Question

Match List I with List II.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ & \text{(Anion)}& & \text{(gas evolved on)}\\ & & & \text{reaction with dil.}{H}_{2}S{O}_4)\\ \text{A.}& C{O}_3^{2-}& \text{I.}& \text{Colourless gas which}\\ & & & \text{turns lead acetate paper}\\ & & & \text{black.}\\ \text{B.}& S^{2-}& \text{II}.& \text{Colourless gas which}\\ & & & \text{turns acidified potassium}\\ & & & \text{dichromate solution}\\ & & & \text{green}\\ \text{C.}& S{O}_3^{2-}& \text{III.}& \text{Brown fumes which turns}\\ & & & \text{acidified KI solution}\\ & & & \text{containing starch blue}\\ \text{D.}& N{O}_2^{-}& \text{IV}& \text{Colourless gas evolved}\\ & & & \text{with brisk effervescence,}\\ & & & \text{which turns lime water}\\ & & & \text{milky}\end{array}$

Choose the correct answer from the options given below:

Answer 1

$C{O}_3^{2-}$ : On action of dil sulphuric acid, $C{O}_2$ gas is released which turns lime water milky.
$S^{2-}$ : On action of dil sulphuric acid, $H_{2}S$ gas is released which turns lead acetate paper black.
$S{O}_3^{2-}$ : On action of dil $H_{2}S{O}_4,S{O}_2$ gas is evolved which turns acidified potassium dichromate solution green.
$N{O}_2^{-}$: On action of dil $H_{2}S{O}_4$, brown coloured $N{O}_2$ gas is evolved which turns Kl solution containing starch, blue.