In qualitative analysis, $\ce{NH4+}$ ions are confirmed using Nessler's reagent:-
$$\ce{NH4+ + 2[HgI4]^{2−} + 4OH− → HgO·Hg(NH2)I ↓ + 7I^{−} + 3H2O}$$
The brown precipitate formed is called iodide of Millon's base. It is also written as $\ce{3HgO.Hg(NH3)2I2}$ and as $\ce{NH2.Hg2I3}$. (wikipedia).
But its structure is present differently in different sites:-
- Source 1 - ionic form
- Source 2 - Here the reaction is given differently:-
$$\ce{([K2]HgI4 <=> 2KI + HgI2) * 2}$$ $$\ce{HgI2 + NH3 -> Hg[NH2]I + NH4I}$$ $$\ce{Hg[NH2]I + HgI2 -> NH2Hg2I3 }$$ $$\ce{[K2]HgI4 + 2NH3 -> NH2Hg2I3 + 4KI + KH4I }$$
structure -
- Source 3- non ionic form
- Source 4 - Hydrate form; Here also the reaction is different:-
$$\ce{[K2]HgI4 -> 2KI + HgI2}$$
$$\ce{HgI2 + NH3 -> Hg[NH2]I + HI}$$
This site also says that its structure is still uncertain. Various other structure has been proposed:-
So, what is the correct reaction, correct formula and correct structure of iodide of Millon's base?