Given below are two statements : Statement (I) : The gas liberated on warming a salt with dil 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 :

Statement I is true but Statement II is false

Both Statement I and Statement II are false

Both Statement I and Statement II are true

Statement I is false but Statement II is true

P Block Elements PYQ — JEE Main 2024 · 30 Jan · Shift 1 · Q3 with Solution

View written solutionFree

Correct answer: B

When a salt containing sulphide ion $\left(\mathrm{S^{2-}}\right)$ is warmed with dilute $\mathrm{H_2SO_4}$ , the gas evolved is hydrogen sulphide:

\mathrm{S^{2-} + 2H^+ \rightarrow H_2S \uparrow}

Example:

\mathrm{FeS + H_2SO_4 \rightarrow FeSO_4 + H_2S \uparrow}

Now, $\mathrm{H_2S}$ turns paper dipped in lead acetate black due to formation of black lead sulphide:

\mathrm{Pb(CH_3COO)_2 + H_2S \rightarrow PbS \downarrow + 2CH_3COOH}

Since blackening of lead acetate paper is a standard confirmatory test for sulphide ion, Statement (I) is true.

Statement (II) says that the black colour is due to formation of lead sulphite.

This is incorrect. The black colour is due to lead sulphide, $\mathrm{PbS}$ , not lead sulphite.

Therefore, Statement (II) is false.

Hence, the correct option is:

\boxed{\text{B}}