Vibrational Spectroscopy
The gross selection rule for infrared activity, and the normal modes of CO2.
The diatomic molecule as a simple harmonic oscillator: energy levels, zero-point energy, and the single fundamental line.
The anharmonic oscillator: the Morse potential, overtones, hot bands and dissociation energy.
The vibrating rotator: \(P\) and \(R\) branches, the band origin and the band head.
8.1 Infrared Activity
Vibrational transitions are observed in the infrared: \begin{equation} \nu = 3\times 10^{12}\text{--}3\times 10^{14}\,\mathrm{Hz}, \qquad \lambda = 100\text{--}1\,\mathrm{\mu m}, \qquad \Delta E \approx e4\,\mathrm{J}\,\mathrm{mol}^{-1} \approx e3\,\mathrm{cm}^{-1} . \label{eq:ir-region} \end{equation}
A vibrational mode is infrared active only if the vibration produces a change in the electric dipole moment of the molecule.
For a diatomic molecule this reduces to the same condition as in Chapter 7: the molecule must be heteronuclear.
| Vibrationally active | Vibrationally inactive |
|---|---|
| HCl, CO, HF, NO | H2, N2, O2, Cl2 |
Note the crucial difference from Chapter 7. Rotational activity needs a permanent dipole; vibrational activity needs a changing dipole. A molecule with no permanent dipole can still be IR active if a particular mode makes the dipole moment non-zero — which is exactly what happens in CO2.
8.2 Normal Modes of CO2
A linear molecule of \(N\) atoms has \(3N-5\) vibrational modes, so CO2 has \(3(3)-5 = 4\).
In the symmetric stretch both C=O bonds lengthen and shorten together, so the centres of positive and negative charge always coincide and the dipole moment stays zero throughout. In the asymmetric stretch one bond compresses while the other extends, so the dipole oscillates. The bending mode is doubly degenerate (it can occur in two perpendicular planes), which is why CO2 has four modes but only three distinct frequencies.
This is also why CO2 is a greenhouse gas despite having no permanent dipole moment.
8.3 The Diatomic Molecule as a Harmonic Oscillator
The bond between two atoms is not rigid — it behaves like a spring of force constant \(k\). Writing \(x = r - r_{e}\) for the displacement from equilibrium, \begin{equation} V(r) = \frac{k\left(r-r_{e}\right)^{2}}{2} = \frac{kx^{2}}{2} , \label{eq:sho-potential} \end{equation} so the Hamiltonian is \begin{equation} \hat H = -\frac{\hbar^{2}}{2\mred}\frac{\dd^{2}}{\dd x^{2}} + \frac{kx^{2}}{2} , \label{eq:sho-hamiltonian} \end{equation} with \(\mred\) the reduced mass. Solving the Schrödinger equation gives \begin{equation} \boxed{\; E_{v} = \frac{h}{2\pi}\sqrt{\frac{k}{\mred}}\left(v+\frac12\right) = h\nu_{\text{osc}}\left(v+\frac12\right)\;}, \qquad v = 0,1,2,3,\dots \label{eq:sho-levels-vib} \end{equation} where \(v\) is the vibrational quantum number and \begin{equation} \nu_{\text{osc}} = \frac{1}{2\pi}\sqrt{\frac{k}{\mred}} . \label{eq:osc-freq} \end{equation}
In wavenumbers, \begin{equation} \varepsilon_{v} = \frac{E_{v}}{hc} = \omega_{e}\left(v+\frac12\right) \ \mathrm{cm}^{-1} , \qquad \varepsilon_{v} = \frac{\omega_{e}}{2},\ \frac{3\omega_{e}}{2},\ \frac{5\omega_{e}}{2},\ \dots \label{eq:sho-wavenumber} \end{equation}
8.3.1 Zero-point energy
The lowest energy is not zero: \begin{equation} \varepsilon_{0} = \frac{\omega_{e}}{2} \ \mathrm{cm}^{-1} . \label{eq:zpe} \end{equation}
The vibrational energy of a diatomic molecule can never be zero — the atoms can never be at rest relative to one another, even at absolute zero. This follows from the uncertainty principle: if the particle were stationary, \(\Delta x = 0\) and hence \(\Delta p = \infty\), but a state with \(E=0\) cannot have infinite momentum uncertainty.
8.3.2 Selection rule and the fundamental band
\begin{equation} \Delta v = \pm 1 . \label{eq:sho-selection} \end{equation} For absorption (\(v \to v+1\)), \begin{equation} \Delta\varepsilon = \omega_{e}\left(v+1+\tfrac12\right) - \omega_{e}\left(v+\tfrac12\right) = \omega_{e} \ \mathrm{cm}^{-1} , \label{eq:sho-line} \end{equation} independent of \(v\). So a harmonic oscillator gives a single line at \(\omega_{e}\).
For a typical molecule \(\omega_{e} \approx 2171\,\mathrm{cm}^{-1}\), which corresponds to \[ \Delta E = \omega_{e}hc = 2171 \times 1.24\times 10^{-4}\,\mathrm{eV}\,\mathrm{cm} = 0.27\,\mathrm{eV} , \] squarely in the infrared. Hence vibrational spectroscopy is infrared spectroscopy.
8.3.3 Isotope effect
Since \(\nu_{\text{osc}} \propto 1/\sqrt{\mred}\) and the force constant \(k\) is unchanged by isotopic substitution (the electronic structure is the same), \begin{equation} \frac{\nu'_{\text{osc}}}{\nu_{\text{osc}}} = \sqrt{\frac{\mred}{\mred'}} < 1 \quad\Longrightarrow\quad \nu'_{\text{osc}} < \nu_{\text{osc}} . \label{eq:vib-isotope} \end{equation} The heavier isotopologue vibrates more slowly, so its fundamental band shifts to lower wavenumber — to the left, exactly as in Chapter 7.
The force constant of the CO molecule is \(1870\,\mathrm{N}\,\mathrm{m}^{-1}\). Calculate the vibrational frequency and the spacing between vibrational energy levels in eV. Take \(\mred = 1.14\times 10^{-26}\,\mathrm{kg}\).
