Answers are grounded in this paper and cite their sources.
Product velocity vector distributions show that the reaction O+(H2,H)OH+ and its isotopic variations proceed from ground state reactants to predominantly electronic ground state products via a direct interaction. Spectator stripping is a prominent feature at the lower relative energies of collision, but the stripping peak is lost at energies where the internal excitation of OH+ (3Σ−) makes it unstable to dissociation. Production of forward scattered electronically excited OH+, probably in the 1Δ state, is observable when the initial relative energy is high enough that OH+(3Σ−) formed by spectator stripping is unstable with respect to dissociation. At these relative energies, OH+ from HD is predominantly scattered into barycentric angles less than 90°, while the OD+ appears at approximately 90° at larger angles. The propensity for formation of OD+ at large angles can be understood in terms of general kinematic considerations which apply to collinear collisions.
No takes yet. Share an insight, caveat, or question.
Gillen et al. (1973) studied this question.
Synapse has enriched 10 related papers. Consider them for comparative context: