Paragraph: The reactions of $\mathrm{Cl}_{2}$ gas with cold-dilute and hot-concentrated $\mathrm{NaOH}$ in…

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. SO2Cl2, PCl5 and H3PO4
  2. SO2Cl2, PCl3 and H3PO3
  3. SOCl2, PCl3 and H3PO2
  4. SOCl2, PCl5 and H3PO4

Solution

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

Asked in: JEE Advanced 2013 (Paper 2)

Practice more p Block Elements (Group 15, 16, 17 & 18) questions on Aicharya