denominator
by 1 m and write it with units J m−2 , or more generally, in units energy/length 2 . Thus, expressing k̃ in units
[cm−1 Å−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
V(r) = 1− k (r-re )2 ; μ = μ0
2
1
2
2
V(r) = −2 k ′(r-re ) ; μ = μ0
υ=5
55
V(r) = k ′(r-re ) ; μ = 0.9 μ0
{k ′= 2 k }
{k ′= 2 k }
υ=3
υ=3
υ=4
ωe
υ=2
υ=3
υ=2
ωe
υ=2
υ=1
υ=1
υ=1
ωe
A
υ=0
υ=0
B
C
υ=0
ωe / 2
- 0.2
0.0
(r - re )
0.2
- 0.2
0.0
(r - re )
0.2
- 0.2
0.0
(r - re )
0.2
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
ΔGHO
v+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
HO
ΔEv+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 §3.1.1
For a diatomic molecule, vibrational transitions typically occur in the energy range 100 − 4 000 cm−1 ,
which is in the infrared (IR) region of the electromagnetic spectrum. For example, for the ground electronic
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 ∝
k̃/μ , these differences reflect differences
in both the bond stiffness force constant k̃ 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.
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
56
reduced mass, μ(HgI)/μ(HgH) ≈ 77.5 , and the force constant k̃ for HgH is only about 1.6 times larger than
that for HgI.
This dependence of vibrational level energies and spacings on k̃ 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 1 Σ+ ground
electronic state of 7 LiH, while for the electronically excited A 1 Σ+ state of this same species, ωe = 234.4 cm−1 ;
i.e., the force constant k̃ 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 6 LiH, 7 LiD and 6 LiD (with natural abundances of 7.499%, 0.014% and 0.001%, respectively)
in the X 1 Σ+ ground electronic state, ωe=1 420.048, 1 054.940 and 1 074.309 cm−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
ν̃vib = ωe . In practice, however, three additional factors make vibrational spectra much more interesting.
(i) While a dipole moment does vary linearly for very small amplitude excursions from equilibrium, in
general quadratic, cubic and higher-power terms are required to describe it accurately. For a harmonic oscillator those terms allow |Δv| > 1 transitions at energies ν̃vib = 2ωe , 3ωe , . . . etc. While
those overtone transitions rapidly become much weaker as |Δv| increases, they could be observed at
frequencies corresponding to multiples of ωe .
(ii) The harmonic-oscillator function of Eq. (3.2) is not a very good overall model for a real potential energy
curve. While almost all intermolecular potentials are approximately quadratic near their minima, a
harmonic oscillator function goes to infinity at large |r − re |, so it could only describe bonds that never
break. This is clearly unrealistic, as a real potential function must approach infinity as the nuclei
are pushed together so that r → 0 , while as r → ∞ it must approach an asymptote at the bond
dissociation energy (see Fig. 1.13). Because of this shape asymmetry, the vibrational level spacings of
a realistic potential are not constant (as they are for a harmonic oscillator), but become smaller with
increasing energy, and the simple Δv = ±1 orthogonality selection rule is no longer strictly true (not
even if the dipole moment is a strictly linear function of r).
(iii) There is no angular momentum associated with radial vibrational motion. Thus, since angular momentum must be conserved when a photon is absorbed or emitted, all vibrational spectra must really
be vibration-rotation spectra, with a simultaneous change in rotational quantum number J taking
account of the photon angular momentum.
These three factors mean that vibrational spectra are much richer and more interesting than would be the
case for a pure harmonic oscillator, and allow such spectra to yield a wealth of detailed information about
molecules. Subsequent sections discuss how factors (ii) and (iii) affect IR (i.e., vibrational) spectra.
Exercise (i): If an ab initio quantum chemistry calculation predicts a bond stretching force constant of
k=1 902 N m−1 for the ground electronic state of CO, what do we predict for the v=1 ← 0 pure vibrational
transition energy of 12 C16 O, in cm−1 ?
Answer: Since the given value of k is given in SI units, one way to address this question is to use Eq. (3.8).
We begin by looking up the atomic masses, calculating the reduced mass, and converting to SI units:
−1
1
6.856 392 × 10−3 [kg/mole]
1
[u] =
μ =
+
= 1.138 531 × 10−26 [kg/molec] .
12.0
15.994 915
6.022 136 7 × 1023 [molec/mole]
A careful student will note that these ωe values are not precisely those predicted by multiplying the value for 7 LiH by the
ratios of the relevant μ values (which would yield 1 420.109, 1 054.721 and 1 074.115, respectively). The small discrepancies are
due to “Born-Oppenheimer breakdown” effects, which are currently a hot research topic.
4
3.3. ANHARMONIC VIBRATIONS AND THE MORSE OSCILLATOR
57
Substituting this value into Eq. (3.8), we obtain a vibrational energy spacing of
k
1 902 [N m−1 ]
−34
h νe = = 1.054 572 66 × 10
= 4.310 317 × 10−20 [J] ,
μ
1.138 531 × 10−26 [kg]
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 = 2 169.9 [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=1 902 [N m−1 ]=1 902 [J m−2 ], we can either look up the J → cm−1
energy conversion factor in Table 1.1 on p. 7, or apply the Planck equation conversion factors:
k̃ = 1 902 [J m−2 ]
10−20
= 9.574 892 × 105 [cm−1 Å−2 ] .
102 hc
Substituting this result into Eq. (3.10) then yields the same result obtained before:
2×16.857 629 × 9.574 892×105 /6.856 392 = 2 169.9 [cm−1 ] .
ωe =
Exercise (ii): A more common problem is that we wish to use experimental spectroscopic data to determine the
properties of a molecule. Consider for example the case of LiH. Recent measurements showed that the v=1−0
vibrational spacing of 7 LiH was ΔG1/2 = 1 359.708 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
1
+
7.016 003 0
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 = 1 359.708 cm−1 . Substituting this value and the reduced mass into Eq. (3.10) and solving for k̃ then
yields
k̃ = (1 359.708)2×0.881 238 16/(2×16.857 629) = 48 323.47 [cm−1 Å−2 ] .
3.3
3.3.1
Anharmonic Vibrations and the Morse Oscillator
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 onedimensional 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)
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
58
VMorse(r)
υ=7
υ=6
υ=5
} hot bands
υ=4
D0
third overtone
υ=3
De
second overtone
υ=2
first overtone
υ=1
fundamental
υ=0
zero point energy
r→
↑
re
Figure 3.4: Vibrational levels and transitions of a Morse potential energy function.
in which De is the bond dissociation energy, the depth of the potential energy well, re is the equilibrium bond
length, the position of the minimum of the potential, and β is a parameter (related to the force constant k)
defining the stiffness of small amplitude vibrations near the potential minimum. The characteristic shape of
a Morse potential and some of its properties are illustrated by Fig. 3.4. In the second version of Eq. (3.12)
we can see that the (squared) positive first exponential term dominates the short-range repulsive behaviour,
the negative middle term is responsible for the attractive potential well and long-range behaviour, while
(constant) the third term defines the potential asymptote. Although this form is not ideal,5 the radial
Schrödinger equation cannot be solved analytically for most more sophisticated potentials, and the Morse
function is sufficiently realistic for the purpose of the following discussion.
As was implied above, the Morse potential is another one of those accidentally special functions for which
the radial Schrödinger equation may be solved analytically to give an exact closed-form expression for its
vibrational level energies:
GMorse (v) = ωe (v + 1/2) − ωe xe (v + 1/2)2 ,
(3.13)
in which6
!
ωe
ωe xe
= = β2
1
2 De β 2
μ
102 hc
2
1
2μ 102 hc
=
=
#
2
$
$ 4 C D [cm−1 ] β [Å−1 ]
u e
%
μ [u]
Cu (β [Å−1 ])2
[cm−1 ] .
μ [u]
[cm−1 ]
(3.14)
(3.15)
Squaring both sides of Eq. (3.14) and dividing by Eq. (3.15) cancels out the factor Cu β 2 and yields the
relation
= (ωe )2 /4 ωe xe .
DMorse
e
(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 − C /r 6 , and
6
not the rapidly-dying exponential tail of a Morse function V (r) ∼ D − 2De e−β(r−re ) .
6 Note that the constant “ω x ” should always be treated as a single symbol, and not as the product ω ×x .
e e
e
e
3.3. ANHARMONIC VIBRATIONS AND THE MORSE OSCILLATOR
59
the Morse function as an expansion about the point r=re . In particular, using a Taylor series to expand the
exponential function(s) of Eq. (3.12),
e
−x
∞
(−x)n
1
1
1 4
= 1 − x + x2 − x3 +
x + ... ,
= 1+
n!
2
6
24
n=1
our Morse potential energy function may be expanded as
7
1
2
3
4
5
[β(r − re )] − [β(r − re )] + . . .
.
V (r) ≈ De [β(r − re )] − [β(r − re )] +
12
4
(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 k̃ = k̃ Morse = 2 De β 2 .
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
ΔGMorse
v+1/2
=
GMorse (v + 1) − GMorse (v)
ωe (v + 1 + 1/2) − ωe xe (v + 1 + 1/2)2 − ωe (v + 1/2) − ωe xe (v + 1/2)2
=
ωe − 2 ωe xe (v + 1) .
=
(3.18)
This expression shows that as the vibrational quantum number v (and hence the energy) increases, the level
spacings become systematically smaller, and go to zero at the dissociation limit.
Another way of thinking about this result is to recognize that the energy level expression Eq. (3.13) is a
parabolic function of v whose maximum lies at De . Taking the first derivative of Eq. (3.13) with respect to
v and setting it equal to zero, yields
dG(v)
= ωe − 2 ωe xe (v + 1/2) = 0 .
dv
(3.19)
Solving this equation yields the effective vibrational index associated with the dissociation limit
vD = ωe /2ωe xe − 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 v D = 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
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
60
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 − ωe xe /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 7 LiH, 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 ωe xe , 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 − 2ωe xe (0 + 1) = ωe − 2ωe xe
ΔG(2 ← 0)
=
=
2 674.560 = GMorse (v = 2) − GMorse (v = 0)
ωe (2 + 1/2) − ωe xe (2 + 1/2)2 − ωe (0 + 1/2) − ωe xe (0 + 1/2)2
=
2 ωe − 6 ωe x e .
(3.21)
(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
ωe xe = (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
=
1 404.564/(2×22.428) − 1/2 = 30.813
for De and vD . The total number of bound vibrational levels of 7 LiH (counting upward from v = 0 ) is therefore
estimated to be v D + 1 = 30 + 1 = 31 .
3.3.3
Higher-Order Anharmonicity and the Dunham Expansion
While the Morse potential function has a qualitatively realistic shape, and the analytic expression for its
level energies is certainly simple to work with, it does not tell the whole truth. The observed vibrational
energies and vibrational level spacings of real molecules are not precisely described by the simple quadratic
3.3. ANHARMONIC VIBRATIONS AND THE MORSE OSCILLATOR
61
function of Eq. (3.13) and the linear vibrational spacing function of Eq. (3.18). A common way of treating
such data is to use the following straightforward empirical generalization of Eq. (3.13):
G(v) = ωe (v + 1/2) − ωe xe (v + 1/2)2 + ωe ye (v + 1/2)3 + ωe ze (v + 1/2)4 + . . . .
This in turn yields the vibrational spacing equation7
ΔGv+1/2 = ωe − 2 ωe xe (v + 1) + ωe ye 3v 2 + 6v + 13/4 + ωe ze 4v 3 + 12v 2 + 13v + 5 + . . . .
(3.23)
(3.24)
It is clear that Eqs. (3.23) and (3.24) are becoming sufficiently cluttered that using such expressions and
solving multiple equations in multiple unknowns would rapidly become a rather unwieldy way of attempting to analyze experimental data. Moreover, that approach of determining Np parameters from Nd data
essentially assumes that all of those input data are ‘perfect’, which is never really true. A more practical
approach when one has data for multiple vibrational levels is to perform least-squares fits to our polynomial
expressions. This may readily be done using either a spreadsheet or a simple computer program. When
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:
2 2
3
r − re
r − re
r − re
r − re
VDun (r) = a0
+ a2
+ a3
+ ...
