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Let us first recall that heavy water is nothing but deuterium oxide, written as $$D_2O$$. Wherever ordinary water $$H_2O$$ would react to give a hydrogen-containing product, heavy water behaves in exactly the same manner but now every hydrogen is replaced by deuterium $$D$$. Using this guiding idea we can test each statement.
Option A: In ordinary water the well-known reaction is
$$H_2O + SO_3 \rightarrow H_2SO_4.$$
Replacing $$H_2O$$ by $$D_2O$$ we obtain
$$D_2O + SO_3 \rightarrow D_2SO_4.$$
This merely changes every $$H$$ to $$D$$, so formation of deuterated sulphuric acid $$D_2SO_4$$ is perfectly correct. Hence Option A is a correct statement.
Option B: Heavy water is famous for its ability to slow down (moderate) fast neutrons in a nuclear reactor. A moderator is a substance that decreases the energy of neutrons so that the nuclear chain reaction can be sustained. Although ordinary (light) water can serve as both coolant and moderator, heavy water is employed almost exclusively as a moderator; the coolant in most reactor designs is usually ordinary light water or some other fluid such as pressurised CO2 or liquid metal. Therefore the sentence “It is used as a coolant in nuclear reactors’’ is not generally correct for heavy water. So Option B is very likely to be the wrong statement.
Option C: With ordinary water the calcium carbide reaction is
$$CaC_2 + 2\,H_2O \rightarrow C_2H_2 + Ca(OH)_2.$$
Writing $$H_2O$$ as $$D_2O$$ and replacing each hydrogen by deuterium we get
$$CaC_2 + 2\,D_2O \rightarrow C_2D_2 + Ca(OD)_2.$$
This is exactly what Option C states, so Option C is correct.
Option D: For aluminium carbide with ordinary water the reaction is
$$Al_4C_3 + 12\,H_2O \rightarrow 3\,CH_4 + 4\,Al(OH)_3.$$
Substituting $$D_2O$$ for $$H_2O$$ and again changing every $$H$$ to $$D$$ we obtain
$$Al_4C_3 + 12\,D_2O \rightarrow 3\,CD_4 + 4\,Al(OD)_3.$$
Thus Option D is also fully correct.
From the above examination, only Option B makes an incorrect claim about the role of heavy water in a nuclear reactor.
Hence, the correct answer is Option B.
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