Q.Rationalise the given statements and give chemical reactions:
• Lead(II) chloride reacts with Cl2 to give PbCl4.
• Lead(IV) chloride is highly unstable towards heat.
• Lead is known not to form an iodide, PbI4.

Step 1 - Background: the inert pair effect for Pb

Down group 14, the $+2$ oxidation state becomes progressively more stable relative to $+4$ because of the inert pair effect (poor shielding of the $6s^2$ electron pair by the filled $4f$/$5d$ shells). This effect is strongest for the heaviest member, lead - so $\text{Pb}^{2+}$ is the thermodynamically more stable oxidation state, and $\text{Pb}^{4+}$ compounds are comparatively unstable, strongly oxidising species.

Step 2 - (i) $\text{PbCl}_2 + \text{Cl}_2 \rightarrow \text{PbCl}_4$

This reaction can be made to occur only under suitably forcing conditions (e.g. with excess/liquid chlorine); it is not facile the way $\text{SnCl}_2 + \text{Cl}_2 \rightarrow \text{SnCl}_4$ is, because oxidising $\text{Pb}^{2+}$ all the way to the less-favoured $\text{Pb}^{4+}$ works against the inert pair effect.

Step 3 - (ii) $\text{PbCl}_4$ unstable to heat

Once formed, $\text{PbCl}_4$ is thermodynamically unstable relative to the $+2$ state, so even mild heating causes it to decompose, releasing chlorine and reverting to the more stable lead(II) chloride:

Step 4 - (iii) $\text{PbI}_4$ does not form …

Apply the inert pair effect for lead (Pb2+ more stable than Pb4+, so Pb4+ is strongly oxidising) to explain: (i) PbCl4 forms only under forcing conditions, (ii) it decomposes thermally back to PbCl2 + Cl2, and (iii) PbI …

• Explaining the instability purely in terms of "size" rather than specifically the inert pair effect (6s2 electrons).
• Forgetting to give the actual redox reaction (Pb4+ + 2I- -> Pb2+ + I2) to justify why PbI4 specifically cannot exist, while PbCl4 (though unsta …