IB Chemistry HLTopic 3 — Classifying the ElementsPaper 1 & 2Core skill~12 min read
Variable Oxidation States in Transition Elements (HL)
Sodium has one oxidation state. Manganese has seven. The difference is not that manganese has more electrons — it is that its outermost electrons all cost about the same to remove, so the atom has no strong preference about where to stop. This page is about the configurations that make that possible.
📘 What you need to know
First-row transition elements fill 4s before 3d, giving configurations of the form [Ar]3dx4s2.
Chromium and copper are exceptions: Cr is [Ar]3d54s1 and Cu is [Ar]3d104s1, because a half-filled or fully filled d sublevel is more stable.
When ions form, 4s electrons are removed first, because once occupied the 4s becomes the outermost subshell.
+2 is available to all of them by removing the two 4s electrons.
Further states come from removing 3d electrons as well, which costs little because 3d and 4s are so close in energy.
The maximum oxidation state equals the total number of 4s and 3d electrons — reaching +7 at manganese.
Successive ionisation energies for these elements rise gradually with no huge jumps early on, which is the evidence for all of this.
The configurations, and the two rebels
Run through the first row and the pattern is boringly regular: scandium [Ar]3d14s2, titanium [Ar]3d24s2, vanadium [Ar]3d34s2, and so on. Then chromium breaks it, and later copper breaks it again.
Both exceptions do the same thing for the same reason. If you can explain chromium you can explain copper, and vice versa.
Forming ions: 4s goes first
This is the point that trips people up more than any other in the d-block. The 4s subshell fills before the 3d, but it also empties before the 3d.
There is no contradiction. Once electrons occupy the 4s, it becomes the outermost subshell, and it is pushed slightly higher in energy than the 3d by repulsion from the electrons already there. Ionisation always removes the outermost, highest-energy electrons, so 4s electrons leave first.
The rule to write down
fill 4s first → but remove 4s first
So iron, [Ar]3d64s2, becomes Fe2+ = [Ar]3d6, and then Fe3+ = [Ar]3d5. Notice that Fe3+ ends up with a half-filled d sublevel, which is part of why the 3+ state of iron is so stable.
Write out the atom’s configuration first, every single time, then take electrons off it. Trying to jump straight to the ion’s configuration is where the mistakes live.
Which states each element reaches
Every one of these metals can in principle reach +2 by losing its two 4s electrons. The chart shows the states you actually meet in reactions and in exam questions.
Two more patterns are worth noticing. The higher states are more common on the left of the row and become harder to reach on the right, because the growing nuclear charge holds the 3d electrons more tightly. And ions in states of +3 and above are strongly polarising — small, highly charged, and therefore capable of distorting nearby anions, which gives their compounds noticeable covalent character.
The ionisation energy evidence
How do we know 4s and 3d really are close in energy? Look at the successive ionisation energies of a transition element. For sodium there is an enormous jump after the first electron, because the second must come from a full inner shell. For titanium or vanadium the first four or five values rise steadily with no dramatic jump at all.
That smooth rise means the electrons being removed are all coming from a similar energy level — which is exactly what “4s and 3d are close together” predicts, and it is why several oxidation states are chemically accessible instead of just one.
Worked examples
WORKED EXAMPLE
Write the full electron configuration of Fe3+.
Step 1: the atom first
Iron has 26 electrons.
Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²Step 2: remove the 4s electrons
Two go from 4s, giving Fe2+.
Step 3: remove one more, now from 3dFe³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵[Ar] 3d⁵ — a half-filled d sublevel, which is why Fe3+ is so stable23 electrons in total; check by adding the superscripts
WORKED EXAMPLE
Explain why chromium’s configuration is [Ar]3d54s1 rather than [Ar]3d44s2, and give the configuration of Cr3+.
Step 1: explain the exception
Promoting one 4s electron to 3d gives a half-filled d sublevel with one electron in each of the five orbitals.
Step 2: why that is preferred
The 4s and 3d are so close in energy that the extra stability of the half-filled arrangement more than pays for the promotion.
Step 3: form the ion
Remove the single 4s electron first, then two 3d electrons.
[Ar] 3d⁵ 4s¹ → [Ar] 3d³Cr3+ is [Ar] 3d3chromium only has one 4s electron to lose, so the second and third come from 3d
WORKED EXAMPLE
Explain why manganese can reach an oxidation state of +7 but nickel cannot.
Step 1: count the available electronsMn: [Ar] 3d⁵ 4s² → 5 + 2 = 7 electrons outside the argon coreNi: [Ar] 3d⁸ 4s² → 8 + 2 = 10 electronsStep 2: why nickel has fewer options despite having more electrons
Nickel has a higher nuclear charge, so its 3d electrons are held much more tightly and removing more than two costs far too much energy.
Step 3: the limit
The maximum state is set by how many electrons can realistically be removed, not by how many exist.
Mn uses all seven of its outer electrons to reach +7; Ni is effectively limited to +2this is why the highest oxidation states appear on the left of the row and fade towards the right
💡 Exam tip
Say “4s fills first but is removed first” in any ion question. It shows you know the point that is being tested.
For chromium and copper, the marking phrase is half-filled or fully filled d sublevel is more stable.
Always write the neutral atom’s configuration before forming the ion.
Both orderings (3d before 4s, or 4s before 3d) are accepted in written configurations. Be consistent.
Check your ion configuration by counting electrons: Fe3+ must have 23.
For “why variable oxidation states?”, the answer is the similar energies of 4s and 3d — not simply “because it is a transition metal”.
⚠ Common mix-up
Writing Fe2+ as [Ar]3d44s2. The 4s electrons go first, so it is [Ar]3d6.
Giving chromium [Ar]3d44s2. That is the expected answer, not the real one.
Assuming every element in the row has an exception. Only chromium and copper do.
Thinking more electrons means a higher maximum oxidation state. Nuclear charge matters more.
Confusing 3d5 stability with 3d5 being full. Half filled is five electrons in five orbitals; full is ten.
Removing electrons from the argon core. Those are far too tightly held to be involved.
Up next: Colour in Transition Metal Complexes (HL) — the last piece. Those partly filled d orbitals split apart when ligands arrive, and the gap that opens up turns out to be exactly the size of a photon of visible light.
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