Crystal field splitting energey `(CFSE)` for the complex `[Fe(CN)_(4)]^(Θ)` is when `DeltagtP` .

Options

A. `a)-2.0`
B. `b)Delta+P`
C. `c)2+P`
D. `d)0`

Explanation

To calculate the Crystal Field Splitting Energy (CFSE) for the complex \([Fe(CN)_4]^{2-}\), we will follow these steps:

### Step 1: Determine the oxidation state of Iron
The complex is \([Fe(CN)_4]^{2-}\). Since CN is a unidentate ligand with a charge of -1, we can set up the equation for the oxidation state of iron (let's denote it as \(x\)):
\[ x + (-1) \times 4 = -2 \]
Solving for \(x\):
\[ x - 4 = -2 \implies x = +2 \]
Thus, the oxidation state of iron in this complex is +2.

### Step 2: Determine the electronic configuration of Iron
Iron has an atomic number of 26. The electronic configuration of neutral iron is:
\[ [Ar] 3d^6 4s^2 \]
In the +2 oxidation state, iron loses two electrons, both from the 4s orbital:
\[ Fe^{2+}: [Ar] 3d^6 \]

### Step 3: Identify the geometry of the complex
The coordination number of the complex is 4 (due to 4 CN ligands). For a coordination number of 4, the geometry can be either tetrahedral or square planar. Given that CN is a strong field ligand and the problem states that \(\Delta\) is greater than \(p\), we can conclude that the complex is tetrahedral.

### Step 4: Understand the crystal field splitting in tetrahedral complexes
In tetrahedral complexes, the crystal field splitting results in two sets of d-orbitals:
- The \(e_g\) set (higher energy)
- The \(t_2g\) set (lower energy)

For a tetrahedral field, the energy levels are reversed compared to octahedral complexes.

### Step 5: Fill the d-orbitals according to Hund's rule
For \(Fe^{2+}\) with \(3d^6\), we will fill the orbitals:
- The \(e_g\) level will be filled first, followed by the \(t_2g\) level.
- Since we have 6 electrons, the filling will be:
- \(e_g^2\) (2 electrons in the higher energy level)
- \(t_2g^4\) (4 electrons in the lower energy level)

### Step 6: Calculate the CFSE
The CFSE can be calculated using the formula:
\[ CFSE = (n_{e_g} \times 0.6 \Delta) + (n_{t_2g} \times -0.4 \Delta) \]
Where:
- \(n_{e_g} = 2\)
- \(n_{t_2g} = 4\)

Substituting the values:
\[ CFSE = (2 \times 0.6 \Delta) + (4 \times -0.4 \Delta) \]
\[ CFSE = (1.2 \Delta) + (-1.6 \Delta) = -0.4 \Delta \]

### Final Result
Thus, the CFSE for the complex \([Fe(CN)_4]^{2-}\) is:
\[ CFSE = -0.4 \Delta \]