
When $S{{O}_{3}}$ treated with heavy water the product is /are
A. Deuterium and sulphuric acid
B. Deuterium and sulphurous acid
C. Only deuterium
D. Dideuterio Sulphuric acid
Answer
220.2k+ views
Hint: Water molecule in which deuterium, a heavy isotope of hydrogen, is replaced in the place of hydrogen is known as heavy water (${{D}_{2}}O$). Ordinary water and heavy water have the same physical and chemical properties. Ordinary water (${{H}_{2}}O$) reacts with sulphur trioxide ($S{{O}_{3}}$) and thereby produces sulphuric acid $({{H}_{2}}S{{O}_{4}})$.
Complete Step by Step Answer:
When sulphur trioxide $(S{{O}_{3}})$ is exposed to air, it takes up water from the atmosphere. Hence it produces white fumes. Sulphur trioxide is very corrosive. That’s why it reacts with water to form highly corrosive sulphuric acid (${{H}_{2}}S{{O}_{4}}$). This is one kind of hydrolysis reaction. $S{{O}_{3}}(g)+{{H}_{2}}O(aq.)\to {{H}_{2}}S{{O}_{4}}(aq.)$
Like ordinary water, heavy water also reacts with sulphur trioxide and forms Dideuterosulphuric acid (${{D}_{2}}S{{O}_{4}}$).
$S{{O}_{3}}(g)+{{D}_{2}}O(aq.)\to {{D}_{2}}S{{O}_{4}}(aq.)$
Heavy water reacts with sulphur trioxide very slowly as compared to ordinary water. But they form a stronger bond $S{{O}_{3}}$than the bonding between normal water and sulphur trioxide.
Mechanism:
Here $S{{O}_{3}}$acts as a strong electrophile and reacts rapidly with heavy water which is a relatively weak nucleophile. When nucleophile attacks sulphur trioxide, initially the dissociation of a $S=O$bond. Additional heavy water molecules rapidly abstract deuterium atoms and form the intermediate ($SO_{4}^{2-}$). After deuterium abstraction by the intermediate results, Deuter Sulphuric acid was the final product.
Thus, option (D) is correct.
Note: As heavy water is not radioactive, it does not mean that we can drink it. The molecular mass of heavy water is higher compared to ordinary water. As a result, it has different physiological activities. It is found in several investigations, that drinking too much heavy water can cause low blood pressure, dizziness and many other harmful diseases.
Complete Step by Step Answer:
When sulphur trioxide $(S{{O}_{3}})$ is exposed to air, it takes up water from the atmosphere. Hence it produces white fumes. Sulphur trioxide is very corrosive. That’s why it reacts with water to form highly corrosive sulphuric acid (${{H}_{2}}S{{O}_{4}}$). This is one kind of hydrolysis reaction. $S{{O}_{3}}(g)+{{H}_{2}}O(aq.)\to {{H}_{2}}S{{O}_{4}}(aq.)$
Like ordinary water, heavy water also reacts with sulphur trioxide and forms Dideuterosulphuric acid (${{D}_{2}}S{{O}_{4}}$).
$S{{O}_{3}}(g)+{{D}_{2}}O(aq.)\to {{D}_{2}}S{{O}_{4}}(aq.)$
Heavy water reacts with sulphur trioxide very slowly as compared to ordinary water. But they form a stronger bond $S{{O}_{3}}$than the bonding between normal water and sulphur trioxide.
Mechanism:
Here $S{{O}_{3}}$acts as a strong electrophile and reacts rapidly with heavy water which is a relatively weak nucleophile. When nucleophile attacks sulphur trioxide, initially the dissociation of a $S=O$bond. Additional heavy water molecules rapidly abstract deuterium atoms and form the intermediate ($SO_{4}^{2-}$). After deuterium abstraction by the intermediate results, Deuter Sulphuric acid was the final product.
Thus, option (D) is correct.
Note: As heavy water is not radioactive, it does not mean that we can drink it. The molecular mass of heavy water is higher compared to ordinary water. As a result, it has different physiological activities. It is found in several investigations, that drinking too much heavy water can cause low blood pressure, dizziness and many other harmful diseases.
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