Chapter 7

Spectroscopy polyatomicmolecules

The spectroscopy of polyatomic molecules is conceptually similar to thatof diatomics, with the important difference that the number of thegrees offreedom increases with the number of atoms and so does the number ofpossible transitions that appear in the spectra. Figuring out “by hand”which transitions are allowed and which are not, would be a hopeless task,but for the simplest molecules.

Group theory comes to rescue with two very important theorems:

1. If {fi} is a basis for a representation Γ(a) with character χ(a) and {gj}is a basis for a representation Γ(b) with character χ(b), then the directproduct {figj} is a basis for the representation Γ(a×b) with characterχ(a×b) = χ(a)χ(b). In other words, if f , which could be a function oran operator, transforms like Γ(a), and g transforms like Γ(b), then f · gtransforms like a representation Γ(a×b) with character χ(a×b) = χ(a)χ(b).the importance of this theorem comes when taken together with thesecond

2. The integral (matrix element)⟨f

∣∣∣O∣∣∣ i⟩ vanishes (is strictly equal to

zero) if Γ(f×O×i) does not contain in its decomposition the totally sym-metric representation (typically named A1). Conversely, if Γ(f×O×i)

decomposition contains A1, then the integral is, in principle, nonzero(wheter it is small or large, group theory arguments cannot tell us).

The two theorems are quite general. We are particularly interested in thecase where O is the dipole moment operator and |f〉 , |i〉 the wavefunctionsof some molecular state. Then we can use the above argument to decidewhether the transition i↔ f is allowed or not.

87

Things are particularly simple if we are interested in transitions thatoriginate from the ground electronic and vibrational state of the molecule,as is often the case in absorption spectroscopy. Since the ground state trans-forms like the totally symmetric representation the requirement that 〈f |μ| i〉transforms like A1 simplifies to the requirement that Γ

(f×μ) contains A1 inits decomposition and this is the case only if f and at least one of the com-ponents of μ transform according to the same representation.

7.0.1 Vibrational spectra

To see how the above concepts apply to a polyatomic molecule, we start withthe example of the X3 molecule, which we have already encountered in § 4.4and 6.4, has two vibrational modes, one of symmetry A′1, associated with thenormal coordinate Q1 = (q4+q5+q6)/

√3 and one of symmetry E ′, associated

with the normal coordinates Q2 = (√3q2 −

√3q3 + 2q4 − q5 − q6)/

√12 and

Q3 = (2q1 − q2 − q3 −√3q5 +

√3q6)/

√12. In spectroscopy it is customary

to label these modes ν1 and ν2 (and, if there were more, ν3, ν4 and so on,typically in order of decreasing frequency). The dipole moment operatoris a vector, therefore it transforms like the coordinates x, y, z. From thecharacter table of D3h we see that (x, y) generates the representation E

′ andz the representation A′′2. Therefore, of the two vibrational modes, only forthe degenerate E ′ the 0 → 1 transition is symmetry allowed (in the dipoleapproximation). Another way of saying this is that ν1 is infrared active andν2 infrared inactive, meaning that in the infrared spectrum of X3, we expectto see the ν1 but not the ν2 fundamental band. The X3 molecule is smallenough that we could have figured all of this out by hand, simply by noticingthat the ν1 vibration corresponds to a uniform expansion and contractionof the whole molecule, which cannot cause a change of the molecular dipolemoment: μ(Q1) is zero at equilibrium and remains zero as Q1 changes fromits equilibrium value. Conversely, ν2 corresponds to a vibration that doesnot keep the length of all three bonds the same, which means that the dipolemoment can change during the vibrational cycle.

Try now with a slightly bigger molecule: benzene, C6H6, Fig. 7.1. Themolecule has 12 atoms, each with 3 degrees of freedom, that’s 36 degrees offreedom. Subtracting transalations and rotations of the whole molecule (3each) leaves 30 degrees of freedom for vibrational modes. Sorting these modesby hand and deciding whether or not they are infrared active is difficult, time-consuming, and not particularly insightful. Let’s see how far we can go bysymmetry arguments.

First of all, we need to identify the molecule symmetry group (D6h) andthe proper group character table, at hand (Tab. 7.1). Second, we need to see

88

C

C

CC

C

C

H

H

HH

H

H

1

2

3

5

6

4

Figure 7.1: Schematic representation of the benzene molecule.

