1. We identify the gas evolved when each anion reacts with dilute acid.

For dilute acid, the relevant reactions are:

• $\mathrm{CO_3^{2-}}$ gives $\mathrm{CO_2}$
• $\mathrm{S^{2-}}$ gives $\mathrm{H_2S}$
• $\mathrm{SO_3^{2-}}$ gives $\mathrm{SO_2}$
• $\mathrm{NO_2^-}$ gives products of nitrous acid decomposition, mainly $\mathrm{NO}$ which is oxidized in air to $\mathrm{NO_2}$ (brown fumes)

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2. Match each with the description in List II.

(A) $\mathrm{CO_3^{2-}}$

On reaction with dilute acid: $\mathrm{CO_3^{2-} + 2H^+ \to CO_2 + H_2O}$ $\mathrm{CO_2}$ is a colourless gas, evolved with brisk effervescence, and it turns lime water milky.

So, $A \to IV$

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(B) $\mathrm{S^{2-}}$

On reaction with dilute acid: $\mathrm{S^{2-} + 2H^+ \to H_2S}$ $\mathrm{H_2S}$ is a colourless gas and turns lead acetate paper black due to formation of black $\mathrm{PbS}$.

So, $B \to I$

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(C) $\mathrm{SO_3^{2-}}$

On reaction with dilute acid: $\mathrm{SO_3^{2-} + 2H^+ \to SO_2 + H_2O}$ $\mathrm{SO_2}$ is a colourless gas and reduces acidified potassium dichromate from orange to green.

So, $C \to II$

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(D) $\mathrm{NO_2^-}$

On reaction with dilute acid: $\mathrm{NO_2^- + H^+ \to HNO_2}$ Nitrous acid is unstable: $\mathrm{2HNO_2 \to NO + NO_2 + H_2O}$ The characteristic observed gas is brown fumes of $\mathrm{NO_2}$.

$\mathrm{NO_2}$ oxidizes iodide to iodine, and iodine turns starch blue: $\mathrm{2I^- + 2NO_2 + 4H^+ \to I_2 + 2NO + 2H_2O}$

So, $D \to III$

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3. Final matching: $A-IV,\; B-I,\; C-II,\; D-III$

This corresponds to Option D.