The CFSE for octahedral `[CoCl_(6)]^(4-)` is `18,000 cm^(-1)`. The CFSE for tetrahedral `[CoCl_(4)]^(2-)` will be

Options

A. `18,000 cm^(-1)`
B. `16,000 cm^(-1)`
C. `8,000 cm^(-1)`
D. `20,000 cm^(-1)`

Explanation

To find the Crystal Field Stabilization Energy (CFSE) for the tetrahedral complex \([CoCl_4]^{2-}\), we can use the relationship between the CFSE of octahedral and tetrahedral complexes.

### Step-by-Step Solution:

1. **Understanding CFSE in Octahedral Complexes:**
The CFSE for an octahedral complex can be calculated using the formula:
\[ \text{CFSE} = (n_t \cdot \Delta_t) - (n_e \cdot \Delta_e) \]
where \(n_t\) is the number of electrons in the lower energy t2g orbitals, \(n_e\) is the number of electrons in the higher energy eg orbitals, \(\Delta_t\) is the energy difference between the t2g and eg orbitals, and \(\Delta_e\) is the energy difference in the octahedral field.

2. **Given CFSE for Octahedral \([CoCl_6]^{4-}\):**
The CFSE for the octahedral complex \([CoCl_6]^{4-}\) is given as \(18,000 \, \text{cm}^{-1}\).

3. **Relationship Between Octahedral and Tetrahedral CFSE:**
The CFSE for tetrahedral complexes is generally lower than that for octahedral complexes. The relationship can be expressed as:
\[ \text{CFSE}_{\text{tetrahedral}} = \frac{4}{9} \times \text{CFSE}_{\text{octahedral}} \]
This is because in tetrahedral complexes, the splitting of d-orbitals is less pronounced compared to octahedral complexes.

4. **Calculating CFSE for Tetrahedral Complex:**
Using the above relationship, we can calculate the CFSE for \([CoCl_4]^{2-}\):
\[ \text{CFSE}_{\text{tetrahedral}} = \frac{4}{9} \times 18,000 \, \text{cm}^{-1} \]
\[ \text{CFSE}_{\text{tetrahedral}} = \frac{4 \times 18,000}{9} \]
\[ \text{CFSE}_{\text{tetrahedral}} = \frac{72,000}{9} = 8,000 \, \text{cm}^{-1} \]

5. **Final Answer:**
The CFSE for the tetrahedral complex \([CoCl_4]^{2-}\) is \(8,000 \, \text{cm}^{-1}\).