While triiodide ion is commonly known, why is that trichloride ion (and its salts) almost unheard of? While it does exist according to literature, why is the stability of trichloride ion in aqueous environment much worse than its bromine and iodine counterparts? Aren't elements down the group bulkier and hence forming weaker covalent bonds with longer bond lengths?
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1$\begingroup$ chemistry.stackexchange.com/questions/27920/… $\endgroup$– Nilay GhoshOct 22, 2016 at 10:17
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As @nilay points out, we can make "polychlorides" with "nonclassical" stoichiometries. But for more typical laboratory conditions it is true that heavier halogens are more favorable for forming polyatomic ions.
This is because polyhalide ions involve two competing forces. On the one hand, the delocalized covalent bonding gives more stability to a polyatomic ion than we would have with the separate monoatomic ions and diatomic molecules. On the other hand, surrounding counterions or dipoles are more strongly attracted to compact, spherical monoatomic anions than to bulky polyatomic ones. With lighter halogens forming the most compact monoatomic anions, the electrostatic effect wins out and we need severe conditions to overcome it. Heavier halogens, with larger atoms and more polarizable electrons, make the polyatomic ions more competitive, as in $\ce{C_5H_6N+Br_3−}$ (http://www.organic-chemistry.org/chemicals/oxidations/pyridiniumhydrobromideperbromide.shtm).
answered Oct 22, 2016 at 13:13
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$\begingroup$ Does it also have to do with the "softness" of iodide ion in a sense that it mingles better with typically "soft" diatomic halogen molecule? Also, have you heard of things like dichloroiodide ion? $\endgroup$– SunnyOct 23, 2016 at 2:49
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$\begingroup$ With multihalogen species things are more complex; the neutral diatomic species (in this case $ICl$) is polar and has an ion-dipole attraction with more anions (such as $Cl^-$). This favors association even with smaller "outer" atoms. Note that you still need a larger halogen ion in the middle to set up the dipole. $\endgroup$ Nov 28, 2016 at 1:45