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.
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