We now use the virial theorem to examine the changes in electronic kinetic and potential energy that occur when a covalent chemical bond is formed in a diatomic molecule. For formation of a stable bond, the $U(R)$ curve must have a substantial minimum. At this minimum we have
\(
\begin{equation}
\left.\frac{d U}{d R}\right|{R{e}}=0 \tag{14.96}
\end{equation}
\)
and Eqs. (14.93) to (14.95) become
\(
\begin{align}
\left.2\left\langle T{\mathrm{el}}\right\rangle\right|{R{e}} & =-\left.\langle V\rangle\right|{R{e}} \tag{14.97}\
\left.\left\langle T{\mathrm{el}}\right\rangle\right|{R{e}} & =-U\left(R{e}\right) \tag{14.98}\
\left.\langle V\rangle\right|{R{e}} & =2 U\left(R{e}\right) \tag{14.99}
\end{align}
\)
These equations resemble those for atoms [Eqs. (14.75) and (14.76)]. At $R=\infty$ we have the separated atoms, and the atomic virial theorem gives
\(
\begin{equation}
\left.2\left\langle T{\mathrm{el}}\right\rangle\right|{\infty}=-\left.\langle V\rangle\right|{\infty},\left.\quad\left\langle T{\mathrm{el}}\right\rangle\right|{\infty}=-U(\infty),\left.\quad\langle V\rangle\right|{\infty}=2 U(\infty) \tag{14.100}
\end{equation}
\)
$U(\infty)$ is the sum of the energies of the two separated atoms. Equations (14.98)-(14.100) give
\(
\begin{align}
& \left.\left\langle T{\mathrm{el}}\right\rangle\right|{R{e}}-\left.\left\langle T{\mathrm{el}}\right\rangle\right|{\infty}=U(\infty)-U\left(R{e}\right) \tag{14.101}\
& \left.\langle V\rangle\right|{R{e}}-\left.\langle V\rangle\right|{\infty}=2\left[U\left(R{e}\right)-U(\infty)\right] \tag{14.102}
\end{align}
\)
For bonding, we have $U\left(R{e}\right)<U(\infty)$. Therefore, Eqs. (14.101) and (14.102) show that the average molecular potential energy at $R{e}$ is less than the sum of the potential energies of the separated atoms, whereas the average molecular kinetic energy is greater at $R{e}$ than at $\infty$. The decrease in potential energy is twice the increase in kinetic energy, and results from allowing the electrons to feel the attractions of both nuclei and perhaps from an increase in orbital exponents in the molecule (see Section 13.5). The equilibrium dissociation energy (13.9) is $D{e}=\frac{1}{2}\left(\left.\langle V\rangle\right|{\infty}-\left.\langle V\rangle\right|{R_{e}}\right)$.
Consider the behavior of the average potential and kinetic energies for large $R$. The forces between uncharged atoms or molecules (other than those due to bond formation) are called van der Waals forces. For two neutral atoms, at least one of which is in an $S$ state, quantum-mechanical perturbation theory shows that the van der Waals force of attraction is proportional to $1 / R^{7}$, and the potential energy behaves like
\(
\begin{equation}
U(R) \approx U(\infty)-\frac{A}{R^{6}}, \quad R \text { large } \tag{14.103}
\end{equation}
\)
where $A$ is a positive constant. (See Kauzmann, Chapter 13.) This expression was first derived by London, and van der Waals forces between neutral atoms are called London forces or dispersion forces. (Recall the discussion near the end of Section 13.7.)
