Comprehensive Study Notes
We shall see in Section 13.1 that to an excellent approximation one can treat separately the motions of the electrons and the motions of the nuclei of a molecule. (This is due to the much heavier mass of the nuclei.) One first imagines the nuclei to be held stationary and solves a Schrödinger equation for the electronic energy $U$. ( $U$ also includes the energy of nuclear repulsion.) For a diatomic (two-atom) molecule, the electronic energy $U$ depends on the distance $R$ between the nuclei, $U=U(R)$, and the $U$ versus $R$ curve has the typical appearance of Fig. 13.1.
After finding $U(R)$, one solves a Schrödinger equation for nuclear motion, using $U(R)$ as the potential energy for nuclear motion. For a diatomic molecule, the nuclear Schrödinger equation is a two-particle equation. We shall see in Section 6.3 that, when the potential energy of a two-particle system depends only on the distance between the particles, the energy of the system is the sum of (a) the kinetic energy of translational motion of the entire system through space and (b) the energy of internal motion of the particles relative to each other. The classical expression for the two-particle internalmotion energy turns out to be the sum of the potential energy of interaction between the particles and the kinetic energy of a hypothetical particle whose mass is $m{1} m{2} /\left(m{1}+m{2}\right)$ (where $m{1}$ and $m{2}$ are the masses of the two particles) and whose coordinates are the coordinates of one particle relative to the other. The quantity $m{1} m{2} /\left(m{1}+m{2}\right)$ is called the reduced mass $\mu$.
The internal motion of a diatomic molecule consists of vibration, corresponding to a change in the distance $R$ between the two nuclei, and rotation, corresponding to a change in the spatial orientation of the line joining the nuclei. To a good approximation, one can usually treat the vibrational and rotational motions separately. The rotational energy levels are found in Section 6.4. Here we consider the vibrational levels.
The Schrödinger equation for the vibration of a diatomic molecule has a kineticenergy operator for the hypothetical particle of mass $\mu=m{1} m{2} /\left(m{1}+m{2}\right)$ and a potential-energy term given by $U(R)$. If we place the origin to coincide with the minimum point of the $U$ curve in Fig. 13.1 and take the zero of potential energy at the energy of this minimum point, then the lower portion of the $U(R)$ curve will nearly coincide with
FIGURE 4.6 Potential energy for vibration of a diatomic molecule (solid curve) and for a harmonic oscillator (dashed curve). Also shown are the boundstate vibrational energy levels for the diatomic molecule. In contrast to the harmonic oscillator, a diatomic molecule has only a finite number of bound vibrational levels.
the potential-energy curve of a harmonic oscillator with the appropriate force constant $k$ (see Fig. 4.6 and Prob. 4.28). The minimum in the $U(R)$ curve occurs at the equilibrium distance $R{e}$ between the nuclei. In Fig. 4.6, $x$ is the deviation of the internuclear distance from its equilibrium value: $x \equiv R-R{e}$.
The harmonic-oscillator force constant $k$ in Eq. (4.26) is obtained as $k=d^{2} V / d x^{2}$, and the harmonic-oscillator curve essentially coincides with the $U(R)$ curve at $R=R{e}$, so the molecular force constant is $k=d^{2} U /\left.d R^{2}\right|{R=R_{e}}$ (see also Prob. 4.28). Differences in nuclear mass have virtually no effect on the electronic-energy curve $U(R)$, so different isotopic species of the same molecule have essentially the same force constant $k$.
