The reaction that does NOT define calcination is:

First, let us recall the meaning of calcination. The definition we use in metallurgy is: “Calcination is the process of heating an ore in the absence or limited supply of air to remove volatile impurities such as water of crystallisation or carbon dioxide.” In contrast, roasting is the process of heating an ore in excess of air or oxygen so that sulphide ores are converted into their oxides with liberation of sulphur dioxide.

Now we examine each option one by one and compare it with the above definition.

Option A gives the chemical change $$Fe_2O_3 \cdot xH_2O \xrightarrow{\Delta} Fe_2O_3 + xH_2O.$$ We see that the hydrated ferric oxide ($$Fe_2O_3\cdot xH_2O$$) is only losing its water of crystallisation on heating. No oxygen from outside appears in the equation. This exactly matches the description of calcination.

Option B shows the reaction $$2Cu_2S + 3O_2 \xrightarrow{\Delta} 2Cu_2O + 2SO_2.$$ Here molecular oxygen $$O_2$$ is an explicit reactant. The sulphide ore $$Cu_2S$$ is being heated in the presence of excess air/oxygen to form an oxide and sulphur dioxide. This is the textbook example of roasting, not calcination.

Option C presents $$ZnCO_3 \xrightarrow{\Delta} ZnO + CO_2.$$ Zinc carbonate loses $$CO_2$$ on heating, with no external oxygen involved. This is again calcination.

Option D offers $$CaCO_3 \cdot MgCO_3 \xrightarrow{\Delta} CaO + MgO + 2CO_2.$$ Both carbonates give off $$CO_2$$ on heating in limited or no air. This is also a straightforward case of calcination.

Summarising the discussion, only the reaction in Option B employs free oxygen as a reactant and converts a sulphide to an oxide, satisfying the definition of roasting, not calcination. All the other reactions meet the criteria for calcination.

Hence, the correct answer is Option B.