Substitution of (14.103) for $U$ and $d U / d R$ into (14.94) and (14.95), and use of (14.100) gives
\(
\begin{equation}
\left.\langle V\rangle \approx\langle V\rangle\right|{\infty}+\frac{4 A}{R^{6}},\left.\quad\left\langle T{\mathrm{el}}\right\rangle \approx\left\langle T{\mathrm{el}}\right\rangle\right|{\infty}-\frac{5 A}{R^{6}}, \quad R \text { large } \tag{14.104}
\end{equation}
\)
Hence, as $R$ decreases from infinity, the average potential energy at first increases, while the average kinetic energy at first decreases. The combination of these conclusions with our conclusions about $\left.\langle V\rangle\right|{R{e}}$ and $\left.\left\langle T{\mathrm{el}}\right\rangle\right|{R{e}}$ shows that $\langle V\rangle$ must go through a maximum somewhere between $R{e}$ and infinity and $\left\langle T_{\mathrm{el}}\right\rangle$ must go through a minimum in this region.
Now consider small values of $R$. One can treat a diatomic molecule by applying perturbation theory to the united atom (UA) formed by merging the two atoms of the molecule. The perturbation is the difference between the molecular and the united-atom Hamiltonians: $H^{\prime}=\hat{H}{\text {mol }}-\hat{H}{\text {UA }}$. One finds that the molecular purely electronic energy has the following form at small $R$ [W. A. Bingel, J. Chem. Phys., 30, 1250 (1959);
I. N. Levine, J Chem. Phys., 40, 3444 (1964); 41, 2044 (1965); W. Byers Brown and E. Steiner, J. Chem. Phys. 44, 3934 (1966)]:
\(
\begin{equation}
E{\mathrm{el}}=E{\mathrm{UA}}+a R^{2}+b R^{3}+c R^{4}+d R^{5}+e R^{5} \ln R+\cdots \tag{14.105}
\end{equation}
\)
where $E{\mathrm{UA}}$ is the united-atom energy and $a, b, c, d, e$ are constants. For $R \ll R{e}$, we can use (14.105) and $U=E{\text {el }}+V{N N}$ [Eq. (13.8)] to write
\(
\begin{equation}
U(R) \approx \frac{Z{a} Z{b}}{R}+E_{\mathrm{UA}}+a R^{2}, \quad R \text { small } \tag{14.106}
\end{equation}
\)
The virial theorem then gives (in atomic units)
\(
\begin{aligned}
\left\langle T{\mathrm{el}}\right\rangle & \approx-E{\mathrm{UA}}-3 a R^{2}, \quad R \text { small } \
\langle V\rangle & \approx \frac{Z{a} Z{b}}{R}+2 E_{\mathrm{UA}}+4 a R^{2}, \quad R \text { small }
\end{aligned}
\)
Since the virial theorem (14.76) holds for the united atom, we have $\left.\left\langle T{\mathrm{el}}\right\rangle\right|{0}=-E{\mathrm{UA}}$ and $\left.\left\langle V{\text {el }}\right\rangle\right|{0}=2 E{\mathrm{UA}}$. Therefore,
\(
\begin{gather}
\left.\left\langle T{\mathrm{el}}\right\rangle \approx\left\langle T{\mathrm{el}}\right\rangle\right|{0}-3 a R^{2}, \quad R \text { small } \tag{14.107}\
\langle V\rangle \approx \frac{Z{a} Z{b}}{R}+\left.\left\langle V{\mathrm{el}}\right\rangle\right|_{0}+4 a R^{2}, \quad R \text { small } \tag{14.108}
\end{gather}
\)
$\langle V\rangle$ goes to infinity as $R$ goes to zero, because of the internuclear repulsion.
Having found the general behavior of $\langle V\rangle$ and $\left\langle T{\text {el }}\right\rangle$ as functions of $R$, we now draw Fig. 14.1. This figure is not for any particular molecule but resembles the known curves for $\mathrm{H}{2}$ and $\mathrm{H}_{2}^{+}$[W. Kolos and L. Wolniewicz, J. Chem. Phys., 41, 3663 (1964); Slater,
FIGURE 14.1 Variation of the average potential and kinetic energies of a diatomic molecule. The unit of energy is taken as the electronic kinetic energy of the separated atoms.
