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Q.
When $H _{2} S$ gas is passed through the $HCl$ containing aqueous solution of $CuCl _{2}, HgCl _{2}$, $BiCl _{3}$ and $CoCl _{2}$, it does not precipitate out
When $H _{2} S$ gas is passed through the $HCl$ containing aqueous solution of $CuCl _{2}, HgCl _{2,} BiCl _{3}$ and $CoCl _{2}$ then
$CuCl _{2}+ H _{2} S \longrightarrow \underset{\text{black ppt}}{CuS} \downarrow+2 HCl$
$HgCl _{2}+ H _{2} S \longrightarrow HgS \downarrow+ HCl$
$2 BiCl _{3}+3 H _{2} S \longrightarrow Bi _{2} S _{3}+6 HCl$
These are the alkaline radical of II A group. These radical precipitate in the presence of dil. $HCl$ and $H _{2} S$. Co is the alkaline radical of IV group which is precipitated in the presence of $H _{2} S$ gas and $NH _{4} OH$. Its reason that the solubility product of sulphide of II nd group is lesser than the solubility product of sulphide of IV group. So, $CoS$ will not precipitate.