1. For the reaction

$4M(s) + nO_2(g) \rightarrow 2M_2O_n(s)$

the Gibbs free energy change is

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

2. The oxide $M_2O_n$ is thermodynamically stable when its formation is spontaneous, i.e.

$\Delta G < 0$

If $\Delta G > 0$, the oxide is not stable with respect to decomposition into metal and oxygen.

3. Therefore, the temperature limit for stability is the temperature at which

$\Delta G = 0$

Below this temperature, if $\Delta G$ is negative, the oxide is stable.

4. On a graph of $\Delta G$ versus $T$, this corresponds to the point where the free energy curve crosses the zero line.

That is, the relevant point is where $\Delta G$ changes sign from negative to positive as temperature increases.

5. Checking options:

• A: free energy changes from negative to positive value
  This is correct: the crossing point $\Delta G=0$ gives the upper temperature limit of oxide stability.

• B: slope changes from positive to negative
  Not the criterion for stability.

• C: slope changes from negative to positive
  Not the criterion for stability.

• D: slope changes from positive to zero
  Also not the criterion for stability.

Hence, the correct option is A.