Question
Chromium (Z = 24) has a ground state electronic configuration of [Ar] 3d⁵ 4s¹, not the [Ar] 3d⁴ 4s² that Aufbau’s rule would predict. Why?
This is a NCERT Class 11 exception that appears almost every year — JEE Main 2023 Session 2 had a direct MCQ on Cr and Cu exceptions, and NEET has tested it in objective form multiple times.
Solution — Step by Step
Following Aufbau principle strictly, electrons fill in order of increasing value. For Cr (Z = 24), we’d expect:
This is what the rule gives us — 4 electrons in 3d, 2 in 4s.
Here’s why the rule breaks: 3d and 4s orbitals are very close in energy for elements in the first transition series. The energy gap between them is small enough that other factors — specifically exchange energy — can tip the balance.
When electrons are in different orbitals of the same subshell with parallel spins, they can exchange positions. Each such exchange lowers the total energy slightly.
For 3d⁴ 4s²: the 4 electrons in 3d give exchange pairs within 3d.
For 3d⁵ 4s¹: the 5 electrons in 3d (all with parallel spin, half-filled) give exchange pairs.
That’s 4 extra exchange interactions in the half-filled arrangement — each one releasing a small amount of energy. The total exchange energy gained makes [Ar] 3d⁵ 4s¹ the lower energy (more stable) state.
This is summarized as the half-filled and fully-filled subshell stability rule:
- Half-filled: , — extra stability due to maximum exchange energy and symmetrical electron distribution
- Fully-filled: , — extra stability due to symmetry
Chromium promotes one 4s electron to 3d to achieve the half-filled configuration. Cu (Z = 29) does the same to reach (fully-filled), giving [Ar] 3d¹⁰ 4s¹ instead of [Ar] 3d⁹ 4s².
Ground state configuration of Cr (Z = 24):
In expanded form:
Why This Works
Aufbau principle is a simplified model — it works for most elements but assumes the energy ordering of orbitals stays fixed. In reality, orbital energies shift as you add more electrons and change the nuclear charge. For transition metals, the 3d and 4s orbitals are so close in energy that exchange energy effects become decisive.
The half-filled has one electron in each of the five 3d orbitals, all with the same spin. This spherically symmetric electron cloud also reduces electron-electron repulsion compared to the asymmetric arrangement. Two reasons working together — extra exchange energy plus reduced repulsion — both favour the half-filled state.
This isn’t a “random exception to memorize” — it’s the Aufbau model showing its limits. At higher atomic numbers, the 3d orbital drops in energy below 4s (which is why we write transition metal ions like Fe²⁺ as [Ar] 3d⁶, not [Ar] 3d⁴ 4s²).
Alternative Method — Using the Half-Filled Stability Rule Directly
If you don’t want to count exchange pairs in an exam, use the shortcut: whenever a transition metal is one electron short of a half-filled or fully-filled d subshell, it will “steal” an electron from the nearest s orbital.
- Cr is at Z = 24 → expected → 3d is one short of → promotes 1e from 4s → gets ✓
- Cu is at Z = 29 → expected → 3d is one short of → promotes 1e from 4s → gets ✓
In JEE/NEET MCQs, you’ll often see these two together. Mo (Z = 42) and Pd (Z = 46) are similar exceptions from Period 5 — Mo gives [Kr] 4d⁵ 5s¹ and Pd gives [Kr] 4d¹⁰ 5s⁰. Don’t be surprised if these appear in JEE Advanced options.
Common Mistake
The most common error: writing Cr as [Ar] 3d⁴ 4s² in the exam because “Aufbau says so.” Some students also flip it and write [Ar] 3d⁶ 4s⁰ — that’s wrong in a different direction entirely. The promoted electron comes FROM 4s (not adding to 3d from nowhere) and goes TO 3d. Always verify: total electron count for Cr must be 24. [Ar] = 18, then 3d⁵ (5) + 4s¹ (1) = 6. 18 + 6 = 24. ✓
Another trap: students sometimes extend this logic to ALL transition metals and write wrong configurations for elements like V (Z = 23) or Ti (Z = 22). The exception only kicks in when you’re exactly one electron short of half-filled or fully-filled. Ti is — no exception, Aufbau holds.