Chemical Activation of Water Molecule by Collision with Spin-Orbit-State-Selected Vanadium Cation:...

We have obtained absolute integral cross sections (σ's) for the reactions of spin-orbit-state-selected vanadium cations, V⁺[a⁵D(J = 0, 2), a⁵F(J = 1, 2), and a³F(J = 2, 3)], with a water molecule (H₂O) in the center-of-mass collision energy range E_cm = 0.1-10.0 eV. On the basis of these state-selected σ curves (σ versus E_cm plots) observed, three reaction product channels, VO⁺ + H₂, VH⁺ + OH, and VOH⁺ + H, from the V⁺ + H₂O reaction are unambiguously identified. Contrary to the previous guided ion beam study of the V⁺(a⁵D) + D₂O reaction, we have observed the formation of the VO⁺ + H₂ channel from the V⁺(a⁵D) + H₂O ground reactant state at low E_cm's (<3.0 eV). No spin-orbit J-state dependences for the σ curves of individual electronic states are discernible, indicating that spin-orbit interactions are weak with little effect on chemical reactivity of the titled reaction. For the three product channels identified, the triplet σ(a³F) values are overwhelmingly higher than the quintet σ(a⁵D) and σ(a⁵F) values, showing that the reaction is governed by a "weak quintet-triplet spin crossing" mechanism, favoring the conservation of total electron spins. The σ curves for exothermic product channels are found to exhibit a rapid decreasing profile as E_cm is increased, an observation consistent with the prediction of the charge-dipole and induced-dipole orbiting model. This experiment shows that the V⁺ + H₂O reaction can be controlled effectively to produce predominantly the VO⁺ + H₂ channel via the V⁺(a³F) + H₂O reaction at low E_cm's (≤0.1 eV) and that the ion-molecule reaction dynamics can be altered readily by selecting the electronic state of V⁺ cation. On the basis of the measured E_cm thresholds for the σ(a⁵D, a⁵F, and a³F: VH⁺) and σ(a⁵D, a⁵F, and a³F: VOH⁺) curves, we have deduced upper bound values of 2.6 ± 0.2 and 4.3 ± 0.3 eV for the 0 K bond dissociation energies, D₀(V⁺-H) and D₀(V⁺-OH), respectively. After correcting for the kinetic energy distribution resulting from the Doppler broadening effect of the H₂O molecule, we obtain D₀(V⁺-H) = 2.2 ± 0.2 eV and D₀(V⁺-OH) = 4.0 ± 0.3 eV, which are in agreement with D₀ determinations obtained by σ curve simulations.