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JEE Advanced · Chemistry · 21. p Block (G15-18)

Paragraph:
The reactions of \(\mathrm{Cl}_{2}\) gas with cold-dilute and hot-concentrated \(\mathrm{NaOH}\) in water give sodium salts of two (different) oxoacids of chlorine, \(\boldsymbol{P}\) and \(\boldsymbol{Q}\), respectively. The \(\mathrm{Cl}_{2}\) gas reacts with \(\mathrm{SO}_{2}\) gas, in presence of charcoal, to give a product \(\boldsymbol{R}\). \(\boldsymbol{R}\) reacts with white phosphorus to give a compound \(\boldsymbol{S}\). On hydrolysis, \(\boldsymbol{S}\) gives an oxoacid of phosphorus, \(\boldsymbol{T}\).
Question:
\(R, S\) and \(T\), respectively, are

  1. A SO2Cl2, PCl5 and H3PO4
  2. B SO2Cl2, PCl3 and H3PO3
  3. C SOCl2, PCl3 and H3PO2
  4. D SOCl2, PCl5 and H3PO4
Verified Solution

Answer & Solution

Correct Answer

(A) SO2Cl2, PCl5 and H3PO4

Step-by-step Solution

Detailed explanation

Cl 2 + cold dil. NaOH NaOCl + NaCl
Cl 2 + hot conc.  NaOH NaClO 3 + NaCl
NaOCl is salt of hypochlorous acid = P.
NaOCl3 is salt of chloric acid = Q.
Cl 2 + SO 2 Charcoal SO 2 Cl 2 R SO 2 Cl 2 + P 4 PCl 5 S + SO 2 PCl 5 + H 2 O H 3 PO 4 T + HCl
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