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 solve the question regarding the crystal field splitting energy (CFSE) for the complex \([Fe(CN)_4]^{3-}\), we will follow these steps:

### Step 1: Determine the oxidation state of Iron
The complex is \([Fe(CN)_4]^{3-}\). The cyanide ion (CN) has a charge of -1. Since there are 4 cyanide ligands, their total contribution to the charge is -4. Let the oxidation state of iron be \(x\).

\[ x + 4(-1) = -3 \\ x - 4 = -3 \\ x = +1 \]

### Step 2: Write the electronic configuration of Iron
Iron in its elemental state has the electronic configuration of \([Ar] 3d^6 4s^2\). In the +3 oxidation state, it loses two 4s electrons and one 3d electron, leading to the configuration:

\[ 3d^5 \]

### Step 3: Identify the ligand field strength
Cyanide (CN\(^-\)) is a strong field ligand. Strong field ligands tend to cause greater splitting of the d-orbitals and favor low-spin configurations.

### Step 4: Determine the d-orbital splitting in a tetrahedral field
In a tetrahedral field, the d-orbitals split into two sets: \(e_g\) (higher energy) and \(t_{2g}\) (lower energy). The splitting pattern is such that:

- \(e_g\) orbitals: higher energy
- \(t_{2g}\) orbitals: lower energy

For tetrahedral complexes, the splitting energy \(\Delta_t\) is less than that of octahedral complexes.

### Step 5: Fill the d-orbitals with electrons
Since we have 5 electrons (from \(3d^5\)), and given that CN is a strong field ligand, we will fill the orbitals to minimize the energy:

1. Fill the \(e_g\) orbitals first (which can hold 4 electrons).
2. Then, fill the \(t_{2g}\) orbital with the remaining electron.

The filling will look like this:

- \(e_g^4\)
- \(t_{2g}^1\)

### Step 6: Calculate the CFSE
The formula for calculating CFSE is:

\[ CFSE = (-0.4 \Delta_t \times n_{e_g}) + (-0.6 \Delta_t \times n_{t_{2g}}) \]

Where:
- \(n_{e_g}\) = number of electrons in \(e_g\) orbitals
- \(n_{t_{2g}\) = number of electrons in \(t_{2g}\) orbitals

Substituting the values:

\[ CFSE = (-0.4 \Delta_t \times 4) + (-0.6 \Delta_t \times 1) \]
\[ = -1.6 \Delta_t - 0.6 \Delta_t \]
\[ = -2.2 \Delta_t \]

### Step 7: Conclusion
The CFSE for the complex \([Fe(CN)_4]^{3-}\) is \(-2.2 \Delta_t\).