456 12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

change the moments of inertia (Fig. 12.13). The effect of centrifugal distortion on adiatomic molecule is to stretch the bond and hence to increase the moment of inertia.As a result, centrifugal distortion reduces the rotational constant and consequentlythe energy levels are slightly closer than the rigid-rotor expressions predict. The effectis usually taken into account largely empirically by subtracting a term from the energyand writing

( J) = J(J + 1) J J2( J + 1)2 (12.16)

The parameter J is the centrifugal distortion constant. It is large when the bond iseasily stretched. The centrifugal distortion constant of a diatomic molecule is relatedto the vibrational wavenumber of the bond, # (which, as we shall see later, is a meas-ure of its stiffness), through the approximate relation (see Problem 12.21)

J = (12.17)

Hence the observation of the convergence of the rotational levels as J increases can beinterpreted in terms of the rigidity of the bond.

12.5 Rotational transitions

Key points (a) For a molecule to give a pure rotational spectrum, it must be polar. The specic rotational selection rules are J = 1, MJ = 0, 1, K = 0. (b) Bond lengths may be obtained fromanalysis of microwave spectra.

Typical values of for small molecules are in the region of 0.110 cm1 (for example,0.356 cm1 for NF3 and 10.59 cm

1 for HCl), so rotational transitions lie in the micro-wave region of the spectrum. The transitions are detected by monitoring the net absorp-tion of microwave radiation. Modulation of the transmitted intensity, which is usedto facilitate detection and amplication of the absorption, can be achieved by varyingthe energy levels with an oscillating electric eld. In this Stark modulation, an electriceld of about 105 V m1 and a frequency of 10100 kHz is applied to the sample.

(a) Rotational selection rules

We have already remarked (Section 12.2) that the gross selection rule for the observa-tion of a pure rotational spectrum is that a molecule must have a permanent electricdipole moment. That is, for a molecule to give a pure rotational spectrum, it must bepolar. The classical basis of this rule is that a polar molecule appears to possess a uctuating dipole when rotating but a nonpolar molecule does not (Fig. 12.14). Thepermanent dipole can be regarded as a handle with which the molecule stirs the electromagnetic eld into oscillation (and vice versa for absorption). Homonucleardiatomic molecules and symmetrical linear molecules such as CO2 are rotationally inactive. Spherical rotors cannot have electric dipole moments unless they becomedistorted by rotation, so they are also inactive except in special cases. An example of a spherical rotor that does become sufciently distorted for it to acquire a dipole moment is SiH4, which has a dipole moment of about 8.3 D by virtue of its rotationwhen J 10 (for comparison, HCl has a permanent dipole moment of 1.1 D; molecu-lar dipole moments and their units are discussed in Section 17.1). The pure rotationalspectrum of SiH4 has been detected by using long path lengths (10 m) through high-pressure (4 atm) samples.

Centrifugal distortionconstant

43

#2

Rotational terms affectedby centrifugal distortion

012

3

4

5

6

7

MJFieldon

Fieldoff

Fig. 12.12 The effect of an electric eld onthe energy levels of a polar linear rotor. All levels are doubly degenerate except thatwith MJ = 0.

Centrifugalforce

Fig. 12.13 The effect of rotation on amolecule. The centrifugal force arisingfrom rotation distorts the molecule,opening out bond angles and stretchingbonds slightly. The effect is to increase themoment of inertia of the molecule andhence to decrease its rotational constant.

12.5 ROTATIONAL TRANSITIONS 457

Fig. 12.14 To a stationary observer, a rotating polar molecule looks like an oscillating dipole that can stir theelectromagnetic eld into oscillation (and vice versa for absorption). Thispicture is the classical origin of the grossselection rule for rotational transitions.

Photon

Fig. 12.15 When a photon is absorbed by a molecule, the angular momentum of the combined system is conserved. If themolecule is rotating in the same sense asthe spin of the incoming photon, then Jincreases by 1.

A brief illustration

Of the molecules N2, CO2, OCS, H2O, CH2=CH2, and C6H6, only OCS and H2O are

polar, so only these two molecules have microwave spectra.

Self-test 12.3 Which of the molecules H2, NO, N2O, and CH4 can have a pure rotational spectrum? [NO, N2O]

The specic rotational selection rules are found by evaluating the transition dipolemoment between rotational states. We show in Further information 12.2 that, for a linear molecule, the transition moment vanishes unless the following conditions arefullled:

J = 1 MJ = 0, 1 (12.18)

The transition J = +1 corresponds to absorption and the transition J = 1 corres-ponds to emission. The allowed change in J in each case arises from the conservationof angular momentum when a photon, a spin-1 particle, is emitted or absorbed (Fig. 12.15).

When the transition moment is evaluated for all possible relative orientations of the molecule to the line of ight of the photon, it is found that the total J + 1 Jtransition intensity is proportional to

|J+1,J |2 = 02 (12.19)

where 0 is the permanent electric dipole moment of the molecule. The intensity isproportional to the square of the permanent electric dipole moment, so strongly polarmolecules give rise to much more intense rotational lines than less polar molecules.

For symmetric rotors, an additional selection rule states that K = 0. To understandthis rule, consider the symmetric rotor NH3, where the electric dipole moment liesparallel to the gure axis. Such a molecule cannot be accelerated into different statesof rotation around the gure axis by the absorption of radiation, so K = 0. Therefore,for symmetric rotors the selection rules are:

J = 1 MJ = 0, 1 K = 0 (12.20)

(b) The appearance of rotational spectra

When these selection rules are applied to the expressions for the energy levels of a rigidspherical or linear rotor, it follows that the wavenumbers of the allowed J + 1 Jabsorptions are

#( J + 1 J) = ( J + 1) ( J) = 2( J + 1) J = 0, 1, 2, . . . (12.21a)

When centrifugal distortion is taken into account, the corresponding expression obtained from eqn 12.16 is

#( J + 1 J) = 2( J + 1) 4J( J + 1)3 (12.21b)

However, because the second term is typically very small compared with the rst, theappearance of the spectrum closely resembles that predicted from eqn 12.21a.

Rotational selection rulesfor symmetric rotors

DEFJ + 1

2J + 1

ABC

Rotational selectionrules for linear rotors

458 12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

Ener

gy

Tran

smitt

ance

Frequency

Fig. 12.16 The rotational energy levels of alinear rotor, the transitions allowed by theselection rule J = 1, and a typical purerotational absorption spectrum (displayedhere in terms of the radiation transmittedthrough the sample). The intensities reectthe populations of the initial level in eachcase and the strengths of the transitiondipole moments.

Example 12.3 Predicting the appearance of a rotational spectrum

Predict the form of the rotational spectrum of 14NH3.

Method We calculated the energy levels in Example 12.2. The 14NH3 molecule is a polar symmetric rotor, so the selection rules J = 1 and K = 0 apply. For absorption, J = +1 and we can use eqn 12.21a.

Answer Because = 9.977 cm1, we can draw up the following table for the J + 1 J transitions.

J 0 1 2 3 . . .#/cm1 19.95 39.91 59.86 79.82 . . ./GHz 598.1 1197 1795 2393 . . .

The line spacing is 19.95 cm1 (598.1 GHz).

Self-test 12.4 Repeat the problem for C35ClH3 (see Self-test 12.2 for details).[Lines of separation 0.944 cm1 (28.3 GHz)]

The form of the spectrum predicted by eqn 12.21 is shown in Fig. 12.16. The mostsignicant feature is that it consists of a series of lines with wavenumbers 2, 4, 6,. . . and of separation 2. The measurement of the line spacing gives , and hence the moment of inertia perpendicular to the principal axis of the molecule. Because the masses of the atoms are known, it is a simple matter to deduce the bond length ofa diatomic molecule. However, in the case of a polyatomic molecule such as OCS orNH3, the analysis gives only a single quantity, I, and it is not possible to infer bothbond lengths (in OCS) or the bond length and bond angle (in NH3). This difcultycan be overcome by using isotopically substituted molecules, such as ABC and ABC;then, by assuming that R(AB) = R(AB), both AB and BC bond lengths can be extracted from the two moments of inertia. A famous example of this procedure is thestudy of OCS; the actual calculation is worked through in Problem 12.7. The assump-tion that bond lengths are unchanged by isotopic substitution is only an approxima-tion, but it is a good approximation in most cases. Nuclear spin, which differs fromone isotope to another, also affects the appearance of high-resolution rotational spec-tra because spin is a source of angular momentum and can couple with the rotation ofthe molecule itself and hence affect the rotational energy levels.

The intensities of spectral lines increase with increasing J and pass through a max-imum before tailing off as J becomes large. The most important reason for the max-imum in intensity is the existence of a maximum in the population of rotational levels.The Boltzmann distribution (Fundamentals F.5) implies that the population of eachstate decays exponentially with increasing J, but the degeneracy of the levels increases,and these two opposite trends result in the population of the energy levels (as distinctfrom the individual states) passing through a maximum. Specically, the populationof a rotational energy level J is given by the Boltzmann expression

NJ NgJeEJ/kT

where N is the total number of molecules and gJ is the degeneracy of the level J. Thevalue of J corresponding to a maximum of this expression is found by treating J asa continuous variable, differentiating with respect to J, and then setting the resultequal to zero. The result is (see Problem 12.26)

Jmax

1/2

(12.22)12DEF

kT

2hc

ABC

12.6 ROTATIONAL RAMAN SPECTRA 459

For a typical molecule (for example, OCS, with = 0.2 cm1) at room temperature, kT 1000hc, so Jmax 30. However, it must be recalled that the intensity of each transition also depends on the value of J (eqn 12.19) and on the population differencebetween the two states involved in the transition. Hence the value of J correspondingto the most intense line is not quite the same as the value of J for the most highly populated level.

12.6 Rotational Raman spectra

Key points A molecule must be anisotropically polarizable for it to be rotationally Raman active.The specic selection rules are: (i) linear rotors, J = 0, 2; (ii) symmetric rotors, J = 0, 1, 2;K = 0.

The gross selection rule for rotational Raman transitions is that the molecule must beanisotropically polarizable. We begin by explaining what this means. A formal deriva-tion of this rule is given in Further information 12.2.

The distortion of a molecule in an electric eld is determined by its polarizability, (Section 17.2). More precisely, if the strength of the eld is E, then the molecule acquires an induced dipole moment of magnitude

= E (12.23)in addition to any permanent dipole moment it may have. An atom is isotropicallypolarizable. That is, the same distortion is induced whatever the direction of the applied eld. The polarizability of a spherical rotor is also isotropic. However, non-spherical rotors have polarizabilities that do depend on the direction of the eld relative to the molecule, so these molecules are anisotropically polarizable (Fig. 12.17).The electron distribution in H2, for example, is more distorted when the eld is applied parallel to the bond than when it is applied perpendicular to it, and we write|| > .

All linear molecules and diatomics (whether homonuclear or heteronuclear) haveanisotropic polarizabilities, and so are rotationally Raman active. This activity is onereason for the importance of rotational Raman spectroscopy, for the technique can beused to study many of the molecules that are inaccessible to microwave spectroscopy.Spherical rotors such as CH4 and SF6, however, are rotationally Raman inactive as wellas microwave inactive. This inactivity does not mean that such molecules are neverfound in rotationally excited states. Molecular collisions do not have to obey such restrictive selection rules, and hence collisions between molecules can result in thepopulation of any rotational state.

We show in Further information 12.2 that the specic rotational Raman selectionrules are

Linear rotors: J = 0, 2Symmetric rotors: J = 0, 1, 2; K = 0

(12.24)

The J = 0 transitions do not lead to a shift in frequency of the scattered photon in purerotational Raman spectroscopy, and contribute to the unshifted Rayleigh radiation.

We can predict the form of the Raman spectrum of a linear rotor by applying theselection rule J = 2 to the rotational energy levels (Fig. 12.18). When the moleculemakes a transition with J = +2, the scattered radiation leaves the molecule in a higherrotational state, so the wavenumber of the incident radiation, initially #i, is decreased.These transitions account for the Stokes lines in the spectrum:

Rotational Ramanselection rules

Distortion

E

E

(a)

(b)

Fig. 12.17 An electric eld applied to amolecule results in its distortion, and thedistorted molecule acquires a contributionto its dipole moment (even if it is nonpolarinitially). The polarizability may be differentwhen the eld is applied (a) parallel or (b) perpendicular to the molecular axis (or,in general, in different directions relative to the molecule); if that is so, then themolecule has an anisotropic polarizability.

460 12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

En

erg

yS

ign

al

Frequency

Stokeslines

Anti-Stokeslines

Ray

leig

h li

ne

Fig. 12.18 The rotational energy levels of a linear rotor and the transitions allowedby the J = 2 Raman selection rules. The form of a typical rotational Ramanspectrum is also shown. The Rayleigh line is much stronger than depicted in the gure; it is shown as a weaker line toimprove visualization of the Raman lines.

#( J + 2 J) = #i {( J + 2) ( J)} = #i 2(2J + 3) (12.25a)

The Stokes lines appear to low frequency of the incident radiation and at displace-ments 6, 10, 14, . . . from #i for J = 0, 1, 2, . . . . When the molecule makes a transition with J = 2, the scattered photon emerges with increased energy. Thesetransitions account for the anti-Stokes lines of the spectrum:

#( J 2 J) = #i {( J) ( J 2)} = #i + 2(2J 1) (12.25b)

The anti-Stokes lines occur at displacements of 6, 10, 14, . . . (for J = 2, 3, 4, . . . ; J = 2 is the lowest state that can contribute under the selection rule J = 2) to high fre-quency of the incident radiation. The separation of adjacent lines in both the Stokesand the anti-Stokes regions is 4, so from its measurement I can be determined andthen used to nd the bond lengths exactly as in the case of microwave spectroscopy.

Example 12.4 Predicting the form of a Raman spectrum

Predict the form of the rotational Raman spectrum of 14N2, for which = 1.99 cm1,

when it is exposed to 336.732 nm laser radiation.

Method The molecule is rotationally Raman active because end-over-end rotationmodulates its polarizability as viewed by a stationary observer. The Stokes andanti-Stokes lines are given by eqn 12.25.

Answer Because i = 336.732 nm corresponds to #i = 29 697.2 cm1, eqns 12.25aand 12.25b give the following line positions:

J 0 1 2 3Stokes lines#/cm1 29 685.3 29 677.3 29 669.3 29 661.4 /nm 336.868 336.958 337.048 337.139Anti-Stokes lines#/cm1 29 709.1 29 717.1 /nm 336.597 336.507

There will be a strong central line at 336.732 nm accompanied on either side bylines of increasing and then decreasing intensity (as a result of transition momentand population effects). The spread of the entire spectrum is very small, so the incident light must be highly monochromatic.

Self-test 12.5 Repeat the calculation for the rotational Raman spectrum of NH3( = 9.977 cm1).

[Stokes lines at 29 637.3, 29 597.4, 29 557.5, 29 517.6 cm1,anti-Stokes lines at 29 757.1, 29 797.0 cm1]

12.7 Nuclear statistics and rotational states

Key point The appearance of rotational spectra is affected by nuclear statistics, the selective occu-pation of rotational states that stems from the Pauli principle.

If eqn 12.25 is used in conjunction with the rotational Raman spectrum of CO2, therotational constant is inconsistent with other measurements of CO bond lengths. Theresults are consistent only if it is supposed that the molecule can exist in states witheven values of J, so the Stokes lines are 2 0, 4 2, . . . and not 5 3, 3 1, . . . .

12.7 NUCLEAR STATISTICS AND ROTATIONAL STATES 461

The explanation of the missing lines is the Pauli principle and the fact that 16Onuclei are spin-0 bosons: just as the Pauli principle excludes certain electronic states,so too does it exclude certain molecular rotational states. The form of the Pauli prin-ciple given in Section 9.4b states that, when two identical bosons are exchanged, theoverall wavefunction must remain unchanged in every respect, including sign. Whena CO2 molecule rotates through 180, two identical O nuclei are interchanged, so theoverall wavefunction of the molecule must remain unchanged. However, inspectionof the form of the rotational wavefunctions (which have the same form as the s, p, etc.orbitals of atoms) shows that they change sign by (1) J under such a rotation (Fig. 12.19). Therefore, only even values of J are permissible for CO2, and hence theRaman spectrum shows only alternate lines.

The selective occupation of rotational states that stems from the Pauli principle istermed nuclear statistics. Nuclear statistics must be taken into account whenever arotation interchanges equivalent nuclei. However, the consequences are not always assimple as for CO2 because there are complicating features when the nuclei havenonzero spin: there may be several different relative nuclear spin orientations consis-tent with even values of J and a different number of spin orientations consistent withodd values of J. For molecular hydrogen and uorine, for instance, with their twoidentical spin- nuclei, we show in the following Justication that there are three timesas many ways of achieving a state with odd J than with even J, and there is a corres-ponding 3:1 alternation in intensity in their rotational Raman spectra (Fig. 12.20). Ingeneral, for a homonuclear diatomic molecule with nuclei of spin I, the numbers ofways of achieving states of odd and even J are in the ratio

=

(12.26)

For hydrogen, I = , and the ratio is 3:1. For N2, with I = 1, the ratio is 1:2.

Justication 12.1 The effect of nuclear statistics on rotational spectra

Hydrogen nuclei are fermions, so the Pauli principle requires the overall wavefunc-tion to change sign under particle interchange. However, the rotation of an H2molecule through 180 has a more complicated effect than merely relabelling thenuclei, because it interchanges their spin states too if the nuclear spins are paired(; Itotal = 0) but not if they are parallel (, Itotal = 1).

First, consider the case when the spins are parallel and their state is (A)(B),(A)(B) + (B)(A), or (A)(B). The (A)(B) and (A)(B) combinations are unchanged when the molecule rotates through 180 so the rotational wavefunc-tion must change sign to achieve an overall change of sign. Hence, only odd valuesof J are allowed. Although at rst sight the spins must be interchanged in the com-bination (A)(B) + (B)(A) so as to achieve a simple A B interchange of labels(Fig. 12.21), (A)(B) + (B)(A) is the same as (A)(B) + (B)(A) apart fromthe order of terms, so only odd values of J are allowed for it too. In contrast, if the nuclear spins are paired, their wavefunction is (A)(B) (B)(A). This combination changes sign when and are exchanged (in order to achieve a simple A B interchange overall). Therefore, for the overall wavefunction tochange sign in this case requires the rotational wavefunction not to change sign.Hence, only even values of J are allowed if the nuclear spins are paired. In accordwith the prediction of eqn 12.26, there are three ways of achieving odd J but only oneof achieving even J.

12

(I + 1)/I for half-integral spin nuclei

I/(I + 1) for integral spin nuclei

123Number of ways of achieving odd J

Number of ways of achieving even J

12

J = 2

J = 1

J = 0

+

+

+

+

Fig. 12.19 The symmetries of rotationalwavefunctions (shown here, for simplicityas a two-dimensional rotor) under arotation through 180. Wavefunctions withJ even do not change sign; those with J odddo change sign.

