Question

Identify the incorrect statement regarding heavy water:
A: It is used as a coolant in nuclear reactors
B: It reacts with to produce $C{D_4}$ and $Al{\left( {OD} \right)_3}$
C: It reacts with $Ca{C_2}$ to produce ${C_2}{D_2}$ and $Ca{\left( {OD} \right)_2}$
D: It reacts with $S{O_3}$ to form deuterated sulphuric acid (${D_2}S{O_4}$)

Answer 1

Heavy water is denoted by ${D_2}O$ i.e. deuterium oxide. It is a compound made up of deuterium and oxygen. Deuterium is a heavy isotope of hydrogen, denoted by either $^2H$or $D$.

Complete Step by step answer:
${D_2}O$ is one of the two fundamental moderators that allow a nuclear reactor to function using natural uranium as fuel. The use of ${D_2}O$ is significant in nuclear proliferation because it makes provision of one additional route to form plutonium that can be used in weapons, totally circumventing uranium enrichment as well as the related technical infrastructure. In addition to this, ${D_2}O$ moderated reactors can even be utilised to make tritium. Now let us briefly discuss each given statement in the options:

Option A: Heavy water is not used as a coolant in nuclear reactors. Rather, it is used as a moderator in nuclear reactors. It is responsible to slow down the speed of fast moving neutrons and also helps in the control of the nuclear fission process.

Option B: Heavy water (${D_2}O$) can react with aluminium carbide ($A{l_4}{C_3}$) in a double displacement reaction to produce $C{D_4}$ and $Al{\left( {OD} \right)_3}$ as shown below:
$A{l_4}{C_3} + 12{D_2}O \to 2Al{(OD)_3} + 3C{D_4}$

Option C: Heavy water (${D_2}O$) can react with calcium carbide ($Ca{C_2}$) to produce deutero acetylene (${C_2}{D_2}$) and calcium deuteroxide ($Ca{\left( {OD} \right)_2}$) as shown below:
$Ca{C_2} + 2{D_2}O \to {C_2}{D_2} + Ca{(OD)_2}$

Option D: Heavy water (${D_2}O$) can react with $S{O_3}$ to form deuterated sulphuric acid (${D_2}S{O_4}$) as shown below:
$S{O_3} + {D_2}O \to {D_2}S{O_4}$
All the statements in the options are correct except statements in the Option A.

Hence, the correct answer is Option A.

Note: Since deuterium’s atomic mass is greater as compared to that of protium (hydrogen), molar mass of ${D_2}O$ is more than that of ${H_2}O$. Thus, ${D_2}O$ possess slightly different physical and chemical properties in comparison to ${H_2}O$. ${D_2}O$ is non-radioactive owing to the deuterium being a stable isotope. ${D_2}O$ can also be produced via hydrogen sulfide-water chemical exchange method, electrolysis and water distillation.