Which one of the following high-spin complexes has the largest CFSE (Crystal Field stabilization energy ) ?

Options

A. `[Mn(H_(2)O)_(6)]^(2+)`
B. `[Cr(H_(2)O)_(6)]^(2+)`
C. `[Mn(H_(2)O)_(6)]^(3+)`
D. `[Cr(H_(2)O)_(6)]^(3-)`

Explanation

To determine which high-spin complex has the largest Crystal Field Stabilization Energy (CFSE), we will analyze the given complexes step by step.

### Step 1: Identify the complexes and their oxidation states
We have the following complexes to analyze:
1. \( \text{[Mn(H}_2\text{O)}_6]^{2+} \)
2. \( \text{[Cr(H}_2\text{O)}_6]^{2+} \)
3. \( \text{[Mn(H}_2\text{O)}_6]^{3+} \)
4. \( \text{[Cr(H}_2\text{O)}_6]^{3+} \)

### Step 2: Determine the electron configurations
- **For \( \text{Mn}^{2+} \)**:
- Manganese (Mn) has an atomic number of 25. Its electron configuration is \( [Ar] 3d^5 4s^2 \).
- For \( \text{Mn}^{2+} \), it loses 2 electrons (from 4s), resulting in \( [Ar] 3d^5 \).

- **For \( \text{Cr}^{2+} \)**:
- Chromium (Cr) has an atomic number of 24. Its electron configuration is \( [Ar] 3d^5 4s^1 \).
- For \( \text{Cr}^{2+} \), it loses 2 electrons (1 from 4s and 1 from 3d), resulting in \( [Ar] 3d^4 \).

- **For \( \text{Mn}^{3+} \)**:
- From \( [Ar] 3d^5 \), it loses 3 electrons, resulting in \( [Ar] 3d^4 \).

- **For \( \text{Cr}^{3+} \)**:
- From \( [Ar] 3d^5 4s^1 \), it loses 3 electrons (1 from 4s and 2 from 3d), resulting in \( [Ar] 3d^3 \).

### Step 3: Determine the CFSE for each complex
- **For \( \text{[Mn(H}_2\text{O)}_6]^{2+} \)**:
- Configuration: \( 3d^5 \)
- CFSE = \( (3 \times -0.4 \Delta_0) + (2 \times 0.6 \Delta_0) = -1.2 \Delta_0 + 1.2 \Delta_0 = 0 \)

- **For \( \text{[Cr(H}_2\text{O)}_6]^{2+} \)**:
- Configuration: \( 3d^4 \)
- CFSE = \( (3 \times -0.4 \Delta_0) + (1 \times 0.6 \Delta_0) = -1.2 \Delta_0 + 0.6 \Delta_0 = -0.6 \Delta_0 \)

- **For \( \text{[Mn(H}_2\text{O)}_6]^{3+} \)**:
- Configuration: \( 3d^4 \)
- CFSE = \( (3 \times -0.4 \Delta_0) + (0 \times 0.6 \Delta_0) = -1.2 \Delta_0 \)

- **For \( \text{[Cr(H}_2\text{O)}_6]^{3+} \)**:
- Configuration: \( 3d^3 \)
- CFSE = \( (3 \times -0.4 \Delta_0) + (0 \times 0.6 \Delta_0) = -1.2 \Delta_0 \)

### Step 4: Compare the CFSE values
- \( \text{[Mn(H}_2\text{O)}_6]^{2+} \): CFSE = 0
- \( \text{[Cr(H}_2\text{O)}_6]^{2+} \): CFSE = -0.6 Δ0
- \( \text{[Mn(H}_2\text{O)}_6]^{3+} \): CFSE = -1.2 Δ0
- \( \text{[Cr(H}_2\text{O)}_6]^{3+} \): CFSE = -1.2 Δ0

### Conclusion
The complex with the largest CFSE is \( \text{[Mn(H}_2\text{O)}_6]^{2+} \) with a CFSE of 0.

### Final Answer
**The high-spin complex with the largest CFSE is \( \text{[Mn(H}_2\text{O)}_6]^{2+} \).**