Shielding and Effective Nuclear Charge predicts that first ionization energy rises steadily across a period. The data almost agree, with two dips in every period: boron is below beryllium, and oxygen is below nitrogen. Your Orbital Diagrams and Magnetism explain both.
Key idea
Group 13 dip (B < Be, Al < Mg): the electron removed from boron is the first one in a new, higher-energy $p$ subshell, partly shielded by the filled $s$ subshell. Group 16 dip (O < N, S < P): the electron removed from oxygen is the first paired electron in the $p$ subshell, and the repulsion from its partner makes it easier to remove.
Why it works
Beryllium loses a $2s$ electron; boron loses a $2p$ electron. A $2s$ electron spends more time close to the nucleus than a $2p$ electron does (it penetrates the $1s$ core), so the $2p$ electron is higher in energy and the boron atom's extra proton doesn't make up for it. In nitrogen, the three $2p$ electrons are in three separate orbitals (Hund's rule). Oxygen's fourth $2p$ electron must share an orbital, and two electrons in one orbital repel strongly. Removing one of them relieves that repulsion, so it costs less than removing one of nitrogen's unpaired electrons. After each dip, the rise resumes, because $Z_\text{eff}$ keeps growing.
Worked example: Ranking with a dip
Rank Be, B and C by first ionization energy.
- Across period 2 the general trend is up, so C should be highest.
- Boron's electron is the first $2p$ electron, higher in energy than beryllium's $2s$ electrons, so B < Be.
- Data: B 801, Be 900, C 1087 kJ/mol.
Answer: B < Be < C
Worked example: Explaining O against N
Nitrogen's first ionization energy is 1402 kJ/mol and oxygen's is 1314 kJ/mol. Explain.
- N: $2p$ ↑ ↑ ↑, three unpaired electrons. O: $2p$ ↑↓ ↑ ↑.
- The electron removed from oxygen shares an orbital with another electron, and that pair repels.
- The repulsion more than offsets oxygen's extra proton, so its first electron is easier to remove.
Answer: Paired-electron repulsion in oxygen's $2p^4$.
Worked example: Period 3
Which has the higher first ionization energy, phosphorus or sulfur?
- P: $[\text{Ne}]\,3s^2 3p^3$, three unpaired $3p$ electrons. S: $3p^4$, with one pair.
- The same dip as N and O: sulfur's paired electron is easier to remove.
Answer: Phosphorus (1012 against 1000 kJ/mol).
Common mistake
Applying "ionization energy rises across a period" to every pair, so ranking B above Be or O above N. The trend describes the overall climb, not each step. Check every pair that crosses from group 2 to 13 or from group 15 to 16: those two steps go down.