The compound with the formula $\mathrm{TII}_{3}$ is a black solid. Given the following standard reduction potentials,
$\begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & & 8^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & & \mathscr{C}^{\circ} &=0.55 \mathrm{~V} \end{aligned}$
would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?