`CuSO_(4)(aq) overset(KCN(uarr))(to)M(darr)overset("Excess")underset(KCN)(to)N+O`
Then final products 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`

Explanation

To solve the given question, we will go through the reactions step by step.

### Step 1: Identify the initial reaction
We start with copper(II) sulfate in aqueous solution, which is represented as \( \text{CuSO}_4(aq) \). When it reacts with potassium cyanide (\( \text{KCN} \)), it forms cupric cyanide and potassium sulfate.

**Reaction:**
\[ \text{CuSO}_4 + 2 \text{KCN} \rightarrow \text{Cu(CN)}_2 + \text{K}_2\text{SO}_4 \]

### Step 2: Identify the compound M
The product formed from the above reaction, \( \text{Cu(CN)}_2 \), is known as cupric cyanide. This compound is represented as \( M \).

### Step 3: Decomposition of M
Next, the cupric cyanide \( \text{Cu(CN)}_2 \) can decompose into cuprous cyanide and cyanogen gas. The decomposition reaction can be written as:

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

Here, \( \text{Cu}_2\text{CN}_2 \) is our compound \( M \) and \( \text{CN}_2 \) is the by-product \( O \).

### Step 4: Reaction with excess KCN
When excess potassium cyanide is added to the cuprous cyanide \( \text{Cu}_2\text{CN}_2 \), it forms a stable complex known as tetracyanocuprate(II) ion.

**Reaction:**
\[ \text{Cu}_2\text{CN}_2 + 6 \text{KCN} \rightarrow 2 \text{K}_3\text{Cu(CN)}_4 \]

### Step 5: Identify the final products N and O
From the above reaction, the final products are:
- \( N \) is \( \text{K}_3\text{Cu(CN)}_4 \) (tetracyanocuprate(II) ion)
- \( O \) is \( \text{CN}_2 \) (cyanogen gas)

### Final Answer
Thus, the final products \( N \) and \( O \) are:
- \( N = \text{K}_3\text{Cu(CN)}_4 \)
- \( O = \text{CN}_2 \)