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JEE Main & Advanced Sample Paper JEE Main Sample Paper-36

  • question_answer
    A metal gives two chlorides A and B. A gives black precipitate with and B gives white. With KI, B gives a red precipitate soluble in excess of \[KI\]. A and B are respectively

    A)  \[HgC{{l}_{2}}\]and \[H{{g}_{2}}C{{l}_{2}}\]

    B)  \[H{{g}_{2}}C{{l}_{2}}\] and \[HgC{{l}_{2}}\]

    C)  \[HgC{{l}_{2}}\]and \[ZnC{{l}_{2}}\]      

    D)  \[ZnC{{l}_{2}}\]and \[HgC{{l}_{2}}\]

    Correct Answer: B

    Solution :

     \[\underset{(A)}{\mathop{H{{g}_{2}}C{{l}_{2}}}}\,+2N{{H}_{4}}OH\xrightarrow{{}}\]\[[{{H}_{2}}N-\underset{(Black)}{\mathop{Hg-Cl}}\,+Hg]\downarrow +N{{H}_{4}}Cl+2{{H}_{2}}O\] \[\underset{(B)}{\mathop{HgC{{l}_{2}}}}\,+2N{{H}_{4}}OH\xrightarrow{{}}\underset{(White)}{\mathop{Hg(N{{H}_{2}})Cl}}\,+N{{H}_{4}}Cl+{{H}_{2}}O\] \[HgC{{l}_{2}}+2KI\xrightarrow{{}}\underset{(\operatorname{Re}d)}{\mathop{Hg{{I}_{2}}}}\,\downarrow +2KCl\] \[Hg{{I}_{2}}+2KI(excess)\xrightarrow{{}}{{K}_{2}}(Hg{{I}_{4}})\]


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