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?


2 Answers 2


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.

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

The problem boils down to competing stabilization factors. As discussed here, an isolated triiodide ion, or any trihalide ion ($\ce{X_3^-}$), is more strongly stabilized than the separate monatomic halide ion plus the neutral halogen diatomic molecule ($\ce{X^- + X2}$). But in practice, except for plasmas, ions are generally not isolated; they are accompanied by counterions or the oppositely-charged ends of solvating dipoles. These are more favorably attracted to a compact, monatomic ion and so tend to pull the triiodide ion apart. Even where the triiodide is stable, such pulling apart commonly occurs to some extent. From Wikipedia:

In ionic compounds, the bond lengths and angles of triiodide vary depending on the nature of the cation. The triiodide anion is easily polarised and in many salts, one $\ce{I−I}$ bond becomes shorter than the other. Only in combination with large cations, e.g. a quaternary ammonium such as $\ce{[N(CH3)4]^+}$, may the triiodide remain roughly symmetrical.1

A table from the Wikipedia article, citing 2 for the methanol solution, renders results for some singly charged cations. In effect the anion is not purely a triiodide ion, but an intermediate between triiodide and dissociated iodide plus $\ce{I2}$:

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When we try a doubly charged cation, the electrostatic attraction is stronger and with it, there is an even greater tendency to pull the proposed triiodide ion apart completely. Attempting to bind triiodide ion with $\ce{Mg^{2+}}$ or even $\ce{Ba^{2+}}$ is likely to produce instead the more tightly packed monatomic iodide salt plus $\ce{I_2}$. Cutting down the electrostatic attraction, or equivalently making the anion more isolated, by using a large multiatomic counterion offers the best chance for creating a bis-triiodide salt.

Cited Reference

  1. Atkins; et al. (2010). Inorganic Chemistry (5th ed.). Oxford University Press. p. 431. ISBN 9780199236176.

  2. Heo, Jun; Kim, Jong Goo; Choi, Eun Hyuk; Ki, Hosung; Ahn, Doo-Sik; Kim, Jungmin; Lee, Seonggon; Ihee, Hyotcherl (26 January 2022). "Determining the charge distribution and the direction of bond cleavage with femtosecond anisotropic x-ray liquidography". Nature Communications. 13 (1): 522. https://doi.org/10.1038/s41467-022-28168-0. ISSN 2041-1723. PMC 8792042. PMID 35082327.


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