Question

$BaC{{l}_{2}}$ solution gives a white precipitate with a solution of an acid radical which dissolves in dilute HCl with the evolution of a colorless, pungent smelling gas. The acid radical may be:
A. $S{{O}_{4}}^{2-}$
B. ${{S}^{2-}}$
C. $S{{O}_{3}}^{2-}$
D. $C{{O}_{3}}^{2-}$

Answer 1

Hint: Generally inorganic salts composed with acidic and basic radicals. Qualitative analysis is used for the detection and identification of acidic and basic radicals present in inorganic salts. Inorganic salts are formed by the reaction of acids and bases or acidic oxides with a base or basic oxides.

Complete answer: The preliminary tests give important clues about the presence of some anions or cations. The systematic analysis of anions is an integral part of salt analysis. Preliminary test involves the color and smell after that check the action of heat that will happen to the solution after producing heat or by flame test i.e. insert the compound in flame and observe the color of flame.
According to the question $BaC{{l}_{2}}$ solution gives white precipitate which can be explained on the basis of following reaction:
$BaC{{l}_{2}}+{{H}_{2}}S{{O}_{3}}\to BaS{{O}_{3}}+2HCl$
This barium sulphite gives white precipitate and the acid radical from this can be shown as:
$BaS{{O}_{3}}\to B{{a}^{2+}}+S{{O}_{3}}^{2-}$
Acid radical is $S{{O}_{3}}^{2-}$
When $BaS{{O}_{3}}$ dissolved in HCl solution then reaction is shown as:
$BaS{{O}_{3}}+2HCl\to BaC{{l}_{2}}+S{{O}_{2}}+{{H}_{2}}O$
Due to the presence of $S{{O}_{2}}$ gas it produces pungent smell.

Hence we can say that option C is the correct answer.

Note: If $BaC{{l}_{2}}$ forms an insoluble white precipitate then the acid radical produced is $S{{O}_{4}}^{2-}$ if there is absence of precipitates then we can mark the absence of $S{{O}_{4}}^{2-}$ and $S{{O}_{3}}^{2-}$ radicals while HCl soluble white precipitates confirms the presence of $S{{O}_{3}}^{2-}$ radical.