.
(3.25)
1 + a1
re
re
re
re
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)
2
3
4
= Y1,0 (v + 1/2) + Y2,0 (v + 1/2) + Y3,0 (v + 1/2) + Y4,0 (v + 1/2) + . . .
l
Yl,0 (v + 1/2) ,
=
(3.26)
l=1
and he determined explicit analytic expressions for the various Yl,0 coefficients in terms of the potential
expansion parameters {ai }:
(3.27)
Y1,0 = 2 a0 Be
3
Be a2 − 5(a1 )2 /4
(3.28)
Y2,0 =
2
(Be )3/2 10a4 − 35 a1 a3 − 17(a2 )2 /2 + 225(a1 )2 a2 /4 − 705(a1 )4 /9
Y3,0 =
(3.29)
√
4 a0
...
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 ai
potential coefficients. In practice, however, simultaneously taking account of the rotational energy level
pattern resolves this problem (see §3.6), and one can determine the first Np {ai } potential parameters from
empirically determined values of the Np leading Yl,0 and Yl,1 (see §3.6) coefficients. However, while it is
appropriate that you be told about this more sophisticated theory, practical applications of this approach
are beyond the scope of this text.8
So far, we have introduced three models for vibrational motion and the resulting transitions probed by
spectroscopy. There is no set rule as to when each of these models should be applied. Obviously, the Morse
or higher-order anharmonic oscillator provides a more realistic picture of molecular vibrations, and should
be used whenever possible. However, if little or no information concerning the anharmonicity is available,
the harmonic oscillator model is a convenient simplification.
Note that as for ωe xe ,6 the variable names ωe ye and ωe ze should each be treated as a single symbol, and not as a product.
It is clear that these Yl,0 coefficients correspond precisely to the empirical coefficients ωe , ωe xe , ωe ye , . . . etc., of Eq. (3.23),
with the anomalous historical sign convention aberration that defines ωe xe = −Y2,0 .
7
8
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
62
3.4
Bond Dissociation Energies and Birge-Sponer Plots
While the Morse function is certainly more realistic than a harmonic oscillator model for representing a
molecular potential energy curve, as the above discussion indicates, it is far from being exact. In addition to
the reservation raised in footnote 5 on p. 58, its prediction that the vibrational level spacing decreases as an
exactly linear function of v is not correct for real molecules. As a result, Eq. (3.16) may not provide us with
particularly accurate estimates of the molecular dissociation energy. Moreover, while its shape is certainly
more flexible, the Dunham potential of Eq. (3.25) goes to +∞ or −∞ (depending on the mathematical sign
of the last ai coefficient) as r → ∞ , and for most more sophisticated potential energy functions, the radial
Schrödinger equation cannot be solved analytically to give expressions for the level energies that can be
inverted to yield a value of De or D0 (as we were able to do in Exercise (iii)). However, we still very much
want to be able to determine accurate bond dissociation energies; thus, we must devise some alternative way
of doing so.
The nature of the problem is illustrated by panels A and B of Fig. 3.5. This example is a case in which
experimental fundamental, overtone and hot band data have given us the vibrational energies for levels
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 dotdot-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
ΔGv+1/2 = 2442.81 [cm−1 ] .
v=0
If we wish to use straight-line extrapolation to estimate the distance from level v+6 to the dissociation limit,
it is first necessary to determine the value of our linear function at the point v+6 . Using the above slope we
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 D rather than D , since experiment can only tell us directly about energies of one level relative to
e
0
another, so information about the zero point energy De − D0 can only be obtained by extrapolation.
3.4. DISSOCIATION ENERGIES AND BIRGE-SPONER PLOTS
D
V(r)
υ=6
4000
∼
⇑?
⇓
D0
63
500
?
Evib(υ)
Birge-Sponer plot
400
∼
3000
ΔE υ + ½
300
5
2000
4
3
2
1
A
0
B
1000
0
0
5
υ
10
15
200
C
100
0
0
5
υ
10
15
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 v 2 ,
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:
14.314
488.339 − 24.3609 v − 0.68152 v 2 dv = . . . = 1439.98 [cm−1 ] ,
D0 − G(v = 6) =
6
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 Le Roy and Richard Bernstein introduced a beautifully simple theory that gave an explicit
analytic expression for the shape expected for a Birge-Sponer plot near its intercept at v=vD . While that
theory is too complicated to present here, the essential result is that near its intercept, a Birge-Sponer plot
has the shape
(n+2)/(n−2)
,
(3.30)
ΔGv+1/2 X0 (vD − v)
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
7/3
that yields the dotted curve extrapolation. It is interesting to
the functional form ΔGv+1/2 ∝ (vD − v)
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.
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
64
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
cm = (xcm , ycm , zcm) , this leaves 3N−3 degrees of freedom for the internal
position of the centre of mass, R
motion.
The overall rotation of the molecule will account for some of these 3N−3 internal degrees of freedom. The
discussion of §2.1 indicated that for a diatomic molecule, or indeed any linear molecule, only two coordinates
are required to characterize its orientation and rotational behaviour, the angular coordinates θ and φ of
Fig. 1.9. Similarly, for a non-linear molecule, three angular coordinates are required (see §2.5.3). This leads
us to the conclusion that
• a linear N –atom molecule has 3N − 5 vibrational degrees of freedom (the other five being three for
translation plus two for overall rotation)
• a non-linear N –atom molecule has 3N − 6 vibrational degrees of freedom (the other six being three
for translation plus three for overall rotation)
Each of these vibrational degrees of freedom corresponds to a separate type of co-ordinated overall motion
of all of the atoms in the molecule, which is called a normal mode of vibration. Of course, all of the same
considerations concerning harmonic and anharmonic vibrations discussed above apply to each of these vibrational modes. As a result, for a general polyatomic molecule the total vibrational energy is conventionally
written as
ωi (vi + 1/2) +
xi,j (vi + 1/2)(vj + 1/2) + . . .
(3.31)
G(v1 , v2 , v3 , . . . ) =
i
i
j≥i
where vi is the vibrational quantum number for mode i, ωi and xi,j are the harmonic and leading anharmonic
vibrational constants.
Equation (3.31) shows that in order to calculate the actual vibrational spacing associated with any particular mode it is necessary to know not only the harmonic and anharmonic constants ωi and xi,i associated
with that mode (the analog of ωe and ωe xe for a diatomic), but also all of the associated mixed anharmonicity
constants, xi,j for j = i . Conversely, to obtain a knowledge of the complete set of harmonic and leading
anharmonic constants for a species with Nm distinct vibrational modes would require Nm (Nm+3)/2 independent vibrational spacing measurements. It is usually quite difficult to obtain that much information, so it
is customary to characterize the energy associated with a given vibrational mode of a polyatomic molecule
by the value of the observed fundamental vibrational spacing for that mode, denoted here as ν̃i , 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 ν̃i .
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
ν1 C − H symmetric stretch
∼
ν1 = 3374 cm-1
ν2 C ≡ C stretch
-1
∼
ν2 = 1974 cm
ν1
∼
-1
ν1 = 3657 cm
ν2
∼
ν2 = 1595 cm-1
ν3 C − H asymmetric stretch
∼
ν3 = 3287 cm-1
A
ν3
-1
ν∼3 = 3756 cm
−
−
+
+
−
+
+
−
Figure 3.6: Vibrational normal modes of: A. water H2 O and
ν4 trans bend
-1
∼
ν4 = 612 cm
B
ν5 cis bend
-1
∼
ν5 = 729 cm
B. acetylene C2 H2 .
(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 v2a and v2b . 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: H2 O
As in the preceding case, N=3 , but since this molecule is non-linear it will have only the 3 × 3 − 6 = 3
independent vibrational modes shown in Fig. 3.6-A. In this case we note that the symmetric stretch
mode is identified as “ν1 ”, the bending mode as “ν2 ” and the antisymmetric stretch as “ν3 ”. The water
molecule has a permanent dipole moment, and it is clear that for all three of these modes that dipole
will oscillate in magnitude (for ν1 and ν2 ) or direction (for ν3 ), so all three of these modes will be IR
active.
(iv) A linear tetra-atomic molecule: acetylene H–C≡C–H
Since N=4 and C2 H2 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
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
66
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 H2 O, 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 ν̃1 = ν̃1ss = 3 441 cm−1 and for the antisymmetric stretch mode an energy of
ν̃3 = ν̃3as = 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≡N
motion, 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 ν̃i for common chemical function groups.
chemical
group
≡C-H
(phenyl)-C-H
=CH2
–CH3
–CH3
–CH2 –
–CH2 –
–CH2 –
–CH3
–CH3
-O-H
NH2
approximate
frequency [cm−1 ]
3300
3060
3030
2970
2870
2930
2860
1470
1460
1375
3600
3400
(antisym. stretch)
(sym. stretch)
(antisym. stretch)
(sym. stretch)
(deformation)
(antisym. deform.)
(sym. deform.)
chemical
group
approximate
frequency [cm−1 ]
S-H
–C≡N
–C≡C–
–C=O
C=C
C=N
C-C, C-N, C-O
C-Cl
C-Br
C-I
C=S
C-F
2580
2250
2220
1750-1600
1650
1600
1200-1000
725
650
550
1100
1050
3.6. ROTATIONAL STRUCTURE IN VIBRATIONAL SPECTRA OF DIATOMICS
67
Figure 3.7: Room temperature absorption spectrum of D35 Cl (the stronger line of each pair) and D37 Cl.
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
√ double and triple bonds, 1200 : 1650 :
√ single,
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 ∝ k̃ .] 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 D35 Cl (or 2 H35 Cl), for which ωe=2 145.133 cm−1 ,
ωe xe=27.1593 cm−1 , ωe ye=0.079 93 cm−1 and ωe ze=− 0.003 03 cm−1 . A harmonic oscillator model would predict that its vibrational spectrum consisted of a single line (or superposition of lines) at a transition energy of
ν̃=ωe=2 145.133 cm−1 . Allowing for anharmonicity, it would be expected to have a strong fundamental transition at 2 091.059 cm−1 , and weaker first and second overtone transitions at 4 128.430 and 6 112.449 cm−1 ,
while if the sample was sufficiently hot to yield measurable populations in excited vibrational levels, there
could be weak “hot band” transitions at 2 037.372 and 1 984.019 cm−1 . In contrast, Fig. 3.7 shows that its
absorption spectrum at room temperature (under which conditions no hot bands would be observed) consists
of about 30 sharp, roughly equally spaced lines spread across a 320 cm−1 interval.
The reason for this was alluded to earlier when it was pointed out that there is no angular momentum
associated with vibrational stretching. As a result, for angular momentum to be conserved when an IR
photon is absorbed or emitted, a simultaneous rotational transition must occur. Since rotational levels
have relatively small energy level separations, a substantial number of them tend to be populated at most
temperatures, and when a collection of molecules in thermal equilibrium undergo a vibrational transition, we
obtain a “band” or group of spectral lines such as that seen in Fig. 3.7. It is because vibrational transitions
of gas phase molecules always occur with a wealth of rotational fine structure that we normally refer to
vibrational spectra as consisting of “bands”.
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
68
R branch
Δ J = +1
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
⎪⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
P branch
Δ J = −1
⎧
7
⎪
⎪
⎪
6
⎪
⎪
⎪
5
υ′ ⎨
⎪
4
⎪
⎪
3
⎪
⎪
2
⎪ J ′= 0
⎩
↑
energy
⎧
7
⎪
⎪
⎪
6
⎪
⎪
5
υ′′ ⎨
⎪
4
⎪
3
⎪
⎪
2
⎪ J ′′= 0
⎩
spectrum
P(8)
P(7)
P(6)
P(5)
P(4) P(3)
band
gap
←⎯→
P(2)
P(1)
R(2) R(3) R(4)
R(1)
R(5)
R(0)
R(6)
R(7)
↓
∼
ν0
∼
ν / cm−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 = ±1
selection rule. The individual transitions must, of course, be labelled by the vibrational and rotational
quantum numbers of the upper and lower states:
, J ) − E(v
, J )i ,
ν̃(v , J ; v , J ) = E(v
(3.32)
where we use the standard convention of labeling the quantum numbers for the upper levels with a single
prime ( ) and those for the lower level in the transition with double primes ( ). The values of J and J are
linked by the normal angular momentum conservation selection rule, and it is customary to label the allowed
transitions according to the sign of the change in J and the rotational quantum number of the lower level.
