Ionisation Enthalpy Exceptions

Introduction

The broad first-ionisation trend rises across a period, but two familiar kinds of local dip recur. One occurs when the electron removed first enters a p subshell after a filled s subshell. Another occurs when a p orbital first becomes paired. Understanding the configurations makes these exceptions predictable rather than arbitrary facts to memorize.

Core explanation

Compare beryllium and boron. Be has ground-state configuration 1s²2s²; B has 1s²2s²2p¹. Boron has one more proton, so a naive nuclear-charge-only argument predicts a higher first ionisation enthalpy. Yet the electron removed from B is a 2p electron, while from Be it is a 2s electron. The 2p orbital is generally higher in energy and less penetrating than 2s in these many-electron atoms, so the boron electron is easier to remove. The first ionisation enthalpy therefore falls from Be to B.

The analogous third-period comparison is Mg, [Ne]3s², versus Al, [Ne]3s²3p¹. The first 3p electron in Al is removed more readily than an electron from Mg's filled 3s subshell. The dip does not imply Al has fewer protons or more inner shells; it identifies a change in subshell of the electron removed.

Now compare nitrogen and oxygen. Nitrogen has 2p³: under Hund's rule, the three p orbitals are singly occupied with parallel spins before pairing. Oxygen has 2p⁴, which requires one of those p orbitals to contain a pair. Repulsion between the paired electrons raises the energy of one of them relative to the singly occupied pattern. Removing that electron from O can be easier than removing an electron from N, so IE₁(O) is lower than IE₁(N) despite oxygen's greater Z.

Phosphorus and sulfur show the corresponding third-period pattern with 3p³ and 3p⁴. A p³ configuration is not magically invulnerable; it simply avoids a paired p orbital and can be comparatively stable in the context of this local ionisation comparison. The full ionisation energy is the difference between the neutral and cation energies, so electron rearrangement after removal is part of the exact result.

These exceptions should be embedded in the broader trend. From Li to Ne, first ionisation enthalpy rises substantially overall even though B lies below Be and O below N. A graph with small notches tells a richer story than a monotonic arrow. A careful response names the pair, writes outer configurations, identifies the removed electron and then links its energy or pairing to the observed direction.

The same reasoning is not a license to invent every anomaly in every heavy period. Orbital energies, shielding and relativistic effects vary. Use actual measurements for precise comparisons beyond the standard examples, and do not use a single “half-filled stability” phrase as a substitute for showing the electron arrangement.

Step-by-step reasoning

1. Write the ground-state outer configurations of the two atoms. 2. Mark which electron is removed first from each. 3. Check whether the comparison changes from s to p or from unpaired to paired p occupation. 4. Explain how orbital energy or electron repulsion changes removal cost. 5. Reconcile the local dip with the broad across-period increase.

Visual explanation

Draw three p-orbital boxes for N with one arrow in each. For O, draw one box with an up and down arrow plus one arrow in each remaining box. Circle the paired electron removed from O. In a second panel, draw a lower 2s box filled for Be and a higher 2p box with one electron for B.

Real-world analogy

Removing a person from a crowded two-seat desk may be easier than removing the only person from a separate desk, because the crowded pair interferes with one another. The analogy captures repulsion but not quantum spin or the full energy change.

Real-world example

When checking a data table, a student may see aluminium's first ionisation enthalpy below magnesium's and suspect a printing error. Their configurations reveal the 3s-to-3p change and turn the apparent contradiction into evidence about orbital energies.

Why?

Why is the first p electron less tightly held in these examples? A p orbital at the same n generally penetrates less and is higher in energy than the corresponding s orbital in a many-electron atom. Extra nuclear charge does not always overcome that change.

Common misconception

“Exceptions mean the periodic trend is false.” A trend describes the broad direction. Local deviations are real and often explainable by additional structural variables rather than random failures.

Worked example

Compare N: [He]2s²2p³ and O: [He]2s²2p⁴. Fill three p boxes singly for N. In O, the fourth p electron pairs in one box. Removing one from this pair relieves repulsion and yields O⁺ with a p³ arrangement. Thus IE₁(O) is lower than IE₁(N), despite O having an additional proton. State both competing effects to make the reasoning complete.

Quick check

1. Which element has lower first ionisation enthalpy, Mg or Al? Answer: Al, because its first removed electron is 3p rather than Mg's 3s electron.

Exam focus

For Be–B or Mg–Al, say “s-to-p subshell change.” For N–O or P–S, draw or describe p-orbital pairing. Include the actual inequality and avoid claiming that nuclear charge decreases.

Advanced insight

Ionisation measurements compare total energies of two many-electron species. Orbital diagrams are valuable explanatory models, but the exact magnitude includes electron correlation and relaxation of the remaining cloud. This is why simple shielding rules predict directions more reliably than exact energies.

Summary

The Be–B and Mg–Al dips reflect easier removal from a newly occupied p subshell. The N–O and P–S dips reflect repulsion in a paired p orbital. These local effects refine the broad increase in first ionisation enthalpy across periods.

Practice questions

1. Write the outer configurations responsible for the Mg–Al anomaly. Answer: Mg ends 3s²; Al ends 3s²3p¹. The Al 3p electron is easier to remove than a Mg 3s electron. 2. Why is O's first ionisation enthalpy lower than N's? Answer: O has a paired 2p orbital; removing one paired electron relieves repulsion, whereas N has three singly occupied 2p orbitals. 3. Would “oxygen has fewer protons than nitrogen” explain the dip? Answer: No. Oxygen has more protons; the local dip requires electron-pair and total-energy reasoning.