Frequency

Fig. 12.20 The rotational Raman spectrumof a diatomic molecule with two identicalspin- nuclei shows an alternation inintensity as a result of nuclear statistics.The Rayleigh line is much stronger thandepicted in the gure; it is shown as aweaker line to improve visualization of the Raman lines.

12

462 12 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

Different relative nuclear spin orientations change into one another only veryslowly, so an H2 molecule with parallel nuclear spins remains distinct from one withpaired nuclear spins for long periods. The two forms of hydrogen can be separated byphysical techniques, and stored. The form with parallel nuclear spins is called ortho-hydrogen and the form with paired nuclear spins is called para-hydrogen. Becauseortho-hydrogen cannot exist in a state with J = 0, it continues to rotate at very lowtemperatures and has an effective rotational zero-point energy (Fig. 12.22). This energy is of some concern to manufacturers of liquid hydrogen, for the slow conver-sion of ortho-hydrogen into para-hydrogen (which can exist with J = 0) as nuclearspins slowly realign releases rotational energy, which vaporizes the liquid. Techniquesare used to accelerate the conversion of ortho-hydrogen to para-hydrogen to avoidthis problem. One such technique is to pass hydrogen over a metal surface: themolecules adsorb on the surface as atoms, which then recombine in the lower energypara-hydrogen form.

The vibrations of diatomic molecules

In this section, we adopt the same strategy of nding expressions for the energy levels,establishing the selection rules, and then discussing the form of the spectrum. Weshall also see how the simultaneous excitation of rotation modies the appearance ofa vibrational spectrum.

12.8 Molecular vibrations

Key point The vibrational energy levels of a diatomic molecule modelled as a harmonic oscillatordepend on a force constant kf (a measure of the bonds stiffness) and the molecules effective mass.

We base our discussion on Fig. 12.23, which shows a typical potential energy curve (asin Fig. 10.1) of a diatomic molecule. In regions close to Re (at the minimum of thecurve) the potential energy can be approximated by a parabola, so we can write

V = kf x2 x = R Re (12.27)

where kf is the force constant of the bond. The steeper the walls of the potential (thestiffer the bond), the greater the force constant.

To see the connection between the shape of the molecular potential energy curveand the value of kf , note that we can expand the potential energy around its minimumby using a Taylor series, which is a common way of expressing how a function variesnear a selected point (in this case, the minimum of the curve at x = 0):

V(x) =V(0) +0

x + 0

x2 + (12.28)

The notation (. . .)0 means that the derivatives are rst evaluated and then x is setequal to 0. The term V(0) can be set arbitrarily to zero. The rst derivative of V iszero at the minimum. Therefore, the rst surviving term is proportional to the squareof the displacement. For small displacements we can ignore all the higher terms, andso write

V(x) 0

x2 (12.29)DEF

d2V

dx2ABC

12

DEFd2V

dx2ABC

12

DEFdV

dx

ABC

Parabolicpotential energy

12

A

A

B

B

AB

(1)J

Changesign ifantiparallel

Ch

ang

e si

gn

Rotateby 180

Fig. 12.21 The interchange of two identicalfermion nuclei results in the change in signof the overall wavefunction. The relabellingcan be thought of as occurring in two steps:the rst is a rotation of the molecule; thesecond is the interchange of unlike spins(represented by the different colours of thenuclei). The wavefunction changes sign inthe second step if the nuclei haveantiparallel spins.

J = 1

J = 0

Lowest rotational stateof ortho-hydrogen

Lowest rotational stateof para-hydrogen

Thermalrelaxation

Fig. 12.22 When hydrogen is cooled, themolecules with parallel nuclear spinsaccumulate in their lowest availablerotational state, the one with J = 1.They can enter the lowest rotational state (J = 0) only if the spins change their relativeorientation and become antiparallel. This is a slow process under normalcircumstances, so energy is slowly released.