In particular, transitions with ΔJ = J − J = Jupper − Jlower = −1 are called P transitions, and those
with ΔJ = +1 are called R transitions.11 For P transitions in vibrational spectra the upper level has less
rotational energy than the lower one, and for R transitions the opposite is true. (As a result, all transitions in
pure rotational spectroscopy are R transitions.) It is very important that these labeling rules be understood
, J ) transition
, J ) ↔ E(v
and followed, since there would be considerable confusion if a particular E(v
was given one name in an absorption spectrum and another in emission.
In view of the above, the energy of a particular P -branch transition in a given (v , v ) vibrational band
would be
ν̃P (J )
, J ) − E(v
, J ) = E(v
, J − 1) − E(v
, J )
= E(v
= {G(v ) + Fv (J − 1)} − {G(v ) + Fv (J )} = {ignoring centrifugal distortion}
= [G(v ) − G(v )] + [Bv (J − 1)(J ) − Bv J (J + 1)]
= ν̃0 (v , v ) − [Bv + Bv ] (J ) − [Bv − Bv ] (J )2
≈ ν̃0 (v , v ) − [Bv + Bv ] (J )
(3.33)
(for small J )
11 Note that we have introduced another standard convention – that the quantum number change is written as J
upper minus
Jlower .
3.6. ROTATIONAL STRUCTURE IN VIBRATIONAL SPECTRA OF DIATOMICS
69
while for an analogous R-branch transition
ν̃R (J )
, J ) − E(v
, J ) = E(v
, J + 1) − E(v
, J )
= E(v
= {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)]
= ν̃0 (v , v ) + [Bv + Bv ] (J + 1) − [Bv − Bv ] (J + 1)2
≈ ν̃0 (v , v ) + [Bv + Bv ] (J + 1)
(3.34)
(for small J )
where ν̃0 (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 all
the transitions seen here are pure rotational sublevels of vibrational level v , and their thermal population
distribution will be given by Eq. (2.31). Since centrifugal distortion can be neglected at low J, we may
expect Eqs. (2.33) to predict the initial-state population distribution accurately, which leads us to predict
pop
Jmax
(293 K) = 3.85 ≈ 4 as the initial-state rotational quantum number for the most intense P - and Rbranch 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 1300 K emission spectrum of Na35 Cl shown in Fig. 3.9, which shows the high frequency
turnaround of the R-branches of the vibrational fundamental and the first hot band. Moving to higher and
higher frequencies with increasing J the lines get closer and closer together, and then pile up on one another
as they begin to turn around and move off to the red (i.e., to lower frequencies) with further increases in J.
The pile-up of intensity located where the branch turns around is called a “band head”, and will be discussed
12 Small rotational intensity correction factors of J /(2J + 1) for P lines and (J + 1)/(2J + 1) for R lines refine these
prediction at small J.
70
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
Figure 3.9: NaCl emission spectrum showing band heads for the fundamental and first hot bands.
further in Chapter 5. Note that the separation between the heads of the (2 − 1) and (1 − 0) bands of Na35 Cl
seen in Fig. 3.9 is approximately the same anharmonicity frequency shift of ∼ 2 ωe xe predicted by Eqs. (3.18)
−1
or (3.24) ( ωe xe (Na35 Cl) = 1.78 cm ) for the shift in the band origin.
A final feature of Fig. 3.7 which deserves comment here is the small, but roughly constant shift to
lower frequencies of the transitions associated with the minority (24% relative abundance) isotopologue
√
or (3.14) we see that all else being equal, ωe ∝ 1/ μ . This implies that for
D37 Cl. From Eqs. (3.10)
D37 Cl, ωe ≈ 2 145.133 μ(D35 Cl)/μ(D37 Cl) = 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 ν̃0 (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 74 Ge; 27.7% is 72 Ge;
21.2% is 70 Ge; 7.7% is 73 Ge; 7.4% is 76 Ge) whose overlapping Δv = 1 emission contribute to this spectrum,
and another is that in addition to the vibrational fundamental, hot bands are observed for (v , v ) = (2, 1)
up to (8, 7). For each hot band and each isotopologue, the band origin is slightly shifted, as predicted
by Eqs. (3.14), (3.15) and (3.24), and for each vibrational level of each isotopologue, all of the rotational
constants have small isotopic and vibrational shifts. In spite of these complexities, the very high resolution
of modern instruments, and our understanding of isotopologue scaling effects allowed this extremely dense
spectrum to be decomposed into 36 distinct bands characterized by very sharply determined parameters.
To illustrate this point, the right half of Fig. 3.10 shows a small segment of the assigned spectra of the
R-branches of the fundamental band; this illustrates the power of modern IR spectroscopy for untangling
very messy spectra.
As indicated both by the preceding discussion and by that of Chapter 2, the rotational constants Bv and
centrifugal distortion constants Dv , Hv , . . . , etc., will in general all depend upon the vibrational level v.
3.7. WHY ARE VIBRATIONAL LEVEL SPACINGS SO LARGE?
71
Figure 3.10: High temperature (1800 K) emission spectrum of GeO. Left: overview spectrum; Right: dispersed segment of the fundamental band showing isotopologue assignments.
One way of representing this behaviour is based on analogs of Eq. (3.23), namely:
Bv
Dv
= Be − αe (v + 12 ) + γe (v + 12 )2 + . . .
= De + βe (v +
1
2)
+ ... .
(3.35)
(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:
l
J) =
E(v,
Km (v)[J(J + 1)]m =
Yl,m v + 12 [J(J + 1)]m ,
(3.37)
m=0
m=0 l=0
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 , ωe xe , αe , . . .
, etc., the general expression of Eq. (3.37) is becoming more commonly employed in the scientific literature.
3.7
Why Are Vibrational Level Spacings So Much Larger Than
Rotational Level Spacings?
In Chapter 2 we stated that rotational spectra were observed in the microwave (MW) region of the electromagnetic spectrum with transition energies in the 1 − 100 cm−1 region, and in the present chapter we have
stated similarly that vibrational transitions occur in the infrared with transition energies of 100−5 000 cm−1 .
In other words, without presenting any justification we said that for any given molecule, vibrational level
spacings are typically 10−100 times larger than rotational level spacings. Similarly, we shall see in Chapter 5
that electronic transitions typically occur in the visible/ultraviolet region of the spectrum, 104 − 106 cm−1 .
While a ‘civilian’ might be content to remember such facts blindly, a scientist would tend to seek a rational
framework to explain it.
It turns out that the simple particle-in-a-square-well model discussed in Chapter 1 provides exactly the
type of explanation we would like to have. We saw in the early sections of Chapters 2 and 3 that the internal
motion of a diatomic molecule could be described exactly in terms of the motion of a pseudo-particle of
mass μ moving in space. For vibration this is motion along the radial coordinate r, while for rotation it is
orbital motion about the centre of mass of the system at a radius of re . More particularly, the discussion
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
72
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 Å, 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 − 100 u. Within a particle-in-a-square-well-box picture, for a fixed box size this difference
would make the energy level spacings for the electron roughly 104 − 105 times larger. However, the radius
of the orbit for one of the outermost electrons in a molecule (the first one to be excited) would be at
least twice the equilibrium internuclear separation re , and this difference increases the effective box length
relative to that for rotation (which we saw was πre ) by a factor of two and hence reduces the level spacing
by a factor of four. Thus, electronic excitation energies are expected to be of order 2 500 − 25 000 times
larger than rotational level spacings for the same molecule, which would make them 25 − 250 times larger
than its vibrational level spacings. Thus, we see that with very simplistic particle-in-a-box reasoning, we
can rationalize the relative magnitudes of the photon energies required to drive rotational, vibrational and
electronic transitions.
3.8
Problems
1. The hydrogen halides have the following fundamental (v = 1 ← 0) band origins (in cm−1 ):
1
H19 F
4143.3
1
H35 Cl
2988.9
1
H81 Br
2649.7
1
H127 I
2309.5
Using the harmonic oscillator model, calculate the force constants k̃ 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 ωe xe ,
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 1 H35 Cl: 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 r̄ for the upper and
lower vibrational levels of this band.
(c) Determine the value of the force constant k̃ for for HCl.
4. The following lines (in cm−1 ) were observed in the vibration-rotation spectrum of 2 H35 Cl: 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 2 H is substituted
for 1 H.
[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 7 Li2 , 12 C2 and 14 N2 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 1 H35 Cl 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 N m−1 , re = 1.597 Å and m(127 I) = 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 1 H127 I.
8. The vibrational frequency of 12 C16 O 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 13 C16 O.
9. An unknown molecule XY has a vibrational frequency of 2331 cm−1 and a force constant of k =
2245 N m−1 . The diatomic oxide of X, XO, has a vibrational frequency of 1876 cm−1 and a force
constant of k = 1 550 N m−1 . Use this information to identify the molecule XY.
10. The IR spectrum of 23 Na127 I has a strong band with its origin at 284.50 cm−1 , a weaker band at
283.00 cm−1 , and the relative intensity of the latter increases with temperature. Determine the bond
force constant, anharmonicity, zero-point energy, dissociation energies De and D0 of NaI in “spectroscopists’ units”.
11. In the infrared spectrum of 1 H35 Cl, 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 Å).
12. For 2 H35 Cl 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 2 H for 1 H 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, H2 O, HCCH,
H2 CCH2 , H3 CCH3 and C60 ? For the first five of these species, how many of those modes are associated
with bending motion?
14. Assuming that 1 H35 Cl is reasonably well described by a Morse potential with parameters: De =
5.250 eV, ωe = 2990.95 cm−1 , and ωe xe = 52.82 cm−1 , determine D0 for each of 1 H35 Cl and 2 H35 Cl.
15. Which bond would be easier to break, a C-1 H bond or a C-2 H 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 k̃ from the IR spectrum.
17. The infrared spectrum of N2 O 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 23 Na35 Cl
and 23 Na37 Cl under the assumption that their force constants are the same.
19. For a molecule whose vibrational data give ωe = 1387.09 and ωe xe = 83.01 cm−1 , obtain an estimate
for De and for the total number of vibrational levels for that molecule.
74
CHAPTER 3. VIBRATIONAL SPECTROSCOPY
20. Determine extended versions of Eqs. (3.33) and (3.34) that take account of the upper and lower level
centrifugal distortion constants, Dv and Dv .
Chapter 4
Raman Spectroscopy
What Is It? Raman spectroscopy determines vibrational and rotational level spacings from the frequency
shifts of scattered light.
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 measurements on such species, a discovery that earned him both a knighthood and the 1930 Nobel Prize in physics.
In one sense this topic introduces nothing new, since it is just another way of measuring the rotational and
vibrational level spacings: once these spacings are known, the theory and methods presented in Chapters
2 and 3 can be used for determining molecular bond lengths, stretching force constants and dissociation
energies. However, the rules governing such spectra are different than those presented previously, so the
Raman mechanism deserves some attention in its own right.
4.1
“Light-As-A-Wave” Description of Raman Scattering
We begin by describing two perspectives on the Raman effect. The first combines our classical view of light
as a wave phenomenon with the fact that a molecule consists of a distribution of negative electronic charge
centred about tiny, heavy positive nuclei. When a molecule is subjected to a strong external electric field E,
that field will distort the molecule’s diffuse electronic charge distribution to induce a small dipole moment
that is proportional to the strength of the field:1
ind = α E
M
(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
CHAPTER 4. RAMAN SPECTROSCOPY
76
external electric field E
→
16
orientation O
of
molecule 18
O
18
16
O
18
O
16
O
16
O
O
16
18
18
O
O
16
O
18
O
O
dipole
induced
by field
1 / νrot
↑
dipole
induced
by field
0
time →
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 cylindrically 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
0 cos(2π ν0 t + φ0 ) .
