Monday, February 21, 2011

Quantum chemistry of hydrogen bonding

Previously, I posted some notes about an empirical valence bond model for hydrogen bonding.
I have been thinking about whether it is possible to justify such a "simple" picture [and effective Hamiltonian] from high-level quantum chemical calculations. A helpful paper I have been looking at is a 2001 J. Chem. Phys. by Kowal, Roszak, and Leszczynski. They perform complete active space - self-consistent field (CASSCF) calculations on the ground and excited states of the water dimer, (H4O2) and H3O2- and H2O52+. The equilibrium geometry of the latter is pictured below. The three systems correspond to weak, intermediate, and strong hydrogen bonds respectively.

This morning I read slowly read through the paper again. They have 12 electrons in an active space of 7-9 (I think) orbitals, which are localised on the individual atoms.
Below I reproduce the potential energy curves as the O-H bond length with increased with the distance between the oxygen atoms at the equilibrium value.
Notice that in the ground state the potential is very flat (and hence anharmonic).

They also find significant differences in the potential energy surfaces between the three molecular systems they study.

They find that the ground state wavefunction is close to one Slater determinant but usually two determinants are required for the excited states. It is not clear to me why this is? I would have thought that the empirical valence bond (EVB) model would have required two Slater determinants for all the states. Furthermore, the EVB model and the associated four electron-three centre bonding model would suggest that a possibly a four-electron three molecular orbital active space might be sufficient.

No comments:

Post a Comment

A very effective Hamiltonian in nuclear physics

Atomic nuclei are complex quantum many-body systems. Effective theories have helped provide a better understanding of them. The best-known a...