JEE Advanced 2013 · Paper 2 · Chemistry Q32

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ChemistrySingle correct+3 / -1Question 32 of 60

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Paragraph for Questions 31 and 32

The reactions of $\ce{Cl2}$ gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The $\ce{Cl2}$ gas reacts with $\ce{SO2}$ gas, in 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.

R, S and T, respectively, are

\ce{SO2Cl2}

\ce{PCl5}

and

\ce{H3PO4}

\ce{SO2Cl2}

\ce{PCl3}

and

\ce{H3PO3}

\ce{SOCl2}

\ce{PCl3}

and

\ce{H3PO2}

\ce{SOCl2}

\ce{PCl5}

and

\ce{H3PO4}

Answer and worked solution

Official answer: A

R: $\ce{SO2 + Cl2 ->[charcoal] SO2Cl2}$ (sulphuryl chloride).
S: $\ce{P4 + 10SO2Cl2 -> 4PCl5 + 10SO2}$ .
T: $\ce{PCl5 + 4H2O -> H3PO4 + 5HCl}$ .

(A)

Topics:Group 15 Group 16

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JEE Advanced 2013 · Paper 2, Chemistry question 32