D6h E 2C6 2C3 C2 3C ′2 3C ′′2 i 2S3 2S6 σh 3σd 3σvA1g 1 1 1 1 1 1 1 1 1 1 1 1 x2 + y2, z2

A2g 1 1 1 1 −1 −1 1 1 1 1 −1 −1 Rz

B1g 1 −1 1 −1 1 −1 1 −1 1 −1 1 −1B2g 1 −1 1 −1 −1 1 1 −1 1 −1 −1 1E1g 2 1 −1 −2 0 0 2 1 −1 −2 0 0 (Rx, Ry) (xz, yz)E2g 2 −1 −1 2 0 0 2 −1 −1 2 0 0 (x2 − y2, xy)A1u 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1A2u 1 1 1 1 −1 −1 −1 −1 −1 −1 1 1 zB1u 1 −1 1 −1 1 −1 −1 1 −1 1 −1 1B2u 1 −1 1 −1 −1 1 −1 1 −1 1 1 −1E1u 2 1 −1 −2 0 0 −2 −1 1 2 0 0 (x, y)E2u 2 −1 −1 2 0 0 −2 1 1 −2 0 0

Table 7.1: Character table of the D6h symmetry group.

89

which irreducible representation are contained in the decomposition of therepresentation generated by the 36 atomic displacement (Γ(tot)). For this weneed the character of Γ(tot):

D6h E 2C6 2C3 C2 3C ′2 3C ′′2 i 2S3 2S6 σh 3σd 3σvΓ(tot) 36 0 0 0 −4 0 0 0 0 12 0 4

(7.1)

According to

ai =1

h

∑R

χ(i)(R)χ()(R) (7.2)

this representation decomposes as

Γ(tot) = 2A1g ⊕ 2A2g ⊕ 0B1g ⊕ 2B2g ⊕ 2E1g ⊕ 4E2g (7.3)

⊕ 0A1u ⊕ 2A2u ⊕ 2B1u ⊕ 2B2u ⊕ 4E1u ⊕ 2E2u. (7.4)

Third, we need to subtract rotations, which according to the charactertable transform as A2g and E1g, and translations (A2u, E1u). This leaves

Γ(vib) = 2A1g ⊕ 1A2g ⊕ 0B1g ⊕ 2B2g ⊕ 1E1g ⊕ 4E2g (7.5)

⊕ 0A1u ⊕ 1A2u ⊕ 2B1u ⊕ 2B2u ⊕ 3E1u ⊕ 2E2u. (7.6)

Fourth, we need to see how the dipole moment components transform.The character table gives A2u for μz and E1u for (μx, μy). Accordingly, ofthe 19 vibrational modes in Γ(vib), only four are infrared active: one A2u andthree E1u, and we expect only four fundamental bands to show up in theinfrared spectrum.

We can say something more about these bands by some approximate ar-guments based on the separation between vibrational frequencies of differentfunctional groups. The idea is that certain normal modes will be mostly de-scribed by one type of motion only (for example, stretching and compressionof the C-H bonds), and very little by other types of motion (bending, stretch-ing of the CC bonds, etc.). The rationale behind this assumption is that eachkind of motion has its own particular frequency (≈ 3000 cm−1 for CH stretch-ing, ≈ 1600 cm−1 for CC stretching and so on) and that motions with verydifferent characteristic frequencies will contribute to different normal modes.Let’s, for sake of example, consider the stretching of the six CH bonds andtake all other bonds and angles as frozen at their equilibrium value. Withthese six coordinates we can construct properly symmetrized linear combina-tions of various symmetries, which will (approximately) correspond to someof the real normal modes of the molecule. Instead of doing it by hand, we

90

can, again, use group theory, just like we did earlier. The representationgenerated by the six CH stretching coordinates has the character

D6h E 2C6 2C3 C2 3C ′2 3C ′′2 i 2S3 2S6 σh 3σd 3σvΓ(CH) 6 0 0 0 2 0 0 0 0 6 0 2

(7.7)

and decomposes as

Γ(CH) = 1A1g ⊕ 0A2g ⊕ 0B1g ⊕ 0B2g ⊕ 0E1g ⊕ 1E2g (7.8)

⊕ 0A1u ⊕ 0A2u ⊕ 1B1u ⊕ 0B2u ⊕ 1E1u ⊕ 0E2u. (7.9)

Of these, only the 1E1u is infrared active, and we expect, therefore thataround the frequency of the CH stretching vibration (≈ 3000 cm−1) thereshould be only one fundamental vibrational band in the spectrum.

