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?

  • $\begingroup$ 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. $\endgroup$ – Oscar Lanzi Feb 4 '19 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.


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