MCQEasyJEE 2024Salt Analysis (Cations & Anions)

JEE Chemistry 2024 Question with Solution

Given below are two statements: Statement-I: The gas liberated on warming a salt with dilute H2SO4H_2SO_4, turns a piece of paper dipped in lead acetate into black; it is a confirmatory test for sulphide ion. Statement-II: In statement-I the colour of paper turns black because of formation of lead sulphide.

Choose the most appropriate answer from the options below:

  • A

    Both Statement-I and Statement-II are false

  • B

    Statement-I is false but Statement-II is true

  • C

    Statement-I is true but Statement-II is false

  • D

    Both Statement-I and Statement-II are true

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Two statements about the confirmatory test for sulphide ion are to be checked.

Find: Which option correctly describes the truth values of Statement-I and Statement-II.

This is a qualitative analysis question based on the reaction of sulphide salts with dilute acid and the effect of the liberated gas on lead acetate paper.

For a sulphide ion, warming with dilute acid liberates hydrogen sulphide gas:

S2+2H+H2SS^{2-} + 2H^+ \longrightarrow H_2S

The liberated H2SH_2S reacts with lead acetate to form black lead sulphide:

H2S+Pb(CH3COO)2PbS+2CH3COOHH_2S + Pb(CH_3COO)_2 \longrightarrow PbS \downarrow + 2CH_3COOH

So, the blackening of lead acetate paper is indeed a confirmatory test for sulphide ion. Therefore, Statement-I is true.

Now check Statement-II using the product formed in the reaction above. The black compound is lead sulphide, PbSPbS. Hence the explanation that the paper turns black because of formation of lead sulphide is correct. Therefore, Statement-II is true.

So both Statement-I and Statement-II are true.

The solution marks the correct option as C, but the chemical working shown there establishes that both statements are true. Hence there is a discrepancy between the marked option and the stated chemistry. The option most defensible from the working and question text is D.

Detailed Check of Both Statements

Given:

  • A salt is warmed with dilute H2SO4H_2SO_4.
  • The liberated gas turns lead acetate paper black.

Find: Truth values of both statements.

  1. If the salt contains sulphide ion, dilute acid liberates hydrogen sulphide gas:
S2+2H+H2SS^{2-} + 2H^+ \longrightarrow H_2S
  1. Hydrogen sulphide reacts with lead acetate paper to give black lead sulphide:
H2S+Pb(CH3COO)2PbS+2CH3COOHH_2S + Pb(CH_3COO)_2 \longrightarrow PbS + 2CH_3COOH
  1. Since formation of black PbSPbS is the basis of the test, Statement-I is correct.
  2. Statement-II says the paper turns black because of formation of lead sulphide. That is exactly the species formed, so Statement-II is also correct.

Therefore, the chemistry supports both statements being true, corresponding to option D.

Because the source solution explicitly declares option C while also explaining formation of lead sulphide, the answer key and the reasoning are inconsistent.

Common mistakes

  • Confusing lead sulphide with lead sulphite. The black compound formed with H2SH_2S is PbSPbS, not PbSO3PbSO_3. Always identify the ion actually present before naming the product.

  • Assuming the marked option must be correct even when the written reaction contradicts it. In assertion-reason style chemistry questions, verify the product formula from the reaction before choosing the option.

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