[pear]

Hi,

So I'm trying to understand what happens when excess I^- is added to the equilibrium  PbI₂(s)    Pb²⁺(aq) + 2I^-(aq)

What I'm thinking is that first the reaction is governed by solubility product.  And eventually, [I^-] is so high that Q >>>> Ksp and the lead re-dissolves to instantly form covalent bonds with the two outer electrons && the two electrons from the unfilled d shell, thus removing twice the [I^-] as the PbI₂ molecule.

Am I thinking logically/correctly?

[pear]

So here's the answer, for anyone else wondering..

Eventually the amount of energy possessed by a solution with excess negatively charged ions will overcome the high energy input required for the 2 d-shell electrons of Pb to create covalent bonds alongside the 2 outer p-shell electrons.  So instead of the solid PbI₂, which only takes care of 2 I^- ions, the covalent, aqueous molecule PbI₄^2- forms and lessens the amount of excess energy in the solution.

[zeoblade]

Isn't this an application of Le Chatelier's principle where if you add excess product, it drives the reaction to favour the reverse reaction and you will precipitate the reactant?

[Borek]

PbI₄^2- formation is a separate reaction, just applying LeChetelier's principle to half of the system won't get you a correct result.

[pear]

PbI₄^2- formation is a separate reaction, just applying LeChetelier's principle to half of the system won't get you a correct result.

But the explanation I posted is correct, right?