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JEE Main 202430 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Practical Chemistry

JEE Main 2024 · 30 January · Morning Shift · Chemistry · Practical Chemistry

Given below are two statements: Statement-I: The gas liberated on...

Given below are two statements:
Statement-I: The gas liberated on warming a salt with dil H2SO4{H}_{2}{SO}_{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 sulphite.
In the light of the above statements, choose the most appropriate answer from the options given below:

Options

  1. ABoth Statement-I and Statement-II are false
  2. BStatement-I is false but Statement-II is true
  3. CStatement-I is true but Statement-II is false
  4. DBoth Statement-I and Statement-II are true.

Correct answer

C. Statement-I is true but Statement-II is false

Step-by-step solution

With warm dilute H2SO4{H}_{2}{SO}_{4} a sulphide gives hydrogen sulphide gas which smells like rotten eggs. A piece of filter paper dipped in lead acetate solution turns black on exposure to the gas due to the formation of lead sulphide which is black in colour.

As we know, hydrogen sulphide gas is a colourless gas and when it reacts with lead acetate it produces lead sulphide and acetic acid. The chemical reaction of hydrogen sulphide with lead acetate is given as:

Na2S+H2SO4→Na2SO4+H2S{Na}_{2}S + {H}_{2}{SO}_{4} \to {Na}_{2}{SO}_{4} + {H}_{2}S

(CH3COO)2Pb+H2S→PbSBlackleadsulphide+2CH3COOH\mathop{{\left({CH}_{3}COO\right)}_{2}Pb + {H}_{2}S \to \mathop{PbS}_{Blackleadsulphide} + 2{CH}_{3}COOH}_{}

Hence, the answer is option C.

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