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Match List I with  List II.

List I

(Anion)

List II

\text { (gas evolved on reaction with dil. } \mathrm{H}_{2} \mathrm{SO}_{4} \text { ) }
\text { A. } \mathrm{CO}_{3}^{2-} I. Colourless gas which turns lead acetate acetate paper black.
\text { B. } \mathrm{S}^{2-} II. Colourless gas which turns acidified potassium dichromate solution green.
\text { C. } \mathrm{SO}_{3}^{2-} III.Brown fumes which turns acidified KI solution containing starch blue.
\text { D. } \mathrm{NO}_{2}^{-} IV. Colourless gas evolved with brisk effervescence, which turns lime water milky.

Choose the correct answer from the options given below:

Option: 1

\text { A-III, B-I, C-II, D-IV }


Option: 2

\text { A-II, B-I, C-IV, D-III }


Option: 3

\text { A-IV, B-I, C-III, D-II }


Option: 4

\text { A-IV, B-I, C-II, D-III }


Answers (1)

best_answer

The correct matching of the Anion and the gas evolved on reaction with dil. \mathrm{H}_{2} \mathrm{SO}_{4} is

\mathrm{(A)} \mathrm{CO}_{3}{ }^{2-}\mathrm{- } (IV) Colourless gas evolved with brisk effervescence, which turns lime water milky.

\mathrm{(B)} \mathrm{S}^{2-}- (I) Colourless gas which turns lead acetate acetate paper black

\mathrm{(C)}\mathrm{SO}_{3}{ }^{2-} {-} (II) Colourless gas which turns acidified potassium dichromate solution green.

\mathrm{(D)} \mathrm{NO}_{2}^{-}- (IV) Colourless gas evolved with brisk effervescence, which turns lime water milky.

Hence, Option (4) is correct.

Posted by

Ritika Kankaria

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