A) \[KHg{{I}_{4}}\]
B) \[{{K}_{2}}Hg{{I}_{4}}+N{{H}_{4}}OH\]
C) \[{{K}_{2}}Hg{{I}_{4}}+KOH\]
D) \[KHg{{I}_{4}}+N{{H}_{4}}OH\]
Correct Answer: C
Solution :
Nesslers reagent \[={{K}_{2}}Hg{{I}_{4}}+KOH\] \[2{{K}_{2}}Hg{{I}_{4}}+N{{H}_{4}}Cl+4KOH\to \] \[\underset{\begin{smallmatrix} \text{iodide}\,\,\text{of}\,\text{Milion }\!\!\!\!\text{ s}\,\,\text{base}\, \\ \,\,\,\,\,\,\,\,\,\,\,\,\text{(brown}\,\text{ppt}\text{.)} \end{smallmatrix}}{\mathop{N{{H}_{2}}HgOHgI}}\,+7KI+KCl+3{{H}_{2}}O\] Note Nesslers reagent is used tor detection of ammonia. It gives brown ppt. with ammonium salt.
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