Rationalise the given statements and give chemical reactions.
(a) Lead(II) chloride reacts with $C l_{2}$ to give $P b C l_{4}$ .
(b) Lead(IV) chloride is highly unstable towards heat.
(c) Lead is known not to form an iodide, $P b I_{4}$ .

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Solution

(a) Due to inert pair effect, $P b (I I)$ is more stable than Pb(IV). Hence, there is no reaction between $P b C l_{2}$ and $C l_{2}$ .
(b) Due to inert pair effect, $P b (I I)$ is more stable than $P b (I V)$ . Hence, $P b C l_{4}$ is thermally unstable and on heating decomposes to form $P b C l_{2}$ .
$P b C l_{4} Δ - \to P b C l_{2} + C l_{2}$
(c) $P b I_{4}$ does not exist due to strong oxidizing power of $P b (I V)$ and strong reducing power of $I^{-}$ .

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