Given below are two statements:
Statement-I: The gas liberated on warming a salt with dil $$H_2SO_4$$, 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 sulphite.
In the light of the above statements, choose the most appropriate answer from the options given below:

We need to evaluate both statements about the confirmatory test for sulphide ions.

Statement-I Analysis:

When a sulphide salt is warmed with dilute $$H_2SO_4$$, hydrogen sulphide gas ($$H_2S$$) is liberated:

$$Na_2S + H_2SO_4 \rightarrow Na_2SO_4 + H_2S \uparrow$$

When this $$H_2S$$ gas comes in contact with paper dipped in lead acetate solution, it forms lead sulphide ($$PbS$$), which is black in colour:

$$(CH_3COO)_2Pb + H_2S \rightarrow PbS \downarrow (\text{black}) + 2CH_3COOH$$

This is indeed a standard confirmatory test for sulphide ions. So Statement-I is true.

Statement-II Analysis:

Statement-II claims the black colour is due to the formation of lead sulphite ($$PbSO_3$$). This is incorrect. Lead sulphite is white, not black. The black precipitate formed is lead sulphide ($$PbS$$), not lead sulphite.

So Statement-II is false.

Therefore, Statement-I is true but Statement-II is false.

The correct answer is Option 3.