A) \[K{{H}_{g}}{{I}_{4}}\]
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 :
Nessler's reagent is \[HN{{O}_{3}}\] in excess of\[KOH\].
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