Question

A solution of a metal 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 a solution of cobalt(III) thiocyanate gives to a deep blue crystalline precipitate. The metal ion is :

• A. ${Pb}^{2+}$
• B. ${Hg}^{2+}$
• C. ${Cu}^{2+}$
• D. ${Co}^{2+}$

Answer 1

The correct option is B ${Hg}^{2+}$

The cation is mercury (II) ion. It reacts with KI solution to form red precipitate of mercury (II) iodide.

$HgC{l}_2+2KI\to Hg{I}_2\downarrow \left(redppt\right)+2KCl$

Th precipitate dissolves in excess KI to form Nesseler's solution.

$Hg{I}_2+KI\left(excess\right)\to K_{2}Hg{I}_4\left(Nesslersolution\right)$

Mercury(II) ion reacts with cobalt(III) thiocyanate to form deep blue crystalline precipitate.

$HgC{l}_2+2Co(CNS{)}_3\to HgCo(CNS{)}_4\downarrow \left(deepbluecrystallineprecipitate\right)+CoC{l}_{2}CNS$

Option B is correct.