Question

The reaction of `Cl_(2)` gas with cold-dilute and hot-concentrated NaOH in water give sodium salts to two (different) oxoacids of chlorine, P and Q, respectively. The `Cl_(2)` gas reacts with `SO_(2)` gas, in presence of charcoal, to give a product R. R reacts with white phosphours to give a compound S. On hydrolysis. S gives an oxoacid of phosphorus T.
Q. R, S and T, respectively
A. `SO_(2)Cl_(2),PCl_(5) & H_(3)PO_(4)`
B. `SO_(2)Cl_(2),PCl_(3)&H_(3)PO_(3)`
C. `SOCl_(2),PCl_(3)&H_(3)PO_(2)`
D. `SOCl_(2),PCl_(5)&H_(3)PO_(4)`

Answer 1

Correct Answer - A
(a) `Cl_(2)+"cold dil. "NaOHtoNaOCl+NaCl`
`Cl_(2)+"hot conc. "NaOHtoNaClO_(3)+NaCl`
NaOCl is salt of hypochlorous acid=F
`NaOCl_(3)` is salt of chloric acid=Q.
`Cl_(2)+SO_(2) overset("Charcoal")toSO_(2)Cl_(2)(R)`
`PCl_(5)+H_(2)O to H_(3)PO_(4)(T)+HCl`