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
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, PbCl4 is much less stable than PbCl2. However, the formation of PbCl4 takes place when chlorine gas is bubbled through a saturated solution of PlCl2.
`PbCl_(2(s)) + Cl_(2(g)) -> PbCl_(4(l))`
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).
c) Lead is known not to form PbI4. 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– to I2 and itself gets reduced to Pb(II).
`PbI_4 -> PbI_2 + I_2`
PbCl2 + Cl2 ———> PbCl4.
This is because Pb can show +2 oxidation state more easily than +4 due to inert pair effect.
PbCl4 ———> PbCl2 + Cl2
Because Pb2+ is more stable than Pb4+ due to inert pair effect.
Pbl4 does not exist because I- ion being a powerful reducing agent reduces Pb4+ ion to Pb2+ ion in solution.
Pb4+ + 2I– ——-> Pb2+ + l2