Raman Scattering
The Raman effect: Rayleigh, Stokes and anti-Stokes lines, and the fixed ordering of their energies, wavelengths and intensities.
The classical polarizability theory — what it explains, and the one thing it cannot.
The quantum picture via virtual states, which fixes the intensity ordering.
Vibrational and rotational Raman spectra, their selection rules (\(\Delta v = \pm1\), \(\Delta J = \pm2\)) and the \(6B\), \(4B\) pattern.
The rule of mutual exclusion, and nuclear-spin intensity alternation.
Magnetic resonance: NMR and ESR, and why one is a radio-frequency technique and the other a microwave one.
9.1 Introduction
When a strong beam of monochromatic ultraviolet or visible light of frequency \(\nu\) falls on a transparent solid, a dust-free liquid or a gas, the scattered light contains not only the incident frequency but also a set of weak shifted frequencies. This is the Raman effect, discovered by C. V. Raman in 1928 and the subject of the 1930 Nobel Prize in Physics.
The scattered spectrum contains three kinds of line.
Rayleigh line. Scattering with no change in frequency, at \(\nu\). This is elastic scattering and is by far the strongest component.
Stokes lines. Lines at \(\nu - \nu_m\), shifted to lower frequency. The molecule has absorbed energy \(h\nu_m\) from the photon.
Anti-Stokes lines. Lines at \(\nu + \nu_m\), shifted to higher frequency. The molecule has given energy \(h\nu_m\) to the photon.
Here \(\nu_m\) is a molecular frequency — a vibrational or rotational frequency of the scatterer. It is the quantity the experiment measures; the incident frequency merely acts as a carrier.
\[ \text{Energy:}\quad E_{AS} > E_{R} > E_{S}, \qquad \text{Wavelength:}\quad \lambda_{S} > \lambda_{R} > \lambda_{AS}, \qquad \text{Intensity:}\quad I_{R} > I_{S} > I_{AS}. \]
The intensity ordering \(I_R > I_S > I_{AS}\) has been asked verbatim (NET Dec 2019). The reason for \(I_S > I_{AS}\) is population, not selection rules: producing an anti-Stokes photon requires the molecule to start in an excited vibrational or rotational level, and at room temperature very few molecules are there.
9.2 Classical Theory of Raman Scattering
A molecule is an assembly of positively charged nuclei embedded in a cloud of negative charge. In a static electric field \(E\) the centres of positive and negative charge are pulled towards opposite poles of the field, so the molecule acquires an induced dipole moment — a polarization \(P\) — proportional to the field: \begin{equation} P \propto E, \qquad\text{that is}\qquad P = \alpha E , \label{eq:polarization} \end{equation} where \(\alpha\) is the polarizability of the molecule. A large \(\alpha\) means the molecule is easily distorted; a small \(\alpha\) means it is stiff.
In a light wave the field oscillates, \begin{equation} E = E_{0}\sin 2\pi\nu t , \label{eq:incident-field} \end{equation} with \(\nu\) the frequency of the incident radiation, so from \(\eqref{eq:polarization}\) \begin{equation} P = \alpha E_{0}\sin 2\pi\nu t . \label{eq:P-simple} \end{equation}
If \(\alpha\) were a constant this would be the whole story: the induced dipole would oscillate at \(\nu\), radiate at \(\nu\), and only a Rayleigh line would appear. But a molecule vibrates and rotates, and both motions change the shape of the electron cloud. The polarizability is therefore itself modulated at the molecular frequency \(\nu_m\): \begin{equation} \alpha = \alpha_{0} + \beta\sin 2\pi\nu_{m}t , \label{eq:alpha-modulated} \end{equation} where \(\beta\) measures how much the polarizability changes during the motion.
Substituting \(\eqref{eq:alpha-modulated}\) into \(\eqref{eq:P-simple}\), \begin{align} P &= \left(\alpha_{0} + \beta\sin 2\pi\nu_{m}t\right) E_{0}\sin 2\pi\nu t \notag\\ &= \alpha_{0}E_{0}\sin 2\pi\nu t + \beta E_{0}\sin 2\pi\nu_{m}t\,\sin 2\pi\nu t . \label{eq:P-expanded} \end{align} Using the product formula \begin{equation} 2\sin A\,\sin B = \cos(A-B) - \cos(A+B), \label{eq:trig} \end{equation} the second term splits into two, and \begin{equation} \boxed{\; P = \underbrace{\alpha_{0}E_{0}\sin 2\pi\nu t}_{\text{Rayleigh}} + \frac{\beta E_{0}}{2} \Big[\underbrace{\cos 2\pi(\nu-\nu_{m})t}_{\text{Stokes}} - \underbrace{\cos 2\pi(\nu+\nu_{m})t}_{\text{anti-Stokes}}\Big] . \;} \label{eq:P-final} \end{equation}
The manuscript quotes the identity as \(2\sin A\sin B = \sin(A+B) + \sin(A-B)\), which is not correct — the product of two sines gives cosines. The working that follows it uses the right result, so only the quoted line needed fixing.
An oscillating dipole radiates at its own frequency of oscillation. Equation \(\eqref{eq:P-final}\) therefore predicts scattered light at three frequencies: \(\nu\) (Rayleigh) and \(\nu \pm \nu_{m}\) (Raman). It also gives the gross selection rule directly: if the motion does not change the polarizability then \(\beta = 0\), the bracket vanishes, and there is no Raman line at all.
A vibration or rotation is Raman active only if it produces a change in the polarizability of the molecule.
Contrast this with infrared activity (Chapter 8), which requires a change in the dipole moment. This difference is the whole reason Raman spectroscopy is worth doing.
Failure of the classical theory
Classical theory predicts the Stokes and anti-Stokes terms in \(\eqref{eq:P-final}\) with equal amplitudes \(\beta E_0/2\). Experiment says otherwise: \(I_S\) is always larger than \(I_{AS}\), and the ratio depends on temperature. Explaining the relative intensities requires the quantum treatment.
9.3 Quantum Theory of Raman Scattering
Let \(E_{1}\) and \(E_{2}\) be two energy levels of the molecule in its ground electronic state, with \(E_{1} < E_{2}\). By the Boltzmann distribution, at room temperature (\(\approx300\,\mathrm{K}\)) almost all molecules sit in \(E_{1}\) and very few in \(E_{2}\): \begin{equation} \frac{N_{2}}{N_{1}} = \frac{g_2}{g_1}\,e^{-(E_{2}-E_{1})/k_{B}T} \ll 1 . \label{eq:boltz-raman} \end{equation}
An incident photon \(h\nu\) lifts the molecule to a short-lived virtual state — not a real stationary state of the molecule, which is why Raman scattering works at any incident wavelength. Call the virtual states reached from \(E_1\) and \(E_2\) by \(E_1'\) and \(E_2'\). Their lifetime is extremely short (\(\sime-14\,\mathrm{s}\)), and the molecule immediately falls back down.
| Process | Transition | Scattered wavenumber |
|---|---|---|
| Rayleigh | \(E_1' \to E_1\) and \(E_2' \to E_2\) | \(\nu/c\) |
| Stokes | \(E_1' \to E_2\) | \((\nu-\nu_m)/c\) |
| Anti-Stokes | \(E_2' \to E_1\) | \((\nu+\nu_m)/c\) |
The intensity ratio follows immediately from the populations of the two starting levels: \begin{equation} \frac{I_{S}}{I_{AS}} = e^{\,h\nu_m/k_{B}T} = e^{\,hc\bar\nu_m/k_{B}T} . \label{eq:stokes-ratio} \end{equation} Since the exponent is positive, \(I_S > I_{AS}\) always. Raising the temperature populates \(E_2\) and so strengthens the anti-Stokes lines — which is the basis of Raman thermometry.
Equation \(\eqref{eq:stokes-ratio}\) is a standing favourite: given a Stokes / anti-Stokes intensity ratio and a Raman shift, find the temperature. The useful conversion is \(1\,\mathrm{cm}^{-1} \equiv 1.44\,\mathrm{K}\), so \[ T = \frac{1.44\,\bar\nu_m\,[\mathrm{cm}^{-1}]}{\ln(I_S/I_{AS})}\ \mathrm{K} . \]
9.4 Rotational Raman Spectra
Homonuclear molecules such as H2, N2 and O2 have no permanent dipole moment and therefore show no pure rotational (microwave) spectrum at all. They do, however, have a polarizability that changes as the molecule tumbles — an end-on view of H2 is not the same as a side-on view — so they give a pure rotational Raman spectrum. This is exactly the gap that Raman spectroscopy fills.
For a rigid rotator (Chapter 7), \begin{equation} \varepsilon_{J} = BJ(J+1)\ \mathrm{cm}^{-1}, \qquad J = 0,1,2,\dots \label{eq:rigid-rot-8} \end{equation}
\[ \Delta J = 0, \pm 2 \] \[ \Delta J = 0 \ (\text{Rayleigh}), \qquad \Delta J = +2 \ (\text{Stokes},\ S\text{ branch}), \qquad \Delta J = -2 \ (\text{anti-Stokes},\ O\text{ branch}). \] The rule is \(\pm 2\), not \(\pm 1\), because the polarizability returns to its original value twice per revolution.
The line positions follow from \(\eqref{eq:rigid-rot-8}\): \begin{align} (\bar\nu_r)_{\text{Raman}} &= (\bar\nu)_{\text{Rayleigh}} \pm \{\varepsilon_{J+2}-\varepsilon_{J}\} \notag\\ &= (\bar\nu)_{\text{Rayleigh}} \pm \{B(J+2)(J+3) - BJ(J+1)\} \notag\\ &= (\bar\nu)_{\text{Rayleigh}} \pm 4B\left(J+\tfrac32\right). \label{eq:rot-raman} \end{align} Hence \begin{align} (\bar\nu_r)_{\text{Raman}} &= (\bar\nu)_{\text{Rayleigh}} + 4B(J+\tfrac32) && (J \to J+2;\ \text{Stokes side of the shift}),\\ (\bar\nu_r)_{\text{Raman}} &= (\bar\nu)_{\text{Rayleigh}} - 4B(J+\tfrac32) && (J+2 \to J;\ \text{anti-Stokes side}). \end{align}
\[ \left|\Delta\bar\nu\right| = 4B\left(J+\tfrac32\right) = 6B,\ 10B,\ 14B,\ 18B,\dots \qquad J = 0,1,2,\dots \] First line at \(6B\) from the Rayleigh line; successive lines \(4B\) apart.
Three numbers do most of the work in exam questions:
first Raman line lies at \(6B\) from the exciting line;
successive Raman lines are \(4B\) apart;
the closest Stokes and anti-Stokes lines are separated by \(12B\).
Compare the microwave spectrum of the same molecule, where the first line is at \(2B\) and the spacing is \(2B\). Mixing up the two is the single commonest error in this topic.
The non-rigid rotator
Including centrifugal distortion, \(\varepsilon_J = B_0J(J+1) - D_0J^2(J+1)^2\), and the same \(\Delta J = \pm2\) rule gives \begin{align} \varepsilon_{J+2}-\varepsilon_{J} &= 2B_{0}(2J+3) - 4D_{0}(J^{2}+3J+3)(2J+3), \end{align} so the Raman lines lie at \begin{equation} (\bar\nu)_{\text{Raman}} = (\bar\nu)_{\text{Rayleigh}} \pm \Big[2B_{0}(2J+3) - 4D_{0}(J^{2}+3J+3)(2J+3)\Big]. \label{eq:nonrigid-raman} \end{equation} Since \(D_0 \sim 10^{-4}B_0\), the correction only matters at large \(J\), where it pulls the high-\(J\) lines slightly inwards.
9.5 Vibrational Raman Spectra
If the polarizability changes during a vibration, transitions between vibrational states shift the scattered light to either side of the Rayleigh line.
\[ \Delta v = 0, \pm 1 \] \[ \Delta v = 0 \ \ (\text{Rayleigh}), \qquad \Delta v = +1 \ \ (\text{Stokes}), \qquad \Delta v = -1 \ \ (\text{anti-Stokes}). \]
Using the anharmonic-oscillator levels of Chapter 8, \(\varepsilon_v = \omega_e(v+\tfrac12) - \omega_e x_e(v+\tfrac12)^2\), the fundamental spacing is \(\varepsilon_{v+1}-\varepsilon_v = \omega_e(1-2x_e)\), so the Raman lines sit at \begin{align} (\bar\nu_v)_{\text{Raman}} &= (\bar\nu)_{\text{Rayleigh}} \pm \{\varepsilon_{v+1}-\varepsilon_v\} = (\bar\nu)_{\text{Rayleigh}} \pm \omega_e(1-2x_e), \label{eq:vib-raman} \end{align} that is \begin{align} (\bar\nu_v)_{\text{Raman}} &= (\bar\nu)_{\text{Rayleigh}} - \omega_e(1-2x_e) &&(v: 0 \to 1;\ \text{Stokes}),\\ (\bar\nu_v)_{\text{Raman}} &= (\bar\nu)_{\text{Rayleigh}} + \omega_e(1-2x_e) &&(v: 1 \to 0;\ \text{anti-Stokes}). \end{align}
\[ \Delta\bar\nu = \omega_e(1-2x_e) \] — numerically the same as the infrared fundamental band. The Raman shift and the IR fundamental of a molecule coincide when both are allowed.
The intensity of the anti-Stokes line grows with temperature, by \(\eqref{eq:stokes-ratio}\).
9.6 Rotational Fine Structure
The vibrational Raman lines of 9.5 are not in fact single. Just as an infrared band carries rotational structure (Chapter 8), so does a Raman band — though in practice the structure is rarely resolved except for diatomic molecules, where the moment of inertia is small and \(B\) correspondingly large.
Ignoring centrifugal distortion, the vibration–rotation levels are \begin{equation} \varepsilon_{v,J} = \omega_{e}\!\left(v+\tfrac12\right) - \omega_{e}x_{e}\!\left(v+\tfrac12\right)^{2} + BJ(J+1)\ \mathrm{cm}^{-1}, \qquad \begin{aligned} v &= 0,1,2,\dots\\ J &= 0,1,2,\dots \end{aligned} \label{eq:raman-vibrot-levels} \end{equation} For a diatomic molecule the Raman rotational selection rule is \(\Delta J = 0, \pm2\) — not \(\Delta J = \pm1\) as in the infrared, because Raman scattering is a two-photon process. Combining this with the vibrational change \(v = 0 \to 1\) gives three groups of lines, and it is conventional to label them by the value of \(\Delta J\):
Writing \(\bar\nu_{o} = \omega_{e}(1-2x_{e})\) for the vibrational shift, \[ \begin{aligned} \Delta J = 0 : &\quad \Delta\varepsilon_{Q} = \bar\nu_{o} &&\text{(for all } J)\qquad &&\textbf{Q branch}\\ \Delta J = +2 : &\quad \Delta\varepsilon_{S} = \bar\nu_{o} + B(4J+6) &&(J = 0,1,2,\dots) &&\textbf{S branch}\\ \Delta J = -2 : &\quad \Delta\varepsilon_{O} = \bar\nu_{o} - B(4J+6) &&(J = 2,3,4,\dots) &&\textbf{O branch} \end{aligned} \]
The Stokes lines — those to low frequency of the exciting radiation — then appear at \begin{align} \bar\nu_{Q} &= \bar\nu_{\text{ex}} - \Delta\varepsilon_{Q} = \bar\nu_{\text{ex}} - \bar\nu_{o} &&\text{(for all } J),\label{eq:raman-Q}\\ \bar\nu_{O} &= \bar\nu_{\text{ex}} - \Delta\varepsilon_{O} = \bar\nu_{\text{ex}} - \bar\nu_{o} + B(4J+6) &&(J = 2,3,4,\dots),\label{eq:raman-O}\\ \bar\nu_{S} &= \bar\nu_{\text{ex}} - \Delta\varepsilon_{S} = \bar\nu_{\text{ex}} - \bar\nu_{o} - B(4J+6) &&(J = 0,1,2,\dots).\label{eq:raman-S} \end{align}
Note the letters carefully, because they are easy to invert. The branch label follows \(\Delta J\) measured upward in \(J\): \(O\) for \(\Delta J = -2\), \(Q\) for \(0\), \(S\) for \(+2\). Compare the infrared case, where \(\Delta J = \pm1\) gives only \(P\) and \(R\) branches and the \(Q\) branch is absent for a diatomic. The Raman spectrum does show a strong central \(Q\) branch, and that difference is the quickest way to tell the two kinds of band apart:
| Infrared | Raman | |
|---|---|---|
| Rotational rule | \(\Delta J = \pm1\) | \(\Delta J = 0, \pm2\) |
| Branches | \(P\), \(R\) | \(O\), \(Q\), \(S\) |
| Central line | absent | present (\(Q\)) |
| Line spacing | \(2B\) | \(4B\) |
The spacing doubles because \(\Delta J\) changes in steps of two.
Since all the \(Q\)-branch lines fall at the same wavenumber whatever the value of \(J\), the \(Q\) branch appears as a single intense line at \(\bar\nu_{\text{ex}} - \bar\nu_{o}\), flanked on either side by the much weaker \(O\) and \(S\) lines. This is the characteristic appearance of a vibration–rotation Raman band.
9.7 Worked Examples
A laser operating at \(500\,\mathrm{nm}\) is used to excite a molecule. The Raman shift of the Stokes line is \(770\,\mathrm{cm}^{-1}\). Find the positions of the Stokes and anti-Stokes lines, and the vibrational frequency.
Solution. The exciting wavenumber is \[ \bar\nu_{0} = \frac{1}{\lambda_{0}} = \frac{1}{500\times10^{-7}\ \mathrm{cm}} = 20000\,\mathrm{cm}^{-1} . \] The Stokes line is displaced downwards by the shift: \[ \bar\nu_{S} = 20000 - 770 = 19230\,\mathrm{cm}^{-1} . \] The shift is the same on both sides, so \[ \bar\nu_{AS} = 20000 + 770 = 20770\,\mathrm{cm}^{-1} . \] In wavelength, \[ \lambda_{S} = \frac{1}{19230\ \mathrm{cm}^{-1}} = 520\,\mathrm{nm}, \qquad \lambda_{AS} = \frac{1}{20770\ \mathrm{cm}^{-1}} = 481.5\,\mathrm{nm} . \] The vibrational frequency of the molecule is \[ \nu_{m} = c\,\Delta\bar\nu = 3\times10^{10}\times770 = 2.31\times 10^{13}\,\mathrm{Hz} . \]
The source prints \(20000+770 = 20700\) (and then uses \(20770\) in the very next line), and gives \(770\times3\times10^{10} = 2.32\times10^{13}\) instead of \(2.31\times10^{13}\). Both are corrected above.
The Raman shift of the rotational lines of a molecule is \(\Delta\bar\nu = \pm(52J + 78)\ \mathrm{cm}^{-1}\). Find the rotational constant and the moment of inertia, and the bond length if the reduced mass is \(13.36\times10^{-24}\ \mathrm{g}\).
Solution. Compare with the standard form \[ \Delta\bar\nu = \pm 4B\left(J+\tfrac32\right) = \pm(4BJ + 6B). \] Matching coefficients, \(4B = 52\) and \(6B = 78\) — consistently \[ B = 13\,\mathrm{cm}^{-1} . \] Then, from \(B = h/8\pi^{2}Ic\), \[ I = \frac{h}{8\pi^{2}Bc} = \frac{6.62\times10^{-27}\ \mathrm{erg}\,\mathrm{s}} {8\times(3.14)^{2}\times 13\ \mathrm{cm}^{-1}\times 3\times10^{10}\ \mathrm{cm}\,\mathrm{s}^{-1}} = 2.1\times10^{-40}\ \mathrm{g}\,\mathrm{cm}^2 . \] Finally \(I = \mred r^{2}\) gives \[ r = \sqrt{\frac{I}{\mred}} = \sqrt{\frac{2.1\times10^{-40}}{13.36\times10^{-24}}} = 3.9\times10^{-9}\ \mathrm{cm} = 0.39\,\mathrm{Å}. \]
The question as printed reads \(\pm(52J+88)\) but the solution uses \(\pm(52J+78)\). Only \(78\) is consistent: \(4B = 52 \Rightarrow B = 13 \Rightarrow 6B = 78\). The solution then writes \(\pm48(J+\tfrac32)\), which should be \(\pm52(J+\tfrac32)\). The value of \(B\), and therefore the rest of the answer, is unaffected.
O2 is rotationally Raman active with \(B = 1.99\,\mathrm{cm}^{-1}\). It is exposed to monochromatic radiation of wavelength \(336.32\,\mathrm{nm}\). Find the wavenumber and wavelength of the first two Stokes and anti-Stokes lines.
