[sonicbug]

Hello all, I have a few questions regarding a lab on the oxidation states of tin. In the lab we prepared Tin 4 Iodide which ended up being orange crystals. We prepared the following from a reflux setup, and then filtration to separate the mixture.I did some research on this lab but was unable to understand the chemistry behind the lab, which is disappointing to say the least, anyways. I have some idea of what was going on but would really like to know exactly what happened. First, SnI4 is stable, why are PbBr4,PbI4 unstable compounds? I thought it had to do with their electronic configurations, in that SnI4 has a greater Zeff and a smaller covalent radius opposed to the others. Secondly I have to right out the oxidation and reduction (redox) half reactions for this experiment. For, Sn + 2I2 --> SnI4. And finally, when my group combined SnI4 with acetone, then split the two into equal amounts, and added H20 to one flask, the mixture went clear as opposed to the other one, which we added KI solution to, it turned yellow.

Any information pertaining to this topic would be much appreciated Thank you sonicbug

[JaCh]

I think although SnI₄ is relatively stable it can be decomposed in the presence of KI to SnI₂:  SnI₄  = SnI₂ + I₂ <BR>
It is facilitated as I₂ with KI forms KI₃, so iodine (as such) is removed from the solution. By removal of the product we can force the reaction to proceed.