Solution. From \(\eqref{eq:osc-freq}\), \[ \nu_{\text{osc}} = \frac{1}{2\pi}\sqrt{\frac{1870}{1.14\times10^{-26}}} = 6.45\times 10^{13}\,\mathrm{Hz} . \] The level spacing is \[ \Delta E = h\nu_{\text{osc}} = 6.62\times10^{-34}\times6.45\times10^{13} = 4.27\times 10^{-20}\,\mathrm{J} = \frac{4.27\times10^{-20}}{1.6\times10^{-19}} = 0.267\,\mathrm{eV} . \]
The source's data line read “\(1\,\mathrm{eV} = 1.60\times 10^{-26}\,\mathrm{erg}\)”. The correct conversion is \(1\,\mathrm{eV} = 1.60\times 10^{-12}\,\mathrm{erg}\); the worked arithmetic in fact uses the correct value.
The wavenumbers of the normal modes of CO2 are \(\nu_{1} = 1330\,\mathrm{cm}^{-1}\), \(\nu_{2} = 667\,\mathrm{cm}^{-1}\) and \(\nu_{3} = 2349\,\mathrm{cm}^{-1}\). Calculate the zero-point energy.
Solution. CO2 has four modes, the bending mode \(\nu_{2}\) being doubly degenerate, so it must be counted twice. Each mode contributes \(\tfrac12 hc\,\nu_{i}\): \[ E_{0} = \frac{hc}{2}\left(\nu_{1} + \nu_{2} + \nu_{2} + \nu_{3}\right) = \frac{6.62\times10^{-27}\times3\times10^{10}}{2} \left(1330+667+667+2349\right) \] \[ = 9.93\times10^{-17}\times5013 = 4.98\times 10^{-13}\,\mathrm{erg} = 0.311\,\mathrm{eV} . \]
Forgetting to count the degenerate bending mode twice is the standard error here. For a linear \(N\)-atom molecule the mode count is \(3N-5\); for a non-linear one it is \(3N-6\).
The elastic constant of a Cu atom is \(100\,\mathrm{N}\,\mathrm{m}^{-1}\) and the atomic spacing is \(0.256\,\mathrm{nm}\).
Calculate the kinetic, potential and total energy of Cu atoms at \(300\,\mathrm{K}\).
Calculate the amplitude of vibration at \(300\,\mathrm{K}\) as a percentage of the equilibrium spacing.
Solution. (a) By equipartition each quadratic degree of freedom carries \(\tfrac12 k_{B}T\): \[ \text{K.E.} = \text{P.E.} = \frac{k_{B}T}{2} = \frac{1.38\times10^{-23}\times300}{2} = 2.07\times 10^{-21}\,\mathrm{J} , \] \[ E_{\text{total}} = k_{B}T = 4.14\times 10^{-21}\,\mathrm{J} . \]
(b) For a 1D oscillator of amplitude \(A\), \(E = \tfrac12 kA^{2}\). Equating to \(k_{B}T\): \[ A = \sqrt{\frac{2k_{B}T}{k}} = \sqrt{\frac{2\times1.38\times10^{-23}\times300}{100}} = 9.09\times 10^{-12}\,\mathrm{m} = 0.009\,\mathrm{nm} , \] \[ \frac{A}{a} \times 100 = \frac{0.009}{0.256}\times100 = 3.5\,\% . \]
Lindemann's melting criterion says a solid melts when the thermal vibration amplitude reaches roughly \(10\%\) of the interatomic spacing. At \(3.5\%\), copper at room temperature is comfortably solid — consistent with its melting point near \(1358\,\mathrm{K}\).
Find the amplitude of vibration of the HCl molecule in the second excited vibrational level. Take \(k = 480\,\mathrm{N}\,\mathrm{m}^{-1}\) and \(\mred = 1.62\times 10^{-27}\,\mathrm{kg}\).
Solution. The second excited level is \(v = 2\), so from \(\eqref{eq:sho-levels-vib}\) \[ E_{2} = \left(2+\tfrac12\right)h\nu_{\text{osc}} = \frac{5}{2}h\nu_{\text{osc}} . \] At the classical turning point all the energy is potential, \(\tfrac12 kx_{0}^{2} = E_{2}\), so \[ x_{0} = \sqrt{\frac{5h\nu_{\text{osc}}}{k}} . \] The oscillation frequency is \[ \nu_{\text{osc}} = \frac{1}{2\pi}\sqrt{\frac{480}{1.62\times10^{-27}}} = 8.7\times 10^{13}\,\mathrm{Hz} , \] giving \[ x_{0} = \sqrt{\frac{5\times6.63\times10^{-34}\times8.7\times10^{13}}{480}} = 2.45\times 10^{-11}\,\mathrm{m} = 0.245\,\mathrm{Å} . \]
The source computes \(\nu_{\text{osc}} = 8.7\times 10^{13}\,\mathrm{Hz}\) but then substitutes \(5.7\times 10^{13}\,\mathrm{Hz}\) into the amplitude formula, obtaining \(x_{0} = 0.19\,\mathrm{Å}\). It also labels the step “for \(v=1\)” while using the \(v=2\) coefficient \(\tfrac52\). With the consistent value \(\nu_{\text{osc}} = 8.7\times 10^{13}\,\mathrm{Hz}\) the answer is \(0.245\,\mathrm{Å}\).
8.4 The Anharmonic Oscillator
A real molecule is not a perfect spring. If the bond is stretched far enough it breaks and the molecule dissociates, so the potential must flatten out at large \(r\) rather than rising without limit. The standard empirical fit is the Morse potential: \begin{equation} \boxed{\;V(r) = D_{e}\left[1 - e^{-a(r-r_{e})}\right]^{2}\;} \label{eq:morse} \end{equation} where \(a\) is a constant and \(D_{e}\) is the dissociation energy measured from the bottom of the well.
Expanding \(\eqref{eq:morse}\) in a Taylor series, \begin{equation} V(r) = f\left(r-r_{e}\right)^{2} - g\left(r-r_{e}\right)^{3} = fx^{2} - gx^{3} , \label{eq:anharm-taylor} \end{equation} where \(fx^{2}\) is the harmonic term and \(gx^{3}\) is the leading anharmonic correction. Solving the resulting Schrödinger equation: \begin{equation} \boxed{\; \varepsilon_{v} = \omega_{e}\left(v+\tfrac12\right) - \omega_{e}x_{e}\left(v+\tfrac12\right)^{2} \ \mathrm{cm}^{-1}\;} \label{eq:anharm-energy} \end{equation} where \(x_{e}\) is the dimensionless anharmonicity constant. The zero-point energy becomes \begin{equation} \varepsilon_{0} = \frac{\omega_{e}}{2}\left(1 - \frac{x_{e}}{2}\right) \ \mathrm{cm}^{-1} , \label{eq:anharm-zpe} \end{equation} slightly below the harmonic value.