Solution. The exciting wavenumber is \[ \bar\nu_{0} = \frac{10^{7}}{\lambda_{0}\,[\mathrm{nm}]} = \frac{10^{7}}{336.32} = 29733.6\,\mathrm{cm}^{-1} . \] Stokes lines lie at \(\bar\nu_{S} = \bar\nu_{0} - B(4J+6)\): \[ \begin{array}{llll} J=0: & \bar\nu_{S} = \bar\nu_{0} - 6B = 29733.6 - 11.94 = 29721.6\,\mathrm{cm}^{-1}, & \lambda_{S} = \dfrac{10^{7}}{29721.6} = 336.46\,\mathrm{nm},\\[6pt] J=1: & \bar\nu_{S} = \bar\nu_{0} - 10B = 29733.6 - 19.90 = 29713.7\,\mathrm{cm}^{-1}, & \lambda_{S} = \dfrac{10^{7}}{29713.7} = 336.55\,\mathrm{nm}. \end{array} \] Anti-Stokes lines lie at \(\bar\nu_{AS} = \bar\nu_{0} + B(4J+6)\): \[ \begin{array}{llll} J=0: & \bar\nu_{AS} = \bar\nu_{0} + 6B = 29733.6 + 11.94 = 29745.5\,\mathrm{cm}^{-1}, & \lambda_{AS} = \dfrac{10^{7}}{29745.5} = 336.19\,\mathrm{nm},\\[6pt] J=1: & \bar\nu_{AS} = \bar\nu_{0} + 10B = 29733.6 + 19.90 = 29753.5\,\mathrm{cm}^{-1}, & \lambda_{AS} = \dfrac{10^{7}}{29753.5} = 336.10\,\mathrm{nm}. \end{array} \] Note the whole pattern spans less than half a nanometre — rotational Raman work needs a high-resolution spectrograph.
The first anti-Stokes line is printed as \(29733.6 \times 1.99\); the intended expression is \(29733.6 + 6\times1.99\). The numerical answer quoted (\(29745.54\,\mathrm{cm}^{-1}\)) is the correct one.
When CCl4 is irradiated with the \(435.8\,\mathrm{nm}\) mercury line, Raman lines are obtained at \(439.9\,\mathrm{nm}\), \(444.6\,\mathrm{nm}\) and \(450.7\,\mathrm{nm}\). Calculate the Raman frequencies of CCl4 in wavenumbers.
Solution. The Raman shift is the difference of the two wavenumbers, \[ \Delta\tilde\nu = \frac{1}{\lambda_{\text{ex}}} - \frac{1}{\lambda_{\text{sc}}} , \] and with wavelengths in nanometres it is convenient to write \(\tilde\nu\,[\mathrm{cm}^{-1}] = 10^{7}/\lambda\,[\mathrm{nm}]\).
The exciting line is at \[ \tilde\nu_{\text{ex}} = \frac{10^{7}}{435.8} = 22946.3\,\mathrm{cm}^{-1} . \] Taking each scattered line in turn: \begin{align*} 439.9\,\mathrm{nm}: \quad \Delta\tilde\nu &= \frac{10^{7}}{435.8} - \frac{10^{7}}{439.9} = 22946.3 - 22732.4 = \boxed{214\,\mathrm{cm}^{-1}}\\[4pt] 444.6\,\mathrm{nm}: \quad \Delta\tilde\nu &= \frac{10^{7}}{435.8} - \frac{10^{7}}{444.6} = 22946.3 - 22492.1 = \boxed{454\,\mathrm{cm}^{-1}}\\[4pt] 450.7\,\mathrm{nm}: \quad \Delta\tilde\nu &= \frac{10^{7}}{435.8} - \frac{10^{7}}{450.7} = 22946.3 - 22187.7 = \boxed{759\,\mathrm{cm}^{-1}} \end{align*} All three shifts are positive, so all three are Stokes lines: the scattered wavelengths are longer than the exciting wavelength, meaning the molecule has gained energy.
Two traps. First, the shift must be computed from the reciprocals, not from the wavelength difference — \(\Delta\lambda\) is not proportional to \(\Delta\tilde\nu\) except over very small intervals. Second, note that \(\tilde\nu \propto 1/\lambda\) means a longer scattered wavelength is a smaller wavenumber, hence a positive shift and a Stokes line. The three values here are the fundamental vibrational wavenumbers of CCl4, and they are what the experiment is for: Raman spectroscopy reaches the vibrations of this highly symmetric molecule, several of which are infrared-inactive.
