Question

Phosphorus on reaction with sulphuryl chloride gives a compound X, which on complete hydrolysis gives Y. X and Y are respectively

• A. \( \text{PCl}_3, \text{H}_3\text{PO}_3 \)
• B. \( \text{PCl}_5, \text{POCl}_3 \)
• C. \( \text{PCl}_5, \text{H}_3\text{PO}_4 \)
• D. \( \text{PCl}_3, \text{H}_3\text{PO}_2 \)

Hint:

Key distinction: - \( \text{SO}_2\text{Cl}_2 \) acts as a chlorinating agent converting P to its higher oxidation state (+5). - Hydrolysis of P(+5) halide yields P(+5) acid (\( \text{H}_3\text{PO}_4 \)).

Answer 1

The Correct Option is C

Step 1: Identifying Reaction 1:

Reaction of White Phosphorus (\( \text{P}_4 \)) with Sulphuryl Chloride (\( \text{SO}_2\text{Cl}_2 \)): \[ \text{P}_4 + 10\text{SO}_2\text{Cl}_2 \to 4\text{PCl}_5 + 10\text{SO}_2 \] Product X is Phosphorus Pentachloride (\( \text{PCl}_5 \)). (Note: Reaction with Thionyl Chloride \( \text{SOCl}_2 \) gives \( \text{PCl}_3 \)).
Step 2: Identifying Reaction 2 (Hydrolysis):

Complete hydrolysis of \( \text{PCl}_5 \) (X): \[ \text{PCl}_5 + 4\text{H}_2\text{O} \to \text{H}_3\text{PO}_4 + 5\text{HCl} \] Product Y is Orthophosphoric Acid (\( \text{H}_3\text{PO}_4 \)).
Step 3: Matching Options:

X = \( \text{PCl}_5 \), Y = \( \text{H}_3\text{PO}_4 \). This corresponds to Option (C).
Step 4: Final Answer:

X is \( \text{PCl}_5 \) and Y is \( \text{H}_3\text{PO}_4 \).