# Why doesn't tetraiodine exist?

I've run into a question that says after doing the Lewis structure of $\ce{IF3}$, which I did, explain why $\ce{I4}$ doesn't exist.

I am confused as to how it can be explained. I've thought that maybe it has to do with iodine's valence orbitals, both hybridised or not being too large and as a result the different orbitals would have to overlap and superimpose. But again I'm not sure.

• There is an $\ce{I4^2-}$ anion. – airhuff May 24 '17 at 18:58
• I think due to the large atomic size of Iodine atom, the central Iodine can't accommodate other 3 Iodine atoms (as well as 2 lone pairs of electrons) . There is strong probability for the ligand Iodine atoms to suffer steric repulsion and hence the molecule $I_4$ doesn't exist! – chail10 May 24 '17 at 19:19
• @chail10 en.wikipedia.org/wiki/Polyiodide iodine can coordinate 3 other iodine atoms. Also elements aren't capitalised. – Mithoron May 24 '17 at 19:58
• @Mithoron Yupp, I wasn't sure about that and that's why I didn't post it as answer! Thanks for the reference. – chail10 May 24 '17 at 20:05

While it's true that there are polyiodides, the $\ce{I4^2-}$ ion has a completely different geometry from $\ce{IF3}$ molecule. The ion is most easily regarded as two iodine molecules stuck end on and the ensemble is bound because two additional electrons are popped into the in-phase combination of each side's $\ce{I(p_z)-I(p_z)}$ antibonding orbital. This is bonding between the fragments and antibonding within the fragments. Since the individual molecules can't do anything with the extra two electrons, it's best just to stay together and stabilize them together.
As to why there isn't a $\ce{I4}$ that is isostructural to $\ce{IF3}$, this is because the stability of interhalogens is heavily influenced by the ionic character of bonds. Recent work[1,2] has shown that the bonding in halogens is dominated by "charge-shift" interactions in which the ground covalent state is stabilized by ionic states in which both electrons of the bond have been transfered to either atom (e.g. $\ce{F-F}$ is stabilized by $\ce{F+-F-}$ and $\ce{F^--F+}$ states). This stabilization is due to highly electronegative atoms. The stability of higher interhalogens relies on this "charge-shift" stabilization. Since I is a lot less electronegative than F, this charge-shift stabilization is much weaker (and this is also the reason why nearly all of the higher interhalogens have fluorine as a bonding partner). Since this charge-shift stabilization is absent, the 4 iodines are much happier to break apart into 2 $\ce{I2}$ molecules.
• $\ce{I4}$ certainly is a minimum on the potential energy surface, albeit not isostructural to $\ce{IF3}$, as the former has $D_\mathrm{3h}$ symmetry, not just $C_\mathrm{2v}$ like the latter. On the (admittedly ridiculously low DF-BP86/def2-SVP) level of theory it is even lower in the electronic energy than the dissociated $\ce{I2}$ molecules. Clearly an explanation solely based on electronic effects, like that "charge-shift explanation", but not limited to it, cannot be a sufficient reason. And just because it has not been observed, it does not mean it cannot exist. – Martin - マーチン Jun 5 '17 at 11:05