Ncert Exercises & Solutions

Rationalise the given statements and give chemical reactions:
• Lead (II) chloride reacts with $C I_{2} t o g i v e P b C I_{4} .$ .
• Lead (IV) chloride is highly unstable towards heat.
• Lead is known not to form an iodide, $P b I_{4} .$

a) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is +2 and +4. On moving down the group, the +2 oxidation state becomes more stable and the +4 oxidation state becomes less stable. This is because of the inert pair effect. Hence,

P b C I_{4}

is much less stable than

P b C I_{2}

. However, the formation of

P b C I_{4}

takes place when chlorine gas is bubbled through a saturated solution of

P I C I_{2}

P b C I_{2 (x)} + C I_{2 (g)} \to P b C I_{4 (1)}

(b) On moving down group IV, 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).

P b C I_{4 (1)} \overset{∆}{\to} P b C I_{2 (x)} + C I_{2 (g)}

(c) Lead is known not to form $P b I_{4}$ . Pb (+4) is oxidising in nature and $I^{-}$ is reducing in nature. A combination of Pb(IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV) oxidises $I^{-} t o I_{2}$ and itself gets reduced to Pb(II).

$P b I_{4} \to P b I_{2} + I_{2}$

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