‫االطياف – المحاضرة السادسة – االنتقاالت الدورانية‬
‫د محمد هاشم مطلوب‬
Rotational transitions
Typical values of B for small molecules are in the region of 0.1 to 10 cm - I (for ex ample, 0.356 cm- I for NF 3 and 10.59 cm-I for HCl), so rotational transitions lie in
the microwave region of the spectrum. The transitions are detected by monitoring
the net absorption of microwave radiation. Modulation of the transmitted intensity
can be achieved by varying the energy levels with an oscillating electric field. In
this Stark modulation, an electric field of about 105 V m-1 and a frequency of
between 10 and 100 kHz is applied to the sample.
We have already remarked that the gross selection rule for the observation of a
pure rotational spectrum is that a molecule must have a permanent electric
dipole moment. That is, for a molecule to give a pure rotational spectrum, it must
he polar. The classical basis of this rule is that a polar molecule appears to
possess a fluctuating dipole when rotating, but a nonpolar molecule does not
(Fig.1).
The permanent dipole can be regarded as a handle with which the molecule stirs
the electromagnetic field into oscillation (and vice versa for absorption).
Homonuclear diatomic molecules and symmetrical linear molecules such as CO 2
are rotationally inactive. Spherical rotors cannot have electric dipole moments
unless they become distorted by rotation, so they are also inactive except in
special cases. An example of a spherical rotor that does become sufficiently
distorted for it to acquire a dipole moment is SiH4, which has a dipole moment of
about 8.3 μD by virtue of its rotation when J=10 (for comparison, HCl has a
permanent dipole moment of 1.1μD.
The pure rotational spectrum of SiH4 has been detected by using long path
lengths (10 m) through high pressure (4 atm) samples.
Flg.1 To a stationary observer, a rotating polar molecule looks like an
oscillating dipole that can stir the electromagnetic field into oscillation (and
vice versa for absorption). This picture is the classical origin of the gross
selection rule for rotational transitions.
Illustration
Identifying rotationally active molecules
Of the molecules N2 , CO 2 , OCS, H20, CH2 =CH2 , C6H6 , only OCS and H20 are
polar, so only these two molecules have microwave spectra.
Test Which of the molecules H2, NO, N20, CH4 can have a pure rotational
spectrum?
Answer: [NO, N2O].
The specific rotational selection rules are found by evaluating the transition
dipole moment between rotational states. For a linear molecule, the transition
moment vanishes unless the following conditions are fulfilled:
MJ = ± 1
ΔMJ = 0, ± 1
The transition ΔJ = +1 corresponds to absorption and the transition ΔJ = -1
corresponds to emission. The allowed change in J in each case arises from the
conservation of angular momentum when a photon, a spin-l particle, is emitted or
absorbed (Fig. 2).
Fig. 2 When a photon is absorbed by a molecule, the angular momentum
of the combined system is conserved. If the molecule is rotating in the
same sense as the spin of the incoming photon, then J increases by 1.
When the transition moment is evaluated for all possible relative orientations of
the molecule to the line of flight of the photon, it is found that the total J + I ↔ J
transition intensity is proportional to
where μ0 is the permanent electric dipole moment of the molecule. The intensity
is proportional to the square of the permanent electric dipole moment, so strongly
polar molecules give rise to much more intense rotational lines than less polar
molecules.
For symmetric rotors, an additional selection rule states that ΔK = 0. To
understand this rule, consider the symmetric rotor NH3 , where the electric dipole
moment lies parallel to the figure axis. Such a molecule cannot be accelerated
into different states of rotation around the figure axis by the absorption of
radiation, so ΔK = 0.When these selection rules are applied to the expressions
for the energy levels of a rigid symmetric or linear rotor, it follows that the
wavenumbers of the allowed J + 1 ← J absorptions are
When centrifugal distortion is taken into account, the corresponding expression is
However, because the second term is typically very small compared with the first,
the appearance of the spectrum closely resembles that predicted from the former
equation.
Example: Predicting the appearance of a rotational spectrum
Predict the form of the rotational spectrum of 14NH3
Method: The molecule 14NH3 is a polar symmetric rotor, so the selection rules
ΔJ = ± 1 and ΔK = 0 apply. For absorption, ΔJ=+1. Because B = 9.977 cm-I, we
can draw up the following table for the J + 1 ← J transitions.
Self-test: Repeat the problem for C35ClH3.
[Lines of separation 0.888 cm- 1 (26.6 GHz)]
The form of the spectrum is shown in Fig. 3. The most significant feature is that it
consists of a series of lines with wavenumbers 2B, 4B, 6B, . . . and of separation
2B. The measurement of the line spacing gives B, 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 of a
diatomic molecule.
Fig. 3 The rotational energy levels of a linear rotor, the transitions allowed
by the selection rule ΔJ = ± 1, and a typical pure rotational absorption
spectrum (displayed here in terms of the radiation transmitted through the
sample). The intensities reflect the populations of the initial level in each
case and the strengths of the transition dipole moments.
However, in the case of a polyatomic molecule such as OCS or NH3 , the
analysis gives only a single quantity, IL and it is not possible to infer both bond
lengths (in OCS) or the bond length and bond angle (in NH 3). This difficulty
can be overcome by using isotopically substituted molecules, such as ABC and
A'BC; then, by assuming that R(A-B) =R(A'-B), both A-B and B-C bond lengths
can be extracted from the two moments of inertia. A famous example of this
procedure is the study of OCS. The assumption that bond lengths are unchanged
by isotopic substitution is only an approximation, but it is a good approximation in
most cases.
The intensities of spectral lines increase with increasing J and pass through a
maximum before tailing off as J becomes large. The most important reason for
the maximum in intensity is the existence of a maximum in the population of
rotational levels.
The Boltzmann distribution implies that the population of each state decays
exponentially with increasing J, but the degeneracy of the levels increases.
Specifically, the population of a rotational energy level J is given by the
Boltzmann expression
where N is the total number of molecules and gJ is the degeneracy of the level J.
The value of J corresponding to a maximum of this expression is found by
treating J as a continuous variable, differentiating with respect to J, and then
setting the result equal to zero. The result is
For a typical molecule (for example, OCS, with B = 0.2 cm- I) at room
temperature, k T = 1000 h c B, so Jrnax = 30. However, it must be recalled that the
intensity of each transition also depends on the value of J and on the population
difference between the two states involved in the transition . Hence the
value of J corresponding to the most intense line is not quite the same as the
value of J for the most highly populated level.