Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly?

Roasting is the step in which a metal sulphide $$MS$$ is first converted to its oxide $$MO$$ by heating it in excess air :

$$2\,MS + 3\,O_2 \rightarrow 2\,MO + 2\,SO_2$$

Only afterwards is the oxide reduced (most commonly by carbon/CO). We do not attempt a direct carbon-reduction of the sulphide

$$MS + C \; \longrightarrow \; M + CS_2 \;(\text{or } CS).$$

The feasibility of any pyrometallurgical reaction is judged from the standard Gibbs energies of formation, which are plotted in an Ellingham diagram. Keeping this in mind, let us examine the four statements.

Case 1: Option B — $$CO_2$$ is thermodynamically more stable than $$CS_2$$

The line for $$CO_2$$ lies far below that for $$CS_2$$ on the Ellingham diagram, i.e. $$\Delta G_f^\circ(CO_2)$$ is much more negative. Carbon therefore has a far greater affinity for oxygen than for sulphur. Hence carbon can reduce an oxide easily (because it is itself oxidised to $$CO_2$$/$$CO$$), but it cannot reduce a sulphide (because the oxidation product $$CS_2$$ is much less stable). This factor is directly responsible for preferring roasting over direct carbon reduction, so Option B is significant.

Case 2: Option C — Metal sulphides are less stable than the corresponding oxides

For most metals the $$MO$$ line also lies below the $$MS$$ line. Consequently the conversion $$MS \rightarrow MO$$ (roasting) is spontaneous, often exothermic, and needs no external reducing agent. Thus Option C is again a valid reason for carrying out roasting.

Case 3: Option D — $$CO_2$$ is more volatile than $$CS_2$$

The gaseous product formed during reduction leaves the reaction mixture, driving the equilibrium to the right. $$CO_2$$ (b.p. -78 °C under 1 atm) is far more volatile than $$CS_2$$ (b.p. 46 °C). Therefore the removal of $$CO_2$$ in oxide reduction is much easier than the removal of $$CS_2$$ in a hypothetical sulphide reduction. This volatility factor also favours the oxide route, so Option D is significant.

Case 4: Option A — Metal sulphides are thermodynamically more stable than $$CS_2$$

This statement compares two sulphur-containing species that are both on the sulphur side of the Ellingham diagram. Whether $$MS$$ happens to be more or less stable than $$CS_2$$ does not decide the success of carbon as a reducing agent; what really matters is the relative stabilities of the oxidation products of carbon (i.e. Option B) and of the ores themselves (i.e. Option C), together with the ease of removing the gaseous product (Option D). Hence the fact quoted in Option A is of no practical significance when deciding to roast the sulphide instead of reducing it directly with carbon.

Therefore, the factor that is not relevant for choosing roasting over direct carbon reduction is:

Option A which is: Metal sulphides are thermodynamically more stable than $$CS_2$$.