Question

The Crystal Field Stabilisation Energy (CFSE) for $[CoCl_6]^{4-}$ is $18000\, cm^{-1}$. The CFSE for $[CoCl_4]^{2-}$ will be-

• A. $6000\, cm^{-1}$
• B. $16000\, cm^{-1}$
• C. $18000\, cm^{-1}$
• D. $8000 \,cm^{-1}$

Answer 1

The Correct Option is D

Step-by-step Solution:

To determine the Crystal Field Stabilisation Energy (CFSE) for $[CoCl_4]^{2-}$, we must first understand the crystal field theory and how CFSE is calculated for different complex geometries.

1. Understanding the Geometries:

$[CoCl_6]^{4-}$ is an octahedral complex, while $[CoCl_4]^{2-}$ is a tetrahedral complex.

2. CFSE Calculation:

In octahedral complexes, CFSE is calculated using the formula: \[\text{CFSE}_{octahedral} = (-0.4x + 0.6y)\Delta_o\], where x is the number of electrons in the t_2g orbitals and y is the number of electrons in the e_g orbitals. For high-spin complexes, the electrons are filled such that they maximize unpaired spins.

For tetrahedral complexes, it is known that: \[\Delta_t = \frac{4}{9} \Delta_o\]

3. Given:

The CFSE for $[CoCl_6]^{4-}$ is $18000\, \text{cm}^{-1}$. Therefore, we use this value to find the corresponding \Delta_o, which remains as $18000\, \text{cm}^{-1}$ (assuming complete crystal field splitting).

4. Calculating CFSE for the Tetrahedral Complex:

Using \Delta_t = \frac{4}{9} \Delta_o, we substitute the given value: \[\Delta_t = \frac{4}{9} \times 18000 \, \text{cm}^{-1}\]

Calculating this gives: \[\Delta_t = 8000 \, \text{cm}^{-1}\]

5. Conclusion:

The CFSE for $[CoCl_4]^{2-}$, a tetrahedral complex, is $8000 \, \text{cm}^{-1}$.

Thus, the correct answer is $8000 \, \text{cm}^{-1}$.