Abstract

The photodissociation of S2Cl2 at 248 and 193 nm has been studied by translational spectroscopy. Upon excitation at 248 nm, S2Cl2 undergoes predominantly the dissociation channels to form the products, (1) S2Cl+Cl and (2) SCl+SCl. The relative yield of (1)/(2) was determined at ∼3.0:1.0. The S2Cl+Cl products were detected with two separate product translational energy distributions of 〈Et〉=14 and 41 kcal/mol. The slow S2Cl fragment, which is internally excited, undergoes a secondary dissociation to form S2+Cl. Dissociation mechanisms which are consistent with the isotropic recoiling are proposed: the slow component could arise from the ground electronic state while the fast component would more likely originate from a mixed excitation of 1A and 1B states (C2 symmetry). The SCl product with anisotropy parameter β=1.6, on the other hand, should proceed by an excited 1B state. At 193 nm, a three-body formation of S2+2Cl becomes the more efficient process than the simple S–Cl or S–S bond fission. Assuming that the rupture of both S–Cl bonds occurs in a single kinetic step, a P(Et) distribution with 〈Et〉=30 kcal/mol is determined. Because of the measured β=−0.3 for the formation of triple products, a higher 1B state will be involved in the excitation.

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