What is CFSE of complex ion `[FeF_(6)]^(4-)` in terms of Dq?

Answer

To find the Crystal Field Stabilization Energy (CFSE) of the complex ion \([FeF_6]^{4-}\) in terms of \(Dq\), we can follow these steps:

### Step 1: Determine the oxidation state of Iron in \([FeF_6]^{4-}\)

- Let the oxidation state of Fe be \(x\).
- The charge of each fluorine atom is \(-1\), and there are 6 fluorine atoms, contributing a total charge of \(-6\).
- The overall charge of the complex is \(-4\).

Setting up the equation:
\[ x - 6 = -4 \]
Solving for \(x\):
\[ x = -4 + 6 = +2 \]

### Step 2: Write the electronic configuration of \(Fe^{2+}\)

- The atomic number of Fe is 26.
- The electronic configuration of neutral Fe is \([Ar] 4s^2 3d^6\).
- For \(Fe^{2+}\), we remove two electrons from the 4s orbital:
\[ Fe^{2+} : [Ar] 3d^6 \]

### Step 3: Identify the geometry of the complex

- The complex \([FeF_6]^{4-}\) has a coordination number of 6, indicating it is an octahedral complex.

### Step 4: Determine the splitting of d-orbitals in an octahedral field

- In an octahedral field, the \(d\) orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy).
- The \(d\) orbital splitting can be represented as:
- \(t_{2g}\) = 3 orbitals
- \(e_g\) = 2 orbitals

### Step 5: Distribute the \(d\) electrons in the split orbitals

- For \(Fe^{2+}\) with \(3d^6\), the electrons will fill the orbitals as follows:
- \(t_{2g}\) will have 4 electrons
- \(e_g\) will have 2 electrons

### Step 6: Calculate the CFSE

- The CFSE is calculated using the formula:
\[ CFSE = (n_{t_{2g}} \times -\frac{2}{5}Dq) + (n_{e_g} \times \frac{3}{5}Dq) \]
Where:
- \(n_{t_{2g}} = 4\) (number of electrons in \(t_{2g}\))
- \(n_{e_g} = 2\) (number of electrons in \(e_g\))

Substituting the values:
\[ CFSE = (4 \times -\frac{2}{5}Dq) + (2 \times \frac{3}{5}Dq) \]
\[ = -\frac{8}{5}Dq + \frac{6}{5}Dq \]
\[ = -\frac{2}{5}Dq \]

### Final Result

Thus, the CFSE of the complex ion \([FeF_6]^{4-}\) in terms of \(Dq\) is:
\[ CFSE = -\frac{2}{5}Dq \]

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