Equation \(\eqref{eq:anharm-zpe}\) was printed with the prefactor \(\omega_{e}/1\); putting \(v=0\) into \(\eqref{eq:anharm-energy}\) gives \(\omega_{e}/2 - \omega_{e}x_{e}/4 = (\omega_{e}/2)(1-x_{e}/2)\), so the denominator is \(2\).
8.4.1 Selection rule and the band structure
Anharmonicity relaxes the harmonic selection rule to \begin{equation} \Delta v = \pm1, \pm2, \pm3, \dots \label{eq:anharm-selection} \end{equation} At room temperature \(N_{v=1}/N_{v=0} \approx 0.008\), so fewer than one per cent of molecules are in \(v=1\). To a very good approximation all absorption starts from \(v=0\), and only three transitions matter.
A population ratio of \(0.008\) is \(0.8\,\%\), not the “\(0.001\,\%\)” stated in the source.
| Transition | \(\Delta v\) | Wavenumber | Name / intensity |
|---|---|---|---|
| \(v=0\to1\) | \(+1\) | \(\omega_{e}\left(1-2x_{e}\right)\) | fundamental band — most intense |
| \(v=0\to2\) | \(+2\) | \(2\omega_{e}\left(1-3x_{e}\right)\) | first overtone — weak |
| \(v=0\to3\) | \(+3\) | \(3\omega_{e}\left(1-4x_{e}\right)\) | second overtone — negligible |
| \(v=1\to2\) | \(+1\) | \(\omega_{e}\left(1-4x_{e}\right)\) | hot band — grows with \(T\) |
The hot band is so called because its intensity rises with temperature: it starts from \(v=1\), whose population grows as \(e^{-\Delta E/k_{B}T}\). Its wavenumber is lower than the fundamental, so it appears as a shoulder on the low-wavenumber side.
From the following vibrational wavenumbers in the ground electronic state of CO, evaluate \(\omega_{e}\) and \(\omega_{e}x_{e}\):
| Transition | \(0\to1\) | \(1\to2\) | \(2\to3\) | \(3\to4\) | \(4\to5\) |
|---|---|---|---|---|---|
| Wavenumber (\(\mathrm{cm}^{-1}\)) | 2143.1 | 2116.1 | 2088.9 | 2061.3 | 2033.5 |
Solution. For the transition \(v \to v+1\), \[ \Delta\varepsilon = \varepsilon_{v+1}-\varepsilon_{v} = \omega_{e} - 2\omega_{e}x_{e}\left(v+1\right) . \] Successive differences therefore give \(2\omega_{e}x_{e}\) directly: \[ 27.0,\ 27.2,\ 27.6,\ 27.8 \quad\Longrightarrow\quad 2\omega_{e}x_{e} = 27.4 \quad\Longrightarrow\quad \omega_{e}x_{e} = 13.7\,\mathrm{cm}^{-1} . \] Adding all five equations, \[ 5\omega_{e} - 30\,\omega_{e}x_{e} = 10442.9 , \] \[ 5\omega_{e} = 10442.9 + 30\times13.7 = 10853.9 \quad\Longrightarrow\quad \omega_{e} = 2171\,\mathrm{cm}^{-1} . \]
The wavenumber of the fundamental vibrational transition of \(^{79}\mathrm{Br}^{81}\mathrm{Br}\) is \(323.2\,\mathrm{cm}^{-1}\). Calculate the force constant of the bond, given \(m_{79} = 78.9\,\mathrm{amu}\) and \(m_{81} = 80.9\,\mathrm{amu}\).
Solution. The reduced mass is \[ \mred = \frac{78.9\times80.9}{78.9+80.9}\times1.67\times10^{-27} = 6.67\times 10^{-26}\,\mathrm{kg} . \] Since \(\omega = \sqrt{k/\mred} = 2\pi c\,\bar\nu\), \[ k = 4\pi^{2}c^{2}\bar\nu^{2}\mred = 4\pi^{2}\left(3\times10^{8}\right)^{2} \left(323.2\times10^{2}\right)^{2}\left(6.67\times10^{-26}\right) = 247.6\,\mathrm{N}\,\mathrm{m}^{-1} . \]
The source's version of this example continued with a second calculation using \(k = 516\,\mathrm{N}\,\mathrm{m}^{-1}\), \(\mred = 1.627\times 10^{-27}\,\mathrm{kg}\) and \(\omega_{e} = 2988\,\mathrm{cm}^{-1}\), then a hot-band evaluation with \(x_{e} = 0.0174\). Those are HCl values, not Br2's, and appear to be a fragment of a different problem spliced in. They have been removed.
The equilibrium vibration wavenumber of the iodine molecule is \(215\,\mathrm{cm}^{-1}\) and the anharmonicity constant is \(x_{e} = 0.003\). Calculate the intensity of the first hot band relative to the fundamental band at \(300\,\mathrm{K}\).
Solution. The fundamental band lies at \[ \omega_{e}\left(1-2x_{e}\right) = 215\left(1-0.006\right) = 213.7\,\mathrm{cm}^{-1} . \] The intensity ratio follows the Boltzmann populations of the originating levels: \[ \frac{I_{0}}{I_{1}} = \frac{N_{0}}{N_{1}} = \exp\!\left[\frac{\Delta E}{k_{B}T}\right], \qquad \frac{\Delta E}{k_{B}T} = \frac{hc\,\Delta\bar\nu}{k_{B}T} . \] With \(\Delta\bar\nu = 213.7\,\mathrm{cm}^{-1}\) and \(T = 300\,\mathrm{K}\), \[ \frac{\Delta E}{k_{B}T} = \frac{6.62\times10^{-27}\times3\times10^{10}\times213.7} {1.38\times10^{-16}\times300} = 1.025 , \] \[ \frac{I_{0}}{I_{1}} = e^{1.025} = 2.79 \quad\Longrightarrow\quad \frac{I_{1}}{I_{0}} = 0.36 . \] So the hot band is about a third as strong as the fundamental — easily visible for a molecule as heavy (and hence low-frequency) as iodine.
8.5 Dissociation Energy
The vibrational levels of \(\eqref{eq:anharm-energy}\) do not continue for ever: they converge, and the highest bound level is where \(\dd\varepsilon_{v}/\dd v = 0\): \begin{equation} \omega_{e} - 2\omega_{e}x_{e}\left(v_{\max}+\tfrac12\right) = 0 \quad\Longrightarrow\quad \boxed{\;v_{\max} = \frac{1}{2x_{e}} - \frac12\;} \label{eq:vmax} \end{equation} Substituting back, \begin{equation} \varepsilon_{v_{\max}} = \frac{\omega_{e}}{4x_{e}} \ \mathrm{cm}^{-1} , \label{eq:eps-max} \end{equation} and the dissociation energy measured from the \(v=0\) level is \begin{equation} \boxed{\; D_{0} = \varepsilon_{v_{\max}} - \varepsilon_{0} = \frac{\omega_{e}}{4x_{e}} - \frac{\omega_{e}}{2} + \frac{\omega_{e}x_{e}}{4} \ \mathrm{cm}^{-1}\;} \label{eq:dissociation} \end{equation}
Note that \(D_{0}\) is dominated by the first term \(\omega_{e}/4x_{e}\). Since \(x_{e}\) is typically \(\sim10^{-2}\), this is \(\sim25\,\omega_{e}\) — a large multiple of the vibrational quantum, of order a few eV. Any answer in the meV range is wrong by three orders of magnitude.
The equilibrium vibration frequency of an oscillator is observed at \(2990\,\mathrm{cm}^{-1}\). The ratio of the wavenumbers of the first overtone to the fundamental is \(1.96\). Find the dissociation energy. (\(hc = 1.239\times 10^{-4}\,\mathrm{eV}\,\mathrm{cm}\).)
Solution. Step 1 — find \(x_{e}\). From Table 8.1, \[ \frac{2\omega_{e}\left(1-3x_{e}\right)}{\omega_{e}\left(1-2x_{e}\right)} = 1.96 \quad\Longrightarrow\quad 2 - 6x_{e} = 1.96 - 3.92x_{e} , \] \[ 0.04 = 2.08\,x_{e} \quad\Longrightarrow\quad x_{e} = 0.0192 . \]
Step 2 — apply \(\eqref{eq:dissociation}\). \[ D_{0} = \omega_{e}\left[\frac{1}{4x_{e}} - \frac12 + \frac{x_{e}}{4}\right] = 2990\left[\frac{1}{0.0768} - 0.5 + 0.0048\right] = 2990 \times 12.52 = 37430\,\mathrm{cm}^{-1} . \] Converting, \[ D_{0} = 37430 \times 1.239\times10^{-4} = \boxed{4.64\,\mathrm{eV}} . \]
The source obtained \(7.12\,\mathrm{meV}\), evaluating \[ E_{D} = hc\,\omega_{e}\,x_{e} = 1.239\times10^{-4}\times2990\times0.0192 . \] That multiplies by \(x_{e}\) where \(\eqref{eq:dissociation}\) divides by \(4x_{e}\), so the answer is out by a factor of \(\sim4x_{e}^{2} \approx 1.5\times10^{-3}\) — roughly a thousandfold.
The sanity check is immediate: a bond dissociation energy of \(7\,\mathrm{meV}\) would be weaker than \(k_{B}T\) at room temperature (\(26\,\mathrm{meV}\)), so the molecule could not exist. Real diatomic bond energies are a few eV, and \(4.6\,\mathrm{eV}\) is exactly right for a molecule with \(\omega_{e} = 2990\,\mathrm{cm}^{-1}\) (which is close to HCl).
Please replace this worked answer in the printed text.
If the leading anharmonic correction to the energy of the \(n\)th vibrational level of a diatomic molecule is \(-x_{e}\left(n+\tfrac12\right)^{2}\hbar\omega\) with \(x_{e} = 0.001\), calculate the total number of possible energy levels.
Solution. From \(\eqref{eq:vmax}\), \[ v_{\max} = \frac{1}{2x_{e}} - \frac12 = \frac{1}{0.002} - 0.5 \approx 500 . \] So there are about \(500\) bound vibrational levels.
8.6 The Vibrating Rotator
Rotational quanta are of order \(e-3\,\mathrm{eV}\) and vibrational quanta of order \(e-1\,\mathrm{eV}\). It is therefore possible to observe pure rotational spectra (Chapter 7), but not pure vibrational spectra: any vibrational transition is inevitably accompanied by rotational structure.
By the Born–Oppenheimer approximation the two motions are separable, so the energies simply add: \begin{equation} \varepsilon_{J,v} = \varepsilon_{J} + \varepsilon_{v} = BJ(J+1) + \omega_{e}\left(v+\tfrac12\right) - \omega_{e}x_{e}\left(v+\tfrac12\right)^{2} \ \mathrm{cm}^{-1} . \label{eq:vibrot-energy} \end{equation} Each vibrational level thus carries a stack of rotational levels.
8.6.1 Selection rules and the two branches
\begin{equation} \Delta v = \pm1, \pm2, \dots \qquad\text{and}\qquad \Delta J = \pm 1 . \label{eq:vibrot-selection} \end{equation}
\(\Delta J = 0\) is not allowed for a diatomic molecule: the vibrational change must be accompanied by a simultaneous rotational change. This is why the band origin itself is missing from the spectrum — there is a gap where the line would be, of width \(2B\).
Writing \(B'\) for the rotational constant of the upper vibrational state and \(B\) for the lower:
\(R\) branch (\(\Delta J = +1\), so \(J' = J+1\)): \begin{equation} \Delta\varepsilon = \omega_{e}\left(1-2x_{e}\right) + 2B' + \left(3B'-B\right)J + \left(B'-B\right)J^{2} . \label{eq:R-branch} \end{equation}
\(P\) branch (\(\Delta J = -1\), so \(J' = J-1\), requiring \(J \ge 1\)): \begin{equation} \Delta\varepsilon = \omega_{e}\left(1-2x_{e}\right) - 2B' + \left(B'-B\right)J + \left(B'-B\right)J^{2} . \label{eq:P-branch} \end{equation}
Both are captured by a single formula using the running index \(m\): \begin{equation} \boxed{\; \Delta\varepsilon = \omega_{e}\left(1-2x_{e}\right) + \left(B'+B\right)m + \left(B'-B\right)m^{2}\;} \label{eq:fortrat} \end{equation} with \[ m = J+1 = 1,2,3,\dots \ \text{for } R(0), R(1), R(2),\dots \] \[ m = -J = -1,-2,-3,\dots \ \text{for } P(1), P(2), P(3),\dots \]
If the vibration–rotation interaction is neglected (\(B' = B\)), these simplify to \begin{equation} R:\ \Delta\varepsilon = \omega_{e}\left(1-2x_{e}\right) + 2B(J+1), \qquad P:\ \Delta\varepsilon = \omega_{e}\left(1-2x_{e}\right) - 2B(J+1) . \label{eq:branches-simple} \end{equation}
8.6.2 Intensity maxima
The intensity of each line follows the population of the originating rotational level, which peaks at \(J_{\max}\) from Chapter 7. Since \(m = \pm(J+1)\), \begin{equation} \Delta\varepsilon_{\max} = \omega_{e}\left(1-2x_{e}\right) \pm 2B\left(\sqrt{\frac{k_{B}T}{2hcB}} + \frac12\right) , \label{eq:branch-max} \end{equation} with \(+\) for the \(R\) branch and \(-\) for the \(P\) branch.
