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 each of the given complexes step by step.

### Step 1: Analyze the first complex
**Complex:** [Mn(H2O)6]²⁺
1. **Oxidation State of Manganese (Mn):**
Let the oxidation state be \( x \).
\[ x + 6(0) = +2 \]
Therefore, \( x = +2 \).
2. **Electronic Configuration of Mn²⁺:**
Manganese has an atomic number of 25, so its configuration is \( [Ar] 3d^5 4s^2 \).
After losing 2 electrons, the configuration becomes \( 3d^5 \).
3. **Filling of Electrons in Octahedral Field:**
In a high-spin configuration, the 5 electrons will fill the \( t_{2g} \) and \( e_g \) orbitals as follows:
\( t_{2g}^3 e_g^2 \) (4 electrons in \( t_{2g} \) and 1 in \( e_g \)).
4. **CFSE Calculation:**
\[ \text{CFSE} = (3 \times -0.4\Delta_0) + (2 \times 0.6\Delta_0) \]
\[ \text{CFSE} = -1.2\Delta_0 + 1.2\Delta_0 = 0 \]

### Step 2: Analyze the second complex
**Complex:** [Cr(H2O)6]²⁺
1. **Oxidation State of Chromium (Cr):**
Let the oxidation state be \( x \).
\[ x + 6(0) = +2 \]
Therefore, \( x = +2 \).
2. **Electronic Configuration of Cr²⁺:**
Chromium has an atomic number of 24, so its configuration is \( [Ar] 3d^5 4s^1 \).
After losing 2 electrons, the configuration becomes \( 3d^4 \).
3. **Filling of Electrons in Octahedral Field:**
In a high-spin configuration, the 4 electrons will fill as follows:
\( t_{2g}^3 e_g^1 \).
4. **CFSE Calculation:**
\[ \text{CFSE} = (3 \times -0.4\Delta_0) + (1 \times 0.6\Delta_0) \]
\[ \text{CFSE} = -1.2\Delta_0 + 0.6\Delta_0 = -0.6\Delta_0 \]

### Step 3: Analyze the third complex
**Complex:** [Mn(H2O)6]³⁺
1. **Oxidation State of Manganese (Mn):**
Let the oxidation state be \( x \).
\[ x + 6(0) = +3 \]
Therefore, \( x = +3 \).
2. **Electronic Configuration of Mn³⁺:**
After losing 3 electrons, the configuration becomes \( 3d^4 \).
3. **Filling of Electrons in Octahedral Field:**
In a high-spin configuration, the 4 electrons will fill as follows:
\( t_{2g}^3 e_g^1 \).
4. **CFSE Calculation:**
\[ \text{CFSE} = (3 \times -0.4\Delta_0) + (1 \times 0.6\Delta_0) \]
\[ \text{CFSE} = -1.2\Delta_0 + 0.6\Delta_0 = -0.6\Delta_0 \]

### Step 4: Analyze the fourth complex
**Complex:** [Cr(H2O)6]³⁺
1. **Oxidation State of Chromium (Cr):**
Let the oxidation state be \( x \).
\[ x + 6(0) = +3 \]
Therefore, \( x = +3 \).
2. **Electronic Configuration of Cr³⁺:**
After losing 3 electrons, the configuration becomes \( 3d^3 \).
3. **Filling of Electrons in Octahedral Field:**
All 3 electrons will fill the \( t_{2g} \) orbitals:
\( t_{2g}^3 \).
4. **CFSE Calculation:**
\[ \text{CFSE} = (3 \times -0.4\Delta_0) + (0 \times 0.6\Delta_0) \]
\[ \text{CFSE} = -1.2\Delta_0 \]

### Conclusion
Now, we compare the CFSE values:
- For [Mn(H2O)6]²⁺: CFSE = 0
- For [Cr(H2O)6]²⁺: CFSE = -0.6Δ₀
- For [Mn(H2O)6]³⁺: CFSE = -0.6Δ₀
- For [Cr(H2O)6]³⁺: CFSE = -1.2Δ₀

The complex with the largest CFSE is [Mn(H2O)6]²⁺ with a CFSE of 0. However, if we consider the absolute value, the complex with the least negative CFSE is [Cr(H2O)6]²⁺ and [Mn(H2O)6]³⁺, both having -0.6Δ₀.

Thus, the answer is that the complex with the largest CFSE is **[Mn(H2O)6]²⁺**.