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

(a) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is +2 and| +4. On moving down the group, the +2 oxidation state becomes more stable and the +4 oxidation state becomes less stable. This is because of the inert pair effect. Hence, `PbCl_(4)` is much less stable than `PbCl_(2)`. However, the formation of `PbCl_(4)` takes place when chlorine gas is bubbled through a saturated solution of `PlCl_(2)`.
`PbCl_(2(s))+Cl_(2(g))rarrPbCl_(4)(l)`
(b) On moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. Pb(IV) is highly unstable and when heated, it reduces to Pb(II).
`PbCl_(4(l))overset(Delta)rarrPbCl_(2(s))+Cl_(2(g))`
(c) Lead is known not to form `PbI_(4)`. Pb (+4) is oxidising in nature and `I^(-)` is reducing in nature. A combination of Pb(IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV) oxidises `I^(–)" to "I^(2)` "and itself gets reduced to Pb(II)".
`PbI_(4)rarrPbl_(2)+I_(2)`