Find `CFSE` of `[Fe(H_(2)O)_(6)]^(2+)` and `[NiCl_(4)]^(2-)`

Options

A. `-0.6triangle_(o)` and `-0.8triangle_(t)`
B. `-0.4triangle_(0)` and `-0.8triangle_(t)`
C. `-2.4triangle_(0)` and `-1.2 triangle_(t)`
D. `-0.4triangle_(o)` and `-1.2triangle_(t)`

Answer

To find the Crystal Field Stabilization Energy (CFSE) for the complexes \([Fe(H_2O)_6]^{2+}\) and \([NiCl_4]^{2-}\), we will follow these steps:

### Step 1: Determine the oxidation state and electronic configuration of the metal ions.

1. **For \([Fe(H_2O)_6]^{2+}\)**:
- The oxidation state of iron (Fe) in this complex is +2.
- The electronic configuration of neutral iron (Fe) is \([Ar] 4s^2 3d^6\).
- For \(Fe^{2+}\), we remove two electrons from the 4s orbital:
\[ Fe^{2+} \rightarrow [Ar] 3d^6 \]

2. **For \([NiCl_4]^{2-}\)**:
- The oxidation state of nickel (Ni) in this complex is +2.
- The electronic configuration of neutral nickel (Ni) is \([Ar] 4s^2 3d^8\).
- For \(Ni^{2+}\), we remove two electrons from the 4s orbital:
\[ Ni^{2+} \rightarrow [Ar] 3d^8 \]

### Step 2: Determine the geometry and the type of ligands.

1. **For \([Fe(H_2O)_6]^{2+}\)**:
- The complex is octahedral because it has six ligands (H2O).
- H2O is a weak field ligand, which means it does not cause pairing of electrons.

2. **For \([NiCl_4]^{2-}\)**:
- The complex is tetrahedral because it has four ligands (Cl).
- Cl is also a weak field ligand, which means it does not cause pairing of electrons.

### Step 3: Fill the d-orbitals according to the geometry and the number of electrons.

1. **For \([Fe(H_2O)_6]^{2+}\)**:
- The \(3d^6\) configuration will fill the orbitals as follows in an octahedral field:
- T2g: 4 electrons (1, 2, 3, 4)
- Eg: 2 electrons (5, 6)
- The arrangement is:
```
T2g: ↑↑↑↑
Eg: ↑↑
```

2. **For \([NiCl_4]^{2-}\)**:
- The \(3d^8\) configuration will fill the orbitals as follows in a tetrahedral field:
- Eg: 4 electrons (1, 2, 3, 4)
- T2: 4 electrons (5, 6, 7, 8)
- The arrangement is:
```
Eg: ↑↑↑↑
T2: ↑↑↑↑
```

### Step 4: Calculate CFSE for both complexes.

1. **For \([Fe(H_2O)_6]^{2+}\)**:
- The CFSE formula for octahedral complexes is:
\[ CFSE = (n_{T2g} \times -0.4\Delta_0) + (n_{Eg} \times 0.6\Delta_0) \]
- Here, \(n_{T2g} = 4\) and \(n_{Eg} = 2\):
\[ CFSE = (4 \times -0.4\Delta_0) + (2 \times 0.6\Delta_0) = -1.6\Delta_0 + 1.2\Delta_0 = -0.4\Delta_0 \]

2. **For \([NiCl_4]^{2-}\)**:
- The CFSE formula for tetrahedral complexes is:
\[ CFSE = (n_{T2} \times -0.6\Delta_t) + (n_{Eg} \times 0.4\Delta_t) \]
- Here, \(n_{T2} = 4\) and \(n_{Eg} = 4\):
\[ CFSE = (4 \times -0.6\Delta_t) + (4 \times 0.4\Delta_t) = -2.4\Delta_t + 1.6\Delta_t = -0.8\Delta_t \]

### Final Results:
- The CFSE for \([Fe(H_2O)_6]^{2+}\) is \(-0.4\Delta_0\).
- The CFSE for \([NiCl_4]^{2-}\) is \(-0.8\Delta_t\).