Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their products given in List-II and choose the correct option.

List - I	List - II
(P) $\mathrm{P}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O} \rightarrow$	(1) $\mathrm{P}(\mathrm{O})\left(\mathrm{OCH}_3\right) \mathrm{Cl}_2$
(Q) $\mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \rightarrow$	(2) $\mathrm{H}_3 \mathrm{PO}_3$
(R) $\mathrm{PCl}_5+\mathrm{CH}_3 \mathrm{COOH} \rightarrow$	(3) $\mathrm{PH}_3$
(S) $\mathrm{H}_3 \mathrm{PO}_2+2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{AgNO}_3 \rightarrow$	(4) $\mathrm{POCl}_3$
	(5) $\mathrm{H}_3 \mathrm{PO}_4$

The reaction of $\mathrm{HClO}_{3}$ with $\mathrm{HCl}$ gives a paramagnetic gas, which upon reaction with $\mathrm{O}_{3}$ produces

The reaction of $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ and $\mathrm{NaCl}$ in water produces a precipitate that dissolves upon the addition of $\mathrm{HCl}$ of appropriate concentration. The dissolution of the precipitate is due to the formation of

A colourless aqueous solution contains nitrates of two metals, X and Y. When it was added to an aqueous solution of NaCl, a white precipitate was formed. This precipitate was found to be partly soluble in hot water to give a residue P and a solution Q. The residue P was soluble in aqueous NH₃ and also in excess sodium thiosulphate. The hot solution Q gave a yellow precipitate with KI. The metals X and Y, respectively, are