ν0 Hz, Eq. (1.3) shows that the electric field felt by the molecule oscillates as E(t)
=E
At the same time, because of the rotation of the molecule, its polarizability along the direction of the field
can be written as
αE 0 (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
0 is a unit vector pointing along the direction of the electric field. Upon
polarizability anisotropy, and E
substituting these expressions for E(t)
and αE 0 (t) into Eq. (4.1), we see that the time-dependent induced
dipole may be written as
ind (t)
M
0 {α0 cos(2π ν0 t) + Δα cos(4π νrot t) cos(2π ν0 t) }
= E
(4.3)
1 1 = E0 α0 cos(2π ν0 t) + 2 E0 Δα cos (2π[ν0 + 2νrot ]t) + 2 E0 Δα cos (2π[ν0 − 2νrot ]t) ,
with the second version of Eq. (4.3) being obtained from the first by utilizing the familiar trigonometric
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
external electric field E
→
16
stretching O
of
molecule 18
16
O
O
18
18
O
O
16
O
[equilibrium]
16
16
O
O
18
O
18
O
[equilibrium]
77
16
O
18
O
[equilibrium]
dipole
induced
by field
1 / νvib
↑
dipole
induced
by field
time →
0
Figure 4.2: Oscillating induced dipole moment of a vibrating non-polar molecule in an external electric field.
and one that is ν0 minus twice the rotation frequency of the molecule. Just as the oscillating charges in a
radio station antenna cause emission of electromagnetic radiation, so these oscillating induced dipoles allow
molecules to scatter light at these three frequencies. The scattering of light at the frequency of the incident
beam ν0 is called Rayleigh scattering, that associated with the lower frequency ν0 − 2νrot is called Stokes
scattering, and that associated with ν0 + 2νrot is called anti-Stokes scattering. Since normally |Δα/α0 | 1 ,
we expect that the Stokes and anti-Stokes scattering will be much weaker than Rayleigh scattering. The
source of this naming convention will be explained below.
Most Raman spectroscopy is performed using high-frequency incident light, since historically it was
much easier to find intense, monochromatic short-wavelength light sources,3 and theory (not presented here)
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,
αE 0 (t) = ᾱ + 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 ᾱ 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:
ind (t)
M
0 ᾱ cos(2π ν0 t) +
= E
1
2
0 δα cos (2π[ν0 + νvib ]t) +
E
1
2
0 δα cos (2π[ν0 − νvib ]t)
E
(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.
CHAPTER 4. RAMAN SPECTROSCOPY
78
Stokes
scattering
Rayleigh
scattering
anti-Stokes
scattering
"virtual
level"
ν0
ν0
δE
νs =
ν0
ν0 − δE/h
νs = ν0+ δE/h
νs=ν0
E ′(υ′,J ′)
E ′′(υ′′,J ′′)
Figure 4.3: Incident ν0 and scattered νs light in Raman scattering involving upper and lower vibrationrotation 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 :
Label :
−3 −2 −1
N
O
P
0
Q
+1 +2 +3
R
S
T
If we ignore the effects of centrifugal distortion, rotational Raman transitions yield Stokes scattering at
energies
4 Fluorescence is spontaneous light emission by molecules in an excited state that was populated by an initial absorption
transition.
CHAPTER 4. RAMAN SPECTROSCOPY
80
ν̃SStokes (J)
= ν̃0 − ΔFv (J + 2 ← J) = ν̃0 − [Fv (J + 2) − Fv (J)]
= ν̃0 − [Bv (J + 2)(J + 3) − Bv J(J + 1)]
= ν̃0 − Bv (4J + 6)
(4.7)
and anti-Stokes scattering at
ν̃Santi−S (J)
=
ν̃0 + ΔF (J + 2 ← J) = ν̃0 + [F (J + 2) − F (J)]
=
ν̃0 + 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 ν̃0 . 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
ν̃0 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
= E(v
, J ) − E(v
, J )
as being the same as infrared spectra, apart from the fact that the level spacing ΔE
is observed as a difference rather than directly and the different rotational selection rules. The latter means
that we have three rotational branches in both the Stokes and anti-Stokes regions. If we ignore centrifugal
distortion, the vibrational Stokes lines occur at energies
Stokes
, J − 2) − E(v
, J)
ν̃O
(J) = ν̃0 − E(v
= ν̃0 − {[G(v ) − G(v )] + [Fv (J − 2) − Fv (J)]}
= ν̃0 − {[G(v ) − G(v )] + [Bv (J − 2)(J − 1) − Bv J(J + 1)]}
= ν̃0 − [G(v ) − G(v )] − (Bv + Bv ) (2J − 1) − (Bv − Bv ) (J 2 − J + 1)
Stokes
, J) − E(v
, J)
(J) = ν̃0 − E(v
ν̃Q
(4.9)
= ν̃0 − {[G(v ) − G(v )] + [Fv (J) − Fv (J)]}
= ν̃0 − {[G(v ) − G(v )] − [Bv − Bv ] J(J + 1)}
, J + 2) − E(v
, J)
ν̃SStokes (J) = ν̃0 − E(v
(4.10)
= ν̃0 − {[G(v ) − G(v )] + [Fv (J + 2) − Fv (J)]}
= ν̃0 − {[G(v ) − G(v )] + [Bv (J + 2)(J + 3) − Bv J(J + 1)]}
(4.11)
2
= ν̃0 − [G(v ) − G(v )] + (Bv + Bv ) (2J + 3) − (Bv − Bv ) (J + 3J + 3) .
The analogous expressions for anti-Stokes transitions are identical, except that the mathematical “−” sign
following the symbol ν̃0 in each of these equations is replaced by a “+” sign. Note, however, that while S(J)
4.5. RAMAN SPECTRA OF POLYATOMIC MOLECULES
fundamental
vibrational
band
(Stokes)
81
⎫
⎬
⎭
Rayleigh line
fundamental
vibrational
band
(anti-Stokes)
Q branch
pure rotation
S branch
(anti-Stokes)
Q branch
⎫
⎬
⎭
O branch
↑
νo
S branch
⎫
⎬
⎭
pure rotation
S branch
(Stokes)
⎫
⎬
⎭
O branch
⎫
⎬
⎭
⎫
⎬
⎭
S branch
ν→
Figure 4.4: Schematic illustration of rotational and vibrational Raman spectra.
and Q(J) transitions occur for all non-negative values of J = 0, 1, 2, 3, . . . , etc., O(J) transitions are only
possible for J ≥ 2 .
The vibrational energy spacings are of course defined by exactly the same harmonic or anharmonic
oscillator models discussed in Chapter 3, and the same vibrational selection rules apply: Δv = ±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
CHAPTER 4. RAMAN SPECTROSCOPY
82
For molecules with a centre of symmetry (e.g., N2 , CO2 , SF6 or C2 H2 ),
• IR active vibrational modes are Raman forbidden, and
• Raman active vibrational modes have no infrared spectra.
For example, for the linear molecule C2 H2 (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 H2 O absorbs very strongly in the IR, it
is a very weak Raman scatterer. This makes Raman spectroscopy the preferred technique for studying
molecules in aqueous solution, as solvent signals will not provide much interference. It has therefore become
particularly important for studying biological molecules and for environmental analysis.
4.6
Problems
1. Determine the generalized versions of Eqs. (4.7) and (4.8) for cases in which the leading centrifugal
distortion constant Dv cannot be ignored.
2. When excited by a laser of wavelength 180.000 nm, the P2 molecule is observed to have a vibrational
Raman spectrum with Q-branch Stokes bands centred at wavelengths 182.538 and 185.123 nm, respectively. Determine the vibrational frequency (in Hz), the harmonic and anharmonic vibrational
constants in cm−1 , and estimate the bond dissociation energy for this molecule (in cm−1 ).
3. Draw and label the level energies, the vibration-rotation transitions, and the resulting anti-Stokes
Raman spectrum of the fundamental band of a diatomic molecule, for all lines up to J = 3 .
4. Determine the generalized versions of Eqs. (4.9) – (4.11) for cases in which the leading centrifugal
distortion constant Dv cannot be ignored.
Chapter 5
Electronic Spectroscopy
What Is It? Electronic spectroscopy uses photons with energies ranging from the visible (eV’s) to ultraviolet (tens of eV) to excite molecules into excited electronic states.
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
CHAPTER 5. ELECTRONIC SPECTROSCOPY
84
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 vibrational bands associated with a single lower-state v value and a series of different upper-state v values, or
with a common v value and a series of different v values, will have distinctly different transition energies.
This is illustrated by Fig. 5.2, which shows a number of vibrational bands in a low resolution spectrum of the
A 1 Σ+ − X 1 Σ+ electronic transition of SrS. The separation between the positions of adjacent bands with
a common v value and different v values will be roughly equal to the fundamental vibrational constant
ωe for the upper state, while adjacent bands with a common v value will be separated by the analogous
lower-state constant, ωe . In particular, in Fig. 5.2 the separations between the peaks associated with the
(v , v )=(0, 0) and (1, 0) bands, the (1, 0) and (2, 0) bands, and between the (1, 1) and (2, 1) bands, are all
roughly equal to ωe . This spectrum also shows that bands in a particular “vibrational sequence”, 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 |ωe − ωe |.
One question raised by the spectrum in Fig. 5.2 is why each band has a sharp intensity maximum at
its high energy (short wavelength) edge, rather than the roughly symmetric double-hump shape seen in the
vibration-rotation spectra of Figs. 3.7 or the left hand side of Fig. 3.10, or the broad smoothly changing
pattern seen in Fig. 2.5. This is readily understood if we examine the expressions for the energies of the
individual transitions observed in electronic spectra.
Transition energies in electronic spectroscopy are described in essentially the same manner as were transition energies in vibrational spectroscopy, except that (i) the energies of the upper and lower states each
also contains an additive electronic energy contribution, Te , the energy at the minimum of the potential
energy curve for that state (see Fig. 5.1), and (ii) the vibration-rotation energies of the upper and lower
levels are governed by different sets of vibration-rotation constants. As a result, the energies of individual
transitions are given by the general expression
ν̃
=
(v , J ) = [Te + G (v ) + Fv (J )] − [Te + G (v ) + Fv (J )]
(v , J ) − E
E
=
{[Te + G (v )] − [Te + G (v )]} + [Fv (J ) − Fv (J )]
=
ν̃0 (v , v ) + [Fv (J ) − Fv (J )] ,
(5.1)
in which the energy difference defining the band origin
ν̃0 (v , v ) = [Te + G (v )] − [Te + G (v )] = ΔTe + G (v ) − G (v )
(5.2)
now contains the difference in electronic energies of the two states, ΔTe=Te − Te . Note that in contrast to
Eqs. (3.33) and (3.34) in vibrational spectroscopy and Eqs. (4.9) - (4.11) in Raman spectroscopy, prime ( ) or
double prime ( ) labels are now associated not only with the vibrational and rotational quantum numbers of
the upper and lower level of each transition, but also with the energies (e.g., Fv (J ) vs. Fv (J )) and with
(see below) molecular constants such as Bv or ωe , since both the molecular constants and the level energies
patterns that they define differ from one electronic state to another.
As in vibrational and rotational spectroscopy, conservation of total angular momentum in an absorption
or emission process means that the usual rotational selection rule ΔJ= ± 1 also applies to transitions in
electronic spectra. However, if the molecule has non-zero electronic angular momentum in one or both states,
the electronic degrees of freedom may take up the photon’s angular momentum, in which case the rotational
selection rule must be extended to allow also ΔJ=0 transitions. This means that in addition to having P
and R branches, each vibrational band of such electronic transitions will also have a Q branch consisting of
transitions for which ΔJ=0 (see Table 4.1). Use of the same manipulations applied in §3.6 then yields the
CHAPTER 5. ELECTRONIC SPECTROSCOPY
86
following expressions for the allowed transition energies:
ν̃P (J ) =
=
ν̃R (J ) =
=
ν̃Q (J ) =
=
(v , J ) − E
(v , J ) = E(v
, J − 1) − E
(v , J )
E
ν̃0 (v , v ) − [Bv + Bv ] (J ) − [Bv − Bv ] (J )2
(5.3)
, J + 1) − E (v , J )
E (v , J ) − E (v , J ) = E(v
ν̃0 (v , v ) + [Bv + Bv ] (J + 1) − [Bv − Bv ] (J + 1)2
ν̃0 (v , v ) − [Bv − Bv ] J (J + 1) ,
(5.4)
, J ) − E (v , J )
E (v , J ) − E (v , J ) = E(v
(5.5)
in which the final expression for each case is obtained by ignoring centrifugal distortion terms. These expressions 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 ν̃0 (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 ( Bv < Bv ), 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 Bv > Bv are called “red shaded”. The bands
shown in Fig. 5.2 are clearly red-shaded (note that since wavelength is increasing to the right in this figure,
the wavenumber or transition energy is increasing to the left).
In contrast with the results shown in Fig. 5.2, Fig. 5.3 shows the detailed structure of a band head for
a case in which the lines are all fully resolved up to, at, and beyond the band head. However, it is not
always possible to obtain fully resolved spectra such as this, so in order to determine the pure vibrational
contribution to the level energies, we need to be able to estimate the displacement between the observed
band head and the band origin. To do this, we need to determine the value of J associated with the point
5.2. VIBRATIONAL-ROTATIONAL STRUCTURE IN ELECTRONIC SPECTRA
87
Figure 5.3: Rotational structure near the (0,0) band head in the A 1 Σ+ − X 1 Σ+ 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 ν̃R (J)
dJ
= [Bv + Bv ] − 2[Bv − Bv ](J + 1) = 0
(5.6)
yields an expression for JhR , the value of J at the band head, the point where the R–branch turns around:
&
(JhR + 1) = [Bv + Bv ] 2[Bv − Bv ] .
(5.7)
Substituting this result into Eq. (5.4) then yields an expression for the position of the band head:
2
Bv + Bv Bv + Bv ν̃R (JhR ) = ν̃0 (v , v ) + [Bv + Bv ]
−
B
]
−
[B
v
v
2 (Bv − Bv )
2 (Bv − Bv )
=
ν̃0 (v , v ) +
(Bv + Bv )
4 (Bv − Bv )
2
.
(5.8)
Alternately, for cases in which Bv < Bv it is the P –branch that turns around, and we obtain
JhP
=
− [Bv + Bv ] /2[Bv − Bv ]
and
ν̃P (JhP ) = ν̃0 (v , v ) −
(Bv + Bv )
4 (Bv − Bv )
= [Bv + Bv ] /2[Bv − Bv ]
(5.9)
2
.
(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 infrared and electronic spectra is merely the different magnitudes of the denominators on the right hand sides of
CHAPTER 5. ELECTRONIC SPECTROSCOPY
88
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 particular, for the 1–0 infrared band of Na35 Cl shown there, Bv =0 = 0.217 251 cm−1 and Bv =1 = 0.215 637 cm−1 ,
so Eqs. (5.7) and (5.8) predict that JhR = 133.10 and that ν̃R (Jh ) − ν̃0 (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
[ν̃R (Jh ) − ν̃0 (v , v )] be relatively large are characteristic properties of infrared vibrational-rotational spectra. However, whether or not those band heads can be observed in a particular case will depend on the
temperature of the system being studied.
In contrast with the infrared case, in electronic spectra the upper- and lower-level Bv values are usually
quite different, so that the denominators of the fractions in Eqs. (5.7) – (5.10) are relatively large. As a result,
those fractions will be relatively small and the band heads will lie close to the band origins. For example, for
the 0 – 0 band of the A 1 Σ+ − X 1 Σ+ spectrum of 63 CuD shown in Fig. 5.3, the published rotational constants
are B0A = 3.475 22 cm−1 and B0X = 3.992 52 cm−1 . Utilizing these values in Eq. (5.8) yields the prediction
that JhR = 6.2 and ν̃R (Jh ) − ν̃0 (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 1 Σ+ − X 1 Σ+ electronic transition of SrS
seen in Fig. 5.2 allow us to determine that Bv =0 = 0.120 565 cm−1 and Bv =0 = 0.113 422 cm−1 . We may
then predict that the band head will occur at J = JhR = 15.38 and that [ν̃R (Jh ) − ν̃0 (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 Schrödinger 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
There are, of course, bands in electronic spectra for which by accident Bv ≈ Bv . 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.
2
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(3 Π0+
) − X(1 Σ+
g ) electronic band system of Br2 . The inner and outer turning points for
u
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
KErad
=
v
1
μ (vr )2 = [Ev − V (r)] .
2
(5.11)
CHAPTER 5. ELECTRONIC SPECTROSCOPY
90
25000
3
B Π 0+u
20000
v’= 0
15000
5
15
10
30
20
60
52
energy
/ cm-1
44
36
10000
28
1
X Σ+g
20
16
12
5000
Br2
8
4
v"= 0
0
2.0
2.5
3.0
r/Å
3.5
4.0
4.5
Figure 5.5: Potential curves and turning points for the Br2 B(3 Π0+
) − X(1 Σ+
g ) system
u
As a result, the time that the vibrating molecule spends with its internuclear distance in the tiny interval
between r and r + dr is
1
μ
δtr ≡ fv (r) dr =
dr =
dr ,
(5.12)
vr
2[Ev − V (r)]
in which fv (r) defines the probability of finding the system at a particular value of r.
For a representative vibrational level with energy Ev in a potential energy curve V (r), Fig. 5.6 shows the
characteristic behaviour of the distribution function fv (r). As with Figs. 1.6, 1.7 and 3.3, this figure is a
superposition of two separate types of plot: the first is an energy vs. distance plot showing the vibrational
level energy Ev and the variation of the potential energy with distance, while the second (here) is a plot of
fv (r) vs. r, with the zero of its vertical axis placed at the energy Ev of the vibrational level in question. It is
clear that fv (r) → ∞ at the inner and outer turning point of every level. Elementary calculus tells us that
these are “integrable singularities”, in that the area under the curve is finite. Nonetheless, it is clear that
(within this classical picture) the molecule spends most of its time with its bond length close to one of those
turning points. As a result, although the stationary point for a given (v , v ) transition can in principle occur
at any radial distance, the transition will be most intense if it lies near one of the classical turning points
where the vibrating molecule spends most of its time. This leads to our second selection rule for vibrational
transitions in electronic spectroscopy.
Selection Rule 2: Vibrational transitions in electronic spectra will be most intense when the upper and
lower vibrational levels have turning points that are nearly coincident.
In practice, quantum mechanics blurs the above rules. In particular, the assumption that the probability
distribution defined by our function fv (r) is only qualitatively correct, especially for small v values. Moreover,
the wavefunctions seen in Figs. 1.6, 1.7 and 3.3, show that in its lowest ( v = 0 ) level, the most probable
place to find a molecule is near its equilibrium distance, half-way between the two turning points. As may be
seen in Fig. 5.5, the fact that the v = 0 level lies very near the potential minimum means that those turning
5.3. VIBRATIONAL PROPENSITY RULES IN ELECTRONIC TRANSITIONS
∞
91
∞
fυ(r)
↑
energy
Eυ
inner
turning
point
KEυ
V(r)
outer
turning
point
r/Å →
Figure 5.6: Classical prediction for the amount of time fv (r) δr that a vibrating molecule spends within
incremental distance δr of a particular radius r.
points will lie relatively close to their midpoint. As a result, when applying Selection Rule 2 to transitions
involving a v = 0 vibrational level, we usually think of its turning points as effectively lying at the potential
minimum, re .
Application of the above considerations to the B(3 Π0+
) − X(1 Σ+
g ) band system of Br2 illustrated by
u
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 upperstate 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
CHAPTER 5. ELECTRONIC SPECTROSCOPY
92
(2,0)
A
(1,0)
(3,0)
(5,0)
(0,0)
(0,0)
I /20
(1,0)
(1,0)
B
(2,0)
(3,0)
(2,0)
(3,0)
(0,0)
(3,0)
(4,0)
(4,0)
(4,0)
(5,0)
(5,0)
(6,0)
(6,0)
(8,0)
C
(10,0)
(8,0)
(2,0)
(1,0)
A
D
(10,0)
(12,0)
(0,0)
40000
(12,0)
45000
B
ν / cm−1
50000
55000
C
D
Figure 5.7: Dependence of vibrational band intensities on the relative radial positions of upper- and lowerstate 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 1 Σ+ ) = 0
1
+
v (A Σ ) = 0
1
2
3
4
5
13910
14244
14594
1
2
3
13525
13863
14200
14541 14152
14487 14103
14446
Use these results to determine estimates of ωe and ωe for this band system.
2. (a) Determine the extended versions of Eqs. (5.3) – (5.5) which take account of the leading centrifugal
distortion constant Dv and Dv .
(b) For a case in which Bv > Bv , use your results from part (a) to determine an expression for the
band head quantum number JhR which takes account of Dv and Dv 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 JhR
for this band, and compare your result to that obtained on p. 88.
3. The leading molecular constants for the X 1 Σ+ and A 1 Σ+ 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 1 Σ+ and A 1 Σ+ electronic states of CuH, all in units cm−1 .
state
Te
ωe
ωe xe
Be
αe
1
0.0
23 407.976
1941.610
1717.543
37.887
53.0
7.9448
6.9294
0.2557
0.2576
+
X Σ
A 1 Σ+
94
CHAPTER 5. ELECTRONIC SPECTROSCOPY
Chapter 6
Photoelectron Spectroscopy
What Is It? Photoelectron spectroscopy (PES) uses photons of high energy, from the ultraviolet (tens
of eV) to X-ray (thousands of eV), to dislodge electrons from molecules. It is precisely analogous to
normal electronic spectroscopy, except that the upper state of a transition is one of the electronic states
of the molecular ion formed upon removing an electron.
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 −
95
1
2
me (ve )2 .
(6.1)
CHAPTER 6. PHOTOELECTRON SPECTROSCOPY
96
ionization
threshold
2sA
1
m v2
−
2 e e
2sB
hν0
H atom
Coulomb
potential
σu*
1sA
1sB
σg
H atom A
H2
H atom B
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 1s1 2p1 → 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 − 12 . The middle segment of the figure then shows
how the energies of the associated bonding σg and anti-bonding σu∗ molecular orbitals that are formed when
the atoms come together to form a molecule, split apart to lie below and above the energies of the parent
isolated-atom orbitals. In the ground state of the neutral H2 molecule, both electrons occupy the σg bonding
orbital, and the lowering of the total system energy associated with this bonding is the source of the binding
energy of the H2 molecule.
It is important to avoid confusing ionization energies with dissociation energies. Ionization corresponds to
−
the removal of an electron from a molecule to create a molecular ion, a process such as H2 + hν0 → 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 σ2s
molecular
+
orbitals. In each case, only one electron is removed, and an N2 ion is produced. However, since the three
cases involve the removal of an electron from a different molecular orbital, each case produces an 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 + :
+
− 1
2
H2 (v = 0) + hν0 → H+
.
(6.3)
2 (v ) + e
2 me (ve )
The left-hand side of this figure then shows the resulting photoelectron spectrum. For v + ≥ 10 , and again
for v + ≥ 13 , a separate scan with enhanced amplitude was obtained by increasing the data collection time
in the experiment.
98
CHAPTER 6. PHOTOELECTRON SPECTROSCOPY
H(1s) + H +
KE e−
υ+= 6
υ =3
+
υ+= 0
+
IE(υ = 2)
hν0
IE adi
H(1s) + H(1s)
υ″= 0
Figure 6.2: Left: He(I) photoelectron spectrum of H2 with small peaks due to N2 impurity marked by arrows.
Right: Potential energy curves for H2 and 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
+
1 2
.
G+ (v + ) = ωe+ (v + + 12 ) − ωe x+
e (v + 2 )
(6.5)
Thus, Eqs. (6.4) and (6.5) allow a set of measured KEe− peak positions to be employed to determine the
adiabatic ionization energy IEadi and the vibrational constants ωe+ and ωe x+
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
−1
in eV, the values of ωe+ and ωe x+
, and the value of the vibrational force constant k̃ for H+
e in cm
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 = ωe+ − 2ωe x+
e (0 + 1)
ΔG+
3/2
=
(5.502 − 5.246) × 8065.54465 = 2065 = ωe+ − 2ωe x+
e (1 + 1) .
−1
. Finally, substituting this value of
Solving these equations then yields ωe+ = 2289 cm−1 and ωe x+
e = 56 cm
+
+
ωe and the reduced mass μ(H2 ) = 0.503 775 338 [u] into Eq. (3.10) yields
k̃ =
μ (ωe+ )2
−2
= 7.83×104 [cm−1 Å ] .
2 Cu
The intensity pattern of the vibrational fine structure peaks in a photoelectron spectrum are readily
explained in terms of the classical Franck-Condon principle discussed in § 5.3. In photoelectron spectroscopy,
ionization almost always occurs from the v=0 level of the neutral parent molecule, so we have the type of
simple cases illustrated by Fig. 5.7. As is shown there, the most intense absorption occurs into vibrational
levels of the upper state whose inner (or outer) turning point lies directly above the equilibrium distance
(re ) of the initial neutral-molecule potential. This intensity maximum defines what is called the “vertical
ionization energy”.
For the case of H2 , Fig. 6.2 shows that the most intense ionization occurs for transitions into the v +=2
+
vibrational level of H+
2 . This tells us that the minimum of the H2 potential energy curve is displaced relative
to that for H2 . From the spectrum alone, we might not know whether this radial displacement was inward
or outward. However, the fact that the observed transitions involve a fairly large number of vibrational
levels suggests that it is probably outward, since the steepness of the inner wall of a potential means that
the inner turning points of the various levels will lie relatively close to one another, and hence would all be
accessible for vertical transitions at distances near the ground-state re value. A more definitive conclusion
is provided by the molecular orbital theory description illustrated by Fig. 6.1. We know that H2 has a bond
1
order of 1, since it has two bonding electrons, while the bond order in H+
2 is only 2 , since it has only one
+
bonding electron. As a result, we expect that H2 will have a longer bond length and smaller dissociation
−1
energy than does neutral H2 . The fact that the value of ωe+ (H+
is roughly half as large
2 ) = 2230 cm
+
−1
as ωe (H2 ) = 4161 cm
confirms that the bond in H2 is much weaker than that in H2 . This in turn
tells us that the potential energy curve for 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:
CHAPTER 6. PHOTOELECTRON SPECTROSCOPY
100
σ3p
*
H(1s)
5
Cl(3p )
σ3p
⇑
bonding
σ3p
*
H(1s)
5
Cl(3p )
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 highestenergy 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,
∗
) molecular orbitals
while the upper segment shows the occupancy of the bonding (σ3p ) and anti-bonding (σ3p
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 2 Π
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+ {σ3p 1 , 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 2 Σ+ 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 antiparallel 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 2 Π 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.
a
b
state
ωe /cm−1
re /Å
D0 /eV
T0 /eV
HCl (X 1 Σ+
g)
2990.9
1.2746
4.4336
0.0 a
HCl+ (X 2 Πi )
HCl+ (A 2 Σ+ )
2673.7
1606.5
1.3147
1.5142
4.653
1.169
12.768
16.252
N2 (X 1 Σ+
g)
2358.6
1.0977
9.7537
0.0 a
2 +
N+
2 (X Σg )
+
2 +
N2 (A Πui )
2 +
N+
2 (B Σu )
2207.4
1903.5+
2420.8
1.1164
1.1749
1.0747
8.7127
7.5949
5.5429 b
15.5808
16.6986
18.7506
The reference energy is the zero point level of the neutral parent molecule.
See the discussion in §6.5.
As a second example, let us consider the case of N2 , for which a schematic molecular orbital diagram and
the photoelectron spectrum generated with a He(I) lamp are shown in Fig. 6.4. The presence of three groups
of transitions in this spectrum indicates that three different electronic states of the ion lie within 21.2182 eV
(the He(I) photon energy) of the ground state of neutral N2 . These states are produced by removing an
electron from one or another of the three highest occupied molecular orbitals of the parent molecule; the
absence of a fourth band indicates that the binding energy of an electron in the σ2s valence electron orbital
is greater than 21.2182 eV.
The energies of the molecular ion states associated with these electronic bands increase from right to
left in Fig. 6.4, and as in Figs. 6.2 and 6.3, within each band the vibrational quantum number associated
with the different peaks also increases from right to left. For all three bands, the sharp intensity breakoff at
the low-energy edge of the band indicates that the lowest-energy peak is indeed associated with ionization
into the ground v +=0 level of that electronic state of the ion. This assignment means that those peak
positions define the threshold ionization energy IEadi , and that the spacings to their neighbours can be used
to determine values for the vibrational constants ωe+ and (and for the middle spectrum, also ωe x+
e ) for these
three electronic states.
102
CHAPTER 6. PHOTOELECTRON SPECTROSCOPY
Figure 6.4: Left: Molecular orbital diagram for N2 .
Right: He(I) photoelectron spectrum of N2 .
Two other prominent features of these 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
2 +
with those of the parent N2 molecule. Removal of a σ2p bonding electron to produce N+
2 (X Σg ) 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 21 . However, the fact that these changes are relatively small
indicates that the σ2p orbital makes only a modest contribution to the binding energy. The very sharp drop
in intensity from the v +=0 to 1 peaks for this band is what we expect for cases in which there is little
difference between the equilibrium bond lengths of the initial and final states.
The photoelectron band system of N2 centred near 17 eV is associated with removal of an electron from
the π2p bonding orbital. The relatively large increase in re and decreases in ωe and D0 compared to the values
for the neutral molecule (see Table 6.1), reflect the fact that the π2p electrons contribute much more to the
binding of N2 than do the σ2p electrons (see the MO diagram in Fig. 6.4). As a result, the product A 2 Π+
ui
molecular ion is distinctly less strongly bound than is the X 2 Σ+
g ground-state ion. The resulting larger
outward displacement of the A 2 Π+
ui state potential then gives rise to the more extended type of vibrational
spectrum illustrated by Fig. 5.7 B.
∗
Finally, the molecular parameters in Table 6.1 show that removal of a σ2s
anti-bonding electron from
neutral N2 increases the strength of the bond in the ion, making the vibrational motion a little stiffer, and
the bond length a little shorter than in either the ground state of the molecular ion or the parent neutral
molecule. The fact that the bond length is only slightly shorter than that in the neutral parent molecule
explains why we again have the same type of intensity pattern found for transitions into the X 2 Σ+
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 2 Σ+
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 1 Σ+ ) and those of HCl+ (X 2 Πi ) (see
Table 6.1) indicate that the lone-pair electrons do contribute somewhat to the bonding in this system.
• Loss of an anti-bonding electron increases the bond order, thereby increasing the vibrational spacings
and decreasing the bond length of the cation, by comparison with the properties of the parent molecule.
2 +
The changes for N+
2 (B Σu ) seen in Table 6.1 are relatively modest, since the relevant anti-bonding
orbital is not in the outermost valence electron subshell. These effects are much more pronounced for
2 +
the case of O+
2 (X Σ ), which is discussed below.
Thus, the peak spacings and intensity pattern of the vibrational fine structure in an ionization band provides
information on the contribution to the bonding in the neutral molecule of the orbital from which the electron
was removed.
6.5
Some Complications in Photoelectron Spectra
The preceding discussion of the photoelectron spectrum of H2 , 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 zerothorder 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 2 Π spin-orbit splitting increases dramatically,
+
and becomes substantially larger than the vibrational peak spacing, while for the A 2 Π+
ui state of N2 (middle
band in Fig. 6.4), these splittings are too small to be experimentally resolved.
Complications of another type arise for O2 . In this case, the only orbitals that are energetically accessible
to He(I) excitation are those formed from the 2p electrons of the parent O atoms. However, while the MO
diagram in Fig. 6.5 shows that only three such orbitals are occupied, the PES spectrum shows (at least) four
distinct bands. The lowest ionization energy for this species is readily assigned to removal of an electron
∗
anti-bonding MO; the substantial increases in both the bond strength D0 and the vibrational
from the π2p
+
constant ωe shown in Table 6.2 are clearly consistent with this assignment. As for the A 2 Π+
u state of N2
discussed above, the spin-orbit splittings in this species are too small to be resolved. However, it is not
immediately clear why there should be (at least) three higher-energy bands, since this species has only two
other types of occupied molecular orbitals.
This conundrum is resolved when we realize that when a π2p or σ2p electron is removed, the energy of
the resulting molecular ion state will depend on whether the spin of the remaining unpaired electron in
∗
that orbital is parallel or anti-parallel to the spins of the two parallel-spin electrons in the anti-bonding π2p
3
orbitals. The former case (parallel) would yield a quartet state with total spin quantum number S=2 , and
Table 6.2: Molecular parameters for some states of O2 and O+
2.
state
ωe /cm−1
re /Å
D0 /eV
T0 /eV
O2 (X 3 Σ−
g )
1580.4
1.2077
5.1156
0.0
2
O+
2 (X Πg )
+
O2 (a 4 Πui )
2
O+
2 (A Πu )
+
4 −
O2 (b Σg )
2 −
O+
2 (B Σg )
1906.1
1035.1
899.0
1197.0
1156.
1.1169
1.3814
1.4090
1.2794
1.298
6.662
2.629
1.694
2.530
1.760
12.0717
16.1042
17.0395
18.1706
20.297
104
CHAPTER 6. PHOTOELECTRON SPECTROSCOPY
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
∗ 2
indeed have the same overall electronic structure, which would be {(π2p )1 (σ2p )2 (π2p
) }. These intensity
patterns also suggest that their bond lengths only differ modestly from that for the ground state of the
neutral molecule, as is confirmed in Table 6.2. Since Hund’s rule suggests that the higher-spin state should
be relatively more stable, it is reasonable to find that the highest-energy band is assigned as the B 2 Σ−
g state
and the one centred near 18.5 eV as the b 4 Σ−
state.
g
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 2 Σ and 3 Σ excited states, so that the 2 Π and 4 Π 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
4 −
2
4
spacing and band profile for the 2 Σ−
g and Σg states, and for the Πu and Πu states, and their differences
from the properties of the ground-state molecular ion and the neutral parent molecule. However, while the
bond energies D0 of the three lowest 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.
4 −
In order to account for the fact that the B 2 Σ−
g and b Σg states have larger vibrational spacings, but
similar or smaller dissociation energies than the corresponding 2 Πu and 4 Πu states, we must take account of
two other concepts. One is that the three lowest molecular ion states dissociate to yield atomic fragments
6.6. X-RAY PHOTOELECTRON SPECTROSCOPY (XPS)
105
Table 6.3: Comparison of calculated orbital binding energies εorbital and experimental adiabatic ionization
energies IEadi for CO.
∗
orbital
σ2p
π2p
σ2s
ion state
εorbital / eV
IEadi / eV
X 2 Σ+
15.09
14.02
A 2 Πu
17.40
16.59
B 2 Σ+
u
21.87
19.71
O(3 P ) and O+ (4 S) in their ground electronic states, but the two highest molecular ion states yield one
3
+ 2
fragment or the other in an excited state. In particular, the B 2 Σ−
u state dissociates to yield O( P2 )+O ( D) ,
4 −
1
+ 4
while the b Σu state dissociates to O( D)+O ( S) . The second consideration is the fact that the electronic
character of a given molecular state, that is, the ordering and energies of its orbitals, may change as a bond
stretches from the equilibrium value towards dissociation. These two effects also explain why the relative
well depth D0 of the highest-energy 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 assumption that ionization energies are a direct measure of the orbital binding energy in the parent molecule, is
only an approximation. As mentioned in §6.2, changes in electron-electron interactions and electron shielding
of the nuclei mean that the molecular orbitals of the cation in general are not the same as the molecular
orbitals of the parent neutral molecule. Table 6.3 provides a simple illustration of this point for the case of
CO, whose molecular orbital structure is similar to that for N2 . In this case the “orbital relaxation” which
occurs when an electron is removed from the parent molecule reduces the energies of all three ion states, but
leaves them with the same energy ordering. That is, however, not always the case.
