# Why does Pb(II) promote ligand substitution?

I have heard of an article where divalent lead $\ce{PbL2}$ ($\ce{L}$ is ligand) reacts with something (again not sure, as I dont know the article) and what it reacts with gains a ligand.

In less confusing words:

$\ce{PbL2 + MXn}$ ($\ce{X}$ is another ligand, not necessarily halogen) ==> $\ce{PbL(n-1)X + MX(n-1)L}$ Or something similar. That is what I got from hearing about the article. I was wondering if you guys can share your thoughts as to why happens?

• It is maybe a better question if you yourself try to clean up some of the "not sure", "or something similar" points and clean up your reaction equation (there is 2L on the left, n L on the right). Also, if you can be a little bit more specific than "why happens?"... It seems like a vaguely documented ligand exchange, which can happen in any equilibrium. – Greg May 11 '15 at 5:35