Question

${HgCl}_2+excessofKI⟶\left(A\right)\underrightarrow{{NH}_3/NaOH}\left(B\right).$
(A) and (B) respectively are:

• A. $K_2{HgI}_4$ (Nessler's reagent) and $Hg{I}_2$
• B. $Hg{I}_2$ and $K_2{HgI}_4$ (Nessler's reagent)
• C. both (A) and (B) correct
• D. none of these

Answer 1

Answer: (B)

$Hg{I}_2$ and $K_2{HgI}_4$ (Nessler's reagent)

$HgC{l}_2+2KI\to \underset{A}{Hg{I}_2\downarrow }+2KCl$
$Hg{I}_2+2KI\stackrel{N{H}_3/NaOH}{\to }\underset{B}{K_{2}Hg{I}_4}$
Thus, (A) is scarlet ppt of mercuric iodide and (B) is Potassium tetraiodomercurate, also known as Nessler's reagent.