Rationalize the given statements and provide 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. However,$PbCl_4$ can be formed by passing $Cl_2$ gas through a saturated solution of $PbCl_2$.
$PbCl_{2(s)} + Cl_{2(g)} \rightarrow PbCl_{4(l)}$
$(b)$ Due to the inert pair effect,the $+4$ oxidation state of $Pb$ is highly unstable. Upon heating,$PbCl_4$ decomposes to form the more stable $Pb(II)$ chloride.
$PbCl_{4(l)} \xrightarrow{\Delta} PbCl_{2(s)} + Cl_{2(g)}$
$(c)$ Lead does not form $PbI_4$ because $Pb^{4+}$ is a strong oxidizing agent and $I^-$ is a strong reducing agent. $Pb(IV)$ oxidizes $I^-$ to $I_2$ and itself gets reduced to $Pb(II)$,making $PbI_4$ unstable.
$PbI_4 \rightarrow PbI_2 + I_2$