I read that lead(II) iodide is soluble in excess KI solution due to complexation, but lead(II) cyanide does not dissolve in excess cyanide solution. This seems wrong since cyanide is a much better ligand that iodide, so shouldn't it be able to form a complex more easily?

• Cyanide is not always a much better ligand than iodide. It preferably complexes some things while iodide preferably complexes others. Lead(II) is one of those things that prefers iodide. – Oscar Lanzi Feb 4 at 11:00

Cyanide is generally better than iodide in transition metal complexes because it acts as a pi-acceptor. It has empty orbitals that overlap with the partially filled d-orbitals of the metal, lowering the energy of these electrons.

Iodide is a sigma-donor, meaning that it donates electron density to empty metallic orbitals. The effect is small compared to a pi-acceptor in a transition metal, because the major contribution to stability is the energy of d-orbital electrons.

However in Pb(II) complexes, the d-orbitals are full already, so here the sigma-bonding mechanism for stabilisation is key. As cyanide isn't as good of a donor as iodide, then iodide forms a complex whereas cyanide doesn't.

I can only guess here, but main-group complexes and transition-metal complexes are sometimes a little different. Once you leave the d-orbital chemistry you have to ask yourself where you get low laying, empty orbitals. If you take $$\ce{BiI3}$$ for example. This is a $$\ce{Bi^3+}$$ so there is still one lone-pair. Hence you will get a tetrahedron with three $$\ce{I^-}$$ and one lone-pair. If you add for example $$\ce{NaI}$$ it will dissolve, but how? The next empty d-orbitals would be the 6d-orbitals, here the 7s would be much more favourable but they are still quite high. What's actually happening is that you shift electron density from the extra iodide-ligand into an anti-bonding $$\ce{Bi-I}$$-bond of the $$\ce{BiI3}$$ molecule.

The same thing goes for $$\ce{PbI2}$$. You have the two iodide-ligands and one lone-pair. So there is already a free spot on the metal. That means first you get $$\ce{[PbI3]^-}$$ and then, by a similar mechanism a $$\ce{[PbI4]^2-}$$. We call this a hypervalent compound.

That is at least how Olaf Kühl describes it in his book.

At this point I can only guess. If the approaching Iodide / Cyanide acts as a Lewis-base this means that it will donate electron density into empty orbitals. If those empty orbitals are the anti-bonding σ*-orbitals of a $$\ce{Pb-L}$$-bond, we require those orbitals to be low in energy.

The problem is that I don't really know how to compare an iodide with a cyanide in terms of covalency and electronegativity here. The bond energies in $$\ce{Pb-X}$$ decrease from fluoride to iodide. Hence I expect the $$\ce{Pb-I}$$-bond to be quite weak. The problem is that cyanide is a soft base as well and only a little smaller.