Question

A solution of metallic ion when treated with $KI$ gives a red precipitate which dissolves in excess of $KI$ to give a colourless solution. Moreover, the solution of metal ion on treatment with solution of cobalt(II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is:

• A. $P{b}^{2+}$
• B. $H{g}^{2+}$
• C. $C{u}^{2+}$
• D. $C{o}^{2+}$

Answer 1

The correct option is B $H{g}^{2+}$

Potassium iodide when added slowly to the solution of $H{g}^{2+}$, a scarlet red precipitate of mercury(II) iodide is formed.
$H{g}^{2+}+2I^{-}\to Hg{I}_2\downarrow$
Scarlet red ppt
The precipitate dissolves in excess reagent, when colorless tetraiodomercurate(II) ions are formed.
$H{g}^{2+}+2I^{-}\to [Hg{I}_4{]}^{2-}orHg{I}_2+2KI\to K_2[Hg{I}_4]$
When $\left[Co\right(SCN{)}_4{]}^{2-}$ is added to the solution of $H{g}^{2+}$ and walls of the vessel are stirred with a glass rod, a deep-blue crystalline precipitate of cobalt tetrathiocyanatomercurate(II) is formed.
$H{g}^{2+}+C{o}^{2+}+4SC{N}^{-}\to Co\left[Hg\right(SCN{)}_4]\downarrow$
Deep-blue ppt