Match List I with List II. List-I (Anion) List-II (gas evolved on reaction with dil. H 2 SO 4 ) (A) CO 3 2 -…

Match List I with List II.

  List-I
(Anion)		   List-II
(gas evolved on reaction with dil. H2SO4)
(A) CO32- (I)

Colourless gas which turns lead acetate paper black.
(B) S2- (II)

Colourless gas which turns acidified potassium dichromate solution green.
(C) SO32- (III) Brown fumes which turns acidified KI solution containing starch blue.
(D) NO2- (IV) Colourless gas evolved with brisk effervescence, which turns lime water milky.

Choose the correct answer from the options given below

  1. A-III,B-I,C-II,D-IV
  2. A-II,B-I,C-IV,D-III
  3. A-IV,B-I,C-III,D-II
  4. A-IV,B-I,C-II,D-III

Solution

When small amount of carbonate salt is treated with dilute $H_2SO_4$ or dilute $HCl$, A colourless, odourless gas $CO_2$ is evolved with brisk effervescence. When small amount of the following salts are treated with dilute $H_2SO_4$ the following observations are seen. | Carbonate salt $CO_3^{2-}$ | Brisk effervescence of $CO_2$ gas, $CO_3^{2-} + H_2SO_4 \rightarrow SO_4^{2-} + H_2O + CO_2 \uparrow$ | | Sulphide salt $S^{2-}$ | A colourless gas $H_2S$ with rotten egg smell | $Na_2S + H_2SO_4 \rightarrow Na_2SO_4 + H_2S \uparrow$ $NO_2^-$ on reaction with dilute $H_2SO_4$ will give brown $NO_2(g)$ will turn KI solution blue. $SO_3^{2-}$ on reaction with dilute $H_2SO_4$ will give $SO_2(g)$, which will turn acidified potassium dichromate solution green.

Asked in: JEE Main 2022 (27 Jun Shift 2)

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