[VISUAL] The table above shows the first ionization energy and atomic radius of several elements. Wh — Atomic Structure Chemistry Question

STEPS:

1. Identify the periodic trend for first-ionization energy: Across a period of the periodic table (from left to right, such as Period 2 from boron to neon), the first-ionization energy generally increases.
2. Explain the physical reason for the general trend: As you move from left to right across a period, protons are added to the nucleus while core electron shielding remains relatively constant. This increases the effective nuclear charge ($Z_{\text{eff}}$), pulling the valence electrons closer and holding them more tightly, as shown by the steadily decreasing atomic radii in the table.
3. Pinpoint the specific deviation for oxygen: In the provided data, first-ionization energy increases as expected from boron ($801\text{ kJ/mol}$) to carbon ($1086\text{ kJ/mol}$) to nitrogen ($1400\text{ kJ/mol}$). However, at oxygen, the ionization energy unexpectedly *drops* to $1314\text{ kJ/mol}$ before rising again at fluorine ($1680\text{ kJ/mol}$).
4. Examine the valence electron configurations of nitrogen and oxygen: To understand this deviation, analyze how their valence electrons are arranged in the $2p$ subshell:
* Nitrogen ($1s^2 2s^2 2p^3$): According to Hund's rule, nitrogen's three $2p$ valence electrons occupy three separate $2p$ orbitals singly (one electron per orbital): $\underline{\uparrow}\ \underline{\uparrow}\ \underline{\uparrow}$.
* Oxygen ($1s^2 2s^2 2p^4$): Oxygen has four valence $2p$ electrons, meaning one $2p$ orbital must contain a pair of electrons with opposite spins: $\underline{\uparrow\downarrow}\ \underline{\uparrow}\ \underline{\uparrow}$.
5. Relate electron-electron repulsion to ionization energy: In oxygen's doubly occupied $2p$ orbital, two negatively charged electrons are forced to share the same region of physical space. This physical proximity results in interelectronic repulsion (or spin-pairing repulsion). This mutual repulsion destabilizes the electron pair, making it energetically easier (requiring less energy) to remove one of these paired electrons compared to removing a singly occupied electron from nitrogen's stable, half-filled $2p$ subshell. This makes Option C the correct answer.

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WHY_OTHERS_WRONG:

• A is incorrect: While the atomic radius of oxygen ($73\text{ pm}$) is indeed greater than that of fluorine ($72\text{ pm}$), this is the normal periodic trend and does not explain why oxygen's ionization energy is *lower* than that of nitrogen (which has a larger atomic radius of $75\text{ pm}$).
• B is incorrect: While the atomic radius of oxygen ($73\text{ pm}$) is indeed less than that of nitrogen ($75\text{ pm}$), a smaller radius means the outermost electron is closer to the nucleus and should experience a stronger Coulombic attraction. If radius were the dominant factor, oxygen would have a *higher* first-ionization energy than nitrogen, so this fact cannot explain the deviation.
• D is incorrect: Electrons are negatively charged particles. Because like charges repel, there is electrostatic repulsion—never attraction—between paired electrons sharing an orbital.