Problem 117
Question
A solution of a metal ion when treated with KI gives a red precipitate which dissolves in excess \(\mathrm{KI}\) 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 (a) \(\mathrm{Pb}^{2+}\) (b) \(\mathrm{Hg}^{2+}\) (c) \(\mathrm{Cu}^{2+}\) (d) \(\mathrm{Co}^{2+}\)
Step-by-Step Solution
Verified Answer
The metal ion is (b) \(\mathrm{Hg}^{2+}\).
1Step 1: Analyze the Reaction with KI
When the metal ion is treated with KI, a red precipitate forms. This is indicative of certain metal ions reacting with iodide ions to form insoluble metal iodides. Among the given options, \(\mathrm{Hg}^{2+}\) forms mercuric iodide \(\mathrm{HgI}_2\), which is a red precipitate that dissolves in excess KI to form \(\mathrm{K}_2\mathrm{HgI}_4\), a soluble complex.
2Step 2: Analyze the Reaction with Cobalt(II) Thiocyanate
The reaction with cobalt(II) thiocyanate yielding a deep blue crystalline precipitate relates to the formation of special complex compounds. The metal ion present is likely to form a complex with thiocyanate and cobalt ions. For \(\mathrm{Hg}^{2+}\), such reactions can occur, giving rise to distinct colored complexes indicative of cobalt compounds with metals.
3Step 3: Consolidating Information from Both Reactions
The key observations are the red precipitate with KI which dissolves in excess KI, indicative of \(\mathrm{HgI}_2\) formation, and the reaction forming a deep blue precipitate with cobalt(II) thiocyanate, both characteristic behaviors of \(\mathrm{Hg}^{2+}\). The option \(\mathrm{Hg}^{2+}\) satisfies both reactions' results comprehensively, unlike the other metal ions which either do not form similar compounds or lack reactivity under these conditions.
Key Concepts
Reaction with Potassium IodideFormation of PrecipitatesComplex Ion FormationMetal Ion IdentificationCobalt(II) Thiocyanate Test
Reaction with Potassium Iodide
When metal ions react with potassium iodide (KI), interesting color changes and the formation of precipitates can occur. This can help in identifying the metal ion present in a solution. For instance, when \( \text{Hg}^{2+} \) ions are treated with \( \text{KI} \), a red precipitate of mercuric iodide \( \text{HgI}_2 \) forms. This reaction is distinct because the red precipitate represents a specific interaction between mercury ions and iodide ions. Interestingly, this red precipitate can dissolve in excess \( \text{KI} \), forming a colorless solution of the complex ion \( \text{K}_2\text{HgI}_4 \). This reversible transformation is evident in the case of mercury ions but not in other metal ions like \( \text{Pb}^{2+} \) or \( \text{Cu}^{2+} \), making it a useful identification step.
Formation of Precipitates
Precipitate formation is a common occurrence in qualitative analysis. It involves the formation of a solid from a solution, often due to a chemical reaction. In the case of the \( \text{Hg}^{2+} \) ion reacting with \( \text{KI} \), the sudden appearance of a red solid (the precipitate) is \( \text{HgI}_2 \). Precipitates occur because the ions in the solution combine to form insoluble compounds. Understanding precipitate formation helps in recognizing specific ions based on their unique color and behavior. For metal ion analysis, the color and solubility of precipitates provide valuable clues. The red color of \( \text{HgI}_2 \) is both distinctive and diagnostic for mercury ions, as it dissolves into a colorless complex only in the presence of excess iodide.
Complex Ion Formation
Complex ion formation is an essential concept in identifying metal ions in solution. A complex ion consists of a metal ion at its center with molecules or anions (called ligands) surrounding it. In the presence of excess \( \text{KI} \), the red precipitate \( \text{HgI}_2 \) can dissolve to form a colorless complex ion \( \text{K}_2\text{HgI}_4 \), where the metal ion is surrounded by iodide ions. This transformation from a precipitate into a soluble complex illustrates how ligands stabilize metal ions in solution. It's significant for identifying specific reactions and differentiating metal ions in qualitative analysis. Mercury's ability to form such complexes is not shared by all metal ions, adding to the diagnostic utility of such reactions.
Metal Ion Identification
Identifying metal ions based on their reactions is a key goal of qualitative analysis. Different ions react in distinctive ways when treated with specific reagents, providing chemical fingerprints that help in their identification. For instance:
- The reaction between \( \text{Hg}^{2+} \) and \( \text{KI} \) leads to a red precipitate and later forms a colorless complex in excess \( \text{KI} \).
- Reactions with different reagents, such as cobalt(II) thiocyanate, further confirm the identity of the metal ion by forming specific colored precipitates, such as the blue crystalline precipitate that occurs uniquely in the presence of cobalt and mercury ions.
Cobalt(II) Thiocyanate Test
The cobalt(II) thiocyanate test is another crucial method for metal ion identification. When cobalt(II) thiocyanate interacts with certain metal ions, unique colored precipitates form. In the presence of \( \text{Hg}^{2+} \), this reaction yields a deep blue crystalline precipitate, which indicates the formation of a characteristic complex. This deep blue color is a strong indicator of the specific interactions between cobalt, thiocyanate, and mercury ions. This test is important as it complements the findings from the \( \text{KI} \) reaction, solidifying the identification process of mercury ions. The cobalt(II) thiocyanate test is thus a valuable tool in qualitative analysis, offering clear visual evidence of the presence of specific metal ions.
Other exercises in this chapter
Problem 115
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What are the products obtained respectively when KI solution in excess is added to \(\mathrm{HgCl}_{2}\) and \(\mathrm{CuSO}_{4}\) solution separately? (a) \(\m
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\(\mathrm{CuSO}_{4}\) decolourize on addition of KCN, the product is (a) \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{2-}\) (b) \(\mathrm{Cu}^{2+}\) gets reduce
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