When ${H_2}S$ gas is passed through $HCl$ containing aqueous solution of $CuC{l_2}$, $HgC{l_2}$ , $BiC{l_3}$ ​, and $CoC{l_2}$​, then which of the following precipitate out ?
(A) $CuS$
(B) $HgS$
(C) $B{i_2}{S_3}$
(D) $CoS$

Hint: This problem is based on the Group 2 basic radicals’ analysis therefore, to check whether the compound will precipitate out or not we first figure out whether the ionic product of a compound is exceeding its solubility product or not hence, we will first analyse every salt present in the aqueous solution.

Complete Step by Step Solution:
To detect the presence of Group 2 basic radicals $\left( {Pb,Bi,Cu,Cd,As{\text{ }}\& {\text{ }}Hg{\text{ }}etc.....} \right)$ , ${H_2}S$ gas in presence of $HCl$ is used as a reagent. We know that whenever an ionic product exceeds a solubility product, precipitation takes place.

Let us consider an aqueous solution of $CuC{l_2}$, $HgC{l_2}$ , $BiC{l_3}$ ​, and $CoC{l_2}$ which is given and try to write a balanced chemical reaction of precipitate formation of all the salts present in aqueous solution.
$CuC{l_2}\xrightarrow[{HCl}]{{{H_2}S}} CuS \downarrow \\$
$HgC{l_2}\xrightarrow[{HCl}]{{{H_2}S}}HgS \downarrow \\$
$BiC{l_3}\xrightarrow[{HCl}]{{{H_2}S}}B{i_2}{S_3} \downarrow \\$

But in case of $CoC{l_2}$ ,${H_2}S$ gas in presence of $N{H_4}OH$ is used as a reagent (as it belongs to Group 4) which results in the formation of $CoS$ .
$CoC{l_2}\xrightarrow[{N{H_4}OH}]{{{H_2}S}}CoS \downarrow $
Thus, only $CuS$,$HgS$ and $B{i_2}{S_3}$ will precipitate out when${H_2}S$ gas in presence of $HCl$ is used as a reagent.
Hence, the correct option is (A), (B), and (C).

Note: Since this is a conceptual-based problem hence, it is essential that the options given in the question are analysed very carefully to give an accurate solution. Also, remember the fundamental properties of basic radicals that belong to a different group and use the correct reasons to give a proper explanation.