Orbital Energy Ordering in Multielectron Atoms

Why does potassium's 19th electron go into the 4s orbital? It seems it should go into the empty, seemingly "closer" 3d orbital instead. A hydrogen atom would answer confidently: 3d sits below 4s. That's because in hydrogen, energy depends only on the principal quantum number $n$ . Potassium disagrees. The periodic table is built entirely on that disagreement. Every s-block metal shows this pattern. Every transition series shows it too. So does every anomaly in ionization energy that a student is told to "just memorize." All of this traces back to one fact. In atoms with more than one electron, orbital energy depends on both $n$ and $l$ . It does not depend on $n$ alone. This matters because the entire architecture of the periodic table depends on it. Why period 4 exists where it does, why transition metals appear after the alkaline earths, why lanthanides hide in a footnote row — all of this is a direct consequence. It stems from the $s$ -before- $d$ energy crossover that first appears at $n=3,4$ . Get this ordering wrong and several things become impossible. You cannot predict a single electron configuration. You cannot rationalize why Cr and Cu break the "expected" pattern. And you cannot explain why transition metals lose their $s$ electrons first upon ionization, even though those electrons filled first. The big picture is a competition between two quantum numbers. Both act on the same nuclear charge. The principal quantum number $n$ sets the gross size and energy scale, exactly as in hydrogen. The angular momentum quantum number $l$ sets something different. It determines how effectively an electron penetrates the shielding cloud of inner electrons. This penetration lets the electron feel the bare nuclear charge close to the nucleus. Low- $l$ orbitals (s, then p) have radial density that sneaks in close to the nucleus. They do this despite the centrifugal barrier. High- $l$ orbitals (d, f) are different: they get pushed outward and shielded more thoroughly. The result is that energy is not simply ordered by $n$ . Instead, it follows a combination of $n$ and $l$ . This is captured, with useful but imperfect accuracy, by the empirical Madelung rule (also called the $n+l$ rule). This article builds the physical picture of penetration-driven energy splitting. It states the Madelung ordering rule precisely. It derives why the rule works, using radial probability distributions and effective nuclear charge. It works through real orbital-filling problems (K, Sc, Cu, Pd). It also catalogs exactly where the rule succeeds and where it visibly fails. It assumes you already understand Effective Nuclear Charge and Penetration and Shielding. It sets up the reasoning you will use immediately in Electron Configurations and the Aufbau Principle.

Key takeaways

In hydrogen, energy depends only on $n$ ; in multielectron atoms it depends on both $n$ and $l$ because of penetration.
Lower- $l$ orbitals (s > p > d > f) penetrate closer to the nucleus, feel a higher effective nuclear charge $Z_{\text{eff}}$ , and drop lower in energy for the same $n$ .
The empirical Madelung ( $n+l$ ) rule: orbitals fill in order of increasing $n+l$ ; ties broken by lower $n$ first.
This produces the famous crossovers: 4s fills before 3d, 6s before 4f before 5d, etc.
The rule is a good approximation, not an exact law — it can be violated (Cr, Cu, Pd, and many lanthanides/actinides) once electron-electron repulsion and exchange stabilization are included.
Ionization reverses apparent filling order: once 3d is occupied, it drops below 4s in energy, so 4s electrons are removed first from transition metal atoms.
Key idea: orbital energy in real atoms is set by how deeply an electron's charge density penetrates the shielding cloud, and the Madelung $(n+l)$ rule is the empirical bookkeeping device that encodes this penetration hierarchy.

The core insight

Orbital energy in a multielectron atom is not a property of $n$ alone; it is a property of the radial penetration of that orbital's electron density into the region of poorly-shielded nuclear charge, and $l$ controls that penetration as strongly as $n$ does.

Where this lives today

The $n,l$ -dependent orbital energy ordering is not a historical curiosity confined to introductory chemistry — it is the load-bearing assumption behind essentially every prediction of transition-metal chemistry. Coordination chemistry's entire logic for why transition metals lose $s$ electrons before $d$ electrons upon oxidation (central to redox catalysis, from cytochrome oxidases to industrial hydrogenation catalysts) traces directly to the s/d crossover analyzed here. Crystal- and ligand-field theory, used to explain the colors, magnetism, and reactivity of every transition-metal complex, is built on the premise that d orbitals are valence orbitals in the ion, even though s filled first in the neutral atom — a distinction that only makes sense once the ionization-reversal mechanism is understood. In materials science, the anomalous $4d^{10}5s^0$ configuration of palladium underlies its unique catalytic behavior in hydrogenation and Suzuki-type cross-coupling catalysis, where Pd's readily accessible, fully-populated 4d shell and empty 5s/5p orbitals make oxidative addition and reductive elimination steps energetically favorable in a way that would not occur for a naively "expected" $4d^85s^2$ configuration. Lanthanide near-degeneracies between 4f and 5d orbitals (the same physics driving the Ce/La anomalies) are exploited in phosphors, MRI contrast agents (Gd complexes), and high-strength magnets (Nd–Fe–B), all of which depend sensitively on which orbitals are energetically accessible for bonding and magnetic coupling. Modern electronic-structure software (density functional theory codes used throughout computational chemistry and materials design) must correctly reproduce these penetration-driven orbital energy crossovers to get transition-metal thermochemistry right at all — a persistent, still-active challenge in DFT functional development, since many standard functionals systematically mis-order near-degenerate $s/d$ or $f/d$ states.

