Crystal field stabilization energy for high spin `d^(4)` octahedral complex is ________

Options

A. `-0.6Delta_(0)`
B. `-1.8Delta_(0)`
C. `-1.6Delta_(0)+p`
D. `-1.2Delta_(0)`

Answer

To find the Crystal Field Stabilization Energy (CFSE) for a high spin \(d^4\) octahedral complex, we can follow these steps:

### Step 1: Understand the Electron Configuration
In a high spin \(d^4\) configuration, we have 4 electrons to place in the \(d\) orbitals of an octahedral complex. The \(d\) orbitals split into two sets due to the presence of the ligand field: the lower energy \(t_{2g}\) orbitals and the higher energy \(e_g\) orbitals.

### Step 2: Identify the Splitting Energy
The energy levels in an octahedral field are:
- \(t_{2g}\) (lower energy)
- \(e_g\) (higher energy)

The barycenter (average energy level) is considered as 0. The energy levels can be represented as:
- \(t_{2g}\): -0.4Δo (for each electron)
- \(e_g\): +0.6Δo (for each electron)

### Step 3: Fill the Electrons
For a high spin \(d^4\) configuration, the electrons will occupy the orbitals as follows:
1. First electron in \(t_{2g}\)
2. Second electron in \(t_{2g}\)
3. Third electron in \(t_{2g}\)
4. Fourth electron in \(e_g\)

So, we have:
- \(p\) (number of electrons in \(t_{2g}\)) = 3
- \(q\) (number of electrons in \(e_g\)) = 1

### Step 4: Calculate the CFSE
Using the formula for CFSE:
\[ \text{CFSE} = (-0.4 \Delta_o \times p) + (0.6 \Delta_o \times q) \]
Substituting the values:
\[ \text{CFSE} = (-0.4 \Delta_o \times 3) + (0.6 \Delta_o \times 1) \]
\[ \text{CFSE} = -1.2 \Delta_o + 0.6 \Delta_o \]
\[ \text{CFSE} = -0.6 \Delta_o \]

### Step 5: Conclusion
The CFSE for a high spin \(d^4\) octahedral complex is \(-0.6 \Delta_o\).

The final answer is:
\[ \text{CFSE} = -0.6 \Delta_o \]