Question

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

• A. $CuS$
• B. $HgS$
• C. $Bi_{2}S_{3}$
• D. $CoS$

Answer 1

Answer: (D)

$CoS$

Answer 2

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.