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, \[Pb{{l}_{4}}\].

Answer:

(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.