What is the crystal field stabilization energy for high spin `d^6` octahedral complex?

Explanation

To find the crystal field stabilization energy (CFSE) for a high-spin \(d^6\) octahedral complex, we will follow these steps:

### Step 1: Understand the Electron Configuration
In a \(d^6\) configuration, there are 6 electrons to be placed in the \(d\) orbitals. In an octahedral field, the \(d\) orbitals split into two sets: the lower energy \(t_{2g}\) orbitals (3 orbitals) and the higher energy \(e_g\) orbitals (2 orbitals).

### Step 2: Fill the Electrons in the Orbitals
For a high-spin configuration, we fill the orbitals according to Hund's rule, which states that electrons will occupy degenerate orbitals singly before pairing.

1. Fill the three \(t_{2g}\) orbitals with 3 electrons:
- \(t_{2g}\): ↑ ↑ ↑
2. Then, place the next 3 electrons in the \(e_g\) orbitals:
- \(e_g\): ↑ ↑

So, the distribution of electrons will be:
- \(t_{2g}\): 3 electrons (all unpaired)
- \(e_g\): 3 electrons (all unpaired)

### Step 3: Calculate the CFSE
The CFSE can be calculated using the formula:
\[ \text{CFSE} = \left( \text{number of electrons in } t_{2g} \times -\frac{2}{5} \Delta_0 \right) + \left( \text{number of electrons in } e_g \times \frac{3}{5} \Delta_0 \right) \]

Where:
- \(\Delta_0\) is the crystal field splitting energy.
- For \(t_{2g}\), the contribution is negative because they are lower in energy.
- For \(e_g\), the contribution is positive because they are higher in energy.

### Step 4: Substitute the Values
1. Number of electrons in \(t_{2g}\) = 6 (3 electrons)
2. Number of electrons in \(e_g\) = 0 (3 electrons)

Now substituting into the formula:
\[ \text{CFSE} = \left( 3 \times -\frac{2}{5} \Delta_0 \right) + \left( 3 \times \frac{3}{5} \Delta_0 \right) \]
\[ \text{CFSE} = -\frac{6}{5} \Delta_0 + \frac{9}{5} \Delta_0 \]
\[ \text{CFSE} = \frac{3}{5} \Delta_0 \]

### Step 5: Final Calculation
The total CFSE for a high-spin \(d^6\) octahedral complex is:
\[ \text{CFSE} = 2.5 \Delta_0 \]

### Conclusion
The crystal field stabilization energy (CFSE) for a high-spin \(d^6\) octahedral complex is \(2.5 \Delta_0\).

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