Premise

Moving from N (atomic number 7) to O (atomic number 8) across Period 2, the general trend predicts IE should increase. Yet IE(N) = 1402 kJ/mol > IE(O) = 1314 kJ/mol. Why?

Step 1 — Write Electronic Configurations

• Nitrogen: $1s^2\,2s^2\,2p^3$ → three 2p electrons, one per orbital (Hund's rule: ↑ ↑ ↑)
• Oxygen: $1s^2\,2s^2\,2p^4$ → four 2p electrons, one orbital has a paired electron (↑↓ ↑ ↑)

Step 2 — Identify the Electron Being Removed

• From N: removing one of three equivalent, unpaired 2p electrons
• From O: removing one of the paired 2p electrons (from the doubly occupied orbital)

Step 3 — Assess Repulsion in the Source Orbital

• In O, the two electrons forced into the same 2p orbital repel each other constantly
• This electron–electron repulsion raises the energy of the paired electron
• A higher-energy electron is easier to remove

Step 4 — Assess Subshell Stability in N

• N's half-filled 2$p^{3}$ configuration: each 2p orbital has exactly one electron → maximum symmetry → minimum repulsion
• This half-filled configuration has extra stability → electron is harder to remove

Step 5 — Quantify the Net Effect

• The extra repulsion in O's 2$p^{4}$ configuration more than cancels the increase in Zeff from N to O
• Net result: removal of the paired 2p electron from O requires less energy than removing an unpaired electron from N

Step 6 — Analogous Reasoning for Be > B

• Be ($2s^2$): removing from fully filled 2$s^{2}$ → extra stability → higher IE
• B ($2s^2\,2p^1$): removing lone 2$p^{1}$ electron → no stability premium, easier

Step 7 — General Rule (NEET Shortcut)

Whenever a subshell is fully filled or half-filled, IE is higher than the immediately following element. This applies to Be/B and N/O in Period 2.