JEE Challenger
More from p-Block Elements (Group 15-18)
ChemMedAdv 2023 • Paper 1

Match Inorganic Phosphorus Reactions with Their Specific Products

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-IList-II(P) P2O3+3H2O→(1) P(O)(OCH3)Cl2(Q) P4+3NaOH+3H2O→(2) H3PO3(R) PCl5+CH3COOH→(3) PH3(S) H3PO2+2H2O+4AgNO3→(4) POCl3(5) H3PO4\begin{array}{ll} \text{\textbf{List-I}} & \text{\textbf{List-II}} \\ \text{(P) } \text{P}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow & \text{(1) } \text{P(O)(OCH}_3\text{)Cl}_2 \\ \text{(Q) } \text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow & \text{(2) } \text{H}_3\text{PO}_3 \\ \text{(R) } \text{PCl}_5 + \text{CH}_3\text{COOH} \rightarrow & \text{(3) } \text{PH}_3 \\ \text{(S) } \text{H}_3\text{PO}_2 + 2\text{H}_2\text{O} + 4\text{AgNO}_3 \rightarrow & \text{(4) } \text{POCl}_3 \\ & \text{(5) } \text{H}_3\text{PO}_4 \end{array}

Options

A

P→2;Q→3;R→1;S→5\text{P} \rightarrow 2; \text{Q} \rightarrow 3; \text{R} \rightarrow 1; \text{S} \rightarrow 5

B

P→3;Q→5;R→4;S→2\text{P} \rightarrow 3; \text{Q} \rightarrow 5; \text{R} \rightarrow 4; \text{S} \rightarrow 2

C

P→5;Q→2;R→1;S→3\text{P} \rightarrow 5; \text{Q} \rightarrow 2; \text{R} \rightarrow 1; \text{S} \rightarrow 3

D

P→2;Q→3;R→4;S→5\text{P} \rightarrow 2; \text{Q} \rightarrow 3; \text{R} \rightarrow 4; \text{S} \rightarrow 5

Correct

Topics & Concepts

p-Block Elements (Group 15-18)p-Block Elements

Step-by-Step Solution

To determine the correct matching between the reactions in List-I and their products in List-II, let us analyze each chemical reaction in detail:

  1. Reaction (P): When phosphorus trioxide (P2O3\text{P}_2\text{O}_3, or P4O6\text{P}_4\text{O}_6) reacts with cold water, it undergoes hydrolysis to form orthophosphorous acid (H3PO3\text{H}_3\text{PO}_3): P2O3+3H2O→2H3PO3\text{P}_2\text{O}_3 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{PO}_3 Therefore, (P) →\rightarrow (2).

  2. Reaction (Q): White phosphorus (P4\text{P}_4) undergoes a disproportionation reaction when heated with concentrated sodium hydroxide solution in an inert atmosphere, producing phosphine (PH3\text{PH}_3) and sodium hypophosphite (NaH2PO2\text{NaH}_2\text{PO}_2): P4+3NaOH+3H2O→PH3+3NaH2PO2\text{P}_4 + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow \text{PH}_3 + 3\text{NaH}_2\text{PO}_2 Therefore, (Q) →\rightarrow (3).

  3. Reaction (R): Phosphorus pentachloride (PCl5\text{PCl}_5) reacts with carboxylic acids such as acetic acid (CH3COOH\text{CH}_3\text{COOH}) to produce acetyl chloride (CH3COCl\text{CH}_3\text{COCl}), phosphoryl chloride (POCl3\text{POCl}_3), and hydrogen chloride (HCl\text{HCl}): PCl5+CH3COOH→CH3COCl+POCl3+HCl\text{PCl}_5 + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COCl} + \text{POCl}_3 + \text{HCl} Therefore, (R) →\rightarrow (4).

  4. Reaction (S): Hypophosphorous acid (H3PO2\text{H}_3\text{PO}_2) acts as a strong reducing agent. It reduces aqueous silver nitrate (AgNO3\text{AgNO}_3) to metallic silver, while itself being oxidized to orthophosphoric acid (H3PO4\text{H}_3\text{PO}_4): H3PO2+2H2O+4AgNO3→4Ag↓+4HNO3+H3PO4\text{H}_3\text{PO}_2 + 2\text{H}_2\text{O} + 4\text{AgNO}_3 \rightarrow 4\text{Ag}\downarrow + 4\text{HNO}_3 + \text{H}_3\text{PO}_4 Therefore, (S) →\rightarrow (5).

Thus, the correct matching is: P→2;Q→3;R→4;S→5\text{P} \rightarrow 2; \quad \text{Q} \rightarrow 3; \quad \text{R} \rightarrow 4; \quad \text{S} \rightarrow 5

This matches the combination given in Option D.

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