Pure rotational Raman spectra

Pure rotational Raman spectra arise from the inelastic scattering of light by molecules undergoing transitions between discrete rotational energy levels, without any accompanying vibrational change. These spectra are an important tool for studying molecular structure and dynamics, providing information about moments of inertia, bond lengths and molecular symmetry, particularly for molecules that may be inactive in microwave spectroscopy.

An ideal pure rotational Raman spectrum is obtained by illuminating a sample with monochromatic visible or near-infrared laser radiation and analysing the frequency shifts of the scattered light. The spectrum consists of a series of lines symmetrically distributed about the incident (Rayleigh) line at $\tilde{\nu}_i$ , corresponding to rotational transitions between discrete rotational energy levels. For a rigid linear molecule in the absence of fine or hyperfine splitting, the allowed rotational Raman transitions follow the selection rule $\Delta J=\pm 2$ , reflecting the requirement that the molecular rotation be accompanied by a change in polarisability.

The rotational energy levels are given by eq44 or eq45 and the frequencies of the transitions are directly related to the rotational constant $B$ . As such, the spacing between the spectral lines provides information about the moment of inertia and the molecular structure. Using the definition $\tilde{\nu}=BJ(J+1)$ , the wavenumber of the Stokes and anti-Stokes lines are:

$\tilde{\nu}_s=\tilde{\nu}_i-B[(J+2)(J+3)-J(J+1)]=\tilde{\nu}_i-2B(2J+3)\;\;\;\;\;\;\;\;30$

and

$\tilde{\nu}_{as}=\tilde{\nu}_i+B[J(J+1)-(J-2)(J-1)]=\tilde{\nu}_i+2B(2J-1)\;\;\;\;\;\;\;\;31$

respectively.

Therefore, the Stokes lines appear at lower wavenumbers than $\tilde{\nu}_i$ , shifted by $6B,10B,14B,\cdots$ for $J=0,1,2,\cdots$ . The anti-Stokes lines appear at higher wavenumbers than $\tilde{\nu}_i$ , shifted by $6B,10B,14B,\cdots$ for $J=2,3,4,\cdots$ , since the lowest anti-Stokes transition is from $J=2$ to $J=0$ (see diagram above).

Question

Calculate the moment of inertia and bond length of HCl if the line spacing in the rotational Raman spectrum is 42.4 cm^-1?

Answer

In a pure rotational Raman spectrum of a linear molecule, the spacing between adjacent Stokes (or anti-Stokes) lines is $4B$ . Therefore, we have $B=10.6$ cm^-1. Using the formula $B=\frac{\hbar}{4\pi cI}$ gives $I=2.64\times 10^{-47}$ kg m². Furthermore, $I=\mu r^2=\biggr$\frac{35.5}{1+35.5}\biggr$\frac{1\times 10^{-3}}{N_A}r^2$ . Therefore, the bond length is $r=1.28\times 10^{-10}$ m.

The intensity of each transition depends partly on the population of the initial level $J$ , which is governed by the Boltzmann distribution, with higher- $J$ states being less populated at lower temperatures:

$\frac{n_J}{N}=\frac{e^{-\frac{\varepsilon_J}{kT}}}{\sum_Je^{-\frac{\varepsilon_J}{kT}}}$

where

$k$ is the Boltzmann constant.
$n_J$ is the number of particles in the energy state $\varepsilon_J$ .
$N$ is the total number of particles in the system.

Since $\frac{n_j}{N}$ represents the fraction of particles in the state $\varepsilon_J$ , the intensity of a spectral line is proportional to $e^{-\frac{\varepsilon_J}{kT}}$ , where the value $\varepsilon_J$ corresponds to a specific energy level. However, there are $(2J+1)$ degenerate states associated with each value of $J$ in a rigid linear molecule. This degeneracy increases the statistical weight of higher $J$ levels, redistributing the population towards more degenerate states, which become thermally accessible at typical laboratory temperatures. In other words, at a given temperature, more particles will occupy a higher $J$ state with greater degeneracy at equilibrium than they would if the state were non-degenerate. Therefore, when a sample is exposed to a laser radiation, the intensity $I$ of each allowed transition in the pure rotational spectrum is proportional to the population of the initial state, which is the product of $e^{-\frac{\varepsilon_J}{kT}}$ and $(2J+1)$ . Because the initial $J$ state of an anti-Stokes process is higher than the final state, the intensities of anti-Stokes lines are typically lower than those of Stokes lines.