7.0.2 Electronic spectra

The same type of symmetry arguments apply to the electronic spectroscopyof polyatomic molecules. In this case, the transition matrix elements that weneed to evaluate are of the kind

〈ψe |μ|ψe′〉 〈ψv|ψv′〉 (7.10)

which is nonzero only if both terms are nonzero.The Frank-Condon integral is nonzero only if both vibrational states have

the same symmetry.The orbital integral 〈ψe |μ|ψe′〉 is nonzero if the product Γ(ψe) × Γ(μ) ×

Γ(ψe′ ) contains the totally symmetric representation, in which case, the tran-sition is said to be orbitally allowed. In the great majority of cases, theground electronic state is totally symmetric1 When this is the case, then theorbital selection rule simplifies to the requirement that there exists at leastone of the dipole moment components with the same symmetry as the upperelectronic state.

Spin selection rules require, as for the diatomic molecule, that the upperand lower electronic state have the same spin degeneracy, since there is nooperator acting on the molecular spin in the electric dipole Hamiltonian.Transitions that satisfy this rule are calle d spin allowed.

Transitions that are both orbitally and spin allowed are called fully al-lowed and are, generally very intense.

1the most notable exceptions are given by radicals, molecules with the HOMO orbitalonly partially filled.

91

Transitions that are orbitally allowed but spin forbidden are still possible,though weaker by several orders of magnitude, because the spin is, to a verygood approximation, a good (conserved) quantum number.

Spin-allowed, orbitally forbidden transitions are also observable, and theirintensity is generally intermediate between fully allowed and spin-forbiddedtransitions. The reason for this is that the separation between electronic andvibrational degrees of freedom is only approximate. To a better degree ofapproximation we should be calculating

〈ψe(xe, Qv)ψv(Qv) |μ(xe, Qv)|ψe′(xx, Qe)ψv′(Qe)〉 (7.11)

instead of

〈ψe(xe, Qv = Qeq) |μ(xe, Qv = Qeq)|ψe′(xe, Qv = Qeq)〉 〈ψv(Qv)|ψv′(Qv)〉(7.12)

that is, take into account that there is a slight dependence of the electronicwavefunction on the vibrational coordinate. It is evident that now the inte-gral cannot be separated anymore. The necessary condition for the integralto be nonzero is that

ψe(xe, Qvψv(Qv)μ(xe, Qv)ψe′(xx, Qe)ψv′(Qe) (7.13)

transforms like the totally symmetric representation, which implies that

Γ(ψe)Γ(ψv)Γ(μ)Γ(ψe′ )Γ(ψv′ ) (7.14)

contains the totally symmetric representation. This is clearly a less restrictiverequirement than for orbitally allowed transitions, since it puts restrictionsonly on the symmetry of the combined electronic and vibrational wavefunc-tions, not on each of them separately.

Consider, again, the molecule of benzene. We will now work out the sym-metry of its HOMO and LUMO orbitals, the symmetry of the ground stateand of the excited states that have one electron transferred from the HOMOto the LUMO horbital and whether the corresponding electronic transitionsare allowed or not. It turns out that the HOMO and LUMO orbitals aremostly formed by the six pz orbitals (perpendicular to the molecular plane).The s orbitals of hydrogen are too low in energy to contribute significantly;the carbon s orbitals and in-plane p orbitals overlap more strongly amongthemselves and give rise to lower energy molecular orbitals than the pz. Also,the pz atomic orbitals of carbon are the only antisymmetric ones with respectto σh, therefore they combine only among themselves to form molecular or-bitals.

92

The representation generated by the six pz orbitals has character

D6h E 2C6 2C3 C2 3C ′2 3C ′′2 i 2S3 2S6 σh 3σd 3σvΓ(CH) 6 0 0 0 −2 0 0 0 0 −6 0 2

(7.15)

and decomposes as

Γ(pz) = a2u ⊕ e1g ⊕ e2u ⊕ b2g (7.16)

the corresponding molecular orbitals, which we can construct using the pro-jection opertator of Ch. 6, are shown in Fig. 7.2, ordered by their energy.To find this energy we need an explicit calculation. Still, we can figure out

a2u

e1g

e2u

b2g

Figure 7.2: Molecular orbitals of benzene formed by the carbon pz orbitals.

the energy ordering considering that the orbital energy increases with thenumber of nodes of the resulting wavefunction. Now we have to accomodatesix electrons in these molecular orbitals. There is only one way to do it forthe ground state (all orbitals fully occupied or fully empty) and two for the

