JEE MainChemistryp Block Elements (Group 13 & 14)
Element M belongs to Group 14. Its dioxide MO₂ acts as a strong oxidising agent, and its tetraiodide MI₄ does not exist. Element N is the element placed immediately above M in the same group. Which of the following correctly represents the relationship between their standard reduction potentials ( E^ _ M⁴⁺/M²⁺ and E^ _ N⁴⁺/N²⁺ ) and the redox nature of the dihalide NX₂ ?
Options
- AE^ _ M⁴⁺/M²⁺ < E^ _ N⁴⁺/N²⁺ and NX₂ is an oxidising agent
- BE^ _ M⁴⁺/M²⁺ > E^ _ N⁴⁺/N²⁺ and NX₂ is an oxidising agent
- CE^ _ M⁴⁺/M²⁺ < E^ _ N⁴⁺/N²⁺ and NX₂ is a reducing agent
- DE^ _ M⁴⁺/M²⁺ > E^ _ N⁴⁺/N²⁺ and NX₂ is a reducing agent
Correct answer
D. E^ _ M⁴⁺/M²⁺ > E^ _ N⁴⁺/N²⁺ and NX₂ is a reducing agent
Step-by-step solution
Based on the given properties, element M is Lead (Pb). PbO₂ is a strong oxidising agent because Pb⁴⁺ is less stable than Pb²⁺ due to the inert pair effect. PbI₄ does not exist because the Pb-I bond initially formed releases insufficient energy to unpair the 6s^2 electrons, and Pb⁴⁺ oxidises I^- to I₂ . Element N , which is immediately above Pb in Group 14, is Tin (Sn). For Pb, the reduction Pb⁴⁺ + 2e^- Pb²⁺ is highly favourable, so E^ _ Pb⁴⁺/Pb²⁺ is highly positive. For Sn, the +4 oxidation state is more stable tha