For `Mn^(3+)` ion the electron pairing energy `P` is about `28,000cm^(1-),Delta_(0)` values for the complexes `[Mn(H_(2)O)_(6)]^(3+)` and `[Mn(CN)_(6)]^(3-)` are `15,800cm^(-1)` and `38,500cm^(-1)` respectively which of the following complex is high spin .

Options

A. Both are high spin
B. `[Mn(H_(2)O)_(6)]^(3+)`
C. `[Mn(CN)_(6)]^(3-)`
D. Noen of these

Answer

To determine which of the complexes \([Mn(H_2O)_6]^{3+}\) or \([Mn(CN)_6]^{3-}\) is a high spin complex, we need to analyze the relationship between the electron pairing energy \(P\) and the crystal field splitting energy \(\Delta_0\) for each complex.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Electron pairing energy \(P = 28,000 \, \text{cm}^{-1}\)
- Crystal field splitting energy for \([Mn(H_2O)_6]^{3+}\): \(\Delta_0 = 15,800 \, \text{cm}^{-1}\)
- Crystal field splitting energy for \([Mn(CN)_6]^{3-}\): \(\Delta_0 = 38,500 \, \text{cm}^{-1}\)

2. **Determine the Spin State for \([Mn(H_2O)_6]^{3+}\):**
- Compare \(\Delta_0\) with \(P\):
\[ \Delta_0 (15,800 \, \text{cm}^{-1}) < P (28,000 \, \text{cm}^{-1}) \]
- Since \(\Delta_0 < P\), this indicates that it is energetically favorable for the electrons to remain unpaired rather than pairing up. Thus, \([Mn(H_2O)_6]^{3+}\) is a **high spin complex**.

3. **Determine the Spin State for \([Mn(CN)_6]^{3-}\):**
- Compare \(\Delta_0\) with \(P\):
\[ \Delta_0 (38,500 \, \text{cm}^{-1}) > P (28,000 \, \text{cm}^{-1}) \]
- Since \(\Delta_0 > P\), this indicates that it is energetically favorable for the electrons to pair up. Thus, \([Mn(CN)_6]^{3-}\) is a **low spin complex**.

4. **Conclusion:**
- The complex that is a high spin is \([Mn(H_2O)_6]^{3+}\).

### Final Answer:
The high spin complex is \([Mn(H_2O)_6]^{3+}\).