Concept:Electronegativity is driven by the nucleus's ability to attract shared electrons, which is maximized when the atom is small and the nuclear charge is high without excessive shielding.
Formula:$$ \text{EN} \propto \frac{Z_{\text{eff}}}{\text{Atomic Radius}} $$
Solution:- Fluorine is at the far right of Period 2 (excluding the noble gas Neon).
- It has a high effective nuclear charge (+9 protons pulling on just 2 electron shells).
- Because it only has two shells, its atomic radius is extremely small, meaning the shared bonding electrons sit very close to the positive nucleus.
- This combination of small size and high nuclear charge gives it the strongest pull on electrons (highest electronegativity).
Why other options are incorrect:- Option A: It does not have a complete outermost shell (it needs 1 more electron).
- Option B & Option C: These are consequences of its chemistry, not the fundamental causes of its electronegativity.
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