Rationalise the given statements and give chemical reactions:
$1$. Lead $(II)$ chloride reacts with $Cl_2$ to give $PbCl_4$.
$2$. Lead $(IV)$ chloride is highly unstable towards heat.
$3$. Lead is known not to form an iodide,$PbI_4$.

(N/A) Lead belongs to group $14$ of the periodic table. The two oxidation states displayed by this group are $+2$ and $+4$. On moving down the group,the $+2$ oxidation state becomes more stable and the $+4$ oxidation state becomes less stable due to 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 $PbCl_2$.
$PbCl_{2(s)} + Cl_{2(g)} \longrightarrow PbCl_{4(l)}$
$(b)$ On moving down group $14$,the higher oxidation state becomes unstable because of the inert pair effect. $Pb(IV)$ is highly unstable and when heated,it reduces to $Pb(II)$.
$PbCl_{4(l)} \stackrel{\Delta}{\longrightarrow} PbCl_{2(s)} + Cl_{2(g)}$
$(c)$ Lead is known not to form $PbI_4$. $Pb(IV)$ is oxidising in nature and $I^-$ is reducing in nature. $A$ combination of $Pb(IV)$ and iodide ion is not stable. $Pb(IV)$ oxidises $I^-$ to $I_2$ and itself gets reduced to $Pb(II)$.
$PbI_4 \longrightarrow PbI_2 + I_2$