When excess of aqueous KCN solution is added to an aqueous solution of copper sulphate , the complex `[Cu(CN)_(4)]^(2-)` is formed . On passing `H_(2)S` gas through this solution no precipitate of CuS is formed because

Options

A. sulphide ions cannot replace `CN^(-)` ions
B. `[Cu(CN)_(4)]^(2-)` does not give `Cu^(2+)` ion in the solution
C. sulphide ions from `H_(2)S ` do not form complexes
D. sulphide ions cannot replace sulphate ions from copper sulphate solution .

Answer

To solve the question, we need to understand the reactions that occur when excess aqueous KCN is added to an aqueous solution of copper sulfate and the subsequent interaction with H2S gas.

### Step 1: Reaction of Copper Sulfate with KCN
When copper sulfate (CuSO4) is mixed with potassium cyanide (KCN), a complex ion is formed. The reaction can be represented as follows:

\[ \text{CuSO}_4 + 4 \text{KCN} \rightarrow \text{[Cu(CN)}_4]^{2-} + \text{K}_2\text{SO}_4 \]

In this reaction, the copper ion (Cu²⁺) from copper sulfate coordinates with four cyanide ions (CN⁻) to form the complex ion \([Cu(CN)_4]^{2-}\).

**Hint:** Identify the reactants and products in the reaction to understand the formation of the complex.

### Step 2: Stability of the Complex
The complex \([Cu(CN)_4]^{2-}\) is stable due to the strong field strength of the cyanide ion, which is a strong ligand. This stability means that the copper ion does not easily release the cyanide ligands.

**Hint:** Consider the nature of the ligands involved and their ability to stabilize the metal ion.

### Step 3: Interaction with H2S
When H2S gas is passed through the solution containing the complex \([Cu(CN)_4]^{2-}\), we would expect a reaction that could potentially lead to the formation of copper sulfide (CuS). However, no precipitate of CuS is formed. This is because the cyanide ligands are strongly bound to the copper ion, preventing the sulfide ions (S²⁻) from displacing them.

**Hint:** Analyze why the presence of strong ligands affects the formation of precipitates with other ions.

### Step 4: Conclusion
The reason no precipitate of CuS is formed when H2S is passed through the solution is that the copper in the complex is too strongly bound to the cyanide ligands, which prevents the sulfide ions from replacing the cyanide ions.

**Final Answer:** No precipitate of CuS is formed because the copper tetra cyanide complex does not release its cyanide ions to react with sulfide ions.

### Summary of Options:
1. **Sulfide ion cannot replace cyanide ion** - Incorrect, as sulfide can replace cyanide in some cases.
2. **Copper tetra cyanide does not give cuprazines in the solution** - Correct, due to the strong binding of cyanide.
3. **Sulfide ion from H2S does not form complexes** - Incorrect, as sulfide can form complexes in other contexts.
4. **Sulfide ion cannot replace sulfate ions from copper sulfate solution** - Incorrect, sulfide can replace sulfate in aqueous solutions.

**Correct Option:** The second option is correct.