Raman Spectroscopy Questions and Answers

Answer: C

The rules are Δν = ±1 and ΔJ = 0, ±2 respectively.

Raman scattering involves two photons, so it follows the polarisability rather than the dipole moment. For a harmonic vibration the polarisability changes linearly with displacement, giving Δv = ±1 (Stokes and anti-Stokes lines). For rotation of a linear molecule the polarisability repeats twice per turn, giving ΔJ = 0, ±2; ΔJ = 0 is the unshifted Rayleigh line.

ΔJ = 0, ±1 is the rule for infrared rotation-vibration spectra, and including Δv = 0 in the vibrational rule only adds the Rayleigh line, not a vibrational Raman line.

Answer: D

The Raman shift is still 606 cm\(^{-1}\).

A Raman shift equals the energy of a molecular vibration (or rotation) expressed in wavenumbers. It is a property of the molecule, not of the exciting light, so changing the source from the mercury line to the He-Ne laser changes the absolute wavelengths of the Raman lines but not their shift from the exciting line.

The other values come from wrongly scaling the shift with the excitation wavelength or frequency. The constant given in the question (printed as n, meaning Planck's constant) is not needed.

Answer: C

The molecules are \(O_{2}\) and \(H_{2}\).

A pure rotational (microwave) spectrum needs a permanent dipole moment, and an infrared vibrational band needs a vibration that changes the dipole moment. Homonuclear diatomics such as \(O_{2}\) and \(H_{2}\) have no dipole and their stretching cannot create one, so they are inactive in both. Their polarizability is anisotropic, however, so they show rotational Raman spectra.

HCl, HCN, CO and \(H_{2}O\) are polar and show microwave and IR spectra. \(CO_{2}\) has no microwave spectrum but its bending and asymmetric stretch are IR active, and \(CH_{4}\), a spherical top, is IR active and shows no rotational Raman spectrum.