Quantum Theory of Molecules and Solids, Volume 1, p. 36]. Similar curves hold for other diatomic molecules [see Fig. 1 in J. Hernandez-Trujillo et al., Faraday Discuss., 135, 79 (2007)].
How can we explain the changes in average kinetic and potential energy with $R$ ? Consider $\mathrm{H}_{2}^{+}$. The electronic potential-energy function is in atomic units
\(
\begin{equation}
V{\mathrm{el}}=-\frac{1}{r{a}}-\frac{1}{r_{b}} \tag{14.109}
\end{equation}
\)
If we plot $V{\mathrm{el}}$ for points on the molecular axis for a large value of $R$, we get a curve like Fig. 14.2, which resembles two hydrogen-atom potential-energy curves (Fig. 6.6) placed side by side. We saw that the overlapping of the $1 s$ AOs occurring in molecule formation increases the charge probability density between the nuclei for the ground state. However, Fig. 14.2 shows that the potential energy is relatively high in the region midway between the nuclei when $R$ is large. Thus $\langle V\rangle$ initially increases as $R$ decreases from infinity. Now consider the kinetic energy. The uncertainty principle (5.13) gives $(\Delta x)^{2}\left(\Delta p{x}\right)^{2} \geq \hbar^{2} / 4$. For a stationary state, $\left\langle p{x}\right\rangle$ is zero [see Eq. (3.92) and Prob. 14.31] and (5.11) gives $\left(\Delta p{x}\right)^{2}=\left\langle p{x}^{2}\right\rangle$. Hence a small value of $(\Delta x)^{2}$ means a large value of $\left\langle p{x}^{2}\right\rangle$ and a large value of the average kinetic energy, which equals $\left\langle p^{2}\right\rangle / 2 m$. Thus a compact $\psi{\text {el }}$ corresponds to a large electronic kinetic energy. In the separated atoms, the wave function is concentrated in two rather small regions about each nucleus (Fig. 6.7). In the initial stages of molecule formation, the buildup of probability density between the nuclei corresponds to having a wave function that is less compact than it was in the separated atoms. Thus, as $R$ decreases from infinity, the electronic kinetic energy initially decreases. The energies $E{\text {el }}$ of the two lowest $\mathrm{H}_{2}^{+}$states have been indicated in Fig. 14.2. For large $R$ the region between the nuclei is classically forbidden, but it is accessible according to quantum mechanics (tunneling).
Now consider what happens as $R$ decreases further. Plotting (14.109) for an intermediate value of $R$, we find that now the region between the nuclei is a region of low potential energy, since an electron in this region feels substantial attractions from both nuclei. (See Fig. 14.3.) Hence at intermediate values of $R$, the overlap charge buildup between the nuclei lowers the potential energy. For intermediate values of $R$, the wave function has
FIGURE 14.3 Potential energy along the internuclear axis for electronic motion in $\mathrm{H}{2}^{+}$at an intermediate internuclear distance.
become more compact compared with large $R$, which gives an increase in $\left\langle T{\mathrm{el}}\right\rangle$ as $R$ is reduced. In fact, we see from Fig. 14.1 and Eq. (14.101) that $\left\langle T{\mathrm{el}}\right\rangle$ is greater at $R{e}$ in the molecule than in the separated atoms. Hence the molecular wave function at $R_{e}$ is more compact than the separated-atoms wave functions.
For very small $R$, the average potential energy goes to infinity, because of the internuclear repulsion. However, for $R=R{e}$, Fig. 14.1 shows that $\langle V\rangle$ is still decreasing sharply with decreasing $R$, and it is the increase in $\left\langle T{\mathrm{el}}\right\rangle$, and not the nuclear repulsion, that causes the $U(R)$ curve to turn up as $R$ becomes less than $R{e}$. The squeezing of the molecular wave function into a smaller region with the associated increase in $\left\langle T{\text {el }}\right\rangle$ is more important than the internuclear repulsion in causing the initial repulsion between the atoms.