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After recent question about non-existent pnictide(V) iodides, I stumbled upon a similar issue in oxygen group. For example, $\ce{SI2}$ was reported to be found… but at the temperature as low as $\pu{9 K}$. Well, OK, I heard that sulfur prefers disulfides, but $\ce{S2I2}$ was characterised at $\pu{-90 °C}$. Moreover, situation with selenium iodides is even worse!

One of the reasons for that is weakness of $\ce{S-I}$ and $\ce{Se-I}$ bonds, which causes the compounds to be endothermic.

Klapötke and Passmore [1, p. 234] provide an explanation for the weakness of these bonds:

The instability of the $\ce{S-I}$ and $\ce{Se-I}$ bonds can be attributed to their very low ionic resonance stabilization energies as the electronegativity of iodine is about the same as that of sulfur and selenium.

I think this is a partial explanation, at best - ionic component sure makes bonds stronger, but by this logic all non-polar covalent bonds should be weak. I’m not saying I’d expect bonds created by iodine to be strong, but it’s not like we need cryogenic temperatures to get methyl iodide!

Could anyone provide a good explanation as to why these bonds are so weak, maybe describing stereo-electronic effects, or using computational methods?

References

  1. T. Klapötke, J. Passmore, J. Acc Chem. Res. 1989, 22, 234–240. DOI:10.1021/AR00163A002.
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  • $\begingroup$ I know it's not what you're looking for, but what's wrong with the good ol' "atom size" explanation? They're all pretty large atoms with minute differences in electronegativity. Or are you looking for a fully non-empirical explanation? $\endgroup$ Jun 24, 2023 at 16:59
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    $\begingroup$ Well, phosphorus has similar size and electronegativity, but PI3 is a common reagent. $\endgroup$
    – Mithoron
    Jun 24, 2023 at 19:16
  • $\begingroup$ @Mithoron The values in question are not bond enthalpies. They are not comparable to hydrogen bond energies. While the dissociation of a hydrogen bond needs similar amount of energy, the process described in the reference provided in OP involves more steps: S-I dissociation consumes energy and recombination of S-S and I-I releases energy. All these contributions are added to give -18kJ/mol, unlike hydrogen bond dissociation, which is a single step process. $\endgroup$
    – Paul Kolk
    Apr 3 at 21:57
  • $\begingroup$ @Mithoron The real reason they need cryogenic temperatures is, to prevent slightly exothermic reaction from happening. Not all reactions need activation energy equal to bond dissociation energy from thermal background. I doubt these bonds are unusually weak at all. Stable molecules may form unstable gas. $\endgroup$
    – Paul Kolk
    Apr 3 at 22:52
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    $\begingroup$ @PaulKolk Oops, this was embarrassing misunderstanding on my part. Worse, now I gotta rework the whole question! $\endgroup$
    – Mithoron
    Apr 4 at 19:21

1 Answer 1

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It isn't just sulfur iodide. Sulfur in low oxidation states does not form any $\ce{SX2}$ molecules that are stable under ambient or near-ambient conditions. Such molecules tend to dimerize, with loss of some halogen atoms in the case of chlorine or a heavier halogens, forming disulfur compounds. (In the case of $\ce{S2I2}$, Wikipedia reports stability up to a somewhat higher temperature than suggested in the question, $\pu{-30 °C}).$

What happens is sulfur in the disulfur compounds can act as a pi donor, thereby strengthening the sulfur–sulfur bond while only slightly weakening the sulfur–halogen bond. The specific case of disulfur dichloride is discussed here.

In addition to the halides, the structure of disulfur monoxide, which appears to exist in the Jovian moon Io, allows a similar interaction between the sulfur–sulfur bond and oxygen.

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