The differnece in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is

Options

A. `Mn^(2+)`
B. `Co^(2+)`
C. `Fe^(2+)`
D. `Ni^(2+)`

Explanation

To determine the metal ion for which the difference in the number of unpaired electrons in its high-spin and low-spin octahedral complexes is two, we can analyze the electronic configurations of various metal ions.

### Step-by-Step Solution:

1. **Identify the Metal Ions**: We will consider common transition metal ions that can exhibit both high-spin and low-spin states in octahedral complexes. The relevant ions are Mn²⁺, Co²⁺, Fe²⁺, and Ni²⁺.

2. **Calculate for Mn²⁺**:
- **Electronic Configuration**: Mn²⁺ has an electronic configuration of d⁵.
- **High-Spin Configuration**: In a high-spin octahedral field, the configuration is T₂g³ E₁g². This results in 5 unpaired electrons.
- **Low-Spin Configuration**: In a low-spin octahedral field, the configuration is T₂g⁵. This results in 1 unpaired electron.
- **Difference in Unpaired Electrons**: 5 (high-spin) - 1 (low-spin) = 4.

3. **Calculate for Co²⁺**:
- **Electronic Configuration**: Co²⁺ has an electronic configuration of d⁷.
- **High-Spin Configuration**: In a high-spin octahedral field, the configuration is T₂g⁵ E₁g². This results in 3 unpaired electrons.
- **Low-Spin Configuration**: In a low-spin octahedral field, the configuration is T₂g⁶ E₁g¹. This results in 1 unpaired electron.
- **Difference in Unpaired Electrons**: 3 (high-spin) - 1 (low-spin) = 2.

4. **Calculate for Fe²⁺**:
- **Electronic Configuration**: Fe²⁺ has an electronic configuration of d⁶.
- **High-Spin Configuration**: In a high-spin octahedral field, the configuration is T₂g⁴ E₁g². This results in 4 unpaired electrons.
- **Low-Spin Configuration**: In a low-spin octahedral field, the configuration is T₂g⁶. This results in 0 unpaired electrons.
- **Difference in Unpaired Electrons**: 4 (high-spin) - 0 (low-spin) = 4.

5. **Calculate for Ni²⁺**:
- **Electronic Configuration**: Ni²⁺ has an electronic configuration of d⁸.
- **High-Spin Configuration**: In a high-spin octahedral field, the configuration is T₂g⁶ E₁g². This results in 2 unpaired electrons.
- **Low-Spin Configuration**: In a low-spin octahedral field, the configuration is T₂g⁶ E₁g². This results in 0 unpaired electrons.
- **Difference in Unpaired Electrons**: 2 (high-spin) - 0 (low-spin) = 2.

6. **Conclusion**: The metal ion that has a difference of 2 unpaired electrons between its high-spin and low-spin states in octahedral complexes is **Co²⁺**.

### Final Answer:
The metal ion is **Co²⁺**.