We expect, therefore, that a reasonable approximation to the vibrational energy levels $E_{\text {vib }}$ of a diatomic molecule would be the harmonic-oscillator vibrational energy levels; Eqs. (4.45) and (4.23) give
\(
\begin{equation}
E{\mathrm{vib}} \approx\left(v+\frac{1}{2}\right) h \nu{e}, \quad v=0,1,2, \ldots \tag{4.58}
\end{equation}
\)
\(
\begin{equation}
\nu{e}=\frac{1}{2 \pi}\left(\frac{k}{\mu}\right)^{1 / 2}, \quad \mu=\frac{m{1} m{2}}{m{1}+m{2}}, \quad k=\left.\frac{d^{2} U}{d R^{2}}\right|{R=R_{e}} \tag{4.59}
\end{equation}
\)
$\nu_{e}$ is called the equilibrium (or harmonic) vibrational frequency. This approximation is best for the lower vibrational levels. As $v$ increases, the nuclei spend more time in regions far from their equilibrium separation. For such regions the potential energy deviates substantially from that of a harmonic oscillator and the harmonic-oscillator approximation is poor. Instead of being equally spaced, one finds that the vibrational levels of a diatomic molecule come closer and closer together as $v$ increases (Fig. 4.6). Eventually, the vibrational energy is large enough to dissociate the diatomic molecule into atoms that are not bound to each other. Unlike the harmonic oscillator, a diatomic molecule has only a finite number of bound-state vibrational levels. A more accurate expression for the molecular vibrational energy that allows for the anharmonicity of the vibration is
\(
\begin{equation}
E{\mathrm{vib}}=\left(v+\frac{1}{2}\right) h \nu{e}-\left(v+\frac{1}{2}\right)^{2} h \nu{e} x{e} \tag{4.60}
\end{equation}
\)
where the anharmonicity constant $\nu{e} x{e}$ is positive in nearly all cases.
Using the time-dependent Schrödinger equation, one finds (Section 9.9) that the most probable vibrational transitions when a diatomic molecule is exposed to electromagnetic radiation are those where $v$ changes by $\pm 1$. Furthermore, for absorption or emission of electromagnetic radiation to occur, the vibration must change the molecule's dipole moment. Hence homonuclear diatomics (such as $\mathrm{H}{2}$ or $\mathrm{N}{2}$ ) cannot undergo transitions between vibrational levels by absorption or emission of radiation. (Such transitions can occur during intermolecular collisions.) The relation $E{\text {upper }}-E{\text {lower }}=h \nu$, the approximate equation (4.58), and the selection rule $\Delta v=1$ for absorption of radiation show that a heteronuclear diatomic molecule whose vibrational frequency is $\nu{e}$ will most strongly absorb light of frequency $\nu{\text {light }}$ given approximately by
$\nu{\text {light }}=\left(E{2}-E{1}\right) / h \approx\left[\left(v{2}+\frac{1}{2}\right) h \nu{e}-\left(v{1}+\frac{1}{2}\right) h \nu{e}\right] / h=\left(v{2}-v{1}\right) \nu{e}=\nu{e}$
The values of $k$ and $\mu$ in (4.59) for diatomic molecules are such that $\nu{\text {light }}$ usually falls in the infrared region of the spectrum. Transitions with $\Delta v=2,3, \ldots$ also occur, but these (called overtones) are much weaker than the $\Delta v=1$ absorption.
Use of the more accurate equation (4.60) gives (Prob. 4.27)
\(
\begin{equation}
\nu{\text {light }}=\nu{e}-2 \nu{e} x{e}\left(v_{1}+1\right) \tag{4.62}
\end{equation}
\)
where $v_{1}$ is the quantum number of the lower level and $\Delta v=1$.
The relative population of two molecular energy levels in a system in thermal equilibrium is given by the Boltzmann distribution law (see any physical chemistry text) as
\(
\begin{equation}
\frac{N{i}}{N{j}}=\frac{g{i}}{g{j}} e^{-\left(E{i}-E{j}\right) / k T} \tag{4.63}
\end{equation}
\)
where energy levels $i$ and $j$ have energies $E{i}$ and $E{j}$ and degeneracies $g{i}$ and $g{j}$ and are populated by $N{i}$ and $N{j}$ molecules, and where $k$ is Boltzmann's constant and $T$ the absolute temperature. For a nondegenerate level, $g_{i}=1$.
The magnitude of $\nu=(1 / 2 \pi)(k / \mu)^{1 / 2}$ is such that for light diatomics (for example, $\mathrm{H}{2}, \mathrm{HCl}, \mathrm{CO}$ ) only the $v=0$ vibrational level is significantly populated at room temperature. For heavy diatomics (for example, $\mathrm{I}{2}$, there is significant room-temperature population of one or more excited vibrational levels.
