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I have read in my textbook that $\ce{CI4}$ is more stable than $\ce{PbI4}$ due to inert pair effect. But shouldn't we also consider the effect of steric repulsions? In $\ce{CI4}$, four large iodine atoms surround the small carbon atom, which should make it less stable than $\ce{PbI4}$.

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    $\begingroup$ You are correct up to a certain point: yes, steric repulsions destabilise CI4. But there is nothing that lets you conclude that that factor will destabilise CI4 to the point where it is less stable than PbI4. It is like saying that I earnt some money at my job, so surely I am richer than Bill Gates now? Well, it's certainly made me richer than I was before; but unfortunately, Bill Gates still has more money. So, yes, sterics sure make CI4 less stable than it should otherwise be, but PbI4 is still less stable. Do you see why the reasoning doesn't make sense? $\endgroup$ – orthocresol May 30 at 17:51
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    $\begingroup$ Thank you very much $\endgroup$ – Akshat May 31 at 0:41
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Steric hindrance does indeed interfere with the stability of carbon tetraiodide, as does poor covalent bonding overlap between the very differently sized carbon and iodine atoms. Even so, carbon tetraiodide is definitely established and in use as an iodinating agent. Storage below normal room temperature is recommended.

Carbon tetraiodide is also very red. Ordinarily saturated carbon compounds are white or colorless because the electronic transitions involving single bonds to carbon require ultraviolet radiation to go, but the relatively weak bonds in carbon tetraiodide enable transitions in the visible light region. Iodoform is more lightly colored for a similar reason.

Lead tetraiodide, meanwhile, does not seem to be definitely known to exist.

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  • $\begingroup$ Just curious, why was this unaccepted given there is no alternative answer? (Meaning what's missing?) $\endgroup$ – Oscar Lanzi Jun 5 at 19:38

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