Rationalise the given statements and give chemical reactions:
- Lead(II) chloride reacts with $\mathrm{Cl}_{2}$ to give $\mathrm{PbCl}_{4}$.
- Lead(IV) chloride is highly unstable towards heat.
- Lead is known not to form an iodide, $\mathrm{Pbl}_{4}$.
(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. $\mathrm{Hence}, \mathrm{PbCl}_{4}$ is much less stable than $\mathrm{PbCl}_{2}$. However, the formation of $\mathrm{PbCl}_{4}$ takes place when chlorine gas is bubbled through a saturated solution of $\mathrm{PICl}_{2}$.
$\mathrm{PbCl}_{2(s)}+\mathrm{Cl}_{2(g)} \longrightarrow \mathrm{PbCl}_{4(l)}$
(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).
$\mathrm{PbCl}_{4(l)} \stackrel{\Delta}{\longrightarrow} \mathrm{PbCl}_{2(s)}+\mathrm{Cl}_{2(g)}$
(c) Lead is known not to form $\mathrm{Pb}_{4}$. $\mathrm{Pb}(+4)$ is oxidising in nature and $\mathrm{I}$, is reducing in nature. A combination of $\mathrm{Pb}$ (IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. $\mathrm{Pb}$ (IV) oxidises $\mathrm{I}^{-}$to $\mathrm{I}^{2}$ and itself gets reduced to $\mathrm{Pb}(\mathrm{II})$.
$\mathrm{PbI}_{4} \longrightarrow \mathrm{PbI}_{2}+\mathrm{I}_{2}$