Question
Write the electronic configurations of all elements from atomic number 1 to 30 using the Aufbau principle, Pauli’s exclusion principle, and Hund’s rule. Also explain the anomalous configurations of Chromium (Z=24) and Copper (Z=29).
Solution — Step by Step
Before writing a single configuration, the three rules are non-negotiable for board exams and JEE:
- Aufbau Principle: Fill orbitals in increasing order of energy — 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p
- Pauli’s Exclusion Principle: No two electrons in the same atom can have all four quantum numbers identical. Each orbital holds max 2 electrons with opposite spins.
- Hund’s Rule: Within a subshell, electrons occupy orbitals singly first, then pair up. All singly occupied orbitals have parallel spins.
The energy order for filling: use the (n + l) rule — lower (n + l) fills first. If tie, lower n fills first.
These follow Aufbau without exception:
| Z | Element | Configuration |
|---|---|---|
| 1 | H | 1s¹ |
| 2 | He | 1s² |
| 3 | Li | [He] 2s¹ |
| 4 | Be | [He] 2s² |
| 5 | B | [He] 2s² 2p¹ |
| 6 | C | [He] 2s² 2p² |
| 7 | N | [He] 2s² 2p³ |
| 8 | O | [He] 2s² 2p⁴ |
| 9 | F | [He] 2s² 2p⁵ |
| 10 | Ne | [He] 2s² 2p⁶ |
| 11 | Na | [Ne] 3s¹ |
| 12 | Mg | [Ne] 3s² |
| 13 | Al | [Ne] 3s² 3p¹ |
| 14 | Si | [Ne] 3s² 3p² |
| 15 | P | [Ne] 3s² 3p³ |
| 16 | S | [Ne] 3s² 3p⁴ |
| 17 | Cl | [Ne] 3s² 3p⁵ |
| 18 | Ar | [Ne] 3s² 3p⁶ |
WHY does 4s fill before 3d? Because at these atomic numbers, the 4s orbital has lower energy than 3d. The (n + l) values: 4s gives 4+0=4, while 3d gives 3+2=5. Lower sum fills first.
| Z | Element | Configuration |
|---|---|---|
| 19 | K | [Ar] 4s¹ |
| 20 | Ca | [Ar] 4s² |
| 21 | Sc | [Ar] 3d¹ 4s² |
| 22 | Ti | [Ar] 3d² 4s² |
| 23 | V | [Ar] 3d³ 4s² |
| 24 | Cr | [Ar] 3d⁵ 4s¹ ← anomaly |
| 25 | Mn | [Ar] 3d⁵ 4s² |
| 26 | Fe | [Ar] 3d⁶ 4s² |
| 27 | Co | [Ar] 3d⁷ 4s² |
| 28 | Ni | [Ar] 3d⁸ 4s² |
| 29 | Cu | [Ar] 3d¹⁰ 4s¹ ← anomaly |
| 30 | Zn | [Ar] 3d¹⁰ 4s² |
For Cr (Z=24), the expected configuration would be [Ar] 3d⁴ 4s². But the actual is [Ar] 3d⁵ 4s¹.
WHY? A half-filled d subshell (3d⁵) is extra stable because of:
- Symmetrical charge distribution — all five 3d orbitals have one electron each
- Maximum exchange energy — electrons with parallel spins in different orbitals lower the total energy through exchange interactions
So one electron “jumps” from 4s to 3d to achieve this stability. The energy gained by half-filling outweighs the cost of promoting an electron.
For Cu (Z=29), expected is [Ar] 3d⁹ 4s². Actual is [Ar] 3d¹⁰ 4s¹.
Same logic — a completely filled d subshell (3d¹⁰) is also extra stable for the same symmetry and exchange energy reasons.
Remember it as: half-filled and completely filled subshells are extra stable. This is a high-weightage concept — expect 1-2 MCQs in JEE Main every year directly on Cr and Cu. The 2023 JEE Main Session 1 had a direct question asking the number of unpaired electrons in Cr.
Why This Works
The stability of half-filled (d⁵) and fully filled (d¹⁰) configurations comes from two factors acting together. First, when all orbitals in a subshell are equally occupied, the electron density is spread symmetrically — there’s no “lopsided” repulsion from one overcrowded orbital. Second, exchange energy is maximised when electrons with the same spin are distributed across orbitals — more such pairs means more energy released.
This is not just a trick to memorise. It is a real quantum mechanical phenomenon. The energy difference between [Ar] 3d⁴ 4s² and [Ar] 3d⁵ 4s¹ in chromium has been measured experimentally — the half-filled configuration sits lower in energy.
For NCERT purposes, the explanation “extra stability due to half-filled/fully filled subshells” is sufficient. For JEE Advanced, knowing about exchange energy and symmetrical distribution gives you the full picture.
Alternative Method
Using the shorthand notation efficiently in exams:
Rather than writing out each configuration from scratch, build on noble gas cores and count the remaining electrons:
For Fe (Z=26): Nearest noble gas below it is Ar (Z=18). Remaining electrons = 26 − 18 = 8. Fill 4s first (2 electrons), then 3d (6 electrons). So [Ar] 3d⁶ 4s².
For Ni (Z=28): 28 − 18 = 10 remaining. 4s gets 2, 3d gets 8. So [Ar] 3d⁸ 4s².
This method is faster under exam pressure and reduces counting errors.
In board exams, always write 3d before 4s in the final answer — i.e., write [Ar] 3d⁶ 4s² not [Ar] 4s² 3d⁶. NCERT consistently uses this order and CBSE marking schemes expect it.
Common Mistake
The most common error: writing Cr as [Ar] 3d⁴ 4s² and Cu as [Ar] 3d⁹ 4s². Students know the rule about anomalies but write the wrong anomaly — they show 3d⁴ instead of 3d⁵, or 3d⁹ instead of 3d¹⁰. Always remember: the electron moves FROM 4s TO 3d (not the other way), and it moves to reach exactly half-filled or fully filled. So for Cr: 4s drops from 2 to 1, and 3d rises from 4 to 5. For Cu: 4s drops from 2 to 1, and 3d rises from 9 to 10.
The final configurations to commit to memory: