A) \[\text{Hg}{{\text{I}}_{\text{2}}}\]
B) \[{{\text{K}}_{\text{2}}}\text{Hg}{{\text{I}}_{\text{4}}}\]
C) \[\text{N}{{\text{H}}_{\text{4}}}\text{HgO}\text{.HgI}\]
D) \[\text{N}{{\text{H}}_{\text{4}}}\text{I}\]
Correct Answer: C
Solution :
When \[\text{NH}_{\text{4}}^{\text{+}}\]ions are treated with Nesslers reagent (alkaline solution of\[{{\text{K}}_{\text{2}}}\text{Hg}{{\text{I}}_{\text{4}}}\]), a brown ppt. of \[N{{H}_{2}}-Hg-O-HgI\]or \[H{{g}_{2}}NI.{{H}_{2}}O\](iodide of Millons base) is formed. \[2{{K}_{2}}Hg{{I}_{4}}+N{{H}_{3}}+3KOH\xrightarrow{{}}\] \[\underset{\begin{smallmatrix} \text{iodide}\,\text{of} \\ \text{Millon }\!\!\!\!\text{ s}\,\text{base} \end{smallmatrix}}{\mathop{N{{H}_{2}}.HgO.HgI}}\,+7KI+2{{H}_{2}}O\]
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