Electron pair repulsion in oxygen's p subshell

Steps:

• Nitrogen's electron configuration is 1s² 2s² 2p³, with three unpaired electrons in p orbitals following Hund's rule.
• Oxygen's configuration is 1s² 2s² 2p⁴, placing two electrons in one p orbital, causing repulsion.
• This repulsion in oxygen makes it easier to remove an electron compared to nitrogen's stable half-filled p subshell.
• Thus, oxygen's first ionization energy (1310 kJ/mol) is lower than nitrogen's (1402 kJ/mol).

Why C is correct:

• According to Coulomb's law, paired electrons in the same orbital experience greater repulsion, reducing the effective nuclear attraction and lowering ionization energy.

Why the others are wrong:

• A: Oxygen has atomic number 8 and nitrogen 7, so oxygen's nuclear charge is greater, which would increase ionization energy if other factors were equal.
• B: Both atoms have outer electrons in the 2p subshell; the difference is in pairing, not subshell.
• D: Both have identical inner electron shielding from 1s and 2s orbitals; oxygen has no additional shielding.

Final answer: C