-
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.