Rationalize 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, $\mathrm{Pbl}_{4}$.
Rationalize 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, $\mathrm{Pbl}_{4}$.

Solution:

– Lead is an individual from bunch 14 of the occasional table. The two oxidation situations with bunch shows are $+2$ and $+4$. The $+2$ oxidation state turns out to be more steady when dropping down the gathering, and the oxidation status $+4$ turns out to be less steady. This is because of the idle pair impact. So $\mathrm{PbCl}_{4}$ is essentially less steady than $\mathrm{PbCl}_{2}$. Be that as it may, $\mathrm{PbCl}_{4}$ development happens when the chlorine gas is risen through a $\mathrm{PICl}_{2}$ soaked arrangement.
$\mathbf{P} \mathrm{bCl}_{4(\mathrm{I})}$
– Due to the idle pair impact, the higher oxidation state becomes unsteady when dropping down bunch IV.
$\mathrm{PbCl}_{4(\mathrm{l})} \longrightarrow \mathrm{PbCl}_{2(\mathrm{~s})}+\mathrm{Cl}_{2(\mathrm{~g})}$
– Lead isn’t known for shaping Pbl4. In nature, $\mathrm{Pb}(+4)$ is oxidizing, and I decrease in nature. A $\mathrm{Pb}(\mathrm{IV})$ and iodide particle mix are not steady. lodide particle decreases strongly in nature. Pb(IV) oxidizes 1 t to $\mathrm{I}_{2}$, and decreases itself to $\mathrm{Pb}(\mathrm{II})$.
$P b I_{4} \rightarrow P b I_{2}+I_{2}$