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 \([CoCl_4]^{2-}\) based on the given CFSE for the octahedral complex \([CoCl_6]^{4-}\), we can use the relationship between the CFSE of octahedral and tetrahedral complexes.

### Step-by-Step Solution:

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

2. **Use the relationship between octahedral and tetrahedral CFSE**:
The relationship between the CFSE of tetrahedral and octahedral complexes is given by:
\[ \text{CFSE}_{\text{tetrahedral}} = \frac{4}{9} \times \text{CFSE}_{\text{octahedral}} \]

3. **Substitute the known values**:
Substitute the CFSE of the octahedral complex into the formula:
\[ \text{CFSE}_{\text{tetrahedral}} = \frac{4}{9} \times 18,000 \, \text{cm}^{-1} \]

4. **Calculate the CFSE for the tetrahedral complex**:
First, calculate \(\frac{18,000}{9}\):
\[ \frac{18,000}{9} = 2,000 \, \text{cm}^{-1} \]
Now multiply this result by 4:
\[ \text{CFSE}_{\text{tetrahedral}} = 4 \times 2,000 = 8,000 \, \text{cm}^{-1} \]

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