What is the crystal field stabalisation energy value of `d^(5)` configuration for tetrahedral complex ?

Options

A. 0.4
B. `0.0`
C. `-2`
D. `+2`

Answer

To determine the Crystal Field Stabilization Energy (CFSE) for a \(d^5\) configuration in a tetrahedral complex, we can follow these steps:

### Step 1: Understand the Tetrahedral Crystal Field Splitting
In a tetrahedral complex, the \(d\) orbitals split into two sets:
- \(e\) set (higher energy): consists of \(d_{z^2}\) and \(d_{x^2-y^2}\)
- \(t_2\) set (lower energy): consists of \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\)

The energy levels are arranged such that the \(t_2\) orbitals are lower in energy compared to the \(e\) orbitals.

### Step 2: Write the Electron Configuration
For a \(d^5\) configuration in a tetrahedral field, the electrons will fill the orbitals as follows:
- The \(t_2\) orbitals will be filled first, and since there are 5 electrons, the configuration will be:
- \(t_2^3 e^2\)

### Step 3: Calculate the CFSE
The CFSE can be calculated using the formula:
\[ \text{CFSE} = (n_{t_2} \times \Delta_t) + (n_e \times \Delta_e) \]
Where:
- \(n_{t_2}\) = number of electrons in \(t_2\) orbitals
- \(n_e\) = number of electrons in \(e\) orbitals
- \(\Delta_t\) = energy difference for \(t_2\) orbitals (which is \(-0.4 \Delta\))
- \(\Delta_e\) = energy difference for \(e\) orbitals (which is \(+0.6 \Delta\))

For \(d^5\):
- \(n_{t_2} = 3\)
- \(n_e = 2\)

Substituting the values:
\[ \text{CFSE} = (3 \times -0.4\Delta) + (2 \times 0.6\Delta) \]
Calculating this gives:
\[ \text{CFSE} = -1.2\Delta + 1.2\Delta = 0 \]

### Step 4: Conclusion
The CFSE for a \(d^5\) configuration in a tetrahedral complex is \(0.0\).

### Final Answer
The crystal field stabilization energy value of \(d^5\) configuration for a tetrahedral complex is \(0.0\).

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