The P-Block Elements JEE 2023: Matching Question Solved

What is the correct answer to the p-block elements JEE 2023 matching question?

Option D is correct for the p-block elements JEE 2023 matching question: reaction R forms phosphoryl chloride, not the methoxy-containing compound in List-II. The source is JEE Advanced 2023, Paper 1, Chemistry. The question bank classifies it as medium difficulty; this is not an official exam rating.

For each reaction in List-I, use the specified reactant ratios to identify one product in List-II, then select the matching option. Each match identifies one product, not the entire product mixture. Include the coproducts when balancing each equation.

List-I

P : & P_{2} O_{3} + 3 H_{2} O \to Q : & P_{4} + 3 N a O H + 3 H_{2} O \to R : & P C l_{5} + C H_{3} C O O H \to S : & H_{3} P O_{2} + 2 H_{2} O + 4 A g N O_{3} \to

List-II

1 & : P (O) (O C H_{3}) C l_{2} 2 & : H_{3} P O_{3} 3 & : P H_{3} 4 & : P O C l_{3} 5 & : H_{3} P O_{4}

The options give entries in the order P, Q, R, S:

The official method uses hydrolysis, reaction with alkali, acid-chloride formation and oxidation by silver ions. Write complete reactions rather than match similar-looking formulas: the products and their coefficients justify each choice.

Why does reaction P give phosphorous acid rather than phosphoric acid?

Hydrolysis gives phosphorous acid, because the reactant is an acidic phosphorus(III) oxide. Water forms the corresponding acid without changing the phosphorus oxidation state.

The phosphorus oxidation state in the oxide is:

2 x + 3 (- 2) = 0 \Rightarrow x = + 3

The complete balanced reaction is:

P_{2} O_{3} + 3 H_{2} O \to 2 H_{3} P O_{3}

The atom totals on each side are:

P = 2, H = 6, O = 6

Phosphorous acid and phosphoric acid are different compounds. Their phosphorus oxidation states distinguish them:

H_{3} P O_{3} (phosphorous acid) : & x = + 3 H_{3} P O_{4} (phosphoric acid) : & x = + 5

No oxidation occurs in this hydrolysis, so phosphoric acid is not the product. The match is:

P \to 2

Why does reaction Q give both phosphine and sodium hypophosphite?

White phosphorus reacts with alkali and water to form phosphine and sodium hypophosphite. Phosphine supplies the numbered match, but omitting hypophosphite hides the oxidation step and leaves the equation unbalanced.

The official reaction is:

P_{4} + 3 N a O H + 3 H_{2} O \to P H_{3} + 3 N a H_{2} P O_{2}

The atom totals on each side are:

P = 4, N a = 3, H = 9, O = 6

This is disproportionation: phosphorus starts in its elemental state and undergoes both reduction and oxidation.

In phosphine: & 0 \to - 3 In sodium hypophosphite: & 0 \to + 1

One phosphorus atom gains three electrons. Three other phosphorus atoms each lose one electron:

Electrons gained = 1 \times 3 = 3 = electrons lost = 3 \times 1

Phosphine appears in List-II; sodium hypophosphite remains a genuine coproduct. Thus:

Q \to 3

Why does reaction R produce phosphoryl chloride?

Phosphorus pentachloride converts a carboxylic acid into an acid chloride, producing phosphoryl chloride alongside it. Acetic acid is a carboxylic acid, so this rule determines the products in reaction R.

The organic transformation replaces the carboxyl hydroxyl group with chlorine: $C H_{3} C O O H \to C H_{3} C O C l$

The complete equation includes the organic and phosphorus-containing products:

P C l_{5} + C H_{3} C O O H \to C H_{3} C O C l + P O C l_{3} + H C l

The five chlorine atoms are allocated as follows:

\underset{acetyl chloride}{\underset{⏟}{1}} + \underset{phosphoryl chloride}{\underset{⏟}{3}} + \underset{hydrogen chloride}{\underset{⏟}{1}} = 5

The remaining atom totals on each side are:

P = 1, C = 2, H = 4, O = 2

The phosphorus-containing product is phosphoryl chloride, product 4. This reaction separates options A and D:

R \to 4

Why do four silver ions oxidise hypophosphorous acid to phosphoric acid?

The specified four silver ions accept four electrons, taking phosphorus to the oxidation state found in phosphoric acid. Hypophosphorous acid acts as a reducing agent: it reduces silver ions to metallic silver while itself being oxidised.

The initial phosphorus oxidation state in hypophosphorous acid is:

H_{3} P O_{2} : 3 (+ 1) + x + 2 (- 2) = 0 \Rightarrow x = + 1

The final state in phosphoric acid is:

H_{3} P O_{4} : 3 (+ 1) + x + 4 (- 2) = 0 \Rightarrow x = + 5

Phosphorus loses four electrons. Each silver ion accepts one: $A g^{+} + e^{-} \to A g$

The complete equation is:

H_{3} P O_{2} + 2 H_{2} O + 4 A g N O_{3} \to H_{3} P O_{4} + 4 A g + 4 H N O_{3}

The atom totals on each side are:

P = 1, H = 7, O = 16, A g = 4, N = 4

Stopping at phosphorous acid accounts for only a two-electron oxidation of phosphorus, not the four-electron change required here. The specified stoichiometry fixes the extent of oxidation.

Reaction S therefore matches phosphoric acid, product 5. Combining all four reactions gives option D:

P \to 2, Q \to 3, R \to 4, S \to 5

How does the wrong reaction rule produce option A?

A and D differ only in reaction R. Both assign P to 2, Q to 3 and S to 5; A assigns R to 1, while D assigns R to 4.

Treating the methyl group in acetic acid as evidence for a methoxy group in the phosphorus product can lead to product 1 and therefore option A:

P (O) (O C H_{3}) C l_{2}

That is a structural error. In acetic acid, the methyl carbon is bonded to the carbonyl carbon, not directly to oxygen. The reactant is not methanol and contains no methoxy group.

Use the functional group, not formula resemblance. Identify the carboxylic acid, apply acid-chloride formation, and write every product before selecting the match.

How can you check your method with three related phosphorus questions?

Use reaction Q’s mole ratios, reaction R’s functional-group rule and reaction S’s electron balance to solve these checks. These are original follow-up practice, not additional verified JEE PYQs.

1. How much phosphine and sodium hypophosphite form from 0.50 mol of white phosphorus?

With sufficient sodium hydroxide and water, the products are 0.50 mol of phosphine and 1.50 mol of sodium hypophosphite. Reaction Q gives the amounts directly from its coefficients:

n (P_{4}) = 0.50 m o l

P_{4} : P H_{3} : N a H_{2} P O_{2} = 1 : 1 : 3

n (P H_{3}) & = 0.50 \times 1 = 0.50 m o l n (N a H_{2} P O_{2}) & = 0.50 \times 3 = 1.50 m o l

2. What forms when propanoic acid reacts with phosphorus pentachloride?

The products are propanoyl chloride, phosphoryl chloride and hydrogen chloride. Changing the alkyl group does not change the carboxylic-acid-to-acid-chloride rule.

C H_{3} C H_{2} C O O H + P C l_{5} \to C H_{3} C H_{2} C O C l + P O C l_{3} + H C l

3. How many moles of silver ions are reduced when 0.25 mol of hypophosphorous acid is fully oxidised to phosphoric acid?

1.00 mol of silver ions is reduced. Each phosphorus atom loses four electrons, and each silver ion accepts one:

n (H_{3} P O_{2}) = 0.25 m o l

n (A g^{+}) = 4 \times 0.25 = 1.00 m o l

Before selecting a match, write the complete equation and check its atom totals. For a redox reaction, verify the electron balance too.

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Frequently asked questions

What is the correct answer to the p-block matching question in JEE Advanced 2023?

Option D is correct: P → 2, Q → 3, R → 4 and S → 5. The matched products are phosphorous acid, phosphine, phosphoryl chloride and phosphoric acid, respectively. Each match identifies one listed product, not the entire product mixture.

Why does P2O3 react with water to form H3PO3 instead of H3PO4?

Hydrolysis of phosphorus(III) oxide forms phosphorous acid without changing the phosphorus oxidation state of +3. The balanced equation is P2O3 + 3H2O → 2H3PO3. Forming H3PO4 would require oxidation of phosphorus to +5, which does not occur here.

What happens when white phosphorus reacts with NaOH and water?

White phosphorus forms phosphine and sodium hypophosphite: P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2. This is disproportionation because phosphorus changes from 0 to −3 in phosphine and from 0 to +1 in hypophosphite.

Why does PCl5 react with acetic acid to give POCl3?

PCl5 converts acetic acid into acetyl chloride, with phosphoryl chloride and hydrogen chloride as coproducts. The complete equation is PCl5 + CH3COOH → CH3COCl + POCl3 + HCl. Acetic acid contains no methoxy group, so assigning the methoxy-containing product in List-II leads to the wrong option A.

Why does H3PO2 react with four AgNO3 to form H3PO4?

Phosphorus is oxidised from +1 in H3PO2 to +5 in H3PO4, losing four electrons. Four silver ions each accept one electron and form metallic silver. Stopping at H3PO3 would account for only two electrons, not the four required by the specified stoichiometry.