:::quiz[Why is a Sc²⁺ ion configuration [Ar]3d¹ rather than [Ar]4s¹, even though 4s filled before 3d in neutral Sc?]

Because 4s electrons are always removed first once 3d becomes populated and depresses below 4s in energy ✓
Because the Madelung rule applies only to anions
Because 3d orbitals become unstable upon ionization
Because Hund's rule forbids single 4s occupancy in cations

Ask the tutor: Why does the 4s/3d crossover happen specifically between period 3 and period 4, rather than somewhere else in the periodic table?

The physical picture

Start from the one-electron (hydrogenic) result everyone already knows:

E_n = -\frac{Z^2}{n^2}\,\text{Ry}

This depends only on $n$ ; all orbitals of a given $n$ (2s and 2p, 3s/3p/3d, …) are exactly degenerate for a single electron orbiting a bare nucleus. That degeneracy is a special feature of the pure $1/r$ Coulomb potential — it is not generic quantum mechanics, it is a symmetry accident of hydrogen (a consequence of the extra conserved Laplace–Runge–Lenz vector). The moment you add a second electron, that accident is destroyed. In a multielectron atom, each electron moves in the field of the nucleus plus the averaged, non-uniform charge cloud of every other electron. An electron with quantum numbers $n,l$ therefore does not feel a clean $-Z/r$ potential; it feels an effective potential that looks like $-Z/r$ very close to the nucleus (where other electrons rarely intrude) and like $-1/r$ far away (where the nucleus is almost fully screened by $Z-1$ other electrons). Whether an orbital's energy sits high or low depends on how much of its probability density lives in that inner, poorly-screened region — this is exactly the concept of penetration built on top of Effective Nuclear Charge. Radial probability distributions make this concrete. The radial distribution function $P(r) = r^2 R_{n,l}(r)^2$ for a given $n$ shifts outward and develops more nodes as $l$ decreases, but crucially, lower- $l$ orbitals retain a small inner lobe close to the nucleus that higher- $l$ orbitals of the same $n$ lack entirely. A 3s electron has two inner maxima tucked inside the "main" 3s lobe; a 3p electron has one inner lobe; a 3d electron has none — its density is a single lobe pushed entirely outward, held off the nucleus by the strong centrifugal barrier $l(l+1)\hbar^2/2m_er^2$ that grows with $l$ . So for the same $n$ , the ordering of average distance from the nucleus is

\langle r \rangle_{ns} < \langle r \rangle_{np} < \langle r \rangle_{nd} < \langle r \rangle_{nf}

and the ordering of penetration (time spent very close to the nucleus, inside the shielding cloud of core electrons) is the same: s penetrates most, f penetrates least.

Figure

Radial probability density: penetration difference between 3s, 3p, 3d

The physical consequence: because 3s and 3p have finite density inside the shielding shell formed by inner ( $n=1,2$ ) electrons, they occasionally feel nearly the full nuclear charge $Z$ , not the heavily reduced $Z_{\text{eff}}$ that applies far out. Averaged over the whole orbital, this pulls their energy down substantially relative to what a naive $-Z_{\text{eff}}^2/n^2$ formula using only the outer-lobe screening would predict. 3d, having no inner lobe, is screened almost as effectively as classical electrostatics would suggest, and sits much higher. This penetration-driven splitting within a shell grows with $Z$ (more nuclear charge to differentially feel) and, critically, grows as more electrons are added to inner shells. That is why the s/d energy gap is small and reversible in neutral early transition-metal atoms but becomes large and one-directional once the d-subshell starts filling with real electrons — those new d electrons shield later s electrons only partially (since d orbitals are themselves diffuse and poor screeners), while s electrons penetrate in and shield the d electrons quite effectively. This asymmetric mutual shielding is the mechanistic root of the famous 4s-before-3d, but 3d-below-4s-once-occupied puzzle, worked out explicitly below.

