Rationalise the given statements and give chemical reactions:
• Lead (II) chloride reacts with .
• Lead (IV) chloride is highly unstable towards heat.
• Lead is known not to form an iodide, $Pb{I}_{4}.$

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, $PbC{I}_{4}$ is much less stable than $PbC{I}_{2}$. However, the formation of $PbC{I}_{4}$ takes place when chlorine gas is bubbled through a saturated solution of $PIC{I}_{2}$.$PbC{I}_{2\left(x\right)}+C{I}_{2\left(g\right)}\to PbC{I}_{4\left(1\right)}$
(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).

(c) Lead is known not to form $Pb{I}_{4}$. 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  and itself gets reduced to Pb(II).

$Pb{I}_{4}\to Pb{I}_{2}+{I}_{2}$