The group 14 elements A and B have the first ionisation enthalpy values of 708 and 715 kJ mol⁻¹ respectively. The above values are lowest among their group members. The nature of their ions A²⁺ and B⁴⁺ respectively is

Solution

Chemistry ma’am · 3 steps

Name the two elements from their ionisation enthalpies, then ask which oxidation state each one prefers.

1. Identify A and B. Down group 14 the first ionisation enthalpies are C 1086, Si 786, Ge 761, Sn 708 and Pb 715 kJ mol⁻¹. The two lowest are tin and lead, so A is Sn and B is Pb. Lead is slightly higher than tin because its poorly shielding 4f and 5d electrons let the nucleus hold the outer electrons more tightly.
2. Inert pair effect. Going down the group, the ns² pair holds back more and more from bonding. So the +2 state becomes more stable down the group. For tin, +4 is still the more stable state. For lead, +2 is the more stable state.

   Ion	More stable state of the element	What the ion does	Nature
   Sn²⁺ (A)	+4	Loses 2 electrons to reach Sn⁴⁺	reducing
   Pb⁴⁺ (B)	+2	Gains 2 electrons to reach Pb²⁺	oxidising

3. So. Sn²⁺ is reducing and Pb⁴⁺ is oxidising, which is option (d).

The idea to keep. An ion in its less stable oxidation state acts to reach the stable one. If it has to lose electrons, it is a reducing agent. If it has to gain electrons, it is an oxidising agent. For Sn, +4 wins. For Pb, +2 wins.

Answer from the NTA final answer key. Solution by PrepWiser.