Concept:Although Fluorine is the most electronegative element, its exceptionally small atomic radius causes severe inter-electronic repulsion, lowering its electron affinity compared to Chlorine.
Formula:$$ \text{EA}_{(\text{Cl})} = -349 \text{ kJ/mol} > \text{EA}_{(\text{F})} = -328 \text{ kJ/mol} $$
Solution:- Fluorine has a very tiny 2p valence shell packed tightly with 7 electrons.
- This creates a "thick, small electronic cloud" (high charge density), leading to intense electrostatic repulsion.
- When an incoming 8th electron tries to enter this cramped space, it faces significant resistance from the existing electrons, which reduces the net energy released.
- Chlorine has a larger 3p orbital, accommodating the extra electron much more easily.
Why other options are incorrect:- Option A: High electronegativity normally increases electron affinity; it is the size constraint that causes the anomaly.
- Option C: Both F and Cl have 7 valence electrons, so this doesn't explain the difference.
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