Pblock Elementsii Result Question 27
Passage
The reactions of $\mathrm{Cl}_2$ gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, $P$ and $Q$, respectively. The $\mathrm{Cl}_2$ gas reacts with $\mathrm{SO}_2$ gas in the presence of charcoal, to give a product $R . R$ reacts with white phosphorus to give a compound $S$. On hydrolysis, $S$ gives an oxoacid of phosphorus $T$.
27. $R, S$ and $T$, respectively, are
(a) $\mathrm{SO} _2 \mathrm{Cl} _2, \mathrm{PCl} _5$ and $\mathrm{H} _3 \mathrm{PO} _4$
(b) $\mathrm{SO} _2 \mathrm{Cl} _2, \mathrm{PCl} _3$ and $\mathrm{H} _3 \mathrm{PO} _3$
(c) $\mathrm{SOCl} _2, \mathrm{PCl} _3$ and $\mathrm{H} _3 \mathrm{PO} _2$
(d) $\mathrm{SOCl} _2, \mathrm{PCl} _5$ and $\mathrm{H} _3 \mathrm{PO} _4$
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Answer:
Correct Answer: 27. ( a )
Solution:
$ 2 \hspace{1mm} \mathrm{NaOH}+\mathrm{Cl}_2 \xrightarrow{\text { cold }} \mathrm{NaCl}+\underset{P}{\mathrm{NaOCl}}+\mathrm{H}_2 \mathrm{O} $
$ 6 \hspace{1mm}\mathrm{NaOH}+3 \mathrm{Cl}_2 \xrightarrow{\text { hot }} 5 \hspace{1mm}\mathrm{NaCl}+ \underset{Q} {\mathrm{NaClO}_3} +3 \hspace{1mm} \mathrm{H}_2 \mathrm{O} $
$ \underset{\text{hypochlorous acid}} {\mathrm{HOCl}} \xrightarrow{\mathrm{NaOH}} \underset{P} {\mathrm{NaOCl}}$
$ \underset{\text { chloric acid } } { \mathrm{HClO}_3 } \xrightarrow{\mathrm{NaOH}} \underset{Q} {\mathrm{NaClO}_3} $
$ \mathrm{Cl}_2+\mathrm{SO}_2 \longrightarrow \underset{R} {\mathrm{SO}_2 \mathrm{Cl}_2} $
$ 10 \hspace{1mm}\underset{R}{\mathrm{SO}_2 \mathrm{Cl}_2}+\mathrm{P}_4 \longrightarrow \underset{S}{4 \hspace{1mm} \mathrm{PCl}_5}+10 \hspace{1mm} \mathrm{SO}_2 $
$ \mathrm{PCl}_5+4 \hspace{1mm}\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{T} {\mathrm{H}_3 \mathrm{PO}_4 }+5 \hspace{1mm} \mathrm{HCl} $
BETA
