Exceptions

Exceptions

5 min read Updated Mar 26, 2026

The periodic trends you have learned are powerful, but they are not perfectly smooth. If you graph ionization energy across Period 2, you will see two noticeable dips where the general upward trend stumbles. These dips are not random - they have clear explanations rooted in electron configuration. The MCAT loves testing these exceptions because they separate students who memorized a trend arrow from students who actually understand why the trends exist.

Exception 1: Beryllium (IE = 900 kJ/mol) vs. Boron (IE = 801 kJ/mol)

The general trend says ionization energy should increase from left to right across a period. Beryllium is to the left of boron, so you would expect Be to have a lower IE. But the opposite is true - removing an electron from beryllium requires more energy than removing one from boron.

The explanation lies in their electron configurations:

  • Be: 1s² 2s² - the electron being removed comes from a filled 2s subshell
  • B: 1s² 2s² 2p¹ - the electron being removed comes from a 2p subshell

The 2s orbital is lower in energy and closer to the nucleus than the 2p orbital. Boron’s outermost electron sits in the higher-energy 2p subshell, where it is easier to remove. Additionally, beryllium has a completely filled 2s subshell, which provides a small extra stability. Together, these factors make it easier to ionize boron than beryllium.

Exception 2: Nitrogen (IE = 1402 kJ/mol) vs. Oxygen (IE = 1314 kJ/mol)

Nitrogen sits to the left of oxygen, so the general trend predicts N should have a lower IE. Again, the opposite is true. This exception involves a different principle: the special stability of half-filled subshells.

Compare their electron configurations:

  • N: 1s² 2s² 2p³ - three 2p electrons, each in its own orbital (one in 2px, one in 2py, one in 2pz). This is a half-filled 2p subshell.
  • O: 1s² 2s² 2p⁴ - four 2p electrons, which means one orbital must hold two electrons. That paired electron experiences electron-electron repulsion from its orbital partner.

Nitrogen’s half-filled 2p subshell is unusually stable. Every 2p orbital is singly occupied, minimizing electron-electron repulsion and maximizing exchange energy (a quantum mechanical stabilization). Oxygen, by contrast, has one 2p orbital with two electrons crammed together, and the repulsion between them makes it easier to remove one.

The Underlying Principle: Subshell Stability

Both exceptions stem from the same idea: half-filled and fully-filled subshells are extra stable. Disrupting a half-filled or fully-filled subshell costs more energy than disrupting an incompletely filled one.

This principle also explains the anomalous electron configurations of certain transition metals:

  • Chromium: Expected [Ar] 4s² 3d⁴, actual [Ar] 4s¹ 3d⁵. The atom “borrows” one electron from the 4s to achieve a half-filled 3d subshell.
  • Copper: Expected [Ar] 4s² 3d⁹, actual [Ar] 4s¹ 3d¹⁰. The atom “borrows” one electron from the 4s to achieve a fully-filled 3d subshell.

Visualizing the IE Dips Across Period 2

Here are the first ionization energies for Period 2 elements:

ElementConfigurationIE₁ (kJ/mol)Notes
Li[He] 2s¹520Baseline
Be[He] 2s²900Filled 2s - extra stable
B[He] 2s² 2p¹801Dip - 2p electron easier to remove
C[He] 2s² 2p²1086Resume upward trend
N[He] 2s² 2p³1402Half-filled 2p - extra stable
O[He] 2s² 2p⁴1314Dip - paired electron repulsion
F[He] 2s² 2p⁵1681Resume upward trend
Ne[He] 2s² 2p⁶2081Fully filled - maximum IE

The overall trend is still upward from left to right. The two dips at B and O are local exceptions, not reversals of the entire trend.

Electron Affinity Exceptions

Electron affinity trends are less regular than ionization energy trends, which makes them harder to predict - and less frequently tested in precise comparisons. However, there are a few patterns worth knowing:

Noble gases have electron affinities near zero or positive (endothermic). Their valence shells are full, so an incoming electron would have to enter a new, higher-energy shell. There is no energetic benefit.

Group IIA elements (Be, Mg, Ca) have very low or positive electron affinities. Their s subshells are already filled (ns²), so an incoming electron would need to enter the higher-energy p subshell. This is analogous to the Be vs. B IE exception in reverse.

Nitrogen has a near-zero electron affinity despite being a small, high-Zeff atom. Its 2p subshell is half-filled (2p³), and adding a fourth electron would force pairing in one of the 2p orbitals. The electron-electron repulsion offsets the energy gain from the attractive nuclear charge.

The Big Takeaway

The smooth periodic trend lines you see in textbooks are idealized. Real trends have bumps and dips at predictable locations - wherever a half-filled or fully-filled subshell would be disrupted. On the MCAT, these exceptions are tested more often than the smooth trends themselves, because they require real understanding rather than simple memorization.

When comparing two elements and the trend seems to give the wrong answer, ask yourself: Is a half-filled or fully-filled subshell involved? If yes, the stability of that configuration likely explains the exception.

Which has a higher first ionization energy, beryllium or boron? Explain why this violates the general trend.
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Beryllium (900 kJ/mol) has a higher IE than boron (801 kJ/mol). The general trend predicts IE should increase left to right, making B higher. However, boron's outermost electron occupies the higher-energy 2p subshell, which is easier to remove than beryllium's 2s electron. Additionally, beryllium's filled 2s² subshell has extra stability that resists ionization.
Why does nitrogen have a higher ionization energy than oxygen, even though oxygen is farther to the right in Period 2?
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Nitrogen has a half-filled 2p subshell (2p³) with one electron in each 2p orbital, maximizing exchange energy and minimizing repulsion. Oxygen has a 2p⁴ configuration, forcing two electrons to share one orbital. The electron-electron repulsion from that pairing makes it easier to remove one of oxygen's electrons. Half-filled subshells have extra stability that raises the ionization energy above what the general trend predicts.
Chromium's actual electron configuration is [Ar] 4s¹ 3d⁵ instead of the expected [Ar] 4s² 3d⁴. What principle explains this?
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The extra stability of half-filled subshells. By moving one electron from the 4s to the 3d subshell, chromium achieves a half-filled 3d⁵ configuration. The exchange energy stabilization from having five unpaired d electrons outweighs the cost of leaving the 4s subshell only half-filled. The same principle explains copper's [Ar] 4s¹ 3d¹⁰ configuration (fully-filled 3d).