8.6.3 Vibration–rotation interaction
Because a molecule in a higher vibrational state has a larger average bond length, its rotational constant is smaller: \begin{equation} B_{v} = B_{e} - \alpha_{e}\left(v + \tfrac12\right) , \label{eq:B-v} \end{equation} where \(B_{e}\) is the value at the equilibrium separation and \(\alpha_{e}\) is a small positive constant whose value depends on the shape of the potential curve. Since \(B' < B\), the coefficient \((B'-B)\) in \(\eqref{eq:fortrat}\) is negative, so the \(R\) branch lines gradually crowd together and eventually turn back — the band head.
The turning point follows from \(\dd(\Delta\varepsilon)/\dd m = 0\): \begin{equation} m_{\text{head}} = -\frac{B'+B}{2\left(B'-B\right)}, \qquad \bar\nu_{\text{head}} = \bar\nu_{0} - \frac{\left(B'+B\right)^{2}} {4\left(B'-B\right)} . \label{eq:band-head} \end{equation}
A diatomic molecule has vibrational states with energies \(E_{v} = \hbar\omega\left(v+\tfrac12\right)\) and rotational states with \(E_{J} = BJ(J+1)\), where \(v\) and \(J\) are non-negative integers. Consider transitions in which both the initial and final states are restricted to \(v \le 1\) and \(J \le 3\), subject to \(\Delta v = \pm1\) and \(\Delta J = \pm1\). What is the largest allowed transition energy?
Solution. The total energy is \(E = \hbar\omega\left(v+\tfrac12\right) + BJ(J+1)\). For \(\Delta v = 1\) and \(\Delta J = +1\) starting from \(J\), \[ \Delta E = \hbar\omega + B\left[(J+1)(J+2) - J(J+1)\right] = \hbar\omega + 2B(J+1) . \] With \(J \le 3\) in both states, the possible rotational steps are \(0\to1\), \(1\to2\) and \(2\to3\); the largest is \(J=2\to3\): \[ \Delta E = \hbar\omega + 2B(3) = \boxed{\hbar\omega + 6B} . \]
The fundamental band of a CO-like molecule is represented by \[ \Delta\varepsilon = 25798 + 3.850\,m + 0.068\,m^{2} \ \mathrm{cm}^{-1}, \qquad m = \pm1, \pm2, \pm3, \dots \] Find (i) the rotational constants \(B\) and \(B'\); (ii) the wavenumbers of the first two lines in the \(P\) and \(R\) branches; (iii) the location of the band head; (iv) the value of \(B_{e}\) if \(\alpha_{e} = 0.018\).
Solution. (i) Comparing with \(\eqref{eq:fortrat}\), \[ B' + B = 3.850 , \qquad B' - B = 0.068 , \] \[ B' = 1.959\,\mathrm{cm}^{-1}, \qquad B = 1.891\,\mathrm{cm}^{-1} . \]
(ii) For the \(P\) branch \(m\) is negative, for the \(R\) branch positive: \[ \begin{aligned} P(1),\ m=-1 &: \ 25798 - 3.850 + 0.068 = 25794.2\,\mathrm{cm}^{-1} ,\\ P(2),\ m=-2 &: \ 25798 - 7.700 + 0.272 = 25790.6\,\mathrm{cm}^{-1} ,\\ R(0),\ m=+1 &: \ 25798 + 3.850 + 0.068 = 25801.9\,\mathrm{cm}^{-1} ,\\ R(1),\ m=+2 &: \ 25798 + 7.700 + 0.272 = 25806.0\,\mathrm{cm}^{-1} . \end{aligned} \]
(iii) From \(\eqref{eq:band-head}\), \[ m_{\text{head}} = -\frac{3.850}{2\times0.068} = -28.3 , \] \[ \bar\nu_{\text{head}} = 25798 - \frac{\left(3.850\right)^{2}}{4\times0.068} = 25798 - 54.5 = 25743.5\,\mathrm{cm}^{-1} . \] Since \(m_{\text{head}}\) is negative, the head lies in the \(P\) branch here (because \(B' > B\) for this band).
(iv) From \(\eqref{eq:B-v}\) with the lower state \(v=0\), \[ B = B_{e} - \frac{\alpha_{e}}{2} \quad\Longrightarrow\quad B_{e} = 1.891 + \frac{0.018}{2} = 1.900\,\mathrm{cm}^{-1} . \]
Two problems in the source version of this example:
Midway through the solution the coefficients change from \((3.850m + 0.068m^{2})\) to \((3.60m - 0.75m^{2})\), and the band head is then computed from the second set — giving \(25802\,\mathrm{cm}^{-1}\). The two are inconsistent; the working above uses the coefficients as stated in the question.
Part (iv) was posed but never answered. It is supplied above.
The vibration–rotation spectrum of CO at low resolution is measured at \(T = 300\,\mathrm{K}\). The band is centred at \(2143\,\mathrm{cm}^{-1}\) and the average line separation near the centre is \(3.83\,\mathrm{cm}^{-1}\). Find the wavenumbers at which the \(P\)- and \(R\)-branch intensity maxima are located.