93

b2g

e2u

e1g

a1u

b2g

e2u

e1g

a1u

b2g

e2u

e1g

a1u

X:1A1g

1B2u

, 1B1u

, 1E1u

3B2u

, 3B1u

, 3E1u

Figure 7.3: Lowest electronic states of benzene

excited state, depending on the relative spin of the two unpaired electrons(Fig. 7.3). The ground state configuration has all electrons paired, hence itmust give rise to a fully symmetric singlet state (1A1g). The second configu-ration has also all paired electron spins (singlet) but two unpaired electrons(e1g and e2u). We can find the corresponding terms by decomposing the rep-resentation2 generated by Γ(e1g) × Γ(e2u) and find 1B2u,

1B1u, and1E1u. The

third configuration differs from the second only for the relative spin of theunpaired electrons (triplet) and gives 3B2u,

3B1u, and3E1u

Of the possible transitions, we expect the strongest ones to be thosethat are spin allowed (singlet→singlet), and among these, the orbitally al-lowed 1A1g →1E1u to be stronger then the orbitally forbidden 1A1g →1B1u

and 1A1g →1B2u. These three bands are indeed observed in the electronicspectrum of benzene (Fig. 7.4), respectively at approximately 185, 200 and260 nm.

7.0.3 Chromophores and functional groups

Molecular spectroscopy would not be the very useful tool it is, if every differ-ent molecule had to be treated as a system of its own, without any relation-ship whatsoever with molecules that are somewhat similar. Luckily this isnot the case, and often a given molecular subunit gives rise, in two differentmolecules, to spectral features that are very similar. Two examples of thesemolecular subunits are the carbonyl group (C=O) that gives rise to charac-

2You are encouraged to try and do it yourselves at the beginning. Afterwards, you canuse pre-compiled tables, which exist for all molecular point groups.

94

Wavelength /nm

extin

ctio

n c

oe

ffic

ien

t

1

1

1

Figure 7.4: Electronic spectrum of benzene

95

teristic bands in the electronic absorption spectra, and CH vibrations, thatgive rise to characteristic bands in the infrared spectrum. These subunitsare usually called chromophores (= bearing color) in electronic spectroscopy,and functional groups in infrared spectroscopy.

That the same subunit can give similar spectral features in differentmolecules is not too surprising: most molecular orbitals extend over justa few atoms, somewhat implying that the local molecular structure is notvery sensitive to how the molecule looks like at distances larger than one ortwo bonds. One notable exception is the benzene ring with its delocalizedmolecular orbitals extending over several atoms. However, if we consider thewhole ring as a subunit, it too gives rise to characteristic electronic spec-tral features. The most common chromophores are the C=O, N=O, andN=N, groups that give rise to a non-bonding (n) to antibonding (π∗) tran-sition (Fig. 7.5a, approximately 290 nm for C=O) the C=C double bond,

n

(a) (b)

C

C

C

C

C

C

π∗

π

π∗

O

O

Figure 7.5: Schematic (a) of a n→ π∗ transition in a C=O chromophore and(b) of a π → π∗ transition in a C=C chromophore.

giving rise to a bonding (π) to antibonding (π∗) transition (Fig. 7.5a) andthe above-mentioned benzene ring, giving bands similar to those observedfor the benzene molecule.

In a similar fashion, most vibrational modes are more or less localized toone or a few bonds, and the same functional in two different molecules givessimilar features in the vibrational spectrum. Functional groups involvinga hydrogen atom are the easiest to identify, because of the high frequencyassociated with the stretching vibration of a X−H bond (Table 7.2). Otherimportant group frequencies include the bending of X−H bonds, in the 1200-1400 cm−1 region and stretching of double/triple bonds involving a carbon

96

Bond Frequency /cm−1 IntensityC−H 2800−3000 strong

=C−H 3000−3100 medium≡C−H 3300 strongO−H 3600−3650 variableN−H 3300−3500 medium

Table 7.2: Vibrational frequencies of some X−H functional groups.

atom (Table 7.3).

Bond Frequency /cm−1 IntensityC=C 1620−1680 variableC≡C 2100−2150 variableC≡N 2200−2300 variableC=O 1700−1750 strong

Table 7.3: Vibrational frequencies of some X=C and X≡C functional groups.

