For a reaction $$4M(s) + nO_2(g) \to 2M_2O_n(s)$$. The free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which:

We begin with the general thermodynamic relation that links the Gibbs free energy change $$\Delta G$$ of a reaction with its enthalpy change $$\Delta H$$ and entropy change $$\Delta S$$. The formula is stated as

$$\Delta G \;=\; \Delta H \;-\; T\,\Delta S,$$

where $$T$$ is the absolute temperature in kelvin. This equation tells us how $$\Delta G$$ varies linearly with temperature for any reaction whose $$\Delta H$$ and $$\Delta S$$ remain approximately constant over the temperature range considered.

For the oxidation reaction under discussion,

$$4M(s) + nO_2(g) \;\longrightarrow\; 2M_2O_n(s),$$

we plot $$\Delta G$$ on the vertical axis and temperature $$T$$ on the horizontal axis. According to the above equation, the plot is a straight line with slope

$$\text{slope} \;=\; -\Delta S.$$

Now, the chemical meaning of the sign of $$\Delta G$$ is crucial:

• If $$\Delta G < 0,$$ the reaction is spontaneous in the forward direction; here, that means the oxide $$M_2O_n$$ is thermodynamically stable relative to the metal $$M$$ and oxygen.
• If $$\Delta G > 0,$$ the oxide is not stable; instead, the reverse reduction reaction is favored.

The temperature at which the oxide just ceases to be stable is the temperature where the free-energy change is exactly zero, because at that point the forward and reverse reactions are in equilibrium:

$$\Delta G \;=\; 0.$$

Setting $$\Delta G = 0$$ in the fundamental relation gives the boundary equation

$$0 \;=\; \Delta H \;-\; T_{\text{eq}}\,\Delta S,$$

and solving for the equilibrium temperature $$T_{\text{eq}}$$ yields

$$T_{\text{eq}} \;=\; \dfrac{\Delta H}{\Delta S}.$$

Below this temperature ($$T < T_{\text{eq}}$$) we have

$$\Delta G = \Delta H - T\Delta S \;<\; 0,$$

so the oxide is stable. Above this temperature ($$T > T_{\text{eq}}$$) we find

$$\Delta G > 0,$$

and the oxide becomes unstable.

Therefore, on the $$\Delta G$$-versus-$$T$$ plot, the critical temperature is identified at the point where the plotted straight line crosses the horizontal axis (the line $$\Delta G = 0$$). At that single point the numerical value of $$\Delta G$$ changes sign—from negative (below the axis) to positive (above the axis). The slope itself does not have to change; only the sign of the ordinate changes.

Translating this understanding into the language of the options given in the question, we seek the description that says “the free energy change shows a change from negative to positive value.” That matches exactly with Option B.

Hence, the correct answer is Option B.