• # question_answer 21) Rationalise the given statements and give chemical reactions. (i) Lead (II) chloride reacts with $C{{l}_{2}}$ to give$PbC{{l}_{4}}$. (ii) Lead (IV) chloride is highly unstable towards heat. (iii) Lead is known not to form an iodide, $Pb{{l}_{4}}$.

(i) On account of inert pair effect, $PbC{{l}_{2}}$ is more stable than$PbC{{l}_{4}}$. Thus, $PbC{{l}_{2}}$ does not react with chlorine to form$Pb{{I}_{4}}$. (ii) On account of greater stability of +2 state over +4 state, $PbC{{l}_{4}}$decomposes on heating into$PbC{{I}_{2}}$. $PbC{{l}_{4}}\xrightarrow{Heat}PbC{{l}_{2}}+C{{l}_{2}}$ (iii) As $P{{b}^{4+}}$ is an oxidising agent while ${{I}^{-}}$ ion is a reducing agent, the formation of $Pb{{I}_{4}}$ is not possible. $P{{b}^{4+}}+4{{I}^{-}}\to Pb{{I}_{2}}+{{I}_{2}}$ Thus, $Pb{{I}_{4}}$ does not exist.