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Question #44113
The Question
Consider the stepwise formation of the complex from (aq) and 1,10-phenanthroline () at 298 K. The stepwise formation constants are denoted as , , and . Which of the following correctly represents the trend in these equilibrium constants and identifies the primary electronic/thermodynamic driving force for this specific trend?
Options
- A.
; The third substitution triggers a high-spin to low-spin crossover (), maximizing CFSE and synergic \pi-backbonding into the LUMO.
- B.
; The chelate effect provides a strictly increasing entropic driving force (\Delta S^\circ > 0) for each successive bidentate ligand addition.
- C.
; Successive substitutions are statistically less probable and sterically hindered, which dominates over the strong-field nature of the ligand.
- D.
; The second ligand completes the equatorial plane, maximizing \sigma-donation, while the third ligand suffers from severe steric repulsion.
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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
Successive Equilibrium Constants|Synergic Bonding in Carbonyls
Why This Question Is Tricky
Common Mistakes
Students often memorize that stepwise formation constants strictly decrease () due to statistical and steric factors, or they mistakenly assume the 'Chelate Effect' causes a continuous increase ().
This question tests the 'Chemical Exception' where an abrupt change in the electronic ground state (high-spin to low-spin crossover) driven by cumulative ligand field strength and synergic \pi-backbonding causes a massive, anomalous spike in the final formation constant ().
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Equilibrium - Class 11 Part 1 |
| Concept | Successive Equilibrium Constants |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |