‫االطياف – المرحلة الرابعة – المحاضرة السابعة – اطياف رامان الدورانية‬
‫د محمد هاشم مطلوب‬
Rotational Raman spectra
The gross selection rule for rotational Raman transitions is that the molecule
must be anisotropically polarizable. We begin by explaining what this means.
The distortion of a molecule in an electric field is determined by its polarizability
α.
More precisely, if the strength of the field is 'E, then the molecule
acquires an induced dipole moment of magnitude
μ=αE
in addition to any permanent dipole moment it may have. An atom is isotropically
polarizable. That is, the same distortion is induced whatever the direction of the
applied field. The polarizability of a spherical rotor is also isotropic. However,
non- spherical rotors have polarizabilities that do depend on the direction of the
field rela- tive to the molecule, so these molecules are anisotropically polarizable
(Fig. 1). The electron distribution in H2 for example, is more distorted when the
field is applied parallel to the bond than when it is applied perpendicular to it, and
we write αII > αL
All linear molecules and diatomics (whether homonuclear or heteronuclear) have
anisotropic polarizabilities, and so are rotationally Raman active. This activity is
one reason for the importance of rotational Raman spectroscopy, for the
technique can be used to study many of the molecules that are inaccessible to
microwave spectroscopy.
Fig.1 An electric field applied to a molecule results in its distortion, and the
distorted molecule acquires a contribution to its dipole moment (even if it
is nonpolar initially). The polarizability may be different when the field 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 the
molecule has an anisotropic polarizability.
Spherical rotors such as CH4 and SF6 however, are rotation ally Raman inactive
as well as microwave inactive. This inactivity does not mean that such molecules
are never found in rotation ally excited states. Molecular collisions do not have to
obey such restrictive selection rules, and hence collisions between molecules
can result in the population of any rotational state.
The specific rotational Raman selection rules are
The ΔJ = 0 transitions do not lead to a shift of the scattered photon's frequency in
pure rotational Raman spectroscopy, and contribute to the unshifted Rayleigh
radiation. We can predict the form of the Raman spectrum of a linear rotor by
applying the selection rule ΔJ = ±2 to the rotational energy levels (Fig. 2±). When
the molecule makes a transition with ΔJ = +2, the scattered radiation leaves the
molecule in a higher rotational state, so the wavenumber of the incident radiation
is decreased. These transitions account for the Stokes lines in the spectrum:
The Stokes lines appear to low frequency of the incident radiation and at
displacements 6B, l0B, 14B,... from ṽi for J = 0, 1, 2,. .. . When the molecule
makes a transition with ΔJ = -2, the scattered photon emerges with increased
energy. These transitions account for the anti-Stokes lines of the spectrum:
The anti-Stokes lines occur at displacements of 6B, l0B, 14B, . . . (for J = 2,3,4, .
. . ; J = 2 is the lowest state that can contribute under the selection rule ΔJ = -2)
to high frequency of the incident radiation. The separation of adjacent lines in
both the Stokes and the anti-Stokes regions is 4B, so from its measurement IL
can be determined and then used to find the bond lengths exactly as in the case
of microwave spectroscopy.
Fig. 2 The rotational energy levels of a linear rotor and the transitions
allowed by the ΔJ = ± 2 Raman selection rules. The form of a typical
rotational Raman spectrum is also shown. The Rayleigh line is much
stronger than depicted in the figure; it is shown as a weaker line to improve
visualization of the Raman lines.
Example
Predicting the form of a Raman spectrum
Predict the form of the rotational Raman spectrum of
14N
2
, for which B =
1.99 cm- 1, when it is exposed to monochromatic 336.732 nm laser radiation.
Method The molecule is rotationally Raman active because end-over-end
rotation modulates its polarizability as viewed by a stationary observer. The
Stokes and anti-Stokes lines.
Answer
Because λi = 336.732 nm corresponds to ṽi = 29697.2 cm-l the above
equations give the following line positions:
Self-test Repeat the calculation for the rotational Raman spectrum of NH 3
(B = 9.977 cm- 1). [Stokes lines at 29 637.3, 29 597.4, 29 557.5, 29 517.6 cm- 1.
anti-Stokes lines at 29 757.1, 29 797.0 cm- l].