This question is based on the inert pair effect and the relative stability of oxidation states of Group 14 elements.
For lead (Pb), due to strong inert pair effect, the +2 oxidation state is more stable than the +4 oxidation state.
Reaction (1):
PbO2 contains Pb in +4 oxidation state, while PbO contains Pb in +2 oxidation state.
The reaction converts Pb4+ to Pb2+, which is a more stable oxidation state for lead.
Formation of a more stable oxidation state is thermodynamically favorable.
Therefore,
$$ \Delta_r G^\circ (1) < 0 $$
For tin (Sn), the inert pair effect is much weaker, and the +4 oxidation state is more stable than the +2 oxidation state.
Reaction (2):
SnO2 contains Sn in +4 oxidation state, while SnO contains Sn in +2 oxidation state.
The reaction converts Sn4+ to Sn2+, which is a less stable oxidation state for tin.
Formation of a less stable oxidation state is thermodynamically unfavorable.
Therefore,
$$ \Delta_r G^\circ (2) > 0 $$
Hence, the correct set is:
ΔrG° (1) < 0; ΔrG° (2) > 0
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.