In what form does the respective halides of thallium exist as?

Question

Thallium Halides like $\ce{TlF}$ and $\ce{TlI}$ exist. However, my text says that $\ce{TlI}$ exists as $\ce{Tl+}$ and $\ce{I3-}$. This is understandable with the help of the inert pair effect. Then why does the fluoride exist as $\ce{Tl^{+3}}$ and $\ce{F-}$?

Answer 1

Two factors are in play here.

First, at least in the sixth period the "inert pair" is not completely inert. It can be pulled into chemical bonding if we use an appropriately strong oxidizing agent. Halogens lighter than iodine are better oxidizing agents than iodine itself, so are more likely to draw off that inert pair. In the case of thallium, bromine and chlorine are capable of forming thallium(III) compounds, although these readily decompose upon warming. Wikipedia gives a brief summary of $\ce{TlX3}$ compounds, noting the contrast in structure between the (tri)iodide and the others.

Second, as described in this answer, trihalide ions are stabilized by de-localized bonding but destabilized by the presence of counterions, which tend to break up the trihalide ion. Triiodide is made of bulky atoms that tend to minimize this instability, whereas lighter trihalide ions are less robust and generally exist under ambient conditions only with bulky counterions such as quaternary ammonium or pyridinium cations (all the tribromide and trichloride examples given or referenced in the answer noted above involve such cations).

So the combination of a less stable trihalide ion (in the concentrated salt environment) and greater oxidizing power favors lighter halogens forming $\ce{TlX3}$ as a largely covalent compound of thallium(III) rather than a trihalide-ion salt of thallium(I).