IB Chemistry HL Topic 3 — Classifying the Elements Paper 1 & 2 Trends ~11 min read

Ionisation Energy Trends Across a Period (HL)

Ionisation energy rises across a period. Except twice, where it drops instead. Those two dips are not annoying exceptions to be memorised — they are the experimental evidence that subshells exist, and an examiner who asks about them is really asking whether you understand where electrons actually live.

📘 What you need to know

The graph you should be able to sketch

First ionisation energy, hydrogen to sodium 0 500 1000 1500 2000 2500first ionisation energy / kJ mol H He Li Be B C N O F Ne Na dip 1 dip 2atomic numberPeaks at the noble gases, troughs at the alkali metals, two small dips between. Values are the accepted first ionisation energies in kJ per mole.
The big features are easy: a full shell is hard to break into, and a lone electron in a fresh shell is easy to remove. The two small dips are where the marks are.

Dip 1: beryllium to boron

Beryllium is 1s22s2 and boron is 1s22s22p1. Boron has one more proton, so the trend says its ionisation energy should be higher. It is lower — 801 against 899 kJ mol−1.

The reason is which orbital the electron comes from. Beryllium’s outermost electron is in a 2s orbital; boron’s is in a 2p. A 2p orbital is higher in energy than a 2s and sits slightly further from the nucleus on average, and it is also very slightly shielded by the 2s electrons underneath it. Both effects make boron’s outer electron easier to remove, and together they outweigh the extra proton.

Dip 2: nitrogen to oxygen

Here both electrons come from a 2p orbital, so the previous explanation cannot apply. Something else must be going on — and it is pairing.

Why oxygen is easier to ionise than nitrogen Same subshell, same shell — but not the same company.N: 2p³ O: 2p⁴ ↑↓ all three electrons unpaired no extra repulsion one orbital holds a pair the pair repel each otherso oxygen’s fourth 2p electron is easier to removeElectrons fill orbitals singly first, and only pair up when they must. Removing the paired electron actually relieves the repulsion.
Nitrogen’s half-filled subshell is often called “extra stable”. Be careful with that phrase in an answer — the mark is for the repulsion between the paired electrons in oxygen, which is the mechanism rather than the label.

What the dips prove

Imagine a model of the atom with shells but no subshells. Across period 2 you would be adding eight electrons to one shell while the nuclear charge climbed steadily, so the graph would rise smoothly with no interruptions at all.

It does not. It dips at exactly the two places where something changes about the type of orbital being filled — when the 2p opens, and when pairing starts within it. That pattern is only explicable if the shell is divided into subshells and orbitals, which is why these two dips are quoted as evidence for the existence of sublevels.

The same dips reappear in period 3, from magnesium to aluminium (3s to 3p) and from phosphorus to sulfur (pairing in 3p). If a question gives you period 3 data, apply exactly the same two arguments.

Worked examples

WORKED EXAMPLE

Explain why the first ionisation energy of aluminium is lower than that of magnesium. [3]

Mark 1: give both configurations Mg: [Ne] 3s²   Al: [Ne] 3s² 3p¹ Mark 2: identify the orbital involved The electron removed from aluminium comes from a 3p orbital, which is higher in energy than the 3s. Mark 3: complete the argument It is further from the nucleus on average and slightly shielded by the 3s electrons, so it is held less strongly. Less energy is needed, despite aluminium having one more proton the phrase “despite the higher nuclear charge” shows you know why this is a surprise
WORKED EXAMPLE

Sulfur has a lower first ionisation energy than phosphorus. Explain, using orbital diagrams. [3]

Mark 1: phosphorus P: 3p³ — three orbitals, one electron in each, all unpaired Mark 2: sulfur S: 3p⁴ — one orbital now contains a pair Mark 3: the consequence The two electrons sharing that orbital repel each other, so one of them is less tightly held. Less energy is needed to remove it, so sulfur’s value is lower say “electron–electron repulsion within the same orbital” — that exact idea is the mark
WORKED EXAMPLE

Why is there such a large drop in first ionisation energy from neon to sodium?

Step 1: compare the electron being removed Neon’s comes from the 2nd shell; sodium’s comes from the 3rd. Step 2: distance Sodium’s outer electron is a whole shell further from the nucleus. Step 3: shielding It is also shielded by all ten inner electrons, which neon’s is not. Distance and shielding both jump at once, so the value falls sharply this drop is what defines the start of a new period on the graph

💡 Exam tip

⚠ Common mix-up

Up next: Characteristic Properties of Transition Elements (HL) — we move into the d-block, where the 4s and 3d subshells sit so close in energy that a whole family of unusual behaviour falls out of it.

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