Abstract
In a crossed molecular beam experiment, we have measured angular and time-of-flight (TOF) distributions of products formed in the reaction K + C6H5I → KI + C6H5 at a collision energy of Etr = 1.9 eV. From these data we have extracted the double-differential reaction cross section in the center-of-mass frame. The brute force technique has been applied to orient for the first time an asymmetric top, namely the reagent molecule C6H5I. The effect of molecular orientation on the angular distribution of products has been investigated. We determined the difference of product intensity for (a) preferred encounters with the I end and the phenyl end (parallel steric effect) and (b) side-on attacks with the axis pointing in two opposite directions (perpendicular steric effect). We find (i) preferred sideways scattering of KI with a mean recoil energy of 31.6% of the totally available energy; (ii) an additional reaction channel with a branching ratio of 0.65% by which a flux of slow products, most likely KI, is formed; (iii) the transformation of the data to a more adapted coordinate frame reveals that backward rather than sideways scattering reflects the dynamics of the reaction; (iv) a tight correlation between the direction of the product flux and the orientation of the molecular axis; (v) the experimental results can be rationalized by the direct interaction with product repulsion (DIPR) model; (vi) the ratio of product yield for attacks of the K atoms to the I end (head) and the phenyl end (tail) amounts to ≈28:1; (vii) the full apex angle of the cone-of-acceptance amounts to ≈110°; (viii) the harpooning mechanism and simple molecular orbital arguments rationalize the impulsive reaction mechanism implicit to the DIPR model and offer an explanation for the existence of a channel for slow KI products.
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