Assertion A: Reduction of metal oxide is easier if metal formed is in liquid state than solid state. Reason R: G…

🧠 On the Ellingham diagram, the slope changes at melting — that's where oxide becomes easier to reduce When the metal product changes from solid to liquid, its entropy increases. Higher entropy of products means $\Delta G^\circ$ for the reduction reaction becomes more negative.

🗺️ Evaluating Assertion and Reason Assertion A: Reduction is easier when the metal is in liquid state → True. On Ellingham diagrams, the $\Delta G^\circ$ line for the oxide breaks downward at the metal's melting point, meaning reduction becomes more feasible.

Reason R: $\Delta G^\circ$ becomes more negative because entropy is higher in liquid state → True. $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Higher entropy of liquid metal (compared to solid) increases $\Delta S^\circ$ for the reduction, making $-T\Delta S^\circ$ more negative, hence $\Delta G^\circ$ more negative.

R correctly explains A → option (a).

⚠️ Trap: Don't confuse the slope of the Ellingham line with its absolute value. The slope (= $-\Delta S^\circ$) changes at the phase transition point, causing $\Delta G^\circ$ to dip further negative.

$\boxed{\text{Answer: (a)}}$