Concept:Electron affinity generally decreases down a group. However, Fluorine is an anomaly due to its extremely small atomic size, which causes high inter-electronic repulsion, dropping its EA below that of Chlorine.
Formula:$$ \text{EA Trend: Cl (Highest)} > \text{F} > \text{Br} > \text{I} $$
Solution:- Chlorine has the highest electron affinity because its 3p orbital is large enough to comfortably accept an extra electron without excessive repulsion.
- Fluorine is highly electronegative but tiny; adding an electron to its crowded 2p shell causes repulsion, releasing less energy than Chlorine.
- After Chlorine, the standard trend resumes: as size increases (Br, I), the nucleus is further away, and EA decreases normally.
Why other options are incorrect:- Option A: Assumes a strict, unbroken trend from top to bottom, ignoring the known Fluorine anomaly.
- Option D: Shows an increasing trend, which opposes the fundamental principle of atomic radius expansion down a group.
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