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