(i) $\mathrm{PbCl}_2 + \mathrm{Cl}_2$ ↛ $\mathrm{PbCl}_4$ Lead is more stable in +2 oxidation state than in +4 state due to inert pair effect. Thus, the reaction is not feasible. (ii) $\mathrm{PbCl}_4$ $\xrightarrow{\Delta}$ $\mathrm{PbCl}_2 + \mathrm{Cl}_2$ Pb is more stable in its +2 oxidation state due to inert pair effect. As a result, when subjected to heat, Pb (IV) goes to Pb (II) state. (iii) $\mathrm{Pb} + 2\mathrm{I}_2$ ↛ $\mathrm{PbI}_4$ $\mathrm{I}^{-}$ is a good reducing agent and therefore, reduces Pb (IV) to Pb (II) easily. That is why, $\mathrm{PbI}_4$ does not exist.