Solution. The line separation is \(2B\), so \[ B = \frac{3.83}{2} = 1.915\,\mathrm{cm}^{-1} . \] The maximum population occurs at \(J_{\max} = \sqrt{k_{B}T/2hcB} - \tfrac12\), so from \(\eqref{eq:branch-max}\) \[ \bar\nu_{P} = \bar\nu_{0} - 2B\left(\sqrt{\frac{k_{B}T}{2hcB}} + \frac12\right). \] Evaluating the bracket: \[ \sqrt{\frac{1.38\times10^{-23}\times300} {2\times6.63\times10^{-34}\times3\times10^{10}\times3.83}} + \frac12 = 7.87 , \] so \[ 2B \times 7.87 = 3.83\times7.87 = 30.15\,\mathrm{cm}^{-1} . \] Hence \[ \bar\nu_{P} = 2143 - 30.15 = 2112.9\,\mathrm{cm}^{-1}, \qquad \bar\nu_{R} = 2143 + 30.15 = 2173.2\,\mathrm{cm}^{-1} . \]
The two maxima sit symmetrically about the band origin, separated by \(\approx60\,\mathrm{cm}^{-1}\). This double-humped envelope is the visual signature of a vibration–rotation band at low resolution, and measuring the hump separation is a quick route to \(B\) and hence the bond length.
8.7 Applications
Vibrational spectra are used to determine:
the force constant \(k\) of the bond, from \(\omega_{e}\);
the anharmonicity \(x_{e}\), from the overtone spacings;
the dissociation energy \(D_{0}\), via \(\eqref{eq:dissociation}\);
the bond length, from the rotational fine structure of the band;
isotopic composition, from the shift of the fundamental band;
chemical identification — functional groups have characteristic frequencies, which is the basis of IR spectroscopy as an analytical tool.
Formula Summary
Gross selection rule the vibration must change the dipole moment (not merely have one)
CO2 \(3N-5 = 4\) modes: symmetric stretch inactive; asymmetric stretch active; bend (doubly degenerate) active
Harmonic oscillator \[ V = \tfrac12 kx^{2}, \qquad \nu_{\text{osc}} = \frac{1}{2\pi}\sqrt{\frac{k}{\mred}}, \qquad E_v = h\nu_{\text{osc}}\left(v+\tfrac12\right) \] \[ \varepsilon_v = \omega_e\left(v+\tfrac12\right)\ \mathrm{cm}^{-1}, \qquad \varepsilon_0 = \frac{\omega_e}{2} \ \text{(zero-point)}, \qquad \Delta v = \pm1 \Rightarrow \text{single line at } \omega_e \] Isotope effect \(\nu \propto 1/\sqrt{\mred}\); heavier isotopologue \(\Rightarrow\) lower wavenumber
Anharmonic oscillator Morse potential \(V = D_e\left[1-e^{-a(r-r_e)}\right]^{2}\) \[ \varepsilon_v = \omega_e\left(v+\tfrac12\right) - \omega_e x_e\left(v+\tfrac12\right)^{2}, \qquad \varepsilon_0 = \frac{\omega_e}{2}\left(1-\frac{x_e}{2}\right) \] \[ \begin{array}{ll} \text{fundamental } (0\to1): & \omega_e(1-2x_e)\\ \text{first overtone } (0\to2): & 2\omega_e(1-3x_e)\\ \text{second overtone } (0\to3): & 3\omega_e(1-4x_e)\\ \text{hot band } (1\to2): & \omega_e(1-4x_e) \end{array} \] successive fundamentals differ by \(2\omega_e x_e\)
Dissociation \[ v_{\max} = \frac{1}{2x_e}-\frac12, \qquad \varepsilon_{v_{\max}} = \frac{\omega_e}{4x_e}, \qquad D_0 = \frac{\omega_e}{4x_e} - \frac{\omega_e}{2} + \frac{\omega_e x_e}{4} \] \(D_0\) is a few eV — never meV
Vibrating rotator \[ \varepsilon_{J,v} = BJ(J+1) + \omega_e\left(v+\tfrac12\right) - \omega_e x_e\left(v+\tfrac12\right)^{2} \] \[ \Delta v = \pm1, \quad \Delta J = \pm1 \ \ (\Delta J = 0 \text{ forbidden}) \] \[ \Delta\varepsilon = \omega_e(1-2x_e) + (B'+B)m + (B'-B)m^{2}, \qquad \begin{cases} m = J+1 & R \text{ branch}\\ m = -J & P \text{ branch}\end{cases} \] If \(B'=B\): \ \(R\): \(\omega_e(1-2x_e)+2B(J+1)\); \ \(P\): \(\omega_e(1-2x_e)-2B(J+1)\); \ gap of \(2B\) at the origin
Intensity maxima \(\omega_e(1-2x_e) \pm 2B\left(\sqrt{\dfrac{k_BT}{2hcB}}+\dfrac12\right)\)
Vibration–rotation interaction \(B_v = B_e - \alpha_e\left(v+\tfrac12\right)\) \[ m_{\text{head}} = -\frac{B'+B}{2(B'-B)}, \qquad \bar\nu_{\text{head}} = \bar\nu_0 - \frac{(B'+B)^{2}}{4(B'-B)} \]
Practice Problems
These problems are original to this book. They cover the examinable core of vibrational spectroscopy: anharmonicity and dissociation, zero-point energy in isotope exchange, hot bands, the \(P\) and \(R\) branch structure, and the origin of infra-red intensity. Answers with the key steps follow.
Constants: \(h = 6.626\times 10^{-34}\,\mathrm{J}\,\mathrm{s}\); \(c = 2.998\times 10^{10}\,\mathrm{cm}\,\mathrm{s}^{-1}\); \(k_{B} = 1.381\times 10^{-23}\,\mathrm{J}\,\mathrm{K}^{-1}\); \(4\pi^{2} = 39.478\); \(1\,\mathrm{cm}^{-1} \equiv 11.96\,\mathrm{J}\,\mathrm{mol}^{-1} \equiv 1.2398\times 10^{-4}\,\mathrm{eV}\); \(k_{B}T/hc = 208.5\,\mathrm{cm}^{-1}\) at \(300\,\mathrm{K}\).
The fundamental and first overtone bands of \(^{12}\)C\(^{16}\)O are centred at \(2143.3\,\mathrm{cm}^{-1}\) and \(4260.0\,\mathrm{cm}^{-1}\) respectively. Taking the vibrational term values as \(\varepsilon_{v} = \omega_{e}(v+\tfrac12) - \omega_{e}x_{e}(v+\tfrac12)^{2}\): (a) evaluate \(\omega_{e}\) and the anharmonicity constant \(x_{e}\); (b) find the exact zero-point energy; (c) calculate the force constant, taking \(\mred = 1.139\times 10^{-26}\,\mathrm{kg}\); (d) treating \(v\) as a continuous variable, find the value of \(v\) at which \(\varepsilon_{v}\) is a maximum, and hence estimate the dissociation energy of CO in \(\mathrm{cm}^{-1}\) and in \(\mathrm{eV}\); (e) criticize the method used in (d).
The \(v = 0 \to 1\) wavenumbers of four molecules are HCl \(2885\,\mathrm{cm}^{-1}\), DCl \(1990\,\mathrm{cm}^{-1}\), HD \(3627\,\mathrm{cm}^{-1}\) and D2 \(2990\,\mathrm{cm}^{-1}\). Considering only zero-point energies, calculate the energy change of \[ \text{HCl} + \text{D}_{2} \longrightarrow \text{DCl} + \text{HD} \] in \(\mathrm{kJ}\,\mathrm{mol}^{-1}\), and state whether energy is absorbed or liberated. Explain why a purely isotopic substitution can change a reaction energy at all.
The equilibrium vibration wavenumber of I2 is \(214.5\,\mathrm{cm}^{-1}\) and its anharmonicity constant is \(x_{e} = 0.00296\). At \(300\,\mathrm{K}\), what is the intensity of the “hot band” (\(v=1\to2\)) relative to that of the fundamental (\(v=0\to1\))? Why is this ratio so much larger for I2 than for HCl?
An infra-red band of a linear molecule is recorded at \(300\,\mathrm{K}\) with the rotational structure unresolved, so that only the \(P\) and \(R\) branch maxima are visible. Show that their separation is approximately \(\Delta\tilde\nu \approx \sqrt{8Bk_{B}T/hc}\), and evaluate it for OCS, for which \(B = 0.2039\,\mathrm{cm}^{-1}\).
In the \(v = 0 \to 1\) band of HF the rotational constants of the two vibrational states are found to be \(B_{0} = 20.56\,\mathrm{cm}^{-1}\) and \(B_{1} = 19.79\,\mathrm{cm}^{-1}\). Calculate the percentage increase in bond length on going from \(v=0\) to \(v=1\), and say what effect this has on the spacing of the lines in the \(P\) and \(R\) branches.
For H35Cl take the bond length as \(127.5\,\mathrm{pm}\), the bond force constant as \(516.3\,\mathrm{N}\,\mathrm{m}^{-1}\), and the atomic masses as \(^{1}\)H \(= 1.673\times 10^{-27}\,\mathrm{kg}\), \(^{35}\)Cl \(= 58.066\times 10^{-27}\,\mathrm{kg}\). Giving all answers in \(\mathrm{cm}^{-1}\): (a) calculate the zero-point energy and the wavenumber of the fundamental vibration; (b) calculate the rotational constant \(B\); (c) calculate the wavenumbers of the lines \(P(1)\), \(P(2)\), \(P(3)\), \(R(0)\), \(R(1)\) and \(R(2)\); (d) sketch the expected vibration–rotation spectrum, marking the approximate intensity distribution; (e) suggest two differences you would expect between your sketch and the spectrum actually observed for HCl, giving your reasons.
Explain why the C=O stretching vibration of an aldehyde gives rise to a strong absorption in the infra-red, while the absorption due to the C=C stretch in an alkene is normally very weak — and why the latter is nevertheless easy to see in the Raman spectrum.
Answers
P7.1 (a) The two band centres are \(\tilde\nu_{0\to1} = \omega_{e}(1-2x_{e})\) and \(\tilde\nu_{0\to2} = 2\omega_{e}(1-3x_{e})\). Twice the first minus the second gives \(2\omega_{e}x_{e} = 2(2143.3)-4260.0 = 26.6\,\mathrm{cm}^{-1}\), so \(\omega_{e}x_{e} = 13.3\,\mathrm{cm}^{-1}\), \(\omega_{e} = 2143.3+2(13.3) = 2169.9\,\mathrm{cm}^{-1}\) and \(x_{e} = 13.3/2169.9 = 6.13\times 10^{-3}\). (b) \(\varepsilon_{0} = \tfrac12\omega_{e} - \tfrac14\omega_{e}x_{e} = 1084.95 - 3.33 = 1081.6\,\mathrm{cm}^{-1}\). (c) \(k = 4\pi^{2}c^{2}\omega_{e}^{2}\mred = 1.90\times 10^{3}\,\mathrm{N}\,\mathrm{m}^{-1}\). (d) \(\dd\varepsilon_{v}/\dd v = 0\) at \(v_{\max} = 1/2x_{e} - \tfrac12 = 81.1\), and \(D_{e} = \omega_{e}/4x_{e} = 8.85\times 10^{4}\,\mathrm{cm}^{-1} = 10.97\,\mathrm{eV}\). (e) The method extrapolates from the three lowest levels (\(v = 0,1,2\)) all the way to \(v \approx 81\). It assumes the potential is exactly Morse, i.e.\ that a Birge–Sponer plot of \(\Delta G(v)\) against \(v\) is a straight line. Real plots curve away from linearity at high \(v\) because the true potential flattens differently from Morse near dissociation, and the extrapolation is typically in error by \(10\)–\(30\,\%\). CO is unusually well described by a Morse curve, so the agreement here (about \(2\,\%\) below the accepted \(D_{e} = 11.2\,\mathrm{eV}\)) is better than one should expect.
P7.2 Taking each zero-point energy as \(\tfrac12\tilde\nu\), \[ \Delta E = \tfrac12\left[(1990+3627)-(2885+2990)\right] = \tfrac12(-258) = -129\,\mathrm{cm}^{-1}, \] that is \(-129\times11.96 = -1.54\,\mathrm{kJ}\,\mathrm{mol}^{-1}\): energy is liberated. Isotopic substitution leaves the electronic potential-energy curve untouched — within the Born–Oppenheimer approximation it depends only on the nuclear charges — but it changes the reduced mass, and hence the zero-point energy sitting on that curve, since \(\tilde\nu \propto \sqrt{k/\mred}\). Redistributing the isotopes among bonds of different force constant therefore changes the total zero-point energy even though no bond type is made or broken. This is the origin of all equilibrium isotope effects.
