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The (n + l) rule says 4s fills before 3d, so chromium should be 3d⁴4s² and copper 3d⁹4s². Both are wrong, and examiners know students will write them. This post explains why the real configurations are 3d⁵4s¹ and 3d¹⁰4s¹ using a simple pair count, shows which heavier elements behave the same way and which do not, and covers the ion configurations that trip people up.
What does NCERT say, and is it on the syllabus?
The NCERT Class 11 chapter Structure of Atom says that in Cr and Cu, where 4s and 3d "differ slightly in their energies", an electron shifts from 4s to 3d when that leaves the 3d subshell exactly half filled or completely filled. It gives two reasons for the extra stability: a symmetrical distribution of electrons, and larger exchange energy. It also adds a caution that exceptions do exist.
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list "extra stability of half-filled and completely filled orbitals". Check the syllabus for your exam year.
How does counting exchange pairs explain it?
The NCERT Class 12 chapter on the d- and f-block elements says exchange energy is approximately proportional to the total number of possible pairs of parallel spins in the degenerate orbitals. Electrons in different subshells are not counted together, so count within 3d only. For k parallel electrons, the number of pairs is k(k − 1)/2.
Chromium.
3d⁴4s²: four parallel d electrons give 4 × 3 ÷ 2 = 6 pairs.
3d⁵4s¹: five parallel d electrons give 5 × 4 ÷ 2 = 10 pairs.
Copper.
3d⁹4s²: five spin-up electrons give 10 pairs, four spin-down give 6, total 16.
3d¹⁰4s¹: five up give 10, five down give 10, total 20.
In both cases the shifted configuration has four more exchange pairs. That gain only wins because 3d and 4s are so close in energy that moving one electron costs little. The same count also explains Hund's rule: parallel spins in separate orbitals maximise the pairs.
A useful check is unpaired electrons. Chromium 3d⁵4s¹ has six, not the four that 3d⁴4s² would give. Copper has one.
Which other elements break the rule?
The NIST Atomic Spectra Database ground configurations list the free atoms:
Molybdenum, below chromium: 4d⁵5s¹. Silver and gold, below copper: 4d¹⁰5s¹ and 5d¹⁰6s¹. These fit the pattern.
Palladium: 4d¹⁰ with no 5s electron. NCERT notes this exception too.
Niobium: 4d⁴5s¹ and platinum: 5d⁹6s¹. Neither gives a half-filled or filled d subshell, so the "extra stability" story does not explain them.
Tungsten, below molybdenum: 5d⁴6s², the "normal" order.
So the half-filled rule is a guide for the 3d series, not a law. For JEE and NEET, learn Cr and Cu by heart and know that Mo, Ag, Au and Pd are also exceptions.
What is really going on?
Two papers in the Journal of Chemical Education are worth knowing about. Melrose and Scerri (1996) argue that, defined properly, there is nothing paradoxical about 4s being preferred to 3d: the vacant 3d orbitals are not available at lower energy than the occupied 4s. Schwarz (2010) points out that free atoms mostly have configurations ending in ns¹ or ns², but in compounds the dominant configuration of a transition metal of group G is (n − 1)dᴳ with no s electrons, and that free-atom ground states have little meaning in chemistry. That is why the next section matters more for problems.
How do you write the ions?
The d-block chapter states that when d-block elements form ions, ns electrons are lost before (n − 1)d electrons.
Cr⁺: remove the 4s electron, giving [Ar]3d⁵. NIST lists the Cr⁺ ground configuration as 3d⁵.
Cr³⁺: [Ar]3d³, with three unpaired electrons.
Cu⁺: [Ar]3d¹⁰. Cu²⁺: [Ar]3d⁹, one unpaired electron.
Fe²⁺: iron is [Ar]3d⁶4s², so Fe²⁺ is [Ar]3d⁶, not [Ar]3d⁴4s².
NCERT uses this to explain ionisation enthalpies: Cr⁺ is d⁵ and Mn⁺ is 3d⁵4s¹, so removing the next electron is harder from Cr⁺, which would break up a half-filled set.
Common mistakes
Writing Cr as [Ar]3d⁴4s² or Cu as [Ar]3d⁹4s².
Removing 3d electrons before 4s when forming ions.
Extending the chromium pattern to tungsten.
Counting exchange pairs between 3d and 4s electrons.
Forgetting that the total electron count must still match Z: Cr has 24, Cu 29.
For a refresher on the quantum numbers and the (n + l) rule behind all this, see our post on quantum numbers and electron configuration. If a gap in those basics is what keeps tripping you, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing.
A practice routine
Write the configurations of Sc to Zn from memory and circle Cr and Cu.
Do the pair count for Cr and Cu without looking.
Write Cr²⁺, Cr³⁺, Mn²⁺, Fe³⁺, Cu⁺ and Cu²⁺, and count the unpaired electrons in each.
Explain in two sentences why Pd is 4d¹⁰ and why tungsten does not copy chromium.
Next time you meet a d-block question, write the ion's configuration before anything else.
