The hydrogen molecule ion

The hydrogen molecule ion H$_2^+$ is the only molecule for which we can solve the electronic Schrödinger equation exactly. Note that it has just one electron! In fact, there are no multi-electron molecules we can solve exactly. Thus from H$_2$ on up to more complicated molecules, we only have approximate solutions for the allowed electronic energies and wave functions. Before we discuss these, however, let us examine the exact solutions for H$_2^+$ starting with a brief outline of how the exact solution is carried out.

The figure below shows the geometry of the H$_2^+$ molecule ion and the coordinate system we will use.

Figure: The hydrogen molecule ion: Geometry and coordinate system

The two protons are labeled A and B, and the distances from each proton to the one electron are $r_{\rm A}$ and $r_{\rm B}$, respectively. Let $R$ be the distance between the two protons (this is the only nuclear degree of freedom that is important, and the electronic wave function will depend parametrically only on $R$). The coordinate system is chosen so that the protons lie on the $z$-axis one at a distance $R/2$ above the $xy$ plane and one a distance $-R/2$ below the $xy$ plane (see figure). The classical energy of the electron is

$${p^2 \over 2m_e} - {e^2 \over 4\pi\epsilon_0}\left({1 \over r_{\rm A}} + {1 \over r_{\rm B}}\right) = E$$

The nuclear-nuclear term $V_{nn}(R) = e^2/4\pi\epsilon_0 R$ is a constant, and we can define the potential energy relative to this quantity.

The energy is not a simple function of energies for the $x$, $y$, and $z$ directions, so we try another coordinate system to see if we can simplify the problem. In fact, this problem has a natural cylindrical symmetry (analogous to the spherical symmetry of the hydrogen atom) about the $z$-axis. Thus, we try cylindrical coordinates. In cylindrical coordinates the distance of the electron from the $z$-axis is denoted $\rho$, the angle $\phi$ is the azimuthal angle, as in spherical coordinates, and the last coordinate is just the Cartesian $z$ coordinate. Thus,

$$x = \rho\cos\phi\;\;\;\;\;\;\;\;y=\rho\sin\phi\;\;\;\;\;\;\;\;\;z=z$$

Using right triangles, the distance $r_{\rm A}$ and $r_{\rm B}$ can be shown to be

$$r_{\rm A} = \sqrt{\rho^2 + (R/2-z)^2}\;\;\;\;\;\;\;\;\;\; r_{\rm B} = \sqrt{\rho^2 + (R/2+z)^2}$$

The classical energy becomes

$${p_{\rho}^2 \over 2m_e} + {p_{\phi}^2 \over 2m_e \rho^2} + {... ... (R/2-z)^2}} + {1 \over \sqrt{\rho^2 + (R/2+z)^2}}\right) = E$$

First, we note that the potential energy does not depend on $\phi$, and the classical energy can be written as a sum $E = \varepsilon_{\rho,z} + \varepsilon_{\phi}$ of a $\rho$ and $z$ dependent term and an angular term. Moreover, angular momentum is conserved as it is in the hydrogen atom. However, in this case, only one component of the angular momentum is actually conserved, and this is the $z$-component, since the motion is completely symmetric about the $z$-axis. Thus, we have only certain allowed values of $L_z$, which are $m\hbar$, as in the hydrogen atom, where $m = 0,\pm 1, \pm 2,...$. The electronic wave function (now dropping the ``(elec)'' label, since it is understood that we are discussing only the electronic wave function), can be written as a product

$$\psi(\rho,\phi,z) = G(\rho,z) y(\phi)$$

and $y(\phi)$ is given by

$$y_m(\phi) = {1 \over \sqrt{2\pi}}e^{im\phi}$$

which satisfies the required boundary condition $y_m(0) = y_m(2\pi)$. Note that the angular part of this problem is exactly like the particle on a ring problem from problem set # 4.

Unfortunately, what is left in $\rho$ and $z$ is still not that simple. But if we make one more change of coordinates, the problem simplifies. We introduce two new coordinates $\mu$ and $\nu$ defined by

$$\mu = {r_{\rm A} + r_{\rm B} \over R} \;\;\;\;\;\;\;\;\;\; \nu = {r_{\rm A} - r_{\rm B} \over R}$$

Note that when $\nu=0$, the electron is in the $xy$ plane. Thus, $\nu$ is analgous to $z$ in that it varies most as the electron moves along the $z$ axis. The presence or absence of a node in the $xy$ plane will be an important indicator of wave functions that lead to a chemical bond in the molecule or not. The coordinate $\mu$, on the other hand,is minimum when the electron is on the $z$ axis and grows as the distance of the electron from the $z$ axis increases. Thus, $\mu$ is analogous to $\rho$. The advantage of these coordinates is that the wave function turns out to be a simple product

$$\psi(\mu,\nu,\phi,R) = M(\mu)N(\nu)y(\phi)$$

which greatly simplfies the problem and allows the exact solution.

