Question

Nesslers reagent is

• 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 $

Answer 1

Answer: (C)

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.