Concept:Ionization Energy (IE) is inversely proportional to the atomic radius and shielding effect.
Formula:$$ \text{IE} \propto \frac{1}{\text{Atomic Radius}} $$
Solution:- As you move down a group, new principal quantum shells (electron layers) are added.
- This significantly increases the atomic radius, pushing the outermost valence electrons much further from the positive nucleus.
- The combined effect of increased distance and increased inner-shell shielding heavily outweighs the increased proton number, causing the nucleus's grip to weaken, thereby decreasing the ionization energy.
Why other options are incorrect:- Option A & Option B: Both shielding and atomic radius drastically increase down a group; they do not remain constant.
- Option C: While proton number does increase, by itself this would increase IE. It is the overriding increase in atomic radius/shielding that causes IE to decrease.
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