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JEE CHEMISTRY MCQ: ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES AND TECHNIQUES

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Analyzing +4 | -1
A metal M\mathrm{M} and its compound can give the following observable changes in a consequences of reactions

M→HNO3 dilute [ Colourless  solutions ]→NaOH aqueous [ White  precipitate ]→NaOH(aq.)excess[Colourlesssolutions]→H2S[ White  precipitate ]{\text{M}}\xrightarrow[{{\text{HN}}{{\text{O}}_{\text{3}}}}]{{{\text{ dilute }}}}\left[ {\begin{array}{cccccccccccccccccccc} {{\text{ Colourless }}} \\ {{\text{ solutions }}} \end{array}} \right]\xrightarrow[{{\text{NaOH}}}]{{{\text{ aqueous }}}}\left[ {\begin{array}{cccccccccccccccccccc} {{\text{ White }}} \\ {{\text{ precipitate }}} \end{array}} \right]\xrightarrow[{{\text{NaOH}}\left( {{\text{aq}}{\text{.}}} \right)}]{{{\text{excess}}}}\left[ {\begin{array}{cccccccccccccccccccc} {{\text{Colourless}}} \\ {{\text{solutions}}} \end{array}} \right]\xrightarrow{{{{\text{H}}_2}{\text{S}}}}\left[ {\begin{array}{cccccccccccccccccccc} {{\text{ White }}} \\ {{\text{ precipitate }}} \end{array}} \right]
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Correct option: C
Zn\mathrm{Zn}
Option C (Zn) is correct because Zinc (Zn) exhibits the described reaction sequence with dilute HNO3, NaOH, and H2S.

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