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JEE Advanced 2013Paper 2
Exam
JEE Advanced
Year
2013
Paper
Paper 2
Subject
Chemistry
Chapter
p Block Elements (Group 15, 16, 17 & 18)

JEE Advanced 2013 · Paper 2 · Chemistry · p Block Elements (Group 15, 16, 17 & 18)

Paragraph: The reactions of Cl_ 2 gas with cold-dilute and...

Paragraph: The reactions of Cl2\mathrm{Cl}_{2} gas with cold-dilute and hot-concentrated NaOH\mathrm{NaOH} in water give sodium salts of two (different) oxoacids of chlorine, P\boldsymbol{P} and Q\boldsymbol{Q}, respectively. The Cl2\mathrm{Cl}_{2} gas reacts with SO2\mathrm{SO}_{2} gas, in presence of charcoal, to give a product R\boldsymbol{R}. R\boldsymbol{R} reacts with white phosphorus to give a compound S\boldsymbol{S}. On hydrolysis, S\boldsymbol{S} gives an oxoacid of phosphorus, T\boldsymbol{T}.
Question: R,SR, S and TT, respectively, are

Options

  1. ASO2Cl2,PCl5andH3PO4{SO}_{2}{Cl}_{2}, {PCl}_{5}and{H}_{3}{PO}_{4}
  2. BSO2Cl2,PCl3andH3PO3{SO}_{2}{Cl}_{2}, {PCl}_{3}and{H}_{3}{PO}_{3}
  3. CSOCl2,PCl3andH3PO2{SOCl}_{2}, {PCl}_{3}and{H}_{3}{PO}_{2}
  4. DSOCl2,PCl5andH3PO4{SOCl}_{2}, {PCl}_{5}and{H}_{3}{PO}_{4}

Correct answer

A. SO2Cl2,PCl5andH3PO4{SO}_{2}{Cl}_{2}, {PCl}_{5}and{H}_{3}{PO}_{4}

Step-by-step solution

Cl2+cold dil. NaOH→NaOCl+NaCl{\text{Cl}}_{2} + \text{cold dil. NaOH} \to \text{NaOCl} + \text{NaCl}
Cl2+hot conc.NaOH→NaClO3+NaCl{\text{Cl}}_{2} + \text{hot conc.} \text{NaOH} \to {\text{NaClO}}_{3} + \text{NaCl}
NaOCl is salt of hypochlorous acid = P.
NaOCl3 is salt of chloric acid = Q.
{Cl2+SO2⟶CharcoalSO2Cl2(R)SO2Cl2+P4⟶PCl5(S)+SO2PCl5+H2O⟶H3PO4(T)+HCl\begin{cases}{\text{Cl}}_{2} + {\text{SO}}_{2} & \overset{\text{Charcoal}}{⟶} & {\text{SO}}_{2} {\text{Cl}}_{2} \left(\text{R}\right) \\ {\text{SO}}_{2} {\text{Cl}}_{2} + {\text{P}}_{4} & ⟶ & {\text{PCl}}_{5} \left(\text{S}\right) + {\text{SO}}_{2} \\ {\text{PCl}}_{5} + {\text{H}}_{2} \text{O} & ⟶ & {\text{H}}_{3} {\text{PO}}_{4} \left(\text{T}\right) + \text{HCl}\end{cases}
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