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(II) 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 correct statement regarding the stability of Cu(I) and Cu(II), we can analyze the electronic configurations and the implications of oxidation states.

### Step 1: Identify the Electronic Configurations
- The electronic configuration of Copper (Cu) is:
- Atomic number of Cu = 29
- Ground state configuration: \( [Ar] 3d^{10} 4s^1 \)
- For Cu(I):
- Cu(I) loses one electron from the 4s orbital, resulting in \( 3d^{10} \).
- For Cu(II):
- Cu(II) loses two electrons, one from the 4s and one from the 3d orbital, resulting in \( 3d^9 \).

### Step 2: Stability Consideration
- Cu(I) has the configuration \( 3d^{10} \), which is a fully filled d-subshell.
- Cu(II) has the configuration \( 3d^{9} \), which is not fully filled and is generally less stable than a fully filled subshell.

### Step 3: Charge and Stability
- Cu(I) has a +1 oxidation state while Cu(II) has a +2 oxidation state.
- Generally, higher oxidation states can lead to increased stability due to greater effective nuclear charge, but this is not always the case, especially when comparing configurations.
- The stability of transition metal ions can also depend on the ligand field and the nature of the salts formed.

### Step 4: Conclusion
- Given that Cu(I) has a fully filled d-subshell, it is generally more stable than Cu(II) which has a partially filled d-subshell.
- Therefore, the correct statement is that Cu(I) is more stable than Cu(II).

### Final Answer
- **Cu(I) is more stable than Cu(II)**.

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