Other than the population of the initial level, the transition intensity is also dependent on the fourth power of the scattered light frequency $\nu^4$ . This, together with the Boltzmann distribution, gives:

$I=N'(2J+1)\nu^4e^{-\frac{hcBJ(J+1)}{kT}}\;\;\;\;\;\;\;\;32$

where $N'$ is a proportionality constant and $\varepsilon_J$ is given by eq45.

Question

Explain why the transition intensity is also dependent on the fourth power of the scattered light frequency $\nu^4$ .

Answer

Consider an electric dipole oscillating at frequency $\nu$ with a displacement $x=x_0cos(2\pi\nu t)$ . In the “far-field” (away from the molecule), the strength of the radiating electric field $E$ is directly proportional to the acceleration of the charge $E\propto\frac{d^2x}{dt^2}\propto\nu^2$ . Since the intensity (or power) of an electromagnetic wave is proportional to the square of its electric field, $I\propto\nu^4$ .

The factor $(2J+1)$ in eq32 increases linearly with $J$ , while the exponential term decreases exponentially. In pure microwave rotational spectroscopy, the transition intensity is $I=N'(2J+1)e^{-\frac{hcBJ(J+1)}{kT}}$ (excluding the factor $\nu^4$ ) with the maximum intensity occurring at the value of $J$ closest to $J=\sqrt{\frac{kT}{2hcB}}-\frac{1}{2}$ . For Raman spectroscopy, $\nu=\tilde{\nu}_i-2B(2J+3)$ decreases as $J$ increases for Stokes lines, while $\nu=\tilde{\nu}_i+2B(2J-1)$ increases as $J$ increases for anti-Stokes lines. However, the change in $\nu$ over the visible region of the spectrum is very small (only about 1%) for a typical Raman experiment, because the laser frequency $\tilde{\nu}_i$ is around 20,000 cm^-1 and $B$ is about 1-10 cm^-1. Therefore, the derivative $\frac{dI}{dJ}=0$ using eq32 still results in the maximum intensity on each side of the Rayleigh line at

$J\approx\sqrt{\frac{kT}{2hcB}}-\frac{1}{2}$

For a symmetric rotor, $\tilde{\nu}=\tilde{B}J(J+1)+K^2(\tilde{A}-\tilde{B})$ . In a pure rotational Raman spectrum, the selection rule $\Delta K=0$ means that the $K^2(\tilde{A}-\tilde{B})$ term cancels out when taking differences. As a result, the Raman shifts are identical to those of a linear rotor and form a single series of equally spaced lines with spacing $4B$ . However, unlike a linear rotor, each allowed transition receives contributions from multiple $K$ states. These overlapping contributions increase the intensity of the Raman lines but do not change their positions for an ideal spectrum. Consequently, the overall intensity envelope of the spectrum on each side of the Rayleigh line still resembles a skewed bell curve, with a maximum at the value of $J$ corresponding to the highest population.

In practice, the peaks in a real rotational Raman spectrum are not discrete vertical lines but have finite width and shape, appearing as broadened curves. This broadening arises, under typical laboratory conditions, from several effects:

- Doppler broadening, due to the thermal motion of molecules, causes a spread in observed frequencies as molecules move towards or away from the detector.
- Instrumental broadening, resulting from the finite resolution of the spectrometer itself.
- Lifetime broadening, which is based on the energy-time uncertainty relation.

Beyond broadening, other phenomena affect the detailed structure and spacing of the spectral lines:

- Centrifugal distortion causes deviations from the ideal rigid rotor model. As rotational speed increases at higher $J$ levels, the molecular bond stretches slightly, increasing the moment of inertia and decreasing the rotational constant $B$ . This leads to uneven spacing between lines, especially at higher $J$ , and is accounted for using a correction term.

- Isotopic substitution alters the moment of inertia due to changes in atomic mass, thereby changing the rotational constant $B$ . Different isotopologues of the same molecule produce distinct sets of rotational lines, each with slightly different spacings. If multiple isotopes are present in the sample, the resulting spectrum may show clusters or duplications of lines, corresponding to each isotopologue.

Together, these effects lead to a rotational Raman spectrum that is richer and more complex than the idealised, evenly spaced series of lines predicted by the rigid rotor model.

Pure rotational Raman spectra

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