Question

Iodide of Millon’s base is:
A. \[Hg{I_2}\]
B. ${K_2}Hg{I_4}$
C. $N{H_2}HgO.HgI$
D. $N{H_4}I$

Answer 1

Hint : Here in this question we need to know some information of Nessler’s reagent before knowing about Iodide of Millon’s base. Nesssler’s reagent is a solution of mercury (II) and iodide $(Hg{I_2})$ in potassium iodide and potassium hydroxide.It is used in qualitative analysis in laboratory experiments for the determination of ammonia.

Complete answer:
> When we add this reagent to the solution of ammonia mercuric ions react with ammonia to produce reddish brown complex compounds under alkaline conditions (it is made alkaline by adding potassium hydroxide or sodium hydroxide). When Nessler’s reagent reacts with gaseous ammonia it gives brown fumes while on being passed through the solution of ammonia gives brown precipitate.The brown precipitate is called iodide of millon’s base.
> Hence the formula of iodide of Millon’s base is $N{H_2}HgO.HgI$, thus option C is the correct answer.
> Nessler’s reagent gives brown colour with small traces of ammonia and brown precipitation with larger amounts so it is a sensitive reagent for detection and determination of traces of ammonia. Hence to get brown precipitation of iodide of a millon's base we need to use a larger amount of ammonia in analysis. Iodide of a millon's base is insoluble in water. It is formed on addition of base into mercuric amido chloride $(HgN{H_2}Cl)$.

Note : We have performed qualitative analysis of ammonium radical in laboratory So there we have used Nessler’s reagent for the detection of ammonia. Ammonium ions $N{H^ + }$ ions are confirmed using Nessler’s reagent in qualitative analysis. Thus the following reaction takes place and brown precipitate of iodide of Millon’s base is formed.
$N{H^ + }_4 + 2{[Hg{I_4}]^{2 - }} + 4O{H^ - } \to HgO.Hg(N{H_2})I \downarrow + 7{I^ - } + 3{H_2}O$