We know that stability of higher oxidation state (+3) decreases and stability of lower oxidation state (+1) increases, so thallium is most stable in its +1 oxidation state due to the inert pair effect and paired s electrons.

In the compound $\ce{TlI3}$, since $\ce{I3}$ exists as a linear molecule $\ce{I3-}$ so the oxidation number of $\ce{Tl}$ is +1 which is stable and the compound should exist but I have read in many of the books that it does not exists. Why is it so?

  • 4
    $\begingroup$ Which books? Thallium Tri-Iodide,Tl+(I3)-, does exist. Its Thallium (III) Iodide, Tl3+ (I-)3 to my knowledge, does not. Interestingly the [TlI4]- ion does, and contains Tl3+. $\endgroup$
    – Ian Bush
    Apr 14, 2016 at 8:19
  • 3
    $\begingroup$ Note that, according to IUPAC nomenclature, the name tris(iodide) must be used for compounds containing $\ce{3I-}$ rather than triiodide, which is used for $\ce{I3-}$. Therefore, correct names for $\ce{TlI3}$ are “thallium tris(iodide)”, “thallium(III) iodide”, and “thallium(3+) iodide”. Correct names for $\ce{Tl(I3)}$ are “thallium triiodide(1−)”, “thallium(I) (triiodide)”, and “thallium(1+) (triiodide)”. $\endgroup$
    – user7951
    May 6, 2016 at 14:53

1 Answer 1


Thallium triiodide does exist and it is formulated as $\ce{Tl+I3-}$ and not $\ce{Tl^3+(I^{-})3}$. Thus $\ce{TlI3}$ is a thallium(I) compound and contains the triiodide ion, $\ce{I3^{−}}$. This is confirmed by considering the standard potential which indicate that $\ce{Tl(III)}$ is rapidly reduced to $\ce{Tl(I)}$ by the iodide.

$$\begin{align} \ce{Tl^3+ + 2e− &-> Tl+} & E^\circ &= +1.252~\mathrm{V} \\ \ce{I2 + 2e- &-> 2I-} & E^\circ &= +0.5355~\mathrm{V} \\ \hline \ce{Tl^3+ + 2 I- &-> Tl+ + I2} & E^\circ_\mathrm{cell} &= +0.717~\mathrm{V} > 0 \\ \hline \end{align}$$

(data from Wikipedia).

$\ce{TlI3}$ can be prepared by the evaporation of stoichiometric quantities of $\ce{TlI}$ and $\ce{I2}$ in concentrated aqueous $\ce{HI}$, or by reacting $\ce{TlI}$ with $\ce{I2}$ in ethanol.


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