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Calcination is a thermal treatment carried out in a limited or complete absence of air. During this process substances such as carbonates, bicarbonates, hydroxides and hydrated oxides lose volatile components like $$CO_2$$ or $$H_2O$$ and get converted into simple, anhydrous metal oxides.
An ore or compound that is already present as a simple, anhydrous oxide obviously cannot lose anything further; it is already in its most thermally stable oxide form. Consequently, such an oxide does not need calcination before subsequent metallurgical steps such as reduction or smelting.
Now let us examine each option carefully.
Option A: $$ZnO$$ and $$Fe_2O_3 \cdot xH_2O$$
• $$ZnO$$ is a simple oxide, so it does not require calcination.
• $$Fe_2O_3 \cdot xH_2O$$ is a hydrated oxide; the loosely bound water must be expelled. Heating (calcination) is therefore required to transform it into anhydrous $$Fe_2O_3$$.
Because one member of the pair does need calcination, the whole pair cannot be chosen.
Option B: $$ZnO$$ and $$MgO$$
• $$ZnO$$ is already an anhydrous oxide.
• $$MgO$$ is also an anhydrous oxide.
Neither substance contains water of hydration, hydroxide groups, carbonate groups, or any other volatile component. No liberation of $$CO_2$$ or $$H_2O$$ can occur; hence no calcination step is needed for either compound.
Option C: $$ZnCO_3$$ and $$CaO$$
• $$ZnCO_3$$ is a carbonate. It must be calcined to expel $$CO_2$$ and furnish $$ZnO$$.
• $$CaO$$ is a simple oxide and does not need calcination.
Because one member of this pair requires calcination, the pair as a whole is unsuitable.
Option D: $$Fe_2O_3$$ and $$CaCO_3 \cdot MgCO_3$$ (dolomite)
• $$Fe_2O_3$$ itself does not need calcination.
• The dolomite mixture $$CaCO_3 \cdot MgCO_3$$ is wholly carbonate in nature; it must be calcined to expel $$CO_2$$ and yield the mixed oxides $$CaO + MgO$$.
Thus the pair requires calcination.
Out of the four possibilities, only the second option presents a pair in which neither substance demands the removal of volatile components. Both compounds are already fully converted oxides.
Hence, the correct answer is Option 2.
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