A) \[CuS\]
B) \[HgS\]
C) \[B{{i}_{2}}{{S}_{3}}\]
D) \[CoS\]
Correct Answer: D
Solution :
When \[{{H}_{2}}S\] gas is passed through the \[HCl\] containing aqueous solution of \[CuC{{l}_{2}},\,HgC{{l}_{2}}.\,BiC{{l}_{3}}\] and \[CoC{{l}_{2}}\] then \[CuC{{l}_{2}}+{{H}_{2}}S\xrightarrow{{}}\underset{black\,ppt}{\mathop{CuS\downarrow +}}\,2HCl\] \[HgC{{l}_{2}}+{{H}_{2}}S\xrightarrow{{}}HgS\downarrow +HCl\] \[2BiC{{l}_{3}}+3{{H}_{2}}S\xrightarrow{{}}B{{i}_{2}}{{S}_{3}}+6HCl\] These are the alkaline radical of II A group. These radical precipitate in the presence of dil. \[HCl\] and \[H2S\]. \[Co\] is the alkaline radical of IV group which is precipitated in the presence of \[H2S\]gas and is \[N{{H}_{4}}OH\]. Its reason that the solubility product of sulphide of Und group is lesser than the solubility product of sulphide of IV group. So, \[CoS\] will not precipitate.
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