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)	$$P_2O_3 + 3H_2O \to$$	(1)	$$P(O)(OCH_3)Cl_2$$
(Q)	$$P_4 + 3NaOH + 3H_2O \to$$	(2)	$$H_3PO_3$$
(R)	$$PCl_5 + CH_3COOH \to$$	(3)	$$PH_3$$
(S)	$$H_3PO_2 + 2H_2O + 4AgNO_3 \to$$	(4)	$$POCl_3$$
		(5)	$$H_3PO_4$$

The problem gives four phosphorus-based reactions (List-I) and five possible products (List-II). We have to decide which product is formed in each reaction and then select the option that contains the correct matching.

Case P: $$P_2O_3 + 3H_2O \longrightarrow$$ Phosphorus(III) oxide is the anhydride of phosphorous acid. When it is hydrolysed, the only product is phosphorous acid. $$P_2O_3 + 3H_2O \rightarrow 2\,H_3PO_3$$ Hence, for reaction P the required product is $$H_3PO_3$$ (List-II, item 2).

Case Q: $$P_4 + 3NaOH + 3H_2O \longrightarrow$$ White phosphorus in hot alkali undergoes disproportionation. One part is reduced to phosphine, the other is oxidised to hypophosphite. $$P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3\,NaH_2PO_2$$ Thus one of the products is phosphine $$PH_3$$ (List-II, item 3).

Case R: $$PCl_5 + CH_3COOH \longrightarrow$$ $$PCl_5$$ is a strong chlorinating agent. With a carboxylic acid it replaces the -OH by -Cl forming an acyl chloride and simultaneously gets converted to phosphoryl chloride. $$PCl_5 + CH_3COOH \rightarrow CH_3COCl + POCl_3 + HCl$$ Therefore the phosphorus-containing product is $$POCl_3$$ (List-II, item 4).

Case S: $$H_3PO_2 + 2H_2O + 4AgNO_3 \longrightarrow$$ Hypophosphorous acid is a good reducing agent. In the presence of $$Ag^+$$ it reduces silver(I) to metallic silver and itself is oxidised to phosphoric acid. $$H_3PO_2 + 2H_2O + 4AgNO_3 \rightarrow H_3PO_4 + 4Ag + 4HNO_3$$ So the phosphorus-containing product is $$H_3PO_4$$ (List-II, item 5).

Collecting the matches:
P → 2, Q → 3, R → 4, S → 5.

The only option containing this set is Option D.

Final Answer: Option D which is: P → 2; Q → 3; R → 4; S → 5