Figure

Energy ordering by n+l: the Madelung diagonal rule

The core ideas

The single-electron hydrogenic energy law depends on $n$ only:

E_n^{\text{H}} = -\frac{Z^2}{n^2}\,\text{Ry}, \qquad \text{Ry} = 13.6\ \text{eV}

where: $Z$ is nuclear charge, $n$ is the principal quantum number. This is the baseline that multielectron atoms deviate from. For a multielectron atom, replace $Z$ by an orbital- and shell-dependent effective nuclear charge:

E_{n,l} \approx -\frac{Z_{\text{eff}}(n,l)^2}{n^2}\,\text{Ry}, \qquad Z_{\text{eff}}(n,l) = Z - S(n,l)

where: $S(n,l)$ is the shielding constant, which now depends on $l$ as well as $n$ because penetration (and hence how much of the "true" nuclear charge is felt) depends on the orbital's shape, not just its size. This equation is the whole story compressed into one line: all of the $n+l$ physics is really $l$ -dependence hiding inside $S(n,l)$ . The empirical bookkeeping rule that captures the ordering consequence of this $l$ -dependence, without needing to compute $S(n,l)$ from first principles, is the Madelung rule:

\text{Orbitals fill in order of increasing } (n+l); \text{ for equal } (n+l), \text{ lower } n \text{ fills first.}

Reading this rule: it says energy rises step-wise along diagonals of constant $n+l$ in the $(n,l)$ grid, and within a diagonal, smaller- $n$ (i.e., larger- $l$ , more compact within that diagonal — wait, more precisely, smaller $n$ , hence smaller $l$ too since $l\le n-1$ ) orbitals sit lower. It reproduces the familiar sequence 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, exactly as summarized in the diagonal-arrow diagram above.

Derivation AWhy penetration produces an

l

-dependent, not just

n

-dependent, energy

Goal: show from the radial Schrödinger equation why orbitals of the same $n$ but different $l$ do not stay degenerate once a second electron (or a shielding cloud) is present, and why the splitting has the sign s < p < d < f.

Step 1 — the radial equation and the centrifugal term. The radial part of the hydrogenic wavefunction satisfies

-\frac{\hbar^2}{2m_e}\frac{d^2u}{dr^2} + \left[V(r) + \frac{\hbar^2\, l(l+1)}{2m_e r^2}\right]u = Eu, \qquad u(r) = rR(r)

The bracket is an effective potential: the true potential $V(r)$ plus a repulsive centrifugal barrier that grows with $l(l+1)$ . For pure hydrogen, $V(r)=-Ze^2/4\pi\epsilon_0 r$ , and remarkably the eigenvalue only depends on $n$ — this is the accidental Coulomb degeneracy. But the centrifugal term itself already tells you that higher $l$ physically pushes the wavefunction's inner turning point outward, regardless of what $V(r)$ is doing.

Step 2 — turn on a non-Coulombic potential (as in a real atom). In a multielectron atom, $V(r)$ is no longer pure $-Ze^2/4\pi\epsilon_0r$ ; it is modified by the other electrons' charge density $\rho(r')$ via a Hartree-like term,

V(r) = -\frac{Ze^2}{4\pi\epsilon_0 r} + \int \frac{e^2\,\rho(r')}{4\pi\epsilon_0 |r-r'|}\,d^3r'

Near $r\to 0$ the second (screening) term is small — the nucleus dominates and $V(r)\to -Ze^2/4\pi\epsilon_0 r$ (steep, attractive). Far from the nucleus, the screening term nearly cancels $Z-1$ units of nuclear charge, leaving an effective $-e^2/4\pi\epsilon_0 r$ tail (hydrogen-like, charge $+1$ ). This potential is no longer pure Coulomb, so the Laplace–Runge–Lenz symmetry that forced $n$ -only degeneracy in hydrogen is broken. Energy eigenvalues must now depend on $l$ as well as $n$ .

Step 3 — why the sign of the splitting is s below p below d. Combine the centrifugal barrier with the non-Coulomb $V(r)$ : an orbital only samples the steep, high- $Z$ inner region of $V(r)$ if its centrifugal barrier is low enough to let it approach $r\to0$ . Since the barrier $\propto l(l+1)$ , s orbitals ( $l=0$ , no barrier) penetrate deepest and sample the most negative (most attractive, highest- $|Z|$ ) part of $V(r)$ ; d and f orbitals ( $l=2,3$ ) are excluded from that region and sample mostly the outer, heavily-screened tail. Because the inner region is more attractive than the outer tail, deeper penetration means lower (more negative) energy. Hence, for fixed $n$ :

