A) \[KHg{{I}_{4}}\]
B) \[{{K}_{2}}Hg{{I}_{4}}+N{{H}_{4}}OH\]
C) \[[Ne]3{{s}^{2}}3{{p}^{3}}\]
D) \[KHg{{I}_{4}}+N{{H}_{4}}OH\]
Correct Answer: C
Solution :
Nessler?s reagent \[={{K}_{2}}Hg{{I}_{4}}+KOH\] \[2{{K}_{2}}Hg{{I}_{4}}+N{{H}_{4}}Cl+4KOH\underset{\begin{smallmatrix} iodide\,of\,Million's\,base \\ (brown\,ppt) \end{smallmatrix}}{\mathop{\to N{{H}_{2}}HgOHgI}}\,\] \[+7KI+KCl+3{{H}_{2}}O\] Note: Nessler's reagent is required for detection of ammonia. It gives brown ppt with ammonium salt.
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