Molecular Orbital Description of the Hydrogen Molecule

Setting up the Hamiltonian¶

{H = -\frac{\hbar^2}{2m_e}\left( \Delta_1 + \Delta_2\right) + \frac{e^2}{4\pi\epsilon_0}\left(\frac{1}{R} + \frac{1}{r_{12}} - \frac{1}{r_{A1}} - \frac{1}{r_{A2}} - \frac{1}{r_{B1}} - \frac{1}{r_{B2}}\right)}

The main difficulty in the molecular Hamiltonian is the $1/r_{12}$ term, which couples the two electrons to each other. This means that a simple product wavefunction is not sufficient. No analytic solution has been found for the electronic Schrodinger equation of $H_2$ .
For this reason, we solve the problem approximately by using the LCAO-MO approach used previously. For example, the ground state for $H_2$ is obtained by placing two electrons with opposite spins in the $1\sigma_g$ orbital. This assumes that the wavefunction is expressed as an antisymmetrized product (a Slater determinant).

According to the Pauli principle, two electrons with opposite spins can be assigned to a given spatial orbital. As a first approximation, we assume that the molecular orbitals in $H_2$ remain the same as in $H_2^+$ . Hence both electrons occupy the $1\sigma_g$ orbital (the ground state) and the electronic configuration is denoted ( $1\sigma_g)^2$ . This is similar to the notation used previously for atoms (for example, the He atom is ( $1s)^2$ ).
The molecular orbital for electron 1 in the $1\sigma_g$ molecular orbital is:

{1\sigma_g(1) = \frac{1}{\sqrt{2(1 + S)}}(1s_A(1) + 1s_B(1))}

Previously we found that the total wavefunction must be antisymmetric with respect to exchange of electron indices. This can be achieved by using the Slater determinant:

where $\alpha$ and $\beta$ denote the electron spin. The Slater determinant can be expanded as follows:

{\psi_{MO}^{(1\sigma_g)^2} = \frac{1}{\sqrt{2}} (1\sigma_g(1)1\sigma_g(2)\alpha (1)\beta (2) - 1\sigma_g(1)1\sigma_g(2)\beta (1)\alpha (2))}

{= \frac{1}{2\sqrt{2}(1 + S_{AB})}(1s_A(1) + 1s_B(1))(1s_A(2) + 1s_B(2)) (\alpha (1)\beta (2) - \alpha (2)\beta (1))}

Note that this wavefunction is only approximate and is definitely not an eigenfunction of the $H_2$ electronic Hamiltonian. Thus we must calculate the electronic energy by taking the expectation value of this wavefunction with the Hamiltonian (the actual calculation is not shown):

{E(R) = 2E_{1s} + \frac{e^2}{4\pi\epsilon_0 R} - \textnormal{integrals}}

where $E_{1s}$ is the electronic energy of one hydrogen atom. The second term represents the Coulomb repulsion between the two positively charged nuclei, and the last term (“integrals”) contains a series of integrals describing the interactions of various charge distributions with one another (see P. W. Atkins, Molecular Quantum Mechanics, Oxford University Press). With this approach, the minimum energy is reached at $R$ = 84 pm (experimental 74.1 pm) with a dissociation energy $D_e$ = 255 kJ mol $^{-1}$ (experimental 458 kJ mol $^{-1}$ ).

This simple approach is not very accurate, but it demonstrates that the method works. To improve the accuracy, ionic and covalent terms should be considered separately:

{\underbrace{1s_A(1)1s_A(2)}_{\textnormal{Ionic (H}^- \textnormal{ + H}^+)} + \underbrace{[1s_A(1)1s_B(2) + 1s_A(2)1s_B(1)]}_{\textnormal{Covalent (H + H)}} + \underbrace{1s_B(1)1s_B(2)}_{\textnormal{Ionic (H}^+ \textnormal{ + H}^-)}}

Both covalent and ionic terms can be introduced into the wavefunction with their own variational parameters $c_1$ and $c_2$ :

{\psi = c_1\psi_{\textnormal{covalent}} + c_2\psi_{\textnormal{ionic}}}

{\psi_{\textnormal{covalent}} = 1s_A(1)1s_B(2) + 1s_A(2)1s_B(1)}

{\psi_{\textnormal{ionic}} = 1s_A(1)1s_A(2) + 1s_B(1)1s_B(2)}

Note that the variational constants $c_1$ and $c_2$ depend on the internuclear distance $R$ . Minimization of the energy expectation value with respect to these constants gives $R_e$ = 74.9 pm (experiment 74.1 pm) and $D_e$ = 386 kJ mol $^{-1}$ (experiment 458 kJ mol $^{-1}$ ).
Further improvement can be achieved by adding higher atomic orbitals to the wavefunction. The previously discussed Hartree-Fock method provides an efficient way of solving the problem. Recall that this method is only approximate, as it ignores electron-electron correlation effects completely. The full treatment requires configuration interaction methods, which can yield essentially exact results: $D_e$ = 36117.8 cm $^{-1}$ (CI) versus $36117.3\pm1.0$ cm $^{-1}$ (experiment), and $R_e$ = 74.140 pm versus 74.139 pm (experiment).