`CuSO_(4)(aq)overset(H_(2)Suarr)rarrMdarroverset("Excess of KCN")rarrN+O`
Then final product N and O are respectively.

Options

A. `[Cu(CN)_(4)]^(3-), (CN)_(2)`
B. `CuCN,(CN)_(2)`
C. `[Cu(CN)_(4)]^(2-), (CN)_(2)`
D. `Cu(CN)_(2), K_(2)S`

Answer

To solve the given problem, we will follow the reactions step by step.

### Step 1: Reaction of Copper Sulfate with Hydrogen Sulfide
The first step involves the reaction of copper sulfate (CuSO₄) with hydrogen sulfide (H₂S).

**Reaction:**
\[ \text{CuSO}_4 (aq) + \text{H}_2\text{S} (g) \rightarrow \text{CuS} (s) + \text{H}_2\text{SO}_4 (aq) \]

Here, copper(II) sulfide (CuS) is formed as a precipitate (M) along with sulfuric acid (H₂SO₄).

### Step 2: Reaction of Copper Sulfide with Excess Potassium Cyanide
Next, we take the precipitated copper sulfide (CuS) and react it with excess potassium cyanide (KCN).

**Reaction:**
\[ \text{CuS} + 2 \text{KCN} \rightarrow \text{Cu(CN)}_2 + \text{K}_2\text{S} \]

However, Cu(CN)₂ is not stable and will further dissociate into its components.

### Step 3: Formation of Final Products
When Cu(CN)₂ is formed, it can further react with KCN to produce stable complexes.

**Final Reaction:**
\[ \text{Cu(CN)}_2 + \text{KCN} \rightarrow \text{CuCN} + \text{CN}^- \]

Thus, the final products N and O are:
- N = CuCN (copper(I) cyanide)
- O = CN⁻ (cyanide ion)

### Summary of Final Products
The final products N and O are:
- N = CuCN
- O = CN⁻