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Question #44287
The Question
Consider the hexaaqua complexes and at 298 K. The ionic radius of the central metal undergoes an anomalously large contraction from to . Let be the number of unpaired electrons in , be the magnitude of its Crystal Field Stabilization Energy (CFSE) in terms of (ignoring pairing energy), and be the change in the number of electrons (). Determine the value of .
Options
- A.
0.4
- B.
4.4
- C.
2.4
- D.
1.8
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Exception_Exposure
Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.
Concepts Behind This Question
Atomic vs Ionic Radii|Complex Ion Oxidation States|Color and d-d Transitions
Why This Question Is Tricky
Common Mistakes
This question tests the student's ability to recognize a critical exception in Coordination Chemistry: while is generally a weak field ligand (as seen with ), the increased effective nuclear charge and oxidation state of increase the crystal field splitting energy () such that .
This forces into a low-spin configuration.
Students who blindly apply the spectrochemical series will assume a high-spin configuration, leading to incorrect values for unpaired electrons (), CFSE magnitude (), and occupancy change (), trapping them with the distractor 4.4.
Furthermore, it connects this electronic anomaly directly to the physical phenomenon of ionic radius contraction.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Classification of Elements and Periodicity in Properties - Class 11 Part 1 |
| Concept | Atomic vs Ionic Radii |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |