CHEM0019/22

Molecular Spectroscopy
Professor Helen Fielding

Department of Chemistry, UCL

December 4, 2020
Contents
1 Introduction 3
1.1 Electromagnetic spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Example calculation: unit conversion . . . . . . . . . . . . . . . . . . . 4
1.2 Absorption and emission of radiation . . . . . . . . . . . . . . . . . . . . . . . 4
1.3 Transition dipole moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Born-Oppenheimer approximation . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Rotational spectroscopy 6
2.1 Rigid rotor energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.1.1 Example calculation: centrifugal distortion . . . . . . . . . . . . . . . . 8
2.2 Rotational spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2.1 Example calculation: rotation constant . . . . . . . . . . . . . . . . . . . 10
2.3 Intensities of lines in rotational spectra . . . . . . . . . . . . . . . . . . . . . . . 11
2.3.1 Example calculation: maximum J . . . . . . . . . . . . . . . . . . . . . . 11
2.4 Rotational spectroscopy question . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Vibrational spectroscopy 13
3.1 Harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.1.1 Example calculation: harmonic vibration wavenumber . . . . . . . . . 14
3.2 Anharmonic oscillator energy levels . . . . . . . . . . . . . . . . . . . . . . . . 14
3.3 Vibrational spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.3.1 Example calculation: determining ωe and ωe xe . . . . . . . . . . . . . 17
3.4 Intensities of lines in vibrational spectra . . . . . . . . . . . . . . . . . . . . . . 17
3.4.1 Example calculation: population of vibrational levels . . . . . . . . . . 18
3.5 Dissociation energies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.5.1 Dissociation energy question . . . . . . . . . . . . . . . . . . . . . . . . 20

4 Vibration-rotation spectroscopy 21
4.1 P, Q and R branches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.2 Combination differences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.3 Vibration-rotation spectroscopy question . . . . . . . . . . . . . . . . . . . . . 27

5 Electronic spectroscopy 28
5.1 Term symbols of diatomic molecules . . . . . . . . . . . . . . . . . . . . . . . . 29
5.2 Electronic selection rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
5.2.1 Example: electronic transitions in N2 . . . . . . . . . . . . . . . . . . . 30
5.3 Vibrational structure of electronic transitions . . . . . . . . . . . . . . . . . . . 31
5.4 Dissociation energies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5.4.1 Deslandres table question . . . . . . . . . . . . . . . . . . . . . . . . . . 37
5.5 Rotational fine-structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

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 5.5.1 Example calculation: rotational fine structure . . . . . . . . . . . . . . . 39
5.6 Electronic spectroscopy questions . . . . . . . . . . . . . . . . . . . . . . . . . . 41

6 Term symbols in more detail 42

6.1 Comparing term symbols for atoms and molecules . . . . . . . . . . . . . . . . 42
6.2 g/u and ± . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

7 Tutorial problems 46

8 Fundamental physical constants 49

2
1 Introduction
The origin of spectral lines in molecular spectroscopy is the absorption, emission or scatter-
ing of photons. The difference between molecular spectroscopy and atomic spectroscopy
(CHEM0005) is that a molecule can undergo changes in rotational and vibrational state as
well as electronic state. As a result, molecular spectra are more complex than atomic spectra;
however, molecular spectra contain a wealth of information about bond lengths and strengths
and molecular electronic structure.

The general strategy we apply in molecular spectroscopy is as follows:

• set up expressions for rotational, vibrational and/or electronic energy levels;

• apply transition selection rules;

• consider populations;

• infer the form of the spectrum.

In this course, we will restrict our discussion to diatomic molecules, although the same ideas
can be extended to polyatomic molecules (CHEM0079 - 4th year course).

These lectures are based on Chapters 12 and 13 of ‘Atkins’ Physical Chemistry’ 10th edition
(Peter Atkins and Julio de Paula, Oxford) and many of the figures are taken from the lecturers’
resources that accompany this book. This book is available electronically through BibliU and
is also a good source of additional worked problems. If you are particularly interested in
molecular spectroscopy and would like to learn more, I recommend ‘Modern Spectroscopy’
4th edition (J. Michael Hollas); this has been purchased as an e-book and is accessible through
Explore (the UCL library catalogue).

1.1 Electromagnetic spectrum

Figure 1: The electromagnetic spectrum and its classification into regions.

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Molecular spectroscopists often use wavelengths (1 nm = 10−9 m) for photon energies,
wavenumbers (cm−1 ) for rotational and vibrational levels of a molecule and wavenumbers
or electron volts (1 eV = e J) for electronic energies.

1.1.1 Example calculation: unit conversion

Convert 200 nm to cm−1 and eV.

First, convert nm to m: 200 nm = 200 × 10−9 m.

1 1
ν̃ = = = 5 × 106 m−1 = 50000 cm−1 .
λ (m) 200 × 10−9
hc
By definition, 1 eV = e J, where e is the charge on an electron, and E(J) = .
λ
hc
Therefore, E(eV) = = 6.2 eV.
λe

N OW TRY THE M OODLE QUIZ ON UNIT CONVERSION .

1.2 Absorption and emission of radiation

Figure 2: Absorption and emission of radiation.

In absorption spectroscopy, a molecule undergoes a transition from a state of low energy,

E1 , to one of higher energy, E2 .

In emission spectroscopy, a molecule undergoes a transition from a state of high energy, E2 ,

to one of lower energy, E1 .

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1.3 Transition dipole moment
The intensity of a spectral line arising from a molecular transition between an initial state
with wavefunction ψi and a final state with wavefunction ψ f , depends on the (electric) dipole
moment, µ = −er.

For a transition to be dipole allowed, the expectation value of the dipole moment operator
(CHEM0019 Quantum Mechanics), known as the transition dipole moment R, must be
non-zero. Z
R = ψ∗f µ̂ψi dτ

1.4 Born-Oppenheimer approximation

The Born-Oppenheimer approximation supposes that the nuclei of a molecule, being much
heavier than the electrons, move relatively slowly and may be treated as stationary while the
electrons move in their field. The Born-Oppenheimer approximation allows us to select an
internuclear separation in a diatomic molecule and to solve the Schrödinger equation for the
electrons at that nuclear separation. We then select a different nuclear separation and repeat
the calculation. In this way, we can explore how the energy of a molecule varies with bond
length and obtain a molecular potential energy surface.

Since electron motion is much faster than molecular vibrations, which are in turn much faster
than molecular rotations, we can, as a first approximation, write an approximate molecular
wavefunction as a product of the electronic, vibrational and rotational wavefunctions,

Ψ = ψe ψv ψr ,

and the total molecular energy as the sum of the electronic, vibrational and rotational ener-
gies,
E = Ee + Ev + Er .

Thus, to a first approximation, we can treat all the different types of molecular energy
separately.

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2 Rotational spectroscopy

2.1 Rigid rotor energy levels

A useful model for the end-over-end rotation of a diatomic molecule is the rigid rotor.
The rotational energy levels are quantised and can be obtained by solving the appropriate
Schrödinger equation (CHEM0019 Quantum Mechanics).

Figure 3: Rigid rotor.

The rotational energy levels of a rigid rotor are (in J),

h2
EJ = J ( J + 1),
8π 2 I

where the rotational angular momentum quantum number J = 0, 1, 2, . . ., r is the bond

length and I is the moment of inertia.

The moment of inertia is defined I = ∑i mi ri2 , where mi are the masses of the atoms and ri are
their distances from the centre of mass of the molecule (Figure 3).

The molecule rotates end-over-end about its centre of mass, m1 r1 = m2 r2 . Therefore, the
moment of inertia can be written,

I = m1 r12 + m2 r22
= m2 r1 r2 + m1 r1 r2
= ( m1 + m2 )r1 r2

Using m1 r1 = m2 r2 = m2 (r − r1 ) = m1 (r − r2 ), we can write,

m2 r m1 r
r1 = and r2 = .
m1 + m2 m1 + m2

Substituting these back into the expression for I gives

m1 m2 2
I = ( m1 + m2 )r1 r2 = r = µr2 ,
m1 + m2

6
where µ = m1 m2 /(m1 + m2 ) is known as the reduced mass of the molecule.

The energy levels of rotational states are usually expressed as a rotational term F ( J ) = E J /hc
(in cm−1 ),
F ( J ) = BJ ( J + 1)

where B is the rotation constant and depends on the moment of inertia of the molecule,

h
B=
8π 2 Ic

and has units of cm−1 .

Because F ( J ) ∝ J ( J + 1), the energy separation between adjacent rotational energy levels
increases as J increases (Figure 4).

Figure 4: Rotational energy levels of a rigid rotor.

Typical values of of the rotational constant, B for small molecules are in the region 0.1 −
10 cm−1 . Since the rotational constant, B ∝ 1/I, larger molecules have more closely spaced
rotational energy levels than smaller molecules.

The rigid motor is a good approximation for the energy levels of a rotating diatomic molecule;
however, centrifugal forces stretch the bond and increase the moment of inertia. As a result,

7
centrifugal distortion reduces the rotation constant and the energy levels are slightly closer
together than the rigid rotor expression predicts. The effect of centrifugal distortion is taken
into account empirically by writing the rotational energy term as,

F ( J ) = BJ ( J + 1) − DJ 2 ( J + 1)2 .

D is the centrifugal distortion constant and is related to the vibrational wavenumber of a

diatomic molecule ωe (Section 3),
D = 4B3 /ωe2 .

2.1.1 Example calculation: centrifugal distortion

For 12 C16 O, B = 1.931 cm−1 and ωe = 2170 cm−1 . Therefore,

4 × 1.9313
D= 2
= 6.116 × 10−6 cm−1 .
2170

Because D << B, centrifugal distortion has only a very small effect on the energy levels and
so we will generally neglect it.

2.2 Rotational spectroscopy

Consider a heteronuclear diatomic molecule with a permanent dipole moment. As it rotates,
it generates a fluctuating electric field which can interact with the electric field vector of
electromagnetic radiation.

Figure 5: Rotating polar molecule generating a fluctuating electric field.

The selection rules are found by evaluating the transition dipole moment (Section 1.3),
Z
Rr = ψr0∗ µψr00 dτ,

8
where the double prime 00 and single prime 0 represent the initial and final states, respec-
tively.

This integral is only non-zero if the molecule has a permanent dipole moment and ∆J = ±1,
which are the rotational selection rules for linear rotors.

The ∆J = ±1 selection rule has a physical interpretation. When a photon is absorbed by a

molecule, the angular momentum of the system must be conserved. Since a photon has a spin
of 1, when a molecule absorbs a photon, its rotational angular momentum J must increase by
1 (Figure 6).

Figure 6: Physical interpretation of the ∆J = +1 selection rule for absorption of a photon.

Applying the selection rules to the expression for the energy levels of a rigid rotor, the
wavenumbers of the allowed J + 1 ← J transitions are

ν̃( J + 1 ← J ) = F ( J + 1) − F ( J )
= B( J + 1)( J + 2) − BJ ( J + 1)
= 2B( J + 1).

Therefore, the spectrum consists of a series of equally spaced lines with wavenumbers
2B, 4B, 6B, . . . , that is, they are separated by 2B.

9
Figure 7: The rotational energy levels of a rigid rotor and the transitions allowed by the
selection rule ∆J = +1 (top) and the corresponding rotational absorption spectrum (bottom).

2.2.1 Example calculation: rotation constant

14 N16 O has a bond length of 0.1151 nm. Calculate the rotation constant B in cm−1 .

[Relative isotopic masses: 14 N 14.003, 16 O 15.995]

First calculate the reduced mass, taking care to convert masses in units of g mol−1 into kg
and to use an appropriate number of decimal places (as a general rule, 3 d.p. works well for
spectroscopic accuracy):
m1 m2 14.003 × 15.995
µ= · mu = · 1.661 × 10−27 = 1.240 × 10−26 kg.
m1 + m2 14.003 + 15.995
h 6.626 × 10−34
B= = = 170.4 m−1 = 1.704 cm−1 .
8π 2 µr2 c 8π 2 · 1.240 × 10−26 · (1.151 × 10−10 )2 · 2.998 × 108

N OW TRY THE M OODLE QUIZ ON ROTATIONAL CONSTANT AND BOND LENGTH CALCULA -
TIONS .

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2.3 Intensities of lines in rotational spectra
The intensities of lines in a rotational spectrum increase with increasing J and pass through
a maximum before tailing off as J becomes large (Figure 7). The existence of a maximum
arises because of a maximum in the population of rotational energy levels. The population
of a rotational energy level J is given by the Boltzmann expression (CHEM0019 Statistical
Mechanics),
NJ = Ng J e−(E J /kT )

where N is the total number of molecules in the sample, g J = (2J + 1) is the degeneracy of the
level J, k is the Boltzmann constant and T is the rotational temperature of the sample.

The relative population of rotational level J is,

NJ /N0 = (2J + 1)e−hcBJ ( J +1)/kT .

To find the value of J with the maximum population we need to differentiate (by parts),
 
dNJ /N0 2 hcB
= 2 − (2J + 1) e−hcBJ ( J +1)/kT .
dJ kT

For a maximum,
dNJ /N0
=0
dJ
hcB
2 − (2J + 1)2 =0
kT
2kT
(2J + 1)2 =
hcB
r
2kT
2J + 1 =
r hcB
kT 1
Jmax = −
2hcB 2

2.3.1 Example calculation: maximum J

For a molecule with rotational constant B = 0.2 cm−1 , kT ≈ 1000hcB at room temperature, so
√
Jmax = 500 − 0.5 ≈ 22.

N OW TRY THE M OODLE QUIZ ON ROTATIONAL TEMPERATURE CALCULATIONS .

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2.4 Rotational spectroscopy question
The wavenumbers of the first 5 absorption lines in the microwave spectrum of the 12 C16 O
molecule, together with their relative intensities, are listed below.

ν̃ / cm−1 3.845 7.690 11.535 15.380 19.226

Relative intensity / arb. units 1.0 2.5 3.0 2.6 1.7

1. Assign rotational numbers J 00 and J 0 to each transition.

2. Deduce the rotation constant B (in cm−1 ) and the bond length r (in nm).

3. Estimate the rotational temperature.

Relative isotopic masses: 12 C 12.000, 16 O 15.995

G ET A HINT OR CHECK YOUR ANSWERS ON M OODLE TO ACCESS THE FULL SOLUTION .

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3 Vibrational spectroscopy

3.1 Harmonic oscillator

For small molecular vibrations, the molecular bond obeys Hooke’s law and the restoring
force, dV ( x )/dx = −kx, where k is the force constant of the bond and x = r − re , where re
is the equilibrium bond length. Integrating the expression for the restoring force gives the
potential energy V ( x ) = 12 kx2 (Figure 8).

Figure 8: Potential energy curves for a range of force constants.

The vibrational energy levels are quantised and can be determined by solving the appropriate
Schrödinger equation (CHEM0019 Quantum Mechanics). The energy levels of a harmonic
oscillator are (in J):
Ev = (v + 1/2)h̄ω,

where the vibrational quantum number v = 0, 1, 2, . . . and ω = (k/µ)1/2 is the vibration

angular frequency (in rad s−1 ).

The energy levels of the vibrational states are usually expressed as a vibrational term G (v) =
Ev /hc (in cm−1 ),
G (v) = ωe (v + 1/2)

where ωe is the harmonic vibration wavenumber (in cm−1 ) and depends on the force constant
and reduced mass of the diatomic molecule,
s
1 k
ωe = .
2πc µ

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3.1.1 Example calculation: harmonic vibration wavenumber

The force constant of the bond in HCl, k = 516 N m−1 . The reduced mass, µ = 1.63 × 10−27 kg
(close to that of the H atom).
r
1 516
ωe = = 2987 cm−1
2π × 2.998 × 1010 1.63 × 10−27

N OW TRY THE M OODLE QUIZ ON FORCE CONSTANT CALCULATIONS .

3.2 Anharmonic oscillator energy levels

A real diatomic molecule is not a harmonic oscillator because when r → ∞, the molecule
dissociates into two neutral atoms. At this point, the force dV/dx = 0 so the curve flattens
out. When r → 0, the positive charges on the nuclei cause mutual repulsion and the force,
dV/dx increases, i.e. the curve becomes steeper (Figure 9).

Figure 9: Potential energy curves for a harmonic oscillator (parabola) and a real diatomic
molecule.

The potential energy curve for a diatomic molecule can be expressed in terms of a Taylor’s
expansion about the equilibrium bond length (x = 0),

d2 V d3 V
     
dV 1 2 1
V ( x ) = V (0) + x+ x + x3 + · · ·
dx 0 2! dx2 0 3! dx3 0

For small displacements, the molecule behaves as a harmonic oscillator (the Taylor’s series is
truncated at the x2 term) but for larger displacements, the higher order (anharmonic) terms
become important.

14
One approach to determining the energy levels of an anharmonic oscillator is to use a
potential that resembles the true potential energy (Figure 9) more accurately, such as the
Morse potential,

2
V ( x ) = hcDe 1 − e−ax ,

1/2
where De is the depth of the potential well, x = r − re and a = 2π 2 µcωe2 /hDe .

As required for a real diatomic molecule, when r → 0, V ( x ) → ∞ and when r → ∞,

V ( x ) → hcDe (the well depth).

Solving the Schrödinger equation for the Morse potential leads to the vibrational term
becoming

G (v) = ωe (v + 1/2) − ωe xe (v + 1/2)2 + · · · ,

where ωe xe is the first anharmonic wavenumber. As the vibrational quantum number

increases, the anharmonic term becomes more significant and the vibrational energy levels
become more closely spaced until they converge at the dissociation limit (Figure 10).

Figure 10: The Morse potential energy curve reproduces the general shape of a real molecular
potential energy curve for a diatomic molecule.

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3.3 Vibrational spectroscopy
The selection rules are found by evaluating the transition dipole moment (Section 1.3),
Z
Rv = ψv0∗ µψv00 dτ.

This integral is only non-zero if the molecule has a permanent dipole moment, µ. For a
heteronuclear diatomic molecule, µ varies with x. This variation can be expressed as a
Tayor’s series,
1 d2 µ
   
dµ
µ = µe + x+ x2 + · · ·
dx e 2! dx2 e

where the subscript ’e’ refers to the equilibrium bond length.

Substituting for µ in the expression for the dipole transition moment gives
  Z
dµ
Z
Rv = µe ψv0∗ ψv00 dτ + ψv0∗ xψv00 dτ + · · ·
dx e

Since ψv0 and ψv00 are eigenfunctions of the same Hamiltonian, they are orthogonal; therefore,
the first term, µe ψv0∗ ψv00 dτ = 0 (CHEM0019 Quantum Mechanics).
R

The dipole transition moment then becomes

  Z
dµ
Rv = ψv0∗ xψv00 dτ + · · ·
dx e

and the first term of this expression is non-zero if ∆v = ±1, which is the vibrational selection
rule for a harmonic oscillator.

Higher order terms in the expansion, arising from anharmonicity, modify the vibrational
selection rule to ∆v = ±1, ±2, ±3, . . .. Transitions with ∆v = ±1 are known as fundamental
vibrational transitions and those with ∆v = ±2, ±3, . . . are known as first, second, . . . vibra-
tional overtones. Since the effects of anharmonicity are relatively small, overtones are much
weaker than fundamental transitions.

Applying the selection rule to the expression for the energy levels of an anharmonic oscillator,
the wavenumbers of v + 1 ← v transitions are

ν̃(v + 1 ← v) = G (v + 1) − G (v)
h i
2 2
= ωe (v + 3/2) − ωe xe (v + 3/2) − ωe (v + 1/2) − ωe xe (v + 1/2)
= ωe − ωe xe [3v + 9/4 − v − 1/4]
= ωe − 2ωe xe (v + 1).

16
Thus, the spacings between adjacent vibrational energy levels decreases as v increases (Figure
11).

Figure 11: Vibrational energy levels of an anharmonic oscillator and transitions allowed by
the ∆v = ±1, ±2, ±3, . . . selection rule.

In order to determine ωe and ωe xe , at least two transition wavenumbers are required.

3.3.1 Example calculation: determining ωe and ωe xe

The first two vibrational intervals in HgH are 1203.7 and 965.6 cm−1 . Using ν̃(v + 1 ← v) =
ωe − 2ωe xe (v + 1),

ν̃(1 ← 0) = 1203.7 = ωe − 2ωe xe (1)

ν̃(2 ← 1) = 965.6 = ωe − 4ωe xe (2)

Subtracting, (1) − (2) gives 2ωe xe = 238.1 cm−1 and ωe xe = 119.08 cm−1 . Substituting in (1)
gives, ωe = 1203.7 + 238.15 = 1441.85 cm−1 .

3.4 Intensities of lines in vibrational spectra

If a vibrational spectrum is observed in absorption, which it usually is, the intensities of
the transitions are governed by the populations of the lower vibrational state. The relative
population of vibrational level v is,

Nv /N0 = e−hc[G(v)−G(0)]/kT .

For most diatomic molecules, hc[ G (v) − G (0)] >> kT, so only the ground vibrational state
(v00 = 0) is significantly populated at normal temperatures. Bands with v00 6= 0 are referred to
as hot bands since the intensities of these bands increase with temperature.

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3.4.1 Example calculation: population of vibrational levels

For 14 N2 , the v = 1 ← 0 transition lies at 2329.92 cm−1 . The relative population of the v = 1
level at 300 K is,

6.626 × 10−34 · 2.998 × 1010 · 2329.92

 
N1 /N0 = exp − ∼ 10−5 .
1.381 × 10−23 · 300

N OW TRY THE M OODLE QUIZ ON VIBRATIONAL TEMPERATURE CALCULATIONS .

3.5 Dissociation energies

The dissociation energy of a diatomic molecule, D0 differs from the well-depth, De (Fig-
ure 12) on account of the zero-point energy, G (0) = 12 ωe − 14 ωe xe (CHEM0019 Quantum
Mechanics).

Figure 12: Dissociation energy, D0 and well-depth, De .

18
When several vibrational transitions are observed, a graphical technique known as a Birge-
Sponer plot may be used to determine the dissociation energy, D0 . The basis of the Birge-
Sponer plot is that the sum of successive intervals ∆G (v) = G (v + 1) − G (v) from v = 0 to
the dissociation limit is D0 (Figure 13),

D0 = ∑ ∆G(v).
v

Figure 13: Dissociation energy D0 as sum of separations between vibrational energy levels
from v = 0 to the dissociation limit.

Since ∆G = ωe − 2ωe xe (v + 1), the area under a plot of ∆G against v + 1/2 is equal to the
sum of the separations between the energy levels and thus to D0 (Figure 14).

Figure 14: Birge-Sponer plot showing that the area is the dissociation energy, D0 .

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This method usually overestimates D0 because higher order anharmonic terms result in a
deviation from linearity as v increases (Figure 14).

If we know D0 , it is straightforward to calculate De by adding the zero-point energy (Figure

12),
1 1
De = D0 + G (0) = D0 + ωe − ωe xe .
2 4

Alternatively, we can estimate De if we know ωe and ωe xe . Given G (v) = ωe (v + 1/2) −

ωe xe (v + 1/2)2 and recognising that the vibrational levels converge towards the dissociation
limit (Figure 10),

dG (v)
= ωe − 2ωe xe (v + 1/2) = 0.
dv

So, (v + 1/2) = ωe /2ωe xe . Substituting this back into the expression for G (v) gives,

ωe ωe2 ωe2
G (vmax ) = ωe . − ωe xe = .
2ωe xe 4ωe xe2 4ωe xe

Thus,
ωe2
De =
4ωe xe

and it is straightforward to calculate D0 by subtracting the zero-point energy (Figure 12),

 
1 1
D0 = De − G (0) = De − ωe − ωe xe .
2 4

N OW TRY THE M OODLE QUIZ ON DISSOCIATION ENERGY CALCULATIONS .

3.5.1 Dissociation energy question

1. Observed vibrational intervals in H2+ lie at 2191, 2064, 1941, 1821, 1705, 1591, 1479, 1368,
1257, 1145, 1033, 918, 800, 677, 548 and 411 cm−1 . Use a graphical method to estimate
the dissociation energy, D0 .

2. The vibrational constants for H2+ are ωe = 2321.7 cm−1 and ωe xe = 66.2 cm−1 . Estimate
the well-depth, De and the dissociation energy, D0 .

G ET A HINT OR CHECK YOUR ANSWERS ON M OODLE TO ACCESS THE FULL SOLUTION .

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4 Vibration-rotation spectroscopy
Each line in a high resolution vibrational spectrum of a gas-phase heteronuclear diatomic
molecule consists of many closely spaced lines, e.g. Figure 15. They arise from many rotational
transitions accompanying each vibrational transition and are referred to as rovibrational
transitions. Consequently, these spectra are often referred to as band spectra.

Figure 15: High resolution vibration-rotation spectrum of H35 Cl and H37 Cl (ratio 3:1).

For most heteronuclear diatomic molecules, the rovibrational selection rules are ∆v = ±1, ±2, . . .
and ∆J = ±1. The fact that ∆J = 0 is forbidden means that the pure vibrational transition is
not observed. The position at which it would occur is known as the band centre. Exceptions
to this are molecules having orbital angular momentum about their internuclear axis (Section
5.1) for which the rovibrational selection rules are ∆v = ±1, ±2, . . . and ∆J = 0, ±1.

The energy levels of the vibration-rotation terms are the sum of the vibrational and rotational
terms,
S(v, J ) = G (v) + F ( J ).

If we ignore anharmonicity and centrifugal distortion, we can write the vibration-rotation

terms,
S(v, J ) = ωe (v + 1/2) + BJ ( J + 1).

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4.1 P, Q and R branches
When a vibrational transition v + 1 ← v occurs, J changes by ±1 and by 0 (if ∆J = 0 is
allowed). The absorptions fall into groups called branches:

∆J -1 0 +1
Branch P Q R

The transition wavenumbers of the lines in each branch can be determined by applying the
appropriate ∆J selection rule and calculating the differences between the vibration-rotation
terms, ν̃ [S(v0 J 0 ) − S(v00 J 00 )].

ν̃Q ( J ) = ν̃ [S(v + 1, J ) − S(v, J )]

= ωe (v + 3/2) + BJ ( J + 1) − [ωe (v + 1/2) + BJ ( J + 1)]
= ωe

This shows that the Q branch consists of lines at the harmonic wavenumber, ωe .

ν̃R ( J ) = ν̃ [S(v + 1, J + 1) − S(v, J )]

= ωe (v + 3/2) + B( J + 1)( J + 2) − [ωe (v + 1/2) + BJ ( J + 1)]
= ωe + B( J 2 + 3J + 2) − B( J 2 + J )
= ωe + 2B( J + 1)

This shows that the R branch consists of a series of equally spaced lines at higher wavenum-
bers than the Q branch, ωe + 2B, ωe + 4B, . . . .

ν̃P ( J ) = ν̃ [S(v + 1, J − 1) − S(v, J )]

= ωe (v + 3/2) + B( J − 1) J − [ωe (v + 1/2) + BJ ( J + 1)]
= ωe + B ( J 2 − J ) − B ( J 2 + J )
= ωe − 2BJ

This shows that the P branch consists of a series of equally spaced lines at lower wavenumbers
than the Q branch, ωe − 2B, ωe − 4B, . . . .

The formation of P, Q and R branches in a vibration-rotation spectrum is illustrated in

Figure 16. The intensity distribution reflects the populations of the rotational levels associated
with the lower vibrational state, determined by the Boltzmann distribution (Section 2.3).

22
 Figure 16: Formation of P, Q and R branches in a vibration-rotation spectrum.

However, it is clear from the vibration-rotation spectrum of HCl (Figure 15) that P and
R branch lines are not exactly equally spaced but that the R branch lines converge and
the P branch lines diverge, away from the band centre. This can be explained if we take
anharmonicity into account.

For an anharmonic oscillator, the vibration-rotation terms are,

S(v, J ) = ωe (v + 1/2) − ωe xe (v + 1/2)2 + Bv J ( J + 1) + · · · .

The rotation constant is written with a subscript ‘v’ because it depends on the vibrational
state. As the vibrational quantum number v increases, so does the internuclear separation, r
(e.g. Figure 10) and therefore, so does the moment of inertia, I = µr2 . Because B ∝ 1/r2 , B
then decreases as v increases.

The vibrational dependence of the rotational constant is, to a good approximation,

Bv = Be − αe (v + 1/2),

where Be is the hypothetical value of the rotation constant at the bottom of the potential well
(Figure 10) and αe is a vibration-rotation interaction constant.

As for the harmonic oscillator, the transition wavenumbers of the lines in each branch of a
vibration-rotation spectrum of an anharmonic oscillator can be determined by applying the

23
appropriate ∆J selection rule and calculating the differences between the vibration-rotation
terms, ν̃ [S(v0 J 0 ) − S(v00 J 00 )] = ν̃ [ G (v0 ) − G (v00 )] + ν̃ [ F ( J 0 ) − F ( J 00 )].

For the R branch, J 0 = J 00 + 1. Writing v00 = v, v0 = v + 1 and J 00 = J,

ν̃R ( J ) = ν̃ [ G (v + 1) − G (v)] + ν̃ [ F ( J + 1) − F ( J )]
= ν̃v0 v00 + B0 ( J + 1)( J + 2) − B00 J ( J + 1)

where ν̃v0 v00 = ωe − 2ωe xe (v + 1) is the wavenumber of the pure vibrational transition v + 1 ←
v, or band head.

For the P branch, J 0 = J 00 − 1. Writing v00 = v, v0 = v + 1 and J 00 = J,

ν̃P ( J ) = ν̃ [ G (v + 1) − G (v)] + ν̃ [ F ( J − 1) − F ( J )]
= ν̃v0 v00 + B0 J ( J − 1) − B00 J ( J + 1)

From these expressions for the wavenumbers of the P and R branches, it is clear that the P
branch lines appear at lower wavenumbers than the pure vibrational transition and the R
branch lines appear at higher wavenumbers than the pure vibrational transition.

4.2 Combination differences

Every transition depends on two rotation constants, B00 and B0 , for the molecule in its lower
vibrational state, v00 and upper vibrational state, v0 . To determine both B00 and B0 , we use
a method called combination differences, which is very common in spectroscopy. The
principle is that, differences in wavenumber between two transitions with a common lower
vibration-rotation level will give us information about energy differences between rotational
levels in the upper vibrational state. Similarly, differences in wavenumber between two
transitions with a common upper vibration-rotation state will give us information about
energy differences between the rotational levels in the lower vibrational state. The method of
combination differences is illustrated in Figure 17.

24
 Figure 17: Method of combination differences.

From Figure 17 it is clear that ν̃R ( J ) and ν̃P ( J ) have a common lower level, J 00 = J. Thus, we
can subtract the expressions for ν̃R ( J ) and ν̃P ( J ) derived above to obtain an expression for
the energy differences between transitions with a common lower level.

∆20 F ( J ) = ν̃R ( J ) − ν̃P ( J )

= ν̃v0 v00 + B0 ( J + 1)( J + 2) − B00 J ( J + 1) − ν̃v0 v00 + B0 J ( J − 1) − B00 J ( J + 1)
 
h i
= B0 J 2 + 3J + 2 − ( J 2 − J )
= 4B0 ( J + 1/2)

Where the superscript 0 and subscript 2 denote a rotational energy difference in the upper
vibrational level of 2. Similarly, ν̃R ( J − 1) and ν̃P ( J + 1) have a common upper level, J 0 =
J. Thus, we can create and subtract expressions for ν̃R ( J − 1) and ν̃P ( J + 1) to obtain an
expression for the energy differences between transitions with a common upper level.

∆200 F ( J ) = ν̃R ( J − 1) − ν̃P ( J + 1)

= ν̃v0 v00 + B0 J ( J + 1) − B00 J ( J − 1) − ν̃v0 v00 + B0 J ( J + 1) − B00 ( J + 1)( J + 2)
 
h i
= B00 J 2 + 3J + 2 − ( J 2 − J )
= 4B00 ( J + 1/2)

25
Where the superscript 00 and subscript 2 denote a rotational energy difference in the lower
vibrational level of 2.

Example calculation: determining ν̃R (0) and ν̃P (1) from ν̃10 , B0 and B1

For 1 H35 Cl, the rotational constants B0 = 10.44 cm−1 and B1 = 10.13 cm−1 , and the pure
vibrational transition ν̃10 = 2886.04 cm−1 . Determine the wavenumbers of the first members
of the P and R branches in the vibration-rotation spectrum.

From Figure 18 and the equation for the R branch lines ν̃R ( J ) = ν̃v0 v00 + B0 ( J + 1)( J + 2) −
B00 J ( J + 1), we see that for the R(0) line,

ν̃R (0) = ν̃10 + B1 (0 + 1)(0 + 2) − B0 × 0 × (0 + 1)

= ν̃10 + 2B1
= 2886.04 + 2 × 10.13
= 2906.3 cm−1

Again, from Figure 5.3 and the equation for the P branch lines ν̃P ( J ) = ν̃v0 v00 + B0 J ( J − 1) −
B00 J ( J + 1), we see that for the P(1) line,

ν̃P (1) = ν̃10 + B1 × 1 × 0 − B0 × 1 × (1 + 1)

= ν̃10 − 2B0
= 2886.04 − 2 × 10.44
= 2865.16 cm−1

Similarly, if ν̃P (1) and ν̃R (0) are known, it is straightforward to calculate the wavenumber of
the band origin, ν̃10 = ν̃R (0) − 2B0 = ν̃P (1) + 2B00 (Figure 18).

Figure 18: Band origin.

26
4.3 Vibration-rotation spectroscopy question
In the infrared absorption spectrum of the v = 0 → 1 transition in 1 H79 Br, lines were observed
at the following wavenumbers:

2507.734 2525.358 2542.530 2575.518 2591.334 2606.698 cm−1

1. Label each transition.

2. Using the method of combination differences, deduce B0 and B1 , in cm−1 .

3. Deduce the average bond lengths in each of the vibrational states, r0 and r1 , and explain
your answers.

4. Determine the wavenumber of the pure v = 0 → 1 transition, i.e. the band origin.

5. Determine the relative Boltzmann factors of these two levels at a vibrational temperature
of 100 K.

[Relative isotopic masses: 1 H 1.008, 79 Br 78.918]

G ET A HINT OR CHECK YOUR ANSWERS ON M OODLE TO ACCESS THE FULL SOLUTION .

27
5 Electronic spectroscopy
In the lowest vibrational state of the ground electronic state of a molecule, the nuclei are at
their equilibrium positions. As mentioned in Section 1, we consider electronic transitions to
occur within a stationary nuclear framework (Born-Oppenheimer approximation). When an
electronic transition occurs, the electron distribution is changed and the nuclei are no longer
in their equilibrium positions and so they move (vibrate).

Figure 19: Vibration of a molecule following an electronic transition.

The vibrational transitions that accompany an electronic transition give rise to the vibrational
structure of an electronic transition. This structure can be resolved in the gas-phase (e.g. I2
absorption spectrum, Figure 25) but is often unresolved in liquid (e.g. chlorophyl, Figure 20)
or solid samples.

There are no simple analytical expressions for the electronic energy levels of a molecule,
unlike for the H atom (CHEM0005).

Term symbols of diatomic molecules classify the projections of electronic angular momenta
along the molecular axis.

Selection rules allow us to determine whether a particular electronic transition is allowed or

not.

The molar absorption coefficient is used to quantify the intensity of an electronic transition
(recall the Beer-Lambert law from the CHEM0005 lab course).

The Franck-Condon principle provides a basis for explaining the vibrational structure asso-
ciated with an electronic transition (see below).

In the gas-phase, rotational structure can also be visible (Section 5.5).

28
 Figure 20: Absorption spectrum of chlorophyl.

5.1 Term symbols of diatomic molecules

Term symbols of diatomic molecules are simply the molecular analogues of atomic term
symbols, 1 S, 2 P, . . ..

In a molecule, with many electrons, the orbital angular momenta of the electrons are coupled
together to give a resultant angular momentum, L, and all the spins are coupled together to
give a resultant, S.

Recall the angular momentum coupling in the H atom from CHEM0005, where the electron
orbital angular momentum, l, and electron spin angular momentum, s, were coupled together
to give a resultant angular momentum, j = |l − s| ... |l + s|.

In a molecule, the coupling between L and S is usually much less that the coupling between L
and the internuclear axis and between S and the internuclear axis.

Thus, only the components of L and S along the internuclear axis (Λh̄ and Σh̄, respectively)
are well-defined. Note that this is similar to the projection of j along the magnetic field axis
you came across when you studied the Zeeman effect in CHEM0005.

Λ and Σ couple to give a total angular momentum projection along the internuclear axis,
Ω = |Λ + Σ| (Figure 21).

Terms symbols for diatomic molecules are written: 2S+1

Λ±
g/u,Ω

For a detailed, explanation of how electronic term symbols can be derived for diatomic
molecules, see Section 6.

29
 Figure 21: Angular momentum coupling in a diatomic molecule.

5.2 Electronic selection rules

Spin angular momentum ∆S = 0
Orbital angular momentum ∆Λ = 0, ±1
Symmetry g ↔ u, g = g, u = u
If both states are Σ + ↔ +, − ↔ −, + = −
Transitions between multiplet components ∆Σ = 0, ∆Ω = 0, ±1

Examples

Allowed 1Π ↔ 1Σ 3Π ↔ 3Π 1Π ↔ 1 Πg
u
Forbidden 1Π = 3Π 2Σ = 2∆ 1Σ = 1Π 1 Σ+ = 1 Σ−
g g

5.2.1 Example: electronic transitions in N2

The lowest energy electron configuration of N2 is (σg 1s)2 (σu∗ 1s)2 (σg 2s)2 (σu∗ 2s)2 (σg 2p)2 (πu 2p)4
and the corresponding term symbol is 1 Σ+ g.

The molecular term symbols for the lowest energy excited states that result from promoting
an electron from the (σg 2p)2 or (πu 2p)4 molecular orbitals to the (π g∗ 2p) orbital, i.e.

· · · (σg 2p)1 (πu 2p)4 (π g∗ 2p)1 and

· · · (σg 2p)2 (πu 2p)3 (π g∗ 2p)1

are 1,3 Π g , 1,3 Σ±

u and
1,3 ∆ .
u

From the electronic selection rules listed above, the only state accessible by single-photon
absorption from the 1 Σ+
g ground state is the Σu excited state.
1 +

30
5.3 Vibrational structure of electronic transitions
In principle there are no restrictions on the change of vibrational quantum number associated
with an electronic transition (Figure 22).

Figure 22: Vibrational structure of electronic transitions.

Transition dipole moment:

Z
0∗ 00
Rev = ψev µ̂ψev dτev (recall that ψev = ψe ψv )
Z Z
= ψe0∗ µ̂ψe00 dτe ψv0∗ ψv00 dτv
Z
= Re ψv0∗ ψv00 dτv

ψv0∗ ψv00 dτv is the Franck-Condon overlap

R
Re is the electronic transition dipole moment and
integral, or vibrational overlap integral.

As already noted, electronic transitions occur much more rapidly than vibrational transitions.
The requirement for the nuclei to be in the same position before and after an electronic transi-
tion means that the transition takes place between points on the potential energy surfaces
that lie on a vertical line (Figure 23); this is known as the Franck-Condon principle.

31
 Figure 23: Intensities of vibrational components of electronic transitions.

Progressions and sequences

Vibrational transitions accompanying an electronic transition are called vibronic transitions.

These vibronic transitions give rise to bands (Figure 24).

Figure 24: Progressions and sequences.

The set of bands associated with a particular electronic transition is called a band system. A
group of transitions with a common lower or upper vibrational level is called a progression.
A group of transitions with the same value of ∆v is called a sequence.

32
Progressions from v00 > 0 are only visible in absorption when the harmonic vibration
wavenumber is small, such as in the B − X band system of I2 (Figure 25).

Figure 25: Progressions with v00 = 0, 1, 2 in the B3 Π+

u − X Σ g band system of I2 .
1 +

Vibronic transition wavenumber

Figure 26: Vibronic transitions.

The total energy T of a molecule in a particular electronic, vibrational and rotational state
may be written (see Figure 26 for illustration of the definitions of the terms):

T = Te + G (v) + F ( J )

33
The wavenumber of a vibronic transition (excluding rotations) may be written:

ν̃(v0 , v00 ) = Te0 − Te00 + G 0 (v0 ) − G 00 (v00 )

= Te0 − Te00 + ωe0 (v0 + 1/2) − ωe xe0 (v0 + 1/2)2 − [ωe00 (v00 + 1/2) − ωe xe00 (v00 + 1/2)2 ]

There are 5 spectroscopic values to determine which requires a minimum of 5 lines to be

measured.

The illustration of progressions and sequences in Figure 24 suggests that we can use combin-
tation differences to obtain the separations of vibrational levels from the observed transition
wavenumbers. A convenient way to organise the vibration transition energies is a Deslandres
table (Figure 27).

Figure 27: Deslandres table for the A1 Π − X 1 Σ+ system of CO. Differences between
wavenumbers in adjacent columns correspond to vibrational separations in the lower elec-
tronic state. Differences between wavenumbers in adjacent rows correspond to vibrational
separations in the upper electronic state.

Although there is no restriction on the change in the vibrational quantum number, the
intensity distribution of the vibrational components of an electronic transition depends on
the relative positions of the two electronic potential energy curves (Figure 23).

5.4 Dissociation energies

Dissociation energies can be derived from electronic spectra. Recall that the zero-point
dissociation energy, D0 is the energy required to dissociate the molecule in a given electronic
state from the lowest vibrational level of that state. The well-depth, De is the energy from the
equilibrium geometry, i.e. the bottom of the potential energy well, to the dissociation limit
(Figure 28).

34
 Figure 28: Dissociation energies.

0
If we can determine the wavenumber of the v00 = 0 progression limit, ν̃(vmax , 0) and the 0-0
0
transition from a spectrum, we can determine D0 :

0
ν̃(vmax , 0) = ν̃(0, 0) + D00

At the progression limit,

0 0 0 0
∆G (vmax + 1 ← vmax ) = ν̃(vmax + 1, 0) − ν̃(vmax , 0)
= ωe0 − 2ωe xe0 (vmax
0
+ 1)
= 0,

which allows us to deduce that

0 ωe0
vmax = − 1.
2ωe xe0

0
Having determined vmax , it can be substituted in the expression for vibronic transitions:

0
ν̃(vmax , 0) = Te0 − Te00 + ωe0 (vmax
0
+ 1/2) − ωe xe0 (vmax
0
+ 1/2)2 − [ 12 ωe00 − 41 ωe xe00 ].

If the 0 − 0 transition wavenumber is known, we can now determine D00 ,

D00 = ν̃(vmax
0
, 0) − ν̃(0, 0)

35
A more accurate value can be determined from the area under the graph if plotted from
v0 = 0 (analagous to Fig. 14).

To find an accurate value of D000 for the electronic ground state is almost impossible using
the same method as for D00 because few vibrational levels are populated. However, if the
wavenumber difference between the atomic fragments is known, we can use,

D000 = ν̃(vmax
0
, 0) − ∆ν̃atomic .

Figure 29: Determining D000 .

Having obtained values for D000 , it is relatively straightforward to determine De by simply

adding the zero point energy:
De00 = D000 + G 00 (0).

However, obtaining D000 reqiures the determination of vmax

0 first. An alternative method is to
determine De00 using,

2
ωe00
De00 = ,
4ωe xe00

which was derived earlier (Section 3.5). Then, it is relatively straightforward to determine D000
by simply subtracting the zero-point energy from De00 ,

D000 = De00 − G 00 (0).

36
5.4.1 Deslandres table question

The following portion of a Deslandres table for the 31 P14 N molecule lists the wavenumbers in
cm−1 of some transitions observed in the A-X band system (A and X represent electronically
excited and ground states, respectively).

v00 0 1 2 3 4
v0 = 2 41859.1 40536.2 - 37932.9 36652.5

1. Deduce the vibrational constants ωe00 and ωe xe00 and the force constant k00 .

2. Calculate the dissociation energy, D000 and the well-depth, De00 .

3. Explain qualitatively how you would expect these values to change for the 31 P15 N

molecule.

Relative isotopic masses: 14 N 14.003; 15 N 15.000; 31 P 30.970

G ET A HINT OR CHECK YOUR ANSWERS ON M OODLE TO ACCESS THE FULL SOLUTION .

5.5 Rotational fine-structure

For every vibronic transition, there is a set of accompanying transitions between the stack
of rotational levels associated with the upper and lower vibronic states, in a similar way to
infrared rovibrational spectroscopy (Section 4).

The important difference is that the difference between B0 and B00 can be much larger than in
infrared spectroscopy because we are rearranging the electrons.

The rotational selection rules are:

1 Σ −1 Σ transitions ∆J = ±1
1 Π −1 Σ transitions ∆J = 0, ±1

The wavenumbers of individual lines in a rovibronic transition are given by

ν̃( J 0 , J 00 ) = ν̃0 + B0 J 0 ( J 0 + 1) − B00 J 00 ( J 00 + 1),

where the band origin is,

ν̃0 = T 0 − T 00 + G 0 (v0 ) − G 00 (v00 )

P, Q and R-branches

ν̃Q ( J, J ) = ν̃0 + B0 J ( J + 1) − B00 J ( J + 1)

= ν̃0 + ( B0 − B00 ) J ( J + 1)

37
 ν̃P ( J − 1, J ) = ν̃0 + B0 ( J − 1) J − B00 J ( J + 1)
= ν̃0 + ( B0 − B00 ) J 2 − ( B0 + B00 ) J

ν̃R ( J + 1, J ) = ν̃0 + B0 ( J + 1)( J + 2) − B00 J ( J + 1)

= ν̃0 + ( B0 − B00 ) J 2 + (3B0 − B00 ) J + 2B0
= ν̃0 + ( B0 − B00 )( J + 1)2 + ( B0 + B00 )( J + 1)

Since the difference between B0 and B00 can be quite large, one or other of the branches may
end up turning round on itself, forming a band head.

Figure 30: Band heads.

When B0 < B00 , the P-branch diverges and the R-branch converges.

When B0 > B00 , the R-branch diverges and the P-branch converges.

To determine the value of rotational quantum number, J, that corresponds to the band head,
we need to differentiate the expressions for the wavenumbers corresponding to the P and R
branches.

B0 > B00 , P-branch converges:

ν̃P ( J − 1, J ) = ν̃0 + ( B0 − B00 ) J 2 − ( B0 + B00 ) J

dν̃P
= 2J ( B0 − B00 ) − ( B0 + B00 )
dJ

38
 ( B0 + B00 )
Jhead =
2( B0 − B00 )

B0 < B00 , R-branch converges:

ν̃R ( J + 1, J ) = ν̃0 + ( B0 − B00 ) J 2 + (3B0 − B00 ) J + 2B0

dν̃R
= 2J ( B0 − B00 ) + (3B0 − B00 )
dJ
(3B0 − B00 )
Jhead = −
2( B0 − B00 )

For Σ − Σ transitions, ∆J = ±1 so only P and R branches are observed.

Figure 31: Σ − Σ transition in which the P branch diverges and the R branch converges. This
tells us that B0 < B00 , so r 0 > r 00 .

For Π − Σ transitions, ∆J = 0, ±1 so P, Q and R branches are all observed.

5.5.1 Example calculation: rotational fine structure

The values of B0 in the A and X (electronically excited and ground) states of CO are 1.6005
cm−1 and 1.9226 cm−1 , respectively. Calculate which lines in the R branch of the 0-0 band lies
at the band-head. By how many wavenumbers is the band head separated from the band
origin?

B00 < B000 , therefore the R branch converges.

(3B0 − B00 ) 3 × 1.6005 − 1.9226

Jhead = − 0 00
=− = 4.47
2( B − B ) 2 × (1.6005 − 1.9226)

39
Figure 32: Π − Σ transition in which the P branch diverges and the R branch converges. This
tells us that B0 < B00 , so r 0 > r 00 .

Thus, J 00 = 4 is furthest away from the band origin.

Substituting this value into the equation for the R branch gives,

ν̃R ( J + 1, J ) = ν̃0 + B0 J 0 ( J 0 + 1) − B00 J 00 ( J 00 + 1)

= ν̃0 + 1.6005 × 5 × 6 − 1.9226 × 4 × 5
= ν̃0 + 9.563

Therefore, the band head is 9.563 cm−1 away from the band origin.

40
5.6 Electronic spectroscopy questions
1. The spectra of the 3 Π −1 Σ+ electronic system in ICl has been observed in absorption.
The origins of successive bands from the lower 1 Σ+ (v00 = 0) level are found at 17080,
17126, 17167, 17203, 17234 and 17260 cm−1 . The band at 17080 cm−1 corresponds to
v0 = 0.

Deduce the energy of the dissociation limit of the 3 Π state relative to the 1 Σ+ (v00 = 0)
level.

2. The equilibrium distance, re , and vibration-rotation constant, αe , for the 3 lowest elec-
tronic states of CN are:
State X A B
re /nm 0.11718 0.12327 0.11506
αe /cm−1 0.01735 0.01746 0.02215

X is the electronic ground state, A is an electronically excited state and B is a higher-lying

electronically excited state.

Calculate the rotational constants for the lowest vibrational level, B0 , for the X, A and B
states.

Determine the value of the rotational quantum number at which the band head occurs
for the 0-0 band of both A-X and B-X emission spectra, ignoring the effects of centrifugal
distortion.

G ET A HINT OR CHECK YOUR ANSWERS ON M OODLE TO ACCESS THE FULL SOLUTION .

41
6 Term symbols in more detail
This section explains how electronic term symbols can be derived for diatomic molecules. It
is for interest and will not be examined.

Recall that in the H atom, the orbital angular momentum l couples to the spin angular
momentum s to give a resultant angular momentum j.

Angular momentum is a vector and the orbital and spin angular momenta add together to
give a total angular momentum, j = l + s. The total angular momentum quantum number
takes values j = |l + s| . . . |l − s|. That is, for an electron in a p-orbital, l = 1 and s = 12 so
j = 12 , 32 .

The term symbol summarises the information about the angular momenta. The term symbol
is expressed as 2S+1 L J . We use capital letters rather than lower case because this expression is
used for many electron atoms as well as the H atom.

• L is the total angular momentum and L = 0, 1, 2, . . . are given symbols S, P, D, . . ..

• 2S + 1 is the multiplicity and 2S + 1 = 1, 2, 3, . . . are referred to as singlet, doublet,

triplet, . . ..

• J = L + S and so J takes values J = | L + S| . . . | L − S|.

Thus, for the example above, the term symbols are 2 P 1 , 3 .

2 2

6.1 Comparing term symbols for atoms and molecules

In a molecule, the total orbital and spin angular momenta, L and S, couple to the internuclear
axis more strongly than to one another. The projection of L and S on the internuclear axis are
Λ and Σ, respectively.

As with atoms, a term symbol is used to summarise the information about the angular
momenta. The term symbol is expressed as 2S+1 ΛΩ . Note that this is identical to the term
symbol for an atom other than we have exchanged L for Λ and J for Ω.

• Λ = 0, 1, 2, . . . are given symbols Σ, Π, ∆, . . ., i.e. we use capital Greek equivalents of

the capital Roman letters.

• Analogous with atoms, 2S + 1 is the multiplicity and 2S + 1 = 1, 2, 3, . . . are referred to

as singlet, doublet, triplet, . . ..

• Analogous with the total angular momentum in an atom, J, the total angular momentum
along the internuclear axis in a molecule is Ω = |Λ + Σ|.

42
6.2 g/u and ±
In addition to the term symbol shown above, two additional labels are used, so altogether
the term symbol is written as 2S+1 Λ±
g/u,Ω .

The g/u label denotes the parity of the orbital wavefunction in a homonuclear diatomic
molecule, that is whether or not the overall wavefunction is symmetric (g) or antisymmetric
(u) with respect to inversion through the origin. g comes from the German word gerade
meaning even and u comes from the German word ungerade meaning odd. Thus, all totally
symmetry s-orbitals have g symmetry, π-bonding orbitals are antisymmetric with respect to
inversion through the origin so have u symmetry and π-antibonding orbitals are symmetric
with respect to inversion through the origin so have g symmetry (Figure 33).

Figure 33: Schematic representation of π bonding (bottom) and antibonding (top) molecular
orbitals showing that the π bonding orbital has odd parity (antisymmetric with respect to
inversion) and that the π antibonding orbital has even parity (symmetric with respect to
inversion).

The ± superscripts denote the symmetry of the orbital component of the electronic wavefunc-
tion and is distinct from g/u. As you have learnt in the quantum mechanics lectures, the total
electronic wavefunction has to change sign upon exchange of electrons, Ψ(1, 2) = −Ψ(2, 1).
As you also learnt, singlet states have spin wavefunctions that are antisymmetric with respect
to exchange of nuclei and triplet states have spin wavefunctions that are symmetric with
respect to exchange of nuclei. Therefore, since the total wavefunction is the product of
spin and orbital wavefunctions, Ψ = ψspin ψorbital , singlet Σ states have symmetric orbital
wavefunctions and triplet Σ states have antisymmetric orbital wavefunctions. Note that we
only need to worry about the symmetries of the orbital and spin wavefunctions in this way
for Σ states, i.e. Λ = 0, in which we have two identical electrons in a doubly degenerate
π orbital (bonding or anti-bonding). For example, for a . . . (2pπu )2 configuration, the Σ
states have 1 Σ+ g and Σ g term symbols. For molecular configurations in which we have two
3 −

electrons in non-degenerate π molecular orbitals, there are no such restrictions. That is, for a
. . . (2pπu )1 (2pπ g∗ )1 configuration, the Σ states have 1 Σu or 3 Σu term symbols.

43
6.3 Examples
Closed shell configurations

If all the molecular orbitals are full, all the electrons are paired, so S = 0 and Λ = 0. Thus,
the term symbol is 1 Σ+ and, if the molecule has inversion symmetry (e.g. a homonuclear
diatomic molecule) we include the parity so the term symbol is 1 Σ+ g.

One unpaired σ electron

If there is a single unpaired electron in a σ orbital, S = 1/2 and Λ = 0, so the term symbol is
2 Σ+ . Again, if it is a homonuclear diatomic molecule we include the parity, so if the single

electron is in a σg bonding molecular orbital, the term symbol is 2 Σ+g , and if the single electron
is in a σu anti-bonding molecular orbital, the term symbol is 2 Σ+ u.

One or three π electrons

If there is a single unpaired electron in a π orbital, S = 1/2 and Λ = 1, so the term symbol is
2 Π. Note that we do not use the ± label for Π states. Again, if it is a homonuclear diatomic

molecule we include the parity, so the term symbol is 2 Πu or 2 Π g , depending on whether the
single electron is in a πu bonding orbital or a π g antibonding orbital, respectively.

For three π electrons, two will be paired so we only need consider the contribution from
the single unpaired electron. Thus, the term symbols are the same as those for a single π
electron.

Two equivalent π electrons

For two unpaired electrons, we need to consider how the electrons can be arranged between
the pair of degenerate orbitals, φ+1 and φ−1 , where the subscript denotes the value of λ.

First we consider the case where the electrons (labelled 1 and 2) occupy the same orbital:
φ+1 (1)φ+1 (2) or φ−1 (1)φ−1 (2), with Λ = +2 or −2, respectively. Since the electrons are in
the same molecular orbital, their spins must be opposite to satisfy the Pauli principle, so
S = 0. Therefore the term symbol is 1 ∆. If the molecule has a centre of inversion and the
molecular orbital has g or u symmetry, we need to include the parity label. Since the two
electrons are in the same orbital and g × g = u × u = g, the term symbol will be 1 ∆ g .

The other possibility is that the two electrons have different values of λ, i.e. Λ = 0. We have
to be careful to construct the appropriately symmetrised wavefunctions. φ+1 (1)φ+1 (2) is
not appropriate because it is neither symmetric or antisymmetric with respect to exchange
of electrons but the linear combination √1 [φ+1 (1)φ−1 (2) + φ−1 (1)φ+1 (2)] is symmetric with
2
respect to exchange of electrons and √1 [φ+1 (1)φ−1 (2) − φ−1 (1)φ+1 (2)] is antisymmetric with
2
respect to exchange of electrons. The symmetric orbital wavefunction must be paired with
an antisymmetric spin wavefunction and vice versa. Thus the term symbols are 1 Σ+ or 3 Σ− .

44
Again, if the molecule has a centre of inversion, we need to include the parity label and,
since there are two electrons in orbitals with the same parity, the term symbols will be 1 Σ+
g or
3 Σ− .
g

Two non-equivalent π electrons

As with electrons in equivalent π orbitals, electrons in non-equivalent π orbitals can be

arranged to give Λ = 0 or ±2. However, unlike electrons in equivalent π orbitals, electrons
in non-equivalent orbitals satisfy the Pauli principle de facto. Thus, for both Σ and ∆ states,
the electron spins can be parallel or antiparallel, i.e. S = 0 or 1, and the term symbols are
1,3 Σ+ , 1,3 Σ− and 1,3 ∆. For molecules with a centre of inversion, the parity is determined using

g × g = u × u = g and g × u = u × g = u.

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7 Tutorial problems

Tutorial 1
Rotational spectroscopy

1. A typical small diatomic molecule has a rotation constant B = 2.0 cm−1 . What are the
relative populations of the J = 1 and J = 2 rotational states at 100 K?

2. 35 Cl19 F
has a microwave spectrum showing absorption maxima at wavelengths of
1.00 cm, 0.50 cm and 0.33 cm. Calculate its bond length assuming it to be a rigid rotor.

[Relative isotopic masses: 35 Cl 34.969; 19 F 18.998.]

3. Explain what parameters determine the transition energies and intensities in rotation
spectra of diatomic molecules. Why does 1 H127 I possess a microwave spectrum whilst
127 I does not?
2

The following lines, ν̃, were observed in a microwave absorption spectrum of 1 H127 I in
its ground vibrational state: 12.684, 25.368, 38.052, 50.735 cm−1 .

(a) Assign quantum numbers J 00 and J 0 to each transition.

(b) Determine the rotation constant B and hence the bond length of 1 H127 I, in nm.

(c) How would you expect B to differ if the molecule were in an excited vibrational
state?

(d) Which of these lines would you expect to be most intense at a molecular beam
temperature of 85 K?

[Relative isotopic masses: 1 H 1.008; 127 I 126.905.]

Vibrational spectroscopy

1. The 12 C ≡ 14 N group in an organic molecule shows an IR absorption at around

2250 cm−1 . Calculate the force constant in N m−1 and the period of vibration. If
the compound were isotopically substituted with 15 N, at what wavenumber would you
expect the new absorption band?

[Relative isotopic masses: 12 C 12.000; 14 N 14.003; 15 N 15.000. ]

2. Features are observed in the vibrational absorption spectrum of 1 H35 Cl at the following
wavenumbers/cm−1 : 2886 (strong); 5668 (weak), 8347 (very weak). Determine the
well-depth De , and dissociation energy D0 , of 1 H35 Cl.

[Relative isotopic masses: 1 H 1.008; 35 Cl 34.969.]

46
Tutorial 2
Vibration-rotation spectroscopy

1. The wavenumbers of the first few P and R branch transitions associated with the
v = 0 → 1 transition of 12 C16 O are listed in the table below.

J 00 ν̃[ R( J 00 )] ν̃[ P( J 00 )]
0 2147.05
1 2150.82 2139.39
2 2154.56 2135.51

(a) Draw an energy level diagram to illustrate the transitions that take place in vibra-
tion–rotation spectroscopy. Label vibrational and rotational quantum numbers
and P and R branch transitions.

(b) Use the energy level diagram from part (a) or the method of combination differ-
ences to determine the rotational constants B0 and B1 .

(c) Determine the equilibrium bond length re .

(d) Determine the wavenumber of the pure v = 0 → 1 transition. Explain why this
transition is not observed in the IR spectrum.

[Relative isotopic masses: 12 C 12.000; 16 O 15.995.]

Electronic spectroscopy

1. The electronic ground state of N2 is 1 Σ+g . Determine whether the Πu , Σu , Π g and

1 3 + 3
1 Σ− electronically excited states of N can be populated directly from the electronic
u 2
ground state by absorption of a single photon.

2. The following Deslandres table lists some of the wavenumbers of the transitions ob-
served in the d3 Π g ← a3 Πu band system of the C2 molecule arranged by vibrational
quantum numbers v0 and v00 . Deduce the vibrational constants ωe00 and ωe xe00 for the
molecule in the a state. Use these data to calculate the force constant and the dissociation
energies D000 and De00 in kJ mol−1 .

v0 \v00 0 1 2
0 19355 17740 16147
1 21104 19490 17899
2 22812 21202 19611
3 22870 21282

47
3. For the c ← X transition of N2 , Te0 − Te00 = 104476 cm−1 , ωe00 = 2358.57 cm−1 and
ωe xe00 = 14.324 cm−1 . The wavenumbers of the two vibronic transitions are given in the
table below.

v00 v0 ν̃/cm−1
0 1 106555.5
0 2 108688.9

(a) Determine ωe0 and ωe xe0 .

(b) The continuum limit for the c ← X transition, ν̃(vmax 0 , 0) in N2 is 185030 cm−1 .
Draw a schematic diagram of the X and c potential energy curves of N2 and use it
0
to illustrate the continuum limit for the c ← X transition and D00 . Use ν̃(vmax , 0) to
determine D0 . 0

48
8 Fundamental physical constants
Name Symbol Value Unit
Elementary charge e 1.60217733 × 10−19 C
Speed of light in vacuum c 2.99792458 × 108 m s−1
Planck’s constant h 6.6260755 × 10−34 Js
Dirac’s constant h̄ = h/2π 1.0545727 × 10−34 Js
Bohr radius a0 0.52918 Å
Rydberg constant R∞ 13.595 eV
Molar gas constant R 8.31441 J mol−1 K−1
Avogadro’s constant NA 6.0221367 × 1023 mol−1
Boltzmann’s constant k = R/NA 1.380658 × 10−23 J K−1
Electron mass me 9.1093897 × 10−31 kg
Proton mass mp 1.6726231 × 10−27 kg
Neutron mass mn 1.674954 × 10−27 kg
Elementary mass unit mu 1.6605656 × 10−27 kg

49