Abstract

According to the conventional picture, the aqueous or "hydrated" electron, e-(aq), occupies an excluded volume (cavity) in the structure of liquid water. However, simulations with certain one-electron models predict a more delocalized spin density for the unpaired electron, with no distinct cavity structure. It has been suggested that only the latter (non-cavity) structure can explain the hydrated electron's resonance Raman spectrum, although this suggestion is based on calculations using empirical frequency maps developed for neat liquid water, not for e-(aq). All-electron ab initio calculations presented here demonstrate that both cavity and non-cavity models of e-(aq) afford significant red-shifts in the O-H stretching region. This effect is nonspecific and arises due to electron penetration into frontier orbitals of the water molecules. Only the conventional cavity model, however, reproduces the splitting of the H-O-D bend (in isotopically mixed water) that is observed experimentally and arises due to the asymmetric environments of the hydroxyl moieties in the electron's first solvation shell. We conclude that the cavity model of e-(aq) is more consistent with the measured resonance Raman spectrum than is the delocalized, non-cavity model, despite previous suggestions to the contrary. Furthermore, calculations with hybrid density functionals and with Hartree-Fock theory predict that non-cavity liquid geometries afford only unbound (continuum) states for an extra electron, whereas in reality this energy level should lie more than 3 eV below vacuum level. As such, the non-cavity model of e-(aq) appears to be inconsistent with available vibrational spectroscopy, photoelectron spectroscopy, and quantum chemistry.

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