Our discussion of the PES spectrum of N2 was based on the assumption that the orbital ordering shown
on the left-hand side of Fig. 6.4 is correct; that assumption is indeed consistent with a Koopman’s theorem
interpretation of the photoelectron spectrum. However, when molecular orbital calculations are made more
and more accurate and pushed to the Hartree-Fock limit, one finds that the π2p orbital of neutral N2
actually lies slightly above (rather than below, as suggested by Fig. 6.4) the σ2p orbital. In order to reverse
this situation and achieve a level ordering consistent with experiment, the theory has to include ‘configuration
interaction’, which effectively means that we have to give up the simple molecular orbital picture we have
been using. Another way of thinking about this situation is to say that the relaxation of the σ2p orbital on
forming the ion is significantly greater than that of the π2p orbital. Fortunately, however, this re-ordering is
an unusual case, and although Koopman’s theorem definitely has its limitations, in most cases it provides a
useful first-order explanation of the ionization energies observed in photoelectron spectroscopy.
6.6
X-Ray Photoelectron Spectroscopy (XPS)
Up to now we have considered only photo-ionization of valence electrons, whose binding energies are sufficiently small that they can be liberated by ordinary ultraviolet radiation with photon energies of 5 − 100 eV.
However, if a radiation source that produces X-rays (100 − 10 000 eV) is employed, then core electrons, whose
binding energies range from 50 − 10 000 eV, will also be dislodged. These core electrons are not affected much
by chemical bonding, so that their binding energies and other properties are mainly defined by the nature
of the particular atom. As is shown by Table 6.4, those binding energies vary quite dramatically from one
atom to the next. However, most of this variation is explained simply by the increase of the nuclear charge
with atomic number. In particular, the 1s core electrons lie much closer to the nucleus than do any others,
Table 6.4: Ionization energies of the 1s electrons of the first-row elements.
atom
atomic number Z
IE(1s) / eV
{IE(1s)/Z 2 } / eV
He
2
25
6.14
Li
3
55
6.11
Be
4
111
6.94
B
5
188
7.52
C
6
285
7.92
N
7
399
8.14
O
8
532
8.32
F
9
686
8.47
CHAPTER 6. PHOTOELECTRON SPECTROSCOPY
106
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 Z 2 , 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 electronwithdrawing 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+
,
is
called
the
chemical
shift,
ΔIE
,
and
is
given
by
n,
f
Table 6.5: Nitrogen 1s chemical shifts relative to the core ionization energy for N(1s) in gaseous N2 .
Δ{IE(1s)} / eV
compound
NF3
4.3
NO2
3.0
NNO
2.6
ONCl
1.5
NO
0.8
N2
0.0
NNO
−1.3
HCN
−3.1
NH3
−4.3
CH3 NH
−4.8
(CH3 )3 N
−5.2
6.7. AUGER ELECTRON SPECTROSCOPY (AES)
107
valence electron
vacancies
X-ray fluorescence
Auger process
nd
2 electron ejected
ionization threshold
2p
2s
primary core
electron vacancy
1s
Figure 6.7: Schematic level energy diagram illustrating XPS core-hole decay.
ΔIEn, =
'
( IEn, (A+
)
−
IE
(A
)
− IEn, (A+
n,
m
m
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
CHAPTER 6. PHOTOELECTRON SPECTROSCOPY
108
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 Å 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 ωe xe (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 m s−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 1 H2 given in Figure 6.2 as well as any other data
you consider necessary (and there are other data you will need!), predict the energies of the ionizations
of 2 H2 to the first five vibrational energy levels of 2 H+
2.
17. Analyze the vibrational fine structure of HCl (Figure 6.3) to determine ωe and ωe xe 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.
110
CHAPTER 6. PHOTOELECTRON SPECTROSCOPY
Chapter 7
NMR Spectroscopy
What Is It? Nuclear magnetic resonance (or NMR) is based on the interaction between the magnetic
dipoles of nuclei and an external magnetic field.
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
7.1.1
Basics of NMR Spectroscopy
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
all we can ever know about it are its
mechanics tells us that for any given type of angular momentum P,
magnitude and one of its vector components Px , Py or Pz . Quantum mechanics also tells us that the
may be written as
magnitude of the angular momentum P
= R(R + 1) ,
|P|
(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 ,
5
2 , . . . ) number. By convention, we always select the z component of angular momentum as the one which
we know, and its allowed values are Pz = mR , where
mR = − R, − R+1, − R+2, . . . , R−2, R−1, R .
111
(7.2)
112
CHAPTER 7. NMR SPECTROSCOPY
Thus, for any given value of R (i.e., of |P|),
there are (2 R+1) allowed values of mR (i.e., of Pz ). Figure
7.1 illustrates the four possible angular momentum alignment sub-states of a system with total angular
momentum quantum number 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 hydrogenic orbitals ( =0 for s, =1 for p, =2 for d, . . . etc.). The associated
3
−
(2J+1) values of mJ are the source of the rotational degeneracy factor
2
→
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,
R
p, d, . . . , orbitals.
1
The type of angular momentum important for NMR spectroscopy
−
2
is nuclear spin, and transitions between these quantum states yield the
spectra observed in NMR. You have already encountered spin in the
context of electron spin, mS , the last quantum number used when filling
atomic orbitals. This quantum number is actually the z–component or
1
−−
2
projection of the electron spin angular momentum, whose magnitude is
1
defined by the total spin quantum number S = 2 . For an electron,
therefore, the degeneracy rule of Eq. (7.2) tells us that an electron can
only have the 2S+1=2 values: mS=+ 12 or − 21 . We commonly refer to
3
−−
these two states as ‘spin-up’ or α, and ‘spin-down’ or β, respectively. Spin
2
angular momentum has no classical analogue in terms of the mechanical
3 5
−×− →
motion of a particle. While it is tempting to think of it as corresponding
2 3
to the spinning motion of the particle, it really isn’t that simple.
For an atomic nucleus the total angular momentum quantum number Figure 7.1: Space quantization of
3
is denoted I and its z–component mI , and as discussed above, there are angular momentum P for R=2 .
(2I+1) different mI sub states associated with each value of I: mI=− I,
−I+1, −I+2, . . . , I−1, and I. The value of I may be either an integer or half of an odd integer, depending
upon the structure of the nucleus. There are general rules for predicting some nuclear spin properties, but
in most cases we need to refer to a table of isotope properties (such as Table 7.1) to learn their values.
↑
m
P/ h−
⎯
√
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 12 C and 16 O.
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 1 H, 3 H, 13 C, 15 N, 19 F, and 31 P; similarly I = 32 for 11 B and 23 Na, while
I = 52 for 17 O and 27 Al.
General Rule 3. If a nucleus has an odd atomic number and an even mass number, it has integer spin
I > 0. For example, I=1 for 2 H and 14 N, while I=3 for 10 B and 50 V.
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 :
= γ I(I + 1) .
μ = γ |I|
(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
Isotope
Atomic
Number
Nuclear
Spin, I
Natural
Abundance
(%)
Magnetogyric
Ratio, γ
(rad T−1 s−1 )
electron
–
–
−17608.4×107
neutron
–
1
2
1
2
1
2
–
−18.3257×107
1
H
1
2
H
1
3
H
1
Li
3
7
99.985
26.7519×107
1
0.015
4.1066×107
1
2
3
2
3
2
1
2
(radioactive)
28.535×107
92.58
10.3975×107
80.42
8.5843×107
1.108
6.7283×107
11
B
5
13
C
6
14
N
7
1
99.63
1.9338×107
15
N
7
0.37
−2.712×107
17
O
8
1
2
5
2
1
2
5
2
1
2
1
2
7
2
0.037
−3.6279×107
19
27
F
9
Al
13
29
Si
31
59
14
P
15
Co
27
100
25.181×107
100
6.9760×107
4.70
−5.3188×107
100
10.841×107
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
= − (μx Bx + μy By + μz Bz ) .
E = −
μ·B
(7.4)
It is a universal convention to define the coordinate system such that the z axis points in the direction of the
≡ B0 . As a result,
external magnetic field. This means that Bx=By=0 , and it is convenient to write Bz=|B|
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 1 H (or 13 C or 19 F 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=− 12 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)
CHAPTER 7. NMR SPECTROSCOPY
114
10
β(mI = − ½)
5
↑
E
ΔE ( /Bα)
0
-5
α(mI = +½)
-10
↑
/ α
B
/ 0→
B
Figure 7.2: Nuclear spin energy levels in a magnetic field for I =
1
2
.
Since the energy levels for a given nucleus are equally spaced, only one possible transition energy is allowed,
namely,
(7.8)
ΔE = E(mI ) − E(mI + 1) = γ B0 = h ν0 .
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
[s−1 ] .
2π
(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 1 H, 13 C and 19 F nuclei in an 8.0000 T magnetic field ?
26.7519×107 [rad T−1 s−1 ] (8.0000 [T]))
γ B0
1
For H nuclei:
=
ν0 =
2π
2 π [rad]
3.40616×108 [Hz] = 340.62 [MHz]
6.7283×107 [rad T−1 s−1 ] (8.0000 [T])
γ B0
=
=
2π
2 π [rad]
=
For
13
C nuclei:
ν0
=
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 1 H nucleus)
and a 13 C 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 13 C
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 13 C is achieved. Since the γ for 13 C 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 13 C 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 1 H nuclei resonate in a 1.409 T field.
13
(a) What must the field be increased to in order for
C nuclei to resonate?
Firstly, we must determine the Larmor frequency of the proton in this spectrometer.
26.7519×107 [rad T−1 s−1 ] (1.409 [T])
= 60.00 [MHz]
ν0 (1 H) =
2π
If
13
C were to achieve resonance at that frequency, then necessarily
γ(13 C) B0 (13 C)
ν0 (13 C) = 60.00 [MHz] =
2π
Hence, the required field strength would be
2 π ν0
=
B0 (13 C) =
γ(13 C)
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
13
C nuclei to resonate ?
The Larmor equation (7.9) tells us that for a given field strength ν0 ∝ γ . Hence
γ(13 C)
6.7283×107
ν0 (13 C) = ν0 (1 H) ×
= 15.09 [MHz]
= 60.000 [MHz] ×
1
γ( H)
26.7519×107
The effects of increasing the magnetic field and decreasing the radiation frequency are equivalent, in
terms of the NMR spectrum. Because NMR spectra were first measured as a function of magnetic field at
fixed resonance frequency, NMR spectra were plotted with magnetic field increasing to the right, which for a
given pattern of peak positions corresponds to frequency increasing to the left. Even today, this convention,
with frequencies increasing to the left, is used in NMR spectroscopy.
Modern NMR spectrometers use superconducting magnets which are best held at constant magnetic
field strengths. Thus, in practice it is the radiation frequency that is varied, and the NMR spectrum is
collected as a function of frequency, as in the other forms of spectroscopy we have discussed. However,
NMR spectrometers are rarely classified in terms of their magnetic field strengths, but rather in terms of
the Larmor frequencies of the nuclei observed. Since the proton 1 H 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 1 H 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 ?
3
1
19
11
H H F
31
13
2
B P C H
250 MHz
spectrometer
400
300
200
100
0
← ν0 / MHz
3
H
1
H 19F
11
B
31
13
P
C
2
H
600 MHz
spectrometer
600
500
400
300
200
100
0
← ν0 / MHz
Figure 7.3: NMR spectra of various atomic nuclei in 250 MHz and 600 MHz spectrometers.
116
CHAPTER 7. NMR SPECTROSCOPY
Figure 7.4: 1 H 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 (1 H), 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
7.2
7.2.1
117
Chemical Shifts
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 proportionality constant σ is called the chemical shielding constant for that particular type of nucleus in that
specific molecular environment, because it is a measure of the tendency of the local electronic environment
to shield the atomic nucleus from the applied external field.
Because of the presence of this induced field, the strength of the net effective magnetic field Beff seen by
the nucleus becomes
(7.11)
Beff = B0 − Bind = (1 − σ) B0 .
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
γA B0
γA Beff
ν0A =
=
(1 − σA ) .
(7.12)
2π
2π
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
ν0A (bare nucleus) − ν0A (in molecule)
.