The vibrational absorption spectrum of a polar diatomic molecule consists of a $v=0 \rightarrow 1$ band, much weaker overtone bands $(v=0 \rightarrow 2,0 \rightarrow 3, \ldots)$, and, if $v>0$ levels are significantly populated, hot bands such as $v=1 \rightarrow 2,2 \rightarrow 3$. Each band corresponding to a particular vibrational transition consists of several closely spaced lines. Each such line corresponds to a different change in rotational state simultaneous with the change in vibrational state. Each line is the result of a vibration-rotation transition.
The SI unit for spectroscopic frequencies is the hertz $(\mathrm{Hz})$, defined by $1 \mathrm{~Hz} \equiv 1 \mathrm{~s}^{-1}$. Multiples such as the megahertz ( MHz ) equal to $10^{6} \mathrm{~Hz}$ and the gigahertz $(\mathrm{GHz})$ equal to $10^{9} \mathrm{~Hz}$ are often used. Infrared (IR) absorption lines are usually specified by giving their wavenumber $\widetilde{\nu}$ defined as
\(
\begin{equation}
\widetilde{\nu} \equiv 1 / \lambda=\nu / c \tag{4.64}
\end{equation}
\)
where $\lambda$ is the wavelength in vacuum.
In the harmonic-oscillator approximation, the quantum-mechanical energy levels of a polyatomic molecule turn out to be $E{\text {vib }}=\sum{i}\left(v{i}+\frac{1}{2}\right) h \nu{i}$, where the $\nu{i}$ 's are the frequencies of the normal modes of vibration of the molecule and $v{i}$ is the vibrational quantum
number of the $i$ th normal mode. Each $v_{i}$ takes on the values $0,1,2, \ldots$ independently of the values of the other vibrational quantum numbers. A linear molecule with $n$ atoms has $3 n-5$ normal modes; a nonlinear molecule has $3 n-6$ normal modes.
To calculate the reduced mass $\mu$ in (4.59), one needs the masses of isotopic species. Some relative isotopic masses are listed in Table A. 3 in the Appendix.
EXAMPLE
The strongest infrared band of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$ occurs at $\widetilde{\nu}=2143 \mathrm{~cm}^{-1}$. Find the force constant of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$. State any approximation made.
The strongest infrared band corresponds to the $v=0 \rightarrow 1$ transition. We approximate the molecular vibration as that of a harmonic oscillator. From (4.61), the equilibrium molecular vibrational frequency is approximately
\(
\nu{e} \approx \nu{\text {light }}=\widetilde{\nu} c=\left(2143 \mathrm{~cm}^{-1}\right)\left(2.9979 \times 10^{10} \mathrm{~cm} / \mathrm{s}\right)=6.424 \times 10^{13} \mathrm{~s}^{-1}
\)
To relate $k$ to $\nu{e}$ in (4.59), we need the reduced mass $\mu=m{1} m{2} /\left(m{1}+m_{2}\right)$. One mole of ${ }^{12} \mathrm{C}$ has a mass of 12 g and contains Avogadro's number of atoms. Hence the mass of one atom of ${ }^{12} \mathrm{C}$ is $(12 \mathrm{~g}) /\left(6.02214 \times 10^{23}\right)$. The reduced mass and force constant are
\(
\begin{gathered}
\mu=\frac{12(15.9949) \mathrm{g}}{27.9949} \frac{1}{6.02214 \times 10^{23}}=1.1385 \times 10^{-23} \mathrm{~g} \
k=4 \pi^{2} \nu_{e}^{2} \mu=4 \pi^{2}\left(6.424 \times 10^{13} \mathrm{~s}^{-1}\right)^{2}\left(1.1385 \times 10^{-26} \mathrm{~kg}\right)=1855 \mathrm{~N} / \mathrm{m}
\end{gathered}
\)
EXERCISE (a) Find the approximate zero-point energy of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$.
(Answer: $2.1 \times 10^{-20} \mathrm{~J}$.) (b) Estimate $\nu_{e}$ of ${ }^{13} \mathrm{C}^{16} \mathrm{O}$. (Answer: $6.28 \times 10^{13} \mathrm{~s}^{-1}$.)