E_{ns} < E_{np} < E_{nd} < E_{nf}

Step 4 — why this splitting can exceed the $n$ -gap and cause crossovers. The magnitude of the $l$ -splitting within a shell grows with the number of shielding electrons inside it (more structure in $V(r)$ to differentiate penetrating vs. non-penetrating orbitals) and with $Z$ . The $n$ -gap, meanwhile, shrinks like $1/n^2$ for the outer shells once $Z_{\text{eff}}$ itself is reduced by screening. By $n=3\to4$ , the $l$ -splitting of 3d (pushed high because $d$ barely penetrates the now-substantial Ar-like core) becomes comparable to, and then larger than, the bare $n$ -gap between 3d and 4s. This is exactly the crossover visualized in the Madelung diagram: 4s (small $l$ , deep penetration, low energy despite larger $n$ ) drops below 3d (large $l$ , poor penetration, elevated energy despite smaller $n$ ).

Go deeper · Derivation B · 2 minThe Madelung

(n+l)

rule as an approximate mnemonic, and why it can fail

Derivation BThe Madelung

(n+l)

rule as an approximate mnemonic, and why it can fail

Goal: show how the empirical $(n+l, \text{then }n)$ ordering approximates the true penetration-controlled energies, and identify the physical terms it omits — explaining known exceptions (Cr, Cu, Pd, lanthanide/actinide anomalies).

Step 1 — state the rule as an ordering function. Define a priority key

k(n,l) = (n+l)\times 100 + n

Lower $k$ fills first. This reproduces the standard sequence because it sorts primarily by $n+l$ and, for ties, by $n$ . For example, 3d has $k = 5\times100+3 = 503$ ; 4p has $k = 5\times100+4 = 504$ ; 4s has $k=4\times100+4=404$ . So the rule predicts $4s\ (404) < 3d\ (503) < 4p\ (504)$ — precisely the observed neutral-atom filling order.

Step 2 — why $n+l$ tracks real energy at all. This is not derived from first principles; it is a successful empirical regularity (Madelung's rule, also credited independently to Klechkowsky). Its rough justification: energy in real atoms decreases monotonically (for fixed total "size scale") as penetration increases, and penetration is controlled by the number of radial nodes plus angular nodes of the orbital, which for a hydrogenic-like orbital scales like $(n - l - 1) + l = n-1$ radial+angular structure — but more usefully, orbitals with the same $n+l$ turn out empirically to lie close in energy because a decrease in $l$ (more penetration, lower energy) very nearly compensates for the accompanying decrease in $n$ needed to keep $n+l$ fixed... Actually the compensations run the other way: within one $(n+l)$ diagonal, increasing $n$ by 1 while decreasing $l$ by 1 changes both the shell size and the penetration in a way that keeps energies clustered, empirically, close enough that a simple integer sum reproduces the observed order better than sorting by $n$ or $l$ alone.

Step 3 — the missing physics: electron-electron repulsion and exchange. The $(n+l)$ rule treats orbital energy as fixed, one-electron numbers, ignoring that the actual energy of a configuration depends on:

E_{\text{config}} = \sum_i \varepsilon_i + \sum_{i<j}\left(J_{ij} - \tfrac12 K_{ij}\right)

where: $\varepsilon_i$ are one-electron orbital energies (what $(n+l)$ approximates), $J_{ij}$ is Coulomb repulsion between electrons $i,j$ , and $K_{ij}$ is the quantum-mechanical exchange integral, which stabilizes configurations with more parallel-spin electrons in degenerate or near-degenerate orbitals (Hund-type stabilization).

Step 4 — apply this to chromium. The rule predicts $[\text{Ar}]4s^23d^4$ for Cr ( $Z=24$ ). The observed ground configuration is $[\text{Ar}]4s^13d^5$ . Promoting one 4s electron into 3d costs one-electron energy $\varepsilon_{3d}-\varepsilon_{4s} > 0$ , but it converts a $4s^23d^4$ arrangement (with limited exchange stabilization among 4 partially-parallel d electrons plus a filled, no-exchange 4s pair) into a half-filled d-subshell, $3d^5$ , with all five d electrons parallel-spin — maximizing the number of favorable exchange pairs $K_{ij}$ . When the exchange gain exceeds the one-electron promotion cost, the atom rearranges. The same logic, pushed further, explains $3d^{10}4s^1$ for copper (a fully-filled d-subshell plus one 4s electron beats $3d^94s^2$ ).

Step 5 — state the scope of validity. The Madelung rule is reliable for predicting neutral, ground-state, gas-phase filling order across most of the periodic table, and is essentially exact for the first three periods. It becomes only approximate once $(n+l)$ -degenerate or near-degenerate orbitals compete (mid-to-late transition series, lanthanides, actinides), because at that point $J_{ij}$ and $K_{ij}$ terms are no longer small corrections — they are comparable to the one-electron $