Concept:Ionization energy generally increases across a period from left to right due to increasing effective nuclear charge and decreasing atomic radius.
Formula:$$ \text{IE Trend (Period 2): Li} < \text{B} < \text{Be} < \text{C} < \text{O} < \text{N} < \text{F} < \text{Ne} $$
Solution:- Boron (B) is further to the left in Period 2 compared to C, N, and O.
- It has a larger atomic radius and a lower effective nuclear charge than the others.
- Therefore, its outermost electron is held the least tightly, requiring the lowest energy to remove.
Why other options are incorrect:- Option A, Option B, Option C: These elements lie further to the right, possessing smaller radii and stronger nuclear grips on their electrons, hence higher ionization energies.
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