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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.

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    $\begingroup$ There is an $\ce{I4^2-}$ anion. $\endgroup$
    – airhuff
    May 24, 2017 at 18:58
  • $\begingroup$ 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! $\endgroup$
    – chail10
    May 24, 2017 at 19:19
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    $\begingroup$ @chail10 en.wikipedia.org/wiki/Polyiodide iodine can coordinate 3 other iodine atoms. Also elements aren't capitalised. $\endgroup$
    – Mithoron
    May 24, 2017 at 19:58
  • $\begingroup$ @Mithoron Yupp, I wasn't sure about that and that's why I didn't post it as answer! Thanks for the reference. $\endgroup$
    – chail10
    May 24, 2017 at 20:05

1 Answer 1

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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.

  1. Sason Shaik, David Danovich, Wei Wu, and Philippe C. Hiberty, Nature Chemistry 2009, 1, 443 - 449. DOI: 10.1038/nchem.327
  2. Daniel S. Levine, Paul R. Horn, Yuezhi Mao, and Martin Head-Gordon, J. Chem. Theory Comput. 2016, 12 (10), 4812–4820. DOI: 10.1021/acs.jctc.6b00571
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  • $\begingroup$ While it's charge-shift for F2, fluorides are described with 3-center 2-electron bonds. Still your argumentation stays valid, I guess. $\endgroup$
    – Mithoron
    May 24, 2017 at 21:59
  • $\begingroup$ This is a (modern) valence bond theory vs MO theory discrepancy in nomenclature. Both are equivalent and equally valid representations, as you note. $\endgroup$
    – levineds
    May 24, 2017 at 22:03
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    $\begingroup$ $\ce{I4}$ 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. $\endgroup$ Jun 5, 2017 at 11:05
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    $\begingroup$ While that is a local minimum, it has no lifetime. At a much better level of theory ($\omega$B97M-V/cc-pvdz with the Stuttgart Dresden small core relativistic pseudopotential), the $D_{3h}$ $\ce{I4}$ structure is 30 kcal/mol higher in energy than two $\ce{I2}$ molecules. Coupled with the large entropy gain from separating into two molecules, the Gibbs free energy of separation is quite downhill. Also, the lowest vibrational mode of $\ce{I4}$ (13 cm-1) is the reaction path to separation, so it is kinetically facile. The explanation given can fully account for the non-existence of this molecule. $\endgroup$
    – levineds
    Jun 5, 2017 at 22:45
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    $\begingroup$ To clarify, it has no lifetime at any reasonable temperature because the barrier to fragmentation is very small, and it is many 10s of kcal/mol above separated 2$\ce{I2}$ and so not a populated state (per the thermodynamic Boltzmann distribution that $\ce{I4}$ and 2$\ce{I2}$ would obey). $\endgroup$
    – levineds
    Jun 6, 2017 at 0:03

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