Question 11.13 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.
Class XI The p-Block Elements Page 324
• 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.
Class XI The p-Block Elements Page 324
Add an Answer
More Questions From This User See All
Answers & Comments
Verified answer
1. Due to inert pair effect , Pb is more stable in +2 oxidation state than +4 oxidation state. therefore, lead (II) chloride doesn't react with Cl2 to give lead(IV) chloride .2. Lead (IV) chloride on heating decompose to give lead (II) chloride and Clean because Lead in +2 oxidation state is more stable than +4 oxidation state .
e.g., PbCl4 -----> PbCl2 + Cl2
3. Due to strong oxidizing power of Pb+4 ion and reducing power of I- ion , PbI4 doesn't exist. because Pb+4 get reduced and formed +2 oxidation state and I- get oxidised and formed positive oxidation state .
(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) + Cl2(g) → PbCl4(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).
PbCl4(l) → PbCl 2(s) + Cl2(g)
(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 I2and itself gets reduced to Pb(II).
Pbl4 → Pbl 2 + l2