Concept:The second ionization energy involves removing an electron from a \( 1^+ \) cation. If this removal disrupts a highly stable noble gas core configuration, the energy required will be exceptionally large.
Formula:$$ \text{Na}^+ \text{ Electron Configuration: } 1s^2, 2s^2, 2p^6 \text{ (Noble Gas Core)} $$
Solution:- Sodium (Na) is in Group 1. Its first ionization removes its only valence electron, forming \( \text{Na}^+ \), which is isoelectronic with Neon (a stable octet).
- Removing a second electron requires breaking into this incredibly stable, deeply buried, full inner shell.
- The massive increase in effective nuclear charge acting on this inner shell causes Na to have a spectacularly high second ionization energy.
Why other options are incorrect:- Option A, Option B, Option D: For O, F, and N, the second electron is being removed from an already partially filled valence shell, which does not require breaking a stable noble gas core.
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