Before You Begin
Question #44580
The Question
The racemization of the chiral complex (where en = ethylenediamine) in aqueous solution at proceeds via an intramolecular Bailar twist mechanism, passing through a trigonal prismatic () transition state. The purely steric activation barrier () to deform the ligand backbone into the geometry is . However, the total activation energy () must account for the 'hidden' electronic barrier due to the loss of Crystal Field Stabilization Energy (CFSE) during the geometric transition. Given:
- The octahedral crystal field splitting parameter for is .
- The complex is low-spin in its ground state.
- In the transition state, the d-orbital splitting yields a net stabilization energy of (ignoring pairing energy changes).
- Conversion factor: .
Calculate the total activation energy () for the racemization process.
Options
- A.
- B.
- C.
- D.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Numerical_Precision
Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.
Concepts Behind This Question
Optical Isomerism Logic|Reaction Progress Coordinate|Nucleophilic Addition Sterics
Why This Question Is Tricky
Common Mistakes
Students often assume that the activation barrier for conformational changes or isomerizations (like the Bailar twist) is purely dictated by steric hindrance (ligand backbone deformation).
The 'hidden variable' here is the significant loss of Crystal Field Stabilization Energy (CFSE) when the geometry distorts from to .
For low-spin complexes, this electronic penalty is massive and dominates the activation energy, explaining why these complexes are kinetically inert to racemization compared to their or counterparts.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Coordination Compounds - Class 12 |
| Concept | Optical Isomerism Logic |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |