For a reaction 4M(s) + nO2(g) 2M2On(s), the free energy change is plotted as a function of temperature. The temperature…

🧠 Oxide is stable when ΔG is negative — the crossover point marks the stability boundary On an Ellingham-style plot of $\Delta G$ vs $T$: when $\Delta G < 0$, the reaction forming the oxide is spontaneous — the oxide is stable. When $\Delta G > 0$, the oxide breaks down.

🗺️ Reading the plot The temperature below which the oxide is stable is the one below which $\Delta G$ is negative. As temperature rises, $\Delta G$ eventually crosses zero and becomes positive — the oxide is no longer stable.

That crossover is exactly "the free energy change shows a change from negative to positive value" → option (b).

⚠️ Trap: Options (a), (c), (d) describe slope changes. The slope changing means $\Delta S$ is changing (slope = $-\Delta S$). What matters for stability is the sign of $\Delta G$, not the slope.

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