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NEET Sample Paper NEET Sample Test Paper-87

  • question_answer
    An acidic solution of 'X' does not give precipitate on passing \[{{H}_{2}}S\] through it. 'X' gives white precipitate when \[N{{H}_{4}}OH\] is added to it. The white precipitate dissolves in excess of \[NaOH\] solution. Pure 'X' fumes in air and dense white fumes are obtained when a glass rod dipped in \[N{{H}_{4}}OH\] is put in the fumes. Compound 'X' can be

    A) \[ZnC{{l}_{2}}\]           

    B)        \[FeC{{l}_{3}}\]

    C) \[AlC{{l}_{3}}\]                       

    D) \[SnC{{l}_{2}}\]

    Correct Answer: A

    Solution :

    [a] \[X\xrightarrow{N{{H}_{4}}OH}\text{White}\,\text{ppt}.\] \[\xrightarrow[NaOH]{excess}\underset{\text{(No}\text{.}\,\,\text{ppt}\,\,\text{with}\,\,{{\text{H}}_{\text{2}}}\text{S)}}{\mathop{\text{Acidic}\,\text{solution}\,\text{(soluble)}}}\,\] Given reactions (white precipitate with \[{{H}_{2}}S\] in presence of \[N{{H}_{4}}OH\]) indicate that 'X' should be \[ZnC{{l}_{2}}\] which explains all given reactions. \[ZnC{{l}_{2}}+2{{H}_{2}}O\xrightarrow{{}}\]\[Zn{{(OH)}_{2}}+\underset{White\,\,fumes}{\mathop{HCl(g)}}\,\uparrow \]             \[N{{H}_{4}}OH(l)+\underset{White\,\,fumes}{\mathop{HCl(g}}\,)\uparrow \xrightarrow[-{{H}_{2}}O]{}\]\[\underset{Dense\,\,white\,\,fumes}{\mathop{N{{H}_{4}}Cl(g)\uparrow }}\,\]             \[ZnC{{l}_{2}}+2NaOH\xrightarrow{{}}\]\[Zn{{(OH)}_{2}}+2NaCl\]             \[Zn{{(OH)}_{2}}\xrightarrow[Excess]{2NaOH}N{{a}_{2}}Zn{{O}_{2}}+2{{H}_{2}}O\]


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