|MadSci Network: Chemistry|
Because you only mention tin without the form, I will deal with the reaction of iodine with tin metal. I looked in several inorganic chemistry books and found various levels of information. From "Chemistry of the Elements" by Greenwood and Earnshaw (Leeds), supplied some of the following information. To produce SnI2 in anhydrous form, react Sn metal (hot) with I2 at elevated temperatures. No water is used. Also and better, Sn and I2 (excess) can be reacted in a solution of hydrochloric acid to give SnI2. If Sn and I2 are reacted in caustic water, hydrates and hydroxides are formed. The reaction should be done in acid solution, since Sn(OH)x can be formed easily in basic solutions. SnI4 can also be formed if the +4 oxidation form of the metal is used. This is the more common form of Sn iodide and can be formed from the reaction of SnCl4 with I2 or I- (excess). The Sn iodides have the ability of forming molecular polymer structures where one iodide bridges between the Sn. These would be structures like Sn2I3, etc, with larger structure predominating. THe SnI4's do this also. The mechanisms for these reactions will vary. For the anhydrous reaction (no water), the reduction of the iodine and the oxidation of the tin will be a surface phenomenon with the I2 bonding both atoms to the surface. It could also be a single atom attachment with a more complicated mechanism involving iodine radicals. This mechanism has probably been delineated, but would take some library time to find the information. In acid, the mechanism will be similar, except that the oxidized tin (Sn+2) and iodide (I-) may react away from the surface of the metal. For example, electron transfer could take place on the surface, and formation of SnI2 could take place away from the surface (but near) in the form of hydrated atoms.
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