Question

Match List I with List II.

	List I (Anion)		List II (gas evolved on reaction with dil \(H_2SO_4\))
A.	\(CO_3^{ 2−}\)	I.	Colourless gas which turns lead acetate paper black.
B.	\(S^{2–}\)	II.	Colourless gas which turns acidified potassium dichromate solution green
C.	\(SO_3^{ 2−}\)	III.	Brown fumes which turns acidified KI solution containing starch blue.
D.	\(NO_2^{−}\)	IV.	Colourless gas evolved with brisk effervescence, which turns lime water milky.

Choose the correct answer from the options given below:

• A. A-III, B-I, C-II, D-IV
• B. A-II, B-I, C-IV, D-III
• C. A-IV, B-I, C-III, D-II
• D. A-IV, B-I, C-II, D-III

Answer 1

The Correct Option is D

To solve this question, we need to match the anions in List I with the gases evolved upon reaction with dilute \(H_2SO_4\) in List II.

1. When \(CO_3^{2−}\) (carbonate) reacts with dilute \(H_2SO_4\), it produces carbon dioxide gas (CO_2). The carbon dioxide gas is a colorless gas evolved with brisk effervescence, which turns lime water milky. Thus, A matches with IV.
2. The \(S^{2−}\) (sulfide) ion reacts with dilute \(H_2SO_4\) to produce hydrogen sulfide gas (H_2S). Hydrogen sulfide is a colorless gas which turns lead acetate paper black. Thus, B matches with I.
3. When \(SO_3^{2−}\) (sulfite) reacts with dilute \(H_2SO_4\), it produces sulfur dioxide gas (SO_2). Sulfur dioxide is a colorless gas which turns acidified potassium dichromate solution green. Thus, C matches with II.
4. The \(NO_2^{-}\) (nitrite) ion reacts with dilute \(H_2SO_4\) to produce nitrogen dioxide gas (NO_2). Nitrogen dioxide forms brown fumes which turns acidified potassium iodide (KI) solution containing starch blue. Thus, D matches with III.

Based on the above explanations, the correct matching is: A-IV, B-I, C-II, D-III.

Therefore, the correct answer is: A-IV, B-I, C-II, D-III.