7.0.4 Lowering of symmetry

Another very useful concept in molecular spectroscopy is that of approximatesymmetry. The idea is that there are situation in which the symmetry of amolecule may be disrupted by some causes. Think, for example, of a moleculein a liquid environment: the molecule by itself may have a high symmetry, butthe ensemble molecule+environment has none, since the solvent moleculesare randomly oriented. Another possibility is that some of the atoms in themolecule are replaced by isotopes. Yet, if the disruption is not too strong, weexpect that some remnants of the original symmetry of the molecule shouldshow up in the spectrum.

Let’s take as an example, the CH stretching vibrations of the benzenemolecule, (C6H6) where two opposite

12C atoms have been replaced by theslightly heavier 13C. This, in principle, brings down the symmetry of themolecule from D6h to D2h. In practice, the difference in mass is less than10%, and the difference in reduced mass of each CH subunit, about 0.5%. We expect that the isotopic substitution should not have a very drasticinfluence on the vibrational spectroscopy of the molecule (and hardly any onthe electronic spectroscopy).

We had seen earlier that the six CH stretching vibrations give four vibra-

97

tional modes:

Γ(CH) = 1A1g ⊕ 1E2g ⊕ 1B1u ⊕ 1E1u (7.17)

of which only 1E1u is infrared active. What we need to consider is whichsymmetry elements of D6h still exist in D2h and what their name is, there,:

D6h E 2C6 2C3 C2 3C ′2 3C ′′2 i 2S3 2S6 σh 3σd 3σvD2h E − − C2(z) C2(x) C2(y) i − − σ(xy) σ(yz) σv(xz)

(7.18)

next, we need to consider how a given representation, which was irreduciblein D6h decomposes in D2h. Take E1u:

D2h E − − C2(z) C2(x) C2(y) i − − σ(xy) σ(yz) σv(xz)E1u 2 − − −2 0 0 −2 − − 2 0 0

(7.19)

by direct inspection of the character table of D2h we see that it decomposesinto B2u +B3u, both of which are infrared active. Likewise, A1g becomes Ag(inactive), 1E2g gives Ag+B1g, both inactive. Finally B1u, which was inactivein D6h gives B3u, which is active in D2h. Because this mode was inactive inD6h benzene, and because the isotopic substitution does not make a majorchange in the molecule, we expect that the corresponding intensity in thespectrum of D2h benzene should be weak compared to transitions that wereallowed in D6h.

98

7.1 Interaction of molecules with their envi-

ronment

We have, so far, considered molecules as if they were sitting somewhere atrest, without interacting with one another or with their environment. Thisis obviously an idealized situation and, in reality, molecules move around,interacting with whatever they find in their path. This interactions man-ifest themselves in many ways; one of the most important ones, as far asspectroscopy is concerned is the shape and width of spectral lines.

7.1.1 Line shape and width

In absence of any interaction with the outside world, spectral lines wouldhave infinite intensity in an infinitely sharp (zero width) line. In reality, wealways observe lines of finite width and spectral intensity. In part, this is areal effect, related to the sample itself, in part, it is just apparent, due to thefinite resolution of the instrument. In what follows we assume that the in-strument’s resolution is sufficiently good and not limiting the measurement inany way.3 We distinguish two classes of mechanisms of spectral lines broaden-ing, depending on whether the broadening is the same for all molecules in thesample or not. In the first case we speak of homogeneous broadening, whilein the second case of inhomogeneous broadening. Homogeneous broadeningmechanisms include:

1. finite lifetime

(a) natural linewidth

(b) pressure broadening (collisions between molecules)

(c) collisions with cell walls

2. finite-time interaction

3. power broadening.

There are many ways to show the effect of finite lifetime on spectral linewidths.The simplest way is to use Heisemberg’s uncertainty principle:

ΔEΔt ≥ h (7.20)

3Whether or not this is true in reality depends on many factors, including how muchmoney and time we are willing to spend to perform the measurement.

99

The frequency of a photon absorbed during a transition from a level |1〉 withenergy E1 to a level |2〉 with energy E2 is

ν12 = (E2 − E1)/h, (7.21)

but, because of the finite lifetime τ2 of |2〉, the energy E2 is undetermined byan amount

ΔE2 = τ2/h (7.22)

this, in turn implies that it is possible to absorb a photon of a frequencyslightly different than ν12 by an amount

δν12 ≈ 1

2πτ2. (7.23)

The same is true for emission from |2〉 to |1〉. In fact, we can very easilyfind the shape of the spectral line considering the intensity of emission froma sample of identical molecules, all excited to |2〉. This intensity goes as

I(t) = I0e−t/τ2 (7.24)

which implies that the emitted electric field has the form

E(t) = E0e−t/2τ2e−2πiν12t (7.25)

and its Fourier transform gives the power spectrum

I(ν) = I01

(ν − ν12)2 + (Δν/2)2(7.26)

which is a Lorentzian function with full width at half maximum (FWHM)Δν.

When the only mechanism of decay of the excited state is spontaneousemission, the linewidth Δν is called natural linewidth. In a two-level system,the natural linewidth is related to Einstein’s coefficient

A =64π4ν12

3hc|μ12|2 (7.27)

by

τ2 = A−1. (7.28)

100

In a multi-level system we must sum over all the states that |2〉 can decay toby emitting a photon

τ2 = (∑

Ai)−1. (7.29)

Because of the ν3 dependence of A, the natural linewidth is much moreprominent in the visible and UV spectral regions (electronic spectroscopy)than it is in the infrared (vibrational) and microwave (rotational) regions.

We can expect that collision broadening will give a similar result, sincecollisions also affect the lifetime of the excited state. The details are morecomplicated because since, upon collision, a molecule may completely loosememory of its previous state (hard collision) or only partially (soft collision).However, a detail analysis shows that the line profile associated with collisionbroadening is also a Lorentzian whose width depends on the sample pressure:

Δνcoll =1

2πτcoll

= aP. (7.30)

The coefficient a depends on the details of the molecular energy levels andcontains information on the interaction between molecules. The effect of col-lisional broadening is especially important in infrared and microwave spec-troscopy, where the natural linewidth is small, and in liquid environment,where the molecules are “always” in contact with one another.

Finite-time interaction arises, most often, from the molecules moving outof the region occupied by the radiation field. Therefore it becomes importantwhenever molecules have high velocities (high temperature, molecular beams,. . . ) and/or the radiation field is strongly confined (tightly focused laserbeams), and/or the transition being measured is extremely narrow (atomicclocks). Finite-time interaction gives rise to spectral broadening since itlimits the time available for the measurement and, therefore causes an un-certainty on the frequency of the transition measured. The width of thetransition is still determined by the uncertainty principle, while the line-shape is determined by the spatial profile of the radiation field. It is possibleto show that a Gaussian profile (for example, a molecule crossing a tightlyfocused laser beam) gives rise to a Gaussian line-shape.

Inhomogeneous broadening mechanisms arise when molecules in the sam-ple experience a different interaction with their environment or with the ra-diation field; the most important mechanisms are:

1. Doppler broadening

2. Broadening due to the environment.

101

Sometimes the term inhomogeneous broadening is, improperly, used to de-note just the second of these mechanisms.

Doppler broadening arises from the thermal distribution of velocity in thesample. A molecule moving with (non relativistic) velocity v in a monochro-matic electromagnetic field experiences a field of slightly different frequency,depending on the magnitude and orientation of its velocity relative to k, thedirection of propagation of the field. In the frame of reference of the labora-tory, the absorption frequency of the appears shifted from its value at restν0 to a new value

ν = ν0

(1 +

v · kc

). (7.31)

The actual spectrum is the superposition of the spectra of the individualmolecules, each with its own velocity, according to Boltzmann distribution.This gives rise to a inhomogeneous line width, called Doppler profile

I(ν) = I0 exp

[−Mc2

2kT

(ν − ν0

ν0

)2]

(7.32)

this is a Gaussian, centered at ν0, and its FWHM is

Δν =ν

c

√2kT

M2 ln 2 (7.33)

Broadening due to the environment is observed in the solid or liquid phase,where the immediate environment of each molecule in the sample is slightlydifferent, even if on average all molecules experience the same conditions. Asa consequence, the spectrum of each molecule is slightly different and whatwe measure is the superposition of all these different spectra. In general, thisbroadening is sufficient to hide all but the coarsest features of a molecularspectrum (Fig. 7.6). Remark that the spectra are sharper in the crystalenvironment than they are in solution, due to the more uniform environmentprovided by the crystal. Remark also that in the crystal spectrum there aretransitions that do not appear either in the gas phase or in the liquid phasespectrum. This is a consequence of the crystal environment, in which the sitewhere the molecule is located has a different symmetry than the moleculeitself. As a result, the molecular symmetry is lowered, and transitions thatwere forbidden for the more symmetric molecule are now permitted, an effectthat we have already encountered with isotopic substitution.

102

Figure 7.6: Electronic spectra of benzene in different environments.

103