The mathematical structure of the exact solutions is complex and nontransparent, so we will only look at these graphically, where we can gain considerably insight. First, we note that the quantum number $m$ largely determines how the solutions appear. First, let us introduce the nomenclature for designating the orbitals (solutions of the Schrödinger equation, wave functions) of the system

1.

If $m=0$, the orbitals are called $\sigma$ orbitals, analogous to the $s$ orbitals in hydrogen.

2.

If $m=1$, the orbitals are called $\pi$ orbitals, analogous to the $p$ orbitals in hydrogen.

3.

If $m=2$, the orbitals are called $\delta$ orbitals, analogous to the $d$ orbitals in hydrogen.

4.

If $m=3$, the orbitals are called $\phi$ orbitals, analogous to the $f$ orbitals in hydrogen.

These orbitals are known as molecular orbitals because the describe the electronic wave functions of an entire molecule. There are four designators that we use to express each molecular orbital:

I.

A greek letter, $\sigma$, $\pi$, $\delta$, $\phi$, .... depending on the quantum number $m$.

II.

A subscript qualifier g or u depending on how an orbital $\psi$ behaves with respect to a spatial reflection or parity operation ${\bf r}\rightarrow -{\bf r}$. If

$$\psi(-{\bf r}) = \psi(ur)$$

then $\psi({\bf r})$ is an even function of ${\bf r}$, so we use the ``g'' designator, where g stands for the German word gerade, meaning ``even''. If

$$\psi(-{\bf r}) = -\psi({\bf r})$$

then $\psi({\bf r})$ is an odd function of ${\bf r}$, and we use the ``u'' designator, where u stands for the German word ungerade, meaning ``odd''.

III.

An integer $n$ in front of the Greek letter to designate the energy level. This is analogous to the integer we use in atomic orbitals (1s, 2s, 2p,...).

IV.

An asterisk or no asterisk depending on the presence or absence of nodes between the nuclei. If there is significant amplitude between the nuclei, then the orbital favors a chemical bond, and the orbital is called a bonding orbital. If there is a node between the nuclei, the orbital does not favor bonding, and the orbital is called an antibonding orbital.

So, the first few orbitals, in order of increasing energy are:

$$1\sigma_g, 1\sigma_u^*, 2\sigma_g, 2\sigma_u^*, 1\pi_u, 3\sigma_g, 1\pi_g^*, 3\sigma_u$$

These orbitals are depicted in the figure below:

Figure: Molecular orbitals in H$_2^+$.

The $1\sigma _g$ and $1\sigma _u^*$ orbitals are the lowest in energy, however, note that the $1\sigma_u*$ contains one more node than the $1\sigma _g$ orbital, hence it has a higher energy. Similarly for the $2\sigma_g$ and $2\sigma_u^*$ orbitals. The former has two nodes while the latter has three and, therefore, it is of higher energy. The next set of orbitals are displayed in the right panel. In this set of orbitals, the $1\pi_u$ is the lowest energy with a single node. The number of nodes increases as we go up in energy in this subset of orbitals. In addition, in all of the orbitals, the nodal structure reveals the bonding/antibonding character of each orbital. When there is significant amplitude between the nuclei, the orbital is a bonding orbital, otherwise, when there is a node there, it is an antibonding orbital.

What do bonding and antibonding orbitals mean in terms of the corresponding energy levels. Consider just the first two energy levels $\varepsilon_0(R)$ and $\varepsilon_1(R)$ corresponding to the $1\sigma _g$ and $1\sigma _u^*$ orbitals. The ground-state orbital $1\sigma _g$ is bonding and the first excited state $1\sigma _u^*$ is antibonding. In the figure below, we plot the energy levels $\varepsilon_0(R)$ and $\varepsilon_1(R)$ as functions of the internuclear separation $R$:

Figure: First two electronic energy levels as functions of the internuclear separation for the H$_2^+$ molecule ion

The picture reveals that the curve $\varepsilon_0(R)$ (the red curve) has a well-defined minimum, corresponding to the equilibrium bond length. This is due to the bonding character of the orbital. On the other hand, the first excited state $\varepsilon_1(R)$ (the blue curve) has no such minimum. There is no stable bond length in this orbital, which reflects the antibonding character. Thus, exciting the molecule into this electronic state causes it to dissociate. The energy needed to do this depends on the internuclear separation $R$. When $R=R_e$, the equilibrium bond length, an energy of several Ry would be needed. At larger separations, the energy decreases as the two curves approach each other.