Rationalise the given statements and give chemical reactions.
a. Lead(II) chloride does not react with `Cl_(2)` to give `PbCl_(4)`.
b. Lead(IV) chloride is highly unstable towards heat.
c. Lead is known not to form an iodide, `Pbl_(4)`.

Explanation

(i) Due to inert pair effect, Pb shows oxidation states of `+2`and `+4`. Since `Cl_(2)` is a strong oxidising agent, it oxidises `Pb^(2+)` to `Pb^(4+)` and hence `PbCl_(2)` reacts with `Cl_(2)` to form `PbCl_(4)`.
`PbCl_(2) (s) + Cl_(2) (g) rarr PbCl_(4)(l)`
(ii) Due to greater stability of `+2` over `+4` oxidation state because of inert pair effect , lead (IV) chloride on heating decomposes to give lead (II) chloride and `Cl_(2)`.
`PbCl_(4) (l) overset(Delta)rarrPbCl_(2) (s) + Cl_(2) (g)`
(iii) Due to oxidising power of `Pb^(4+)` ions and reducing power of `l^(-)` ion, `PbI_(4)`, does not exist.
Alternatively, `Pb-I` bond initially formed during the reaction does not release enough energy to unpair `6r^(2)` electrons and excite one of them to the higher `6p`-orbital to have four unpaired electrons around lead atom needed for formation of `PbI_(4)`.