The electronic configurationo of Cu(II) is `3d^(9)` whereas that of Cu(I) is `3d^(10)`. Which of the following is correct ?

Options

A. Cu(I) is more stable
B. Cu(II) is less stable
C. Cu(I) and Cu(II) are equally stable
D. Stability of Cu(I) and Cu(II) depends on nature of copper salts

Explanation

To determine the stability of Cu(I) and Cu(II) based on their electronic configurations, we can follow these steps:

### Step 1: Understand the Electronic Configurations
- The electronic configuration of Copper (Cu) in its neutral state is:
- \( [Ar] 3d^{10} 4s^1 \)
- When Copper loses one electron to form Cu(I), the configuration becomes:
- \( [Ar] 3d^{10} \)
- When Copper loses two electrons to form Cu(II), the configuration becomes:
- \( [Ar] 3d^{9} \)

### Step 2: Analyze Stability Based on Electronic Configuration
- A fully filled \( d \) subshell (like \( 3d^{10} \)) is more stable than a partially filled \( d \) subshell (like \( 3d^{9} \)).
- Therefore, Cu(I) with the configuration \( 3d^{10} \) is more stable than Cu(II) with the configuration \( 3d^{9} \).

### Step 3: Evaluate the Given Statements
1. **Copper(I) is more stable, Copper(II) is less stable**: This statement is correct based on the analysis above.
2. **Copper(I) and Copper(II) are equally stable**: This statement is incorrect as we established that Cu(I) is more stable.
3. **Stability of Copper(I) and Copper(II) depends on the nature of copper salts**: This statement is also incorrect since the stability is primarily determined by the electronic configuration, not the nature of the salts.

### Conclusion
Based on the analysis:
- The correct statements are:
- Copper(I) is more stable than Copper(II).
- Copper(II) is less stable than Copper(I).

### Final Answer
The correct option is:
- **Copper(I) is more stable, Copper(II) is less stable.**

---