P7.3 The hot band starts from molecules already in \(v=1\), so the intensity ratio is the population ratio \[ \frac{N_{1}}{N_{0}} = e^{-\Delta\varepsilon_{01}/k_{B}T}, \qquad \Delta\varepsilon_{01} = \omega_{e}(1-2x_{e}) = 213.2\,\mathrm{cm}^{-1} . \] With \(k_{B}T/hc = 208.5\,\mathrm{cm}^{-1}\), \(N_{1}/N_{0} = e^{-213.2/208.5} = e^{-1.02} = 0.36\): the hot band is about \(36\,\%\) as intense as the fundamental. For HCl, \(\tilde\nu \approx 2886\,\mathrm{cm}^{-1}\) gives \(e^{-13.8} \approx 10^{-6}\) — undetectable. Iodine has a very large reduced mass and a weak bond, so its vibrational quantum is smaller than \(k_{B}T\) at room temperature and excited vibrational levels are well populated.
P7.4 The intensity envelope of each branch follows the thermal population of the rotational levels, which peaks at \(J_{\max} \approx \sqrt{k_{B}T/2hcB}-\tfrac12\). The \(R\) maximum lies at \(\tilde\nu_{0}+2B(J_{\max}+1)\) and the \(P\) maximum at \(\tilde\nu_{0}-2B(J_{\max}+1)\), so \[ \Delta\tilde\nu = 4B(J_{\max}+1) \approx 4B\sqrt{\frac{k_{B}T}{2hcB}} = \sqrt{\frac{8Bk_{B}T}{hc}} . \] Since \(k_{B}T/hc = 0.695\,T\ \mathrm{cm}^{-1}\), this is \(\Delta\tilde\nu \approx 2.36\sqrt{BT}\ \mathrm{cm}^{-1}\). For OCS at \(300\,\mathrm{K}\): \(2.36\sqrt{0.2039\times300} = 18.5\,\mathrm{cm}^{-1}\).
P7.5 \(B \propto 1/r^{2}\), so \(r_{1}/r_{0} = \sqrt{B_{0}/B_{1}} = \sqrt{20.56/19.79} = 1.0193\): the bond is \(\mathbf{1.9\,\%}\) longer in \(v=1\). Because \(B_{1} < B_{0}\), the \(R\)-branch lines crowd together as \(J\) rises and eventually turn back on themselves, forming a band head, while the \(P\)-branch lines spread further apart. The band is therefore not symmetric about its centre: it is degraded towards lower wavenumber.
P7.6 \(\mred = 1.6262\times 10^{-27}\,\mathrm{kg}\). (a) \(\tilde\omega = (1/2\pi c)\sqrt{k/\mred} = 2991\,\mathrm{cm}^{-1}\); the zero-point energy is \(\tfrac12\tilde\omega = 1496\,\mathrm{cm}^{-1}\) and the fundamental (harmonic) lies at \(2991\,\mathrm{cm}^{-1}\). (b) \(B = h/8\pi^{2}c\mred r^{2} = 10.588\,\mathrm{cm}^{-1}\), so \(2B = 21.18\,\mathrm{cm}^{-1}\). (c) With \(R(J) = \tilde\nu_{0}+2B(J+1)\) and \(P(J) = \tilde\nu_{0}-2BJ\):
| Line | \(\tilde\nu\ (\mathrm{cm}^{-1})\) | Line | \(\tilde\nu\ (\mathrm{cm}^{-1})\) |
|---|---|---|---|
| \(R(0)\) | \(3012.2\) | \(P(1)\) | \(2969.8\) |
| \(R(1)\) | \(3033.4\) | \(P(2)\) | \(2948.7\) |
| \(R(2)\) | \(3054.5\) | \(P(3)\) | \(2927.5\) |
(d) Two branches symmetric about the missing centre at \(2991\,\mathrm{cm}^{-1}\), with a gap of \(4B = 42.4\,\mathrm{cm}^{-1}\) across the centre and \(2B = 21.2\,\mathrm{cm}^{-1}\) between neighbours within a branch. The intensity envelope peaks at \(J_{\max} = \sqrt{208.5/21.18}-\tfrac12 \approx 2.6\), so around \(P(3)\) and \(R(2)\), falling away on both sides. (e) Two differences. (i) Every line is doubled: natural chlorine is \(76\,\%\) \(^{35}\)Cl and \(24\,\%\) \(^{37}\)Cl, so each line has a companion about \(2\,\mathrm{cm}^{-1}\) lower in wavenumber and about one third as intense. (ii) The real band centre lies at \(2886\,\mathrm{cm}^{-1}\), not \(2991\,\mathrm{cm}^{-1}\): the observed fundamental is \(\omega_{e}-2\omega_{e}x_{e}\), and anharmonicity pulls it down by roughly \(100\,\mathrm{cm}^{-1}\). In addition \(B_{1} < B_{0}\), so the \(R\) branch converges and the \(P\) branch spreads, and the band is not the symmetric picture sketched in (d).
P7.7 Infra-red intensity is proportional to \(\left(\dd\mu/\dd Q\right)^{2}\), the square of the rate of change of the electric dipole moment with the normal coordinate. The C=O bond is strongly polar — oxygen is far more electronegative than carbon — so stretching it changes a large dipole moment substantially, and the aldehyde carbonyl band near \(1720\,\mathrm{cm}^{-1}\) is one of the strongest features in organic infra-red spectroscopy. The C=C bond joins two identical atoms and carries almost no dipole, so stretching it barely changes \(\mu\) and the absorption is weak; in a symmetrically substituted alkene the centre of symmetry makes \(\dd\mu/\dd Q\) vanish exactly and the mode is strictly infra-red inactive.
Raman intensity depends instead on \(\left(\dd\alpha/\dd Q\right)^{2}\), the change in polarizability. The C=C bond has a large, loosely held and highly anisotropic \(\pi\) cloud which distorts easily and changes markedly on stretching, so the mode is strongly Raman active. This is the rule of mutual exclusion of Chapter 9 at work: in a molecule with a centre of symmetry, no mode can be both infra-red and Raman active.
Previous Year Questions
The source carries a single combined previous-year bank headed “Chapter 6 to Chapter 9”. It is typeset in full — GATE, CSIR-NET/JRF and JEST, with worked solutions — at the end of Chapter 10. A set of original practice problems on this chapter's material appears immediately above.
The questions bearing most directly on this chapter are, in that bank: GATE Q7 (second overtone), Q8 (force constant of CO), Q9 (anharmonicity constant), Q10 and Q16 (spectral-region matching), Q12 and Q20 (vibrational energy scaling), Q25 (half-harmonic oscillator) and Q26 (Morse curves compared); NET Q4 (number of vibrational levels), Q5 (largest allowed transition) and Q11 (bond length from the \(P\) and \(R\) branches); and JEST Q2 (amplitude of atomic vibration in copper).