The CFSE for octahedral \([CoCl_6]^{4–}\) is \(18,000 \text{ }cm^{–1}\). The CFSE for tetrahedral \([CoCl_4]^{2–}\) will be
Answer & explanation
Correct answer: option 4
The correct answer is option 4. \(8,000\, \ cm^{–1}\).
CFSE (Crystal Field Stabilization Energy): It's the energy difference between the d orbitals of a metal ion in a complex, caused by the interaction with ligands. It affects the stability and magnetic properties of the complex.
Octahedral vs. Tetrahedral Complexes:
- Octahedral complexes have six ligands arranged around the central metal ion in an octahedral shape.
- Tetrahedral complexes have four ligands arranged in a tetrahedral shape.
- The d orbital splitting patterns differ in these two geometries, leading to different CFSE values.
CFSE Relationship: The CFSE for a tetrahedral complex is approximately 4/9 times the CFSE for an octahedral complex with the same metal ion and ligands.
Calculation:
Given the CFSE for octahedral [CoCl₆]⁴⁻ is 18,000 cm⁻¹, we can calculate the CFSE for tetrahedral [CoCl₄]²⁻: CFSE (tetrahedral) ≈ (4/9) \(×\) CFSE (octahedral) ≈ (4/9) \(×\) 18,000 cm⁻¹ ≈ 8,000 cm⁻¹
Therefore, the CFSE for tetrahedral [CoCl₄]²⁻ is approximately 8,000 cm⁻¹, making option (4) the correct answer.