In general, I found that triiodide ion generally forms ionic bond with cations having +1 charge like $\ce{NH4+I3-}$, $\ce{K+I3-}$, $\ce{Tl+I3-}$. But, I have never seen triiodide ion forming ionic bond with cations having +2 or +3 charge like $\ce{Ba^2+(I3^{-})2}$, $\ce{Pb^2+(I3^{-})2}$, $\ce{Bi^3+(I3^{-})3}$. If properly written, it would look like $\ce{Ba(I3)2}$, $\ce{Pb(I3)2}$, $\ce{Bi(I3)3}$. These compound has never existed and would not likely to exist. But what is the reason behind this?

Recently I went through this question:-

Why isn't thallium triiodide stable?

One of the commenter said:-

Thallium Tri-Iodide $\ce{Tl+(I3)-}$ does exist. Thallium (III) Iodide, $\ce{Tl^3+(I^{-})3}$ to my knowledge, does not.

I was quite surprised at first seeing that $\ce{TlI3}$ exist because I knew that due to inert pair effect, thallium will form Thallium(I) compounds like $\ce{TlCl}$, $\ce{TlBr}$, $\ce{TlI}$ etc. but I realized that the compound contains $\ce{Tl+}$ and not $\ce{Tl^3+}$. But what about the corresponding $\ce{Pb^2+}$ and $\ce{Bi^3+}$ ions? Why does not they form triiodide compounds?


1 Answer 1


Why triiodide ion does not form ionic bond with cation having +2 charge

It does. Specifically, quick googling for "bis-triiodide" gives out of the blue links like this: poke, poke, so the question is not 'why no bivalent triiodides', but 'why only super-large bications can form triiodides.'

The answer, I suspect, lies in charge density of the cations and their ability to form strong ionic bonds.

It is well known, that energy of ionic crystal lattice strongly increases with decrease of the size of ions involved. Triiodide ion, being quite large, is not very good for forming strong ionic lattices, forming only very weak ionic bonds. Iodide ion, on the other hand, is much better for that.

The energy of formation of triiodide anion from iodine molecul and iodide anion is very low. Thus, triiodie will break to form a crystal lattice with higher formation energy at first chance. So, for a solid triiodide to form, an ion with very low charge density is required, so the energy of crystal lattice would not benefit significantly from triiodide anione dissociation. This rules out any 'naked' cation with charge +2 or more and actually most 'naked' cations with charge +1. On the other hand, by covering dication in a shell of organic molecules strongly bound to the cation, we gain a very large cation, and a solid triiodide can form.

A very similar effect guards temperature of decomposition of solid carbonates, for example, as the decomposition involves breaking of a larger carbonate dianion into a much smaller oxide dianion.

  • 1
    $\begingroup$ They are some seriously big compounds :) $\endgroup$ Apr 18, 2016 at 3:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.