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For the question

Why is $\ce{ICl5}$ unstable?

my textbook gave an answer of

$\ce{ICl5}$ has too many highly electronegative chlorine atoms present on the central iodine.

But how does having electronegative atoms around a central atom make a molecule unstable? For example, $\ce{SF6}$ is very stable.

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    $\begingroup$ Steric hindrance, maybe? Fluorine atoms are relatively small. Steric hindrance is less in $SF_6$ than in $ICl_5$. $\endgroup$ – Oscar Lanzi Mar 13 '16 at 0:38
  • $\begingroup$ Could symmetry play a role in the relative stability of these molecules? $\endgroup$ – Kraw Mar 13 '16 at 4:58
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    $\begingroup$ Hypervalent iodine is a common oxidant. Iodine(V) and iodine(III) are both reactive because iodine is most satisfied with a -1 oxidation state. Here, it has an oxidation number of +5. $\endgroup$ – SendersReagent Mar 13 '16 at 10:55
  • $\begingroup$ Actually, iodine is not most satisfied in a $-\mathrm{I}$ oxidation state, especially not in water. Iodinated table salt contains iodate, not iodide, so that it will not get reduced to iodine and cause ugly colouring. $\endgroup$ – Jan Mar 21 '16 at 1:06
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The textbook answer must be wrong, since $\ce{IF5}$ is a sufficiently stable compound and fluorine is more electronegative than chlorine.

Rather, I think that the reason for $\ce{ICl5}$’s lesser stability is the lower electronegativity of chlorine. Creating this type of hypercoordinate compounds requires two sets of four-electron three-centre bonds of the $\ce{X- \bond{...}{I+}-X <-> X-{I+}\bond{...}X-}$ type. This type of bond requires a sufficient electronegativity on the outer atoms to stabilise the partial negative charge better — something fluorine is more capable of than chlorine. $\ce{ICl5}$ should only be possible because of iodine’s low electronegativity and $\ce{BrCl5}$ should be orders of magnitude less stable again.

A second aspect could be the larger size of chlorine atoms compared to fluorine atoms causing greater steric stress, longer bonds and therefore weaker interactions.

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