JEE Challenger
More from The d- and f-Block Elements
ChemMedAdv 2007 • Paper 2

Identification of Mercuric Ion via Qualitative Analysis

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

Options

A

Pb2+\text{Pb}^{2+}

B

Hg2+\text{Hg}^{2+}

Correct
C

Cu2+\text{Cu}^{2+}

D

Co2+\text{Co}^{2+}

Topics & Concepts

The d- and f-Block ElementsQualitative AnalysisMercury SaltsComplex Formation

Step-by-Step Solution

  1. When mercuric ion (Hg2+\text{Hg}^{2+}) is treated with potassium iodide (KI\text{KI}), a scarlet-red precipitate of mercuric iodide (HgI2\text{HgI}_2) is formed:
    Hg2++2I−→HgI2↓  (scarlet red)\text{Hg}^{2+} + 2\text{I}^- \rightarrow \text{HgI}_2\downarrow\; (\text{scarlet red})

  2. With excess KI\text{KI}, the precipitate dissolves completely due to the formation of the soluble, colourless tetraiodomercurate(II) complex:
    HgI2+2I−→[HgI4]2−  (colourless)\text{HgI}_2 + 2\text{I}^- \rightarrow [\text{HgI}_4]^{2-}\; (\text{colourless})

  3. When Hg2+\text{Hg}^{2+} is treated with cobalt(II) thiocyanate, a characteristic deep blue crystalline precipitate of cobalt tetrathiocyanatomercurate(II) is formed:
    Hg2++Co2++4SCN−→Co[Hg(SCN)4]↓  (deep blue)\text{Hg}^{2+} + \text{Co}^{2+} + 4\text{SCN}^- \rightarrow \text{Co}[\text{Hg(SCN)}_4]\downarrow\; (\text{deep blue})

Therefore, the metal ion is Hg2+\text{Hg}^{2+}.

Correct Answer: B

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