# 161: Molecular Orbital Analysis for the Hydrogen Molecule Ion Bond

In this analysis the molecular orbital for the hydrogen molecule ion is formed as a linear combination of scaled hydrogenic 1s orbitals centered on the nuclei, a and b.

$\begin{matrix} \Psi_{MO} = \frac{a+b}{ \sqrt{2+2S}} & \text{where} & a = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} (- \alpha,~r_a ) & b = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( - \alpha,~r_b) & S = \int ab d \tau \end{matrix}$

The molecular energy operator in atomic units:

$H = \frac{-1}{2} \left[ \frac{d}{dr} \left( r^2 \frac{d}{dr} \blacksquare \right) \right] - \frac{1}{r_a} - \frac{1}{r_b} + \frac{1}{R}$

The energy integral to be minimized by the variation method:

$E = \frac{ \int (a+b) H(a+b) d \tau}{2+2S} = \frac{Haa + Hab}{1+S}$

When this integral is evaluated a two-parameter variational expression for the energy (highlighted below) is obtained.

Electron mass:

m = 1

Seed values for the variational parameter and internuclear separation:

$\begin{matrix} \alpha = 1 & R = .1 \end{matrix}$

$E ( \alpha,~R) = \frac{ - \alpha^2}{2m} + \frac{ \frac{ \alpha^2}{m} - \alpha - \frac{1}{R} + \frac{1}{R} (1 + \alpha R) \text{exp} (-2 \alpha R) + \alpha \left( \frac{ \alpha}{m} - 2 \right) (1 + \alpha R) \text{exp} (- \alpha R)}{1 + \text{exp} ( - \alpha R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} + \frac{1}{R}$

Minimization of the energy of the hydrogen molecule ion follows. There are two variational parameters, the orbital scale factor and the internuclear distance.

$\begin{matrix} \text{Given} & \frac{d}{d \alpha} E( \alpha,~R) = 0 & \frac{d}{dR} E ( \alpha,~R) = 0 & \begin{pmatrix} \alpha \\ R \end{pmatrix} = \text{Find} ( \alpha,~R) & \begin{pmatrix} \alpha \\ R \end{pmatrix} = \begin{pmatrix} 1.23803 \\ 2.0033 \end{pmatrix} & E ( \alpha,~R) = -0.58651 \end{matrix}$

The calculation yields a stable molecule ion as is shown here.

$\begin{matrix} ~ & \text{Hydrogen Molecule Ion} = & \text{Hydrogen Atom} + & \text{Hydrogen Ion} \\ \text{Theory} & -0.5865 & -0.5000 & 0 & \text{Bond Energy} = 0.0865 \\ \text{Experiment} & -0.6029 & -0.5000 & 0 & \text{Bond Energy} = 0.1029 \end{matrix}$

The experimental ground state energy is -0.6029 Eh. The error in this calculation is calculate two ways: error in bond energy and error in total ground state energy.

$\begin{matrix} \frac{.1029 - .0865}{.1029} = 15.9378 \% & \left| \frac{E ( \alpha,~R)+ .6029}{-.6029} \right| = 2.71911 \% \end{matrix}$

It is instructive to calculate the kinetic and potential energy contributions to the total energy.

$E = \int \Psi_{MO} (T+V) \Psi_{MO} d \tau = Taa + Tab + Vaa + Vab$

$\begin{matrix} Taa ( \alpha,~R) = \frac{ \frac{ \alpha^2}{2m}}{1 + \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} & Tab ( \alpha,~R) = \frac{ \frac{ \alpha^2}{2m} \text{exp} ( - \alpha,~R) \left( 1 + \alpha R - \frac{ \alpha^2 R^2}{3} \right)}{1 + \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} \\ Vaa ( \alpha,~R) = \frac{ \left( \frac{1}{R} + \alpha \right) \text{exp} (-2 \alpha R) - \alpha}{1 + \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} & Taa ( \alpha,~R) = \frac{ \frac{-1}{3} \text{exp} ( \alpha,~R) \frac{3 \alpha R + 5 \alpha^2 R^2 - 3}{R}}{1 + \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} \end{matrix}$

First we establish that these terms are correct by showing that they sum to the correct ground state energy calculated earlier.

$Taa ( \alpha,~R) + Tab ( \alpha,~R) + Vaa ( \alpha,~R) + Vab ( \alpha,~R) = -0.58651$

Next we establish that the virial theorem is obeyed. Quantum mechanical calculations that violate the appropriate virial theorem are not valid. For atomic and molecular systems the virial theorem is: <E> = -<T> = <V>/2.

$\begin{matrix} \text{Electron Kinetic Energy} & \text{Total Potential Energy} & \text{Virial Theorem Satisfied} \\ Taa ( \alpha,~R) + Tab ( \alpha,~R) = 0.58651 & Vaa ( \alpha,~R) + Vab ( \alpha,~R) = -1.17301 & \frac{|Vaa ( \alpha,~R) + Vab ( \alpha,~R)|}{Taa ( \alpha,~R) + Tab ( \alpha,~R)} = 2 \end{matrix}$

Next we separate the nuclear potential energy into its two components: electron-nucleus and nucleus-nucleus.

$\begin{matrix} \text{Electron Kinetic Energy} & \text{Electron-Nucleus Potential Energy} & \text{Nucleus-Nucleus Potential Energy} \\ T = Taa ( \alpha,~R) + Tab ( \alpha,~R) & Ven = Vaa ( \alpha,~R) + Vab ( \alpha,~R) - \frac{1}{R} & Vnn = \frac{1}{R} \\ T = 0.58651 & Ven = -1.67219 & Vnm = 0.49918 \\ \frac{T}{T + |Ven|+Vnn} = 21.27 \% & \frac{|Ven|}{T + |Ven| + Vnn} = 60.63 \% & \frac{Vnn}{T + |Ven|+Vnn} = 18.10 \% \end{matrix}$

In light of these calculations the following comments on the covalent bond in the H2+ are made. Ven is the largest and only negative term, and might be thought of as the glue holding the H2+ molecule together. However, a ground state (stable molecule) requires "energetic" opposition to the attractive Ven, otherwise molecular collapse occurs. This opposition is provided by T and Vnn. It might surprise those who think that chemical bonding is simply an electrostatic phenomenon, that nuclear repulsion makes a smaller contribution to molecular stability than electron kinetic energy.