(7.13)
σA =
ν0A (bare nucleus)
It is possible to measure transition frequencies with very high precision, especially in the “radio wave”
region of the electromagnetic spectrum associated NMR spectroscopy. However, it is very difficult to determine accurate experimental values of σ A because it is very difficult to get a stable population of bare nuclei
to sit still to allow us to measure the value of ν0A (bare nucleus) in the particular magnetic field of a given
spectrometer. As a result, instead of using the bare nucleus as the reference species it is customary to define
a chemical shift parameter δA by comparison with the Larmor frequency of that nucleus in some chosen
standard chemical environment. Moreover, to avoid having constantly to write down large powers of ten,
this quantity is defined in units of ‘parts-per-million’ as
A
ν0 (in molecule) − ν0A (in reference species)
δA (in molecule) =
(7.14)
× 106 [ppm]
ν0A (in reference species)
The NMR signal of the chosen reference species is, by convention, taken as the zero of the ‘chemical shift
scale’ for the nucleus of interest.
7.2.2
What Determines Chemical Shifts, and The Chemical Shift Scale
The greater the shielding of the nucleus, the smaller the Larmor frequency relative to that for the bare
nucleus. For convenience, reference compounds are chosen to be stable, relatively inert chemical species in
which there is a high degree of shielding. The latter property is a matter of convenience which means that
most measured δA values will be positive. For 1 H and 13 C and 29 Si, the conventional reference compound is
tetramethylsilane (TMS), Si(CH3 )4 , and its NMR signal is, by convention, taken as the zero of the ‘chemical
shift scale’ for these nuclei.
CHAPTER 7. NMR SPECTROSCOPY
118
Figure 7.5: Chemical shifts of 1 H and
13
C 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
δ = 0.23
CH3 Cl
δ = 3.05
CH2 Cl2
δ = 5.33
CHCl3
δ = 7.26
CH3 F
δ = 4.26
CH3 Br
δ = 2.68
CH3 I
δ = 2.16
2. The nature of nearby bonds
Double bonds, triple bonds, and aromatic rings (denoted Ar– ) are increasingly strong electron withdrawing groups. Their presence also reduces the electron density around the nucleus of interest and
causes the resulting transition frequency (and δA value) to be larger.
In larger molecules there are no ‘hard-and-fast’ rules about the size of chemical shifts for a particular type of
atom in a specific local environment. However, Fig. 7.5 illustrates the fact that there are well-defined ranges
of δ values for particular types of bonded atoms.
For protons, δ values typically range from 0 − 12. 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 13 C, 19 F, 31 P, . . . , etc.) are surrounded by many more electrons than is
an H 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 13 C the range of δ values is ∼ 300 (see
Fig. 7.5), while for 31 P it is ∼ 1000. However, 1 H atoms are the most abundant species in many chemical
environments, and hence the remainder of this chapter will focus on proton NMR spectra.
1
7.2.3
Working With Chemical Shifts
The δ scale readily allows us to relate NMR spectra obtained at different magnetic field strengths. However,
we must always work in Hz and ppm interchangeably, so it is necessary to become familiar with switching
between the two sets of units.
Exercise (iii): Consider a molecule containing two types of protons A and B that have chemical shifts of 3.00
ppm and 6.55 ppm, respectively. What are the frequency shifts (from the TMS reference signal) of their NMR
transitions in magnetic fields of 2.35 T and 9.40 T ?
(a) First, we must use the Larmor formula Eq. (7.9) to determine the reference frequencies for these two
spectrometers.
26.7519×107 [rad T−1 s−1 ] (2.35 T )
For the 2.35 T spectrometer: ν0 (reference) =
= 100.0 [MHz]
2π
26.7519×107 [rad T−1 s−1 ] (9.40 T )
For the 9.40 T spectrometer: ν0 (reference) =
= 400.0 [MHz]
2π
(b) Now we rearrange Eq. (7.14) and use it to define the desired frequency shifts.
Δν0A
= ν0 (in molecule) − ν0 (TMS) = δ×10−6 × ν0 (TMS)
Hence, in the 100 MHz spectrometer:
For the 3.00 ppm proton :
Δν0A = 3.00×10−6 × 100.0×106
= 300 [Hz]
For the 6.55 ppm proton :
Δν0A = 6.55×10−6 × 100.0×106
= 655 [Hz]
For the 3.00 ppm proton :
Δν0A = 3.00×10−6 × 400.0×106
= 1200 [Hz]
For the 6.55 ppm proton :
Δν0A = 6.55×10−6 × 400.0×106
= 2620 [Hz]
Similarly, in the 400 MHz 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.
CHAPTER 7. NMR SPECTROSCOPY
120
β(mA = − ½)
½ JAB
ν0A
α(A), β(A)
(mA = ± ½)
ν0A − ½ JAB
½ JAB
α(mA = + ½)
no fields
external
field /B0
ν0A + ½ JAB
nearby
proton α(B)
(mB = + ½)
nearby
proton β(B)
(mB = − ½)
Figure 7.6: Energy level diagram for proton A without (left half) and with (right half) coupling to a
neighbouring proton B of a different type
7.3
7.3.1
Spin-Spin Coupling
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 ν0A 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 ν0A−JAB /2 or ν0A+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 ν0A 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
1 The 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
νA0 − νB0 = (δA− δB) × ν0spectrometer
JAB
JAB
ν0A
←ν
ν0B
Figure 7.7: Simulated NMR spectrum for neighbouring single H atoms A and B with different chemical
shifts.
between A and B changes the NMR spectrum from two lines, one for each type of nucleus, to four lines,
two for each type of proton, due to the two possible relative orientations of the neighboring nucleus. Since
there are equal numbers of the two types of nuclei, the intensities of the original A and B peaks would be
the same, and since each nucleus has an equal probability of being in state α or β, the two components of
each pair have equal intensity. Figure 7.7 shows the nature of the resulting spectrum for this type of case,
which is called an “AX spectrum”.
There are three main features to note.
1. Each type of proton involved in a given type of spin-spin coupling is affected equally; i.e., the splitting
JAB = JBA .
2. Each splitting pattern is symmetric about the chemical shift for that type of nucleus.
3. The coupling constant JAB does not depend on the magnetic field, but only on the natures of the
two types of nuclei and their physical separation. In particular, its magnitude does not depend on
the strength of the external magnetic field B0 . Hence values of JAB are always expressed in absolute
frequency units, Hz, rather than in ppm.
In addition, we add the general point that perturbations due to nuclei that are more than three bonds away
are too small to be seen. Thus, all the splittings we discuss herein are associated with protons attached to
directly bonded atoms.
7.3.2
‘Equivalence’, and Coupling from Multiple Equivalent Nuclei
If a proton of type A has two equivalent nearby protons in a local environment of type B, the energy levels
for proton A take on the pattern shown in Fig. 7.8. Since there are two equivalent protons of type B, they
have four possible relative alignments which give rise to three possible values for the total magnetic moment
which perturbs the energies of proton A :
• α1 (B) α2 (B) with
mtot
B = +1
• α1 (B) β2 (B) with
mtot
B = 0
• β1 (B) α2 (B) with
mtot
B = 0
• β1 (B) β2 (B)
mtot
B = −1
with
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: ν0A−JAB , ν0A and ν0A+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 mtot
B =0 . At the same time, as there is only one nucleus of type A and hence
1
=m
only two possible values of mtot
A=± 2 , the NMR signal for protons of type B will be split into two peaks
A
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
CHAPTER 7. NMR SPECTROSCOPY
122
½ JAB
β(mA = − ½)
½ JAB
ν0A
α(A), β(A)
(mA = ± ½)
ν0A − JAB
ν0A
ν0A + JAB
½ JAB
α(mA = + ½)
no fields
external
field /B0
½ JAB
nearby
protons B
mtot
= +1
B
nearby
protons B
mtot
=0
B
nearby
protons B
mtot
= −1
B
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)–CH2 I .
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 C2 H6 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, ν0A . Thus, although the levels are split, the transitions are not,
νA0 − νB0 = (δA− δB) × ν0spectrometer
JAB
JAB JAB
ν0A
←ν
ν0B
Figure 7.9: Simulated NMR spectrum for proton A interacting with two equivalent neighbouring protons B.
7.3. SPIN-SPIN COUPLING
123
number of
equivalent
atoms
N=1
weights
1
1
N=2
1
N=3
1
N=4
N=5
number
of peaks
1
1
2
3
4
5
1
1
3
6
10 10
1
4
1
5
1
sum of
weights
2
2
3
4
4
8
5
16
6
32
Figure 7.10: Pascal’s triangle and the calculation of weights for split NMR peaks.
and there is only a single NMR line associated with this type of proton (assuming there are not also other
types of proton present).
Consider now the case of the molecule O=(CH)–CH3 . It clearly has two types of proton, the methyl
protons (B) forming one group and the O=(CH)– proton (A) being the other, and its proximity to the
double-bonded and electronegative O atom means that the latter will have a much larger chemical shift
parameter: δA δB . If we now consider all possible alignments of the three methyl protons, we see that
there are eight possible unique arrangements and four possible mtot
B values:
3
tot
• mB = + 2 is obtained one way: α1 (B) α2 (B) α3(B)
1
• mtot
B = + 2 is obtained three ways: α1 (B) α2 (B) β3 (B), α1 (B) β2 (B) α3 (B), β1 (B) α2 (B) α3 (B)
tot
• mB = − 12 is obtained three ways: α1 (B) β2 (B) β3 (B), β1 (B) α2 (B) α3 (B), β1 (B) β2 (B) α3 (B)
3
• mtot
B = − 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 ν0A , 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 mtot
B 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
2 These are also the ‘binomial coefficients’ associated with the different terms xm y N−m when one expands the algebraic
expression (x + y)N .
124
CHAPTER 7. NMR SPECTROSCOPY
Figure 7.11: 1 H 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 spectroscopic tools for structure determination available to chemists. Much of its utility has been due to the
establishment of empirical rules relating the values of chemical shifts and spin-spin coupling constants to the
local environments of the nuclei. For example, a proton near a highly electronegative atom, such as oxygen
or fluorine, will in general have a larger chemical shift. Other empirical rules regarding the strengths of
bonds and local electron density have been used, and you will learn much more about these as you go on in
Chemistry.
Perhaps the most unique aspect of NMR spectra, however, is the phenomenon of J coupling. That
nuclei can communicate with their neighbors, signalling what parts of the molecule they are close to, is
a tremendous tool for determining structures of unknown molecules. NMR spectroscopists have exploited
this feature by developing new experiments that rapidly and readily provide a “fingerprint” for molecular
structure. These experiments have been used to great advantage in biochemistry and molecular biology,
7.5. PROBLEMS
125
where structures of proteins and DNA have been determined solely from their NMR spectra.
Several characteristic examples are given below. Common pieces of organic molecules, composed solely of
carbon and hydrogen and called alkyl groups, give 1 H NMR spectra that act as signatures of their presence
in a molecule. These groups, such as CH3 CH2 (ethyl), CH3 CH2 CH2 (n-propyl), (CH3 )2 CH (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 H2 O, to probe the soft tissue of organisms,
including patients in hospitals. Consider the differences with conventional X-rays, which are only deflected
by hard matter such as bones. MRI can focus on the soft tissue in an organism, without removing the tissue
or harming it, and has been particularly well-developed in the area of brain research, including multiple
sclerosis. By providing pictures of the soft brain tissue, while the brain is still within the patient, physicians
have learned a tremendous amount about the functioning of the brain and the action of various diseases.
7.5
Problems
1. Calculate the magnetic field needed to induce the following nuclei to undergo NMR transitions at a
frequency of 100.000 000 MHz.
(a)
1
H
(b)
15
N
(c)
29
Si
2. Calculate the Larmor frequencies of the following nuclei in a magnetic field B0 = 11.7 T.
(a)
13
C
(b)
19
F
(c)
31
P
3. Two 1 H 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 1 H 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 1 H 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 (CH3 NO2 ) 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 1 H 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 13 C nuclei on
each.
126
CHAPTER 7. NMR SPECTROSCOPY
Figure 7.12: 60 MHz NMR spectrum of an unknown organic compound.
10. The proton NMR spectrum at 60 MHz of an unknown compound (shown above) has been obtained,
and is described as follows. It consists of a triplet (group of three lines) centred at 4.40 ppm, a sextet
(six lines) centred at 2.05 ppm, and another triplet centred at 1.02 ppm. Three structures have been
proposed for this compound: CH3 CH2 CH=CHNO2 , CH3 CH2 CH2 NO2 , and (CH3 )2 CHNO2 . 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 C4 H8 O2 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 C4 H8 O2 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 1 H 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 A2 MX 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.