The first several Raman shifts of 14N2 are \(19.908\), \(27.857\), \(35.812\), \(43.762\), \(51.721\) and \(59.662\,\mathrm{cm}^{-1}\). These lines are due to pure rotational transitions with \(J = 1, 2, 3, 4, 5\) and \(6\). The reduced mass is \(1.162651\times 10^{-26}\,\mathrm{kg}\). What is the internuclear distance?
Solution. Step 1: the line spacing. From \(\eqref{eq:rot-raman}\) the rotational Raman lines lie at \(\Delta\tilde\nu = B(4J+6)\), so consecutive lines are separated by \(4B\). Taking successive differences: \[ \begin{aligned} 27.857 - 19.908 &= 7.949 &\qquad 43.762 - 35.812 &= 7.950\\ 35.812 - 27.857 &= 7.955 &\qquad 51.721 - 43.762 &= 7.959\\ && 59.662 - 51.721 &= 7.941 \end{aligned} \] The five differences average to \[ 4B = \frac{7.949 + 7.955 + 7.950 + 7.959 + 7.941}{5} = 7.951\,\mathrm{cm}^{-1} , \] so \[ B = \frac{7.951}{4} = 1.988\,\mathrm{cm}^{-1} = 198.78\,\mathrm{m}^{-1} . \]
Step 2: the moment of inertia. From \(B = h/8\pi^{2}Ic\), \[ I = \frac{h}{8\pi^{2}Bc} = \frac{6.626\times 10^{-34}\,\mathrm{J}\,\mathrm{s}} {8\pi^{2} \times 198.78\,\mathrm{m}^{-1} \times 3\times 10^{8}\,\mathrm{m}/\mathrm{s}} = 1.407\times 10^{-46}\,\mathrm{kg}\,\mathrm{m}^{2} . \]
Step 3: the bond length. With \(I = \mu r_{\text{eq}}^{2}\), \[ r_{\text{eq}} = \sqrt{\frac{I}{\mu}} = \sqrt{\frac{1.407\times 10^{-46}\,\mathrm{kg}\,\mathrm{m}^{2}}{1.162651\times 10^{-26}\,\mathrm{kg}}} = \sqrt{1.210\times 10^{-20}\,\mathrm{m}^{2}} = 1.100\times 10^{-10}\,\mathrm{m} , \] that is \[ \boxed{\ r_{\text{eq}} = 110\,\mathrm{pm} = 1.10\,\mathrm{Å}\ } \] which is the accepted N\(\equiv\)N bond length.
Averaging the differences rather than using a single pair is not fussiness: it is the standard way of extracting \(B\) from real data, and it also exposes bad points. Note too that this measurement is only possible by Raman scattering — 14N2 is homonuclear, has no permanent dipole moment, and therefore shows no pure rotational microwave spectrum at all (6.4). Its polarizability does change with orientation, so it is rotationally Raman active. This complementarity between infrared and Raman activity is the single most examined idea in the chapter.
9.8 Important Facts and Applications
The Raman effect determines the structure of diatomic and polyatomic molecules, together with bond lengths and force constants.
Besides polar molecules such as HCl, the homonuclear molecules H2, N2, O2 also show Raman spectra although they show no infrared spectra.
Rule of mutual exclusion. If a molecule has a centre of symmetry, then vibrations that are Raman active are infrared inactive, and vice versa. No mode is both.
Half-integer nuclear spin (e.g.\ H2, with \(I = \tfrac12\) for each proton): all rotational levels are present, but the \(J = \text{odd}\) levels carry the larger statistical weight. So \(J: 1\to3\) is more intense than \(J: 0\to2\), in the ratio \(3\!:\!1\).
Integer nuclear spin (e.g.\ N2, with \(I=1\) for each \(^{14}\)N): all levels are present but the \(J = \text{even}\) levels carry the larger weight. So \(J: 1\to3\) is less intense than \(J: 0\to2\).
Zero nuclear spin (e.g.\ \(^{16}\)O2, \(^{12}\)C2): alternate levels are missing altogether. For a \(\Sigma_g^+\) ground state only the even-\(J\) levels survive.
Point 6 is GATE 2023 in one line: for a homonuclear \(X_2\) with even-parity electronic ground state and \(I(X) = 0\), the nuclei are identical bosons, the nuclear spin function is necessarily symmetric, and so only even \(J\) levels exist. The rotational Raman spectrum shows lines from even \(J\) only.
| Molecule | Microwave (rotational) | Infrared | Raman |
|---|---|---|---|
| HCl, CO, HF | active | active | active |
| H2, N2, O2 | inactive | inactive | active |
| CO2 (sym.\ stretch) | inactive | inactive | active |
| CO2 (asym.\ stretch, bend) | inactive | active | inactive |
Applications. Raman spectra are used to obtain molecular parameters — reduced mass, moment of inertia, bond length, force constant, vibrational frequency — for exactly those molecules (homonuclear diatomics, spherical tops) that give no pure rotational or vibrational spectrum. They are also used for chemical fingerprinting, for measuring temperature via \(\eqref{eq:stokes-ratio}\), and, in the surface-enhanced form (SERS), for single-molecule detection.
Formula Summary
Gross selection rule (Raman) the motion must change the
polarizability \(\alpha\)
(infrared needs a changing dipole moment; the two are complementary)
Classical theory \(\alpha = \alpha_{0} + \beta\sin2\pi\nu_{m}t\) \[ P = \alpha_{0}E_{0}\sin2\pi\nu t + \frac{\beta E_{0}}{2}\left[\cos2\pi(\nu-\nu_{m})t - \cos2\pi(\nu+\nu_{m})t\right] \] \(\beta = 0 \Rightarrow\) no Raman line. Classical theory cannot give the relative intensities.
Orderings \(E_{AS} > E_{R} > E_{S}\); \ \(\lambda_{S} > \lambda_{R} > \lambda_{AS}\); \ \(I_{R} > I_{S} > I_{AS}\)
Intensity ratio \(\dfrac{I_{S}}{I_{AS}} = e^{\,hc\bar\nu_{m}/k_{B}T}\); \(T = \dfrac{1.44\,\bar\nu_{m}}{\ln(I_{S}/I_{AS})}\ \mathrm{K}\) with \(\bar\nu_m\) in \(\mathrm{cm}^{-1}\)
Vibrational Raman \(\Delta v = 0, \pm1\) \[ \Delta\bar\nu = \omega_{e}(1-2x_{e}) \qquad\text{(same as the infrared fundamental)} \]
Rotational Raman \(\Delta J = 0, \pm2\) \[ \left|\Delta\bar\nu\right| = 4B\left(J+\tfrac32\right) = 6B,\,10B,\,14B,\dots \] first line at \(6B\); spacing \(4B\); closest Stokes–anti-Stokes gap \(12B\)
Non-rigid rotator \(\Delta\bar\nu = 2B_{0}(2J+3) - 4D_{0}(J^{2}+3J+3)(2J+3)\)
Mutual exclusion centre of symmetry \(\Rightarrow\) Raman active modes are IR inactive, and vice versa
Nuclear spin statistics \(I\) half-integer (H2): odd \(J\) stronger, \(3\!:\!1\); \ \(I\) integer (N2): even \(J\) stronger; \ \(I = 0\) (\(^{16}\)O2): alternate lines missing
NMR \(\mu_{N} = e\hbar/2m_{p} = 5.051\times 10^{-27}\,\mathrm{J}/\mathrm{T}\) \[ E = -g_{N}\mu_{N}B\,m_{I}, \qquad \Delta E = g_{N}\mu_{N}B = h\nu, \qquad 2I+1 \text{ sublevels} \] radio frequency; \(\nu/B \approx 42.6\,\mathrm{MHz}/\mathrm{T}\) for a proton
ESR \(\mu_{B} = e\hbar/2m_{e} = 9.27\times 10^{-24}\,\mathrm{J}/\mathrm{T}\) \[ E = g_{e}\mu_{B}B\,m_{s}, \qquad \Delta E = g_{e}\mu_{B}B = h\nu \] microwave; \(\nu/B \approx 28\,\mathrm{GHz}/\mathrm{T}\); needs unpaired electrons