Q20 / 90 · JEE Main 2022 (27 Jun Shift 2) · Chemistry

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

  • 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
  • Correct answer
    D
    A-IV,B-I,C-II,D-III

Solution

When small amount of carbonate salt is treated with dilute H2SO4 or dilute HCl, A colourless, odourless gas CO2 is evolved with brisk effervescence

When small amount of the following salts are treated with dilute H2SO4  the following observations are seen.

Carbonate salt CO32− Brisk effervescence of CO2 gas, CO32−+H2SO4→SO42−+H2O+CO2↑
Sulphide salt S-2 A colourless gas H2S with rotten egg smell

Na2S+H2SO4→Na2SO4+H2S↑

NO2 –  on reaction with dilute H2SO4 will give brown NO2(g) will turn KI solution blue.

SO32-on reaction with dilute H2SO4 will give SO2g, which will turn acidified potassium dichromate solution green.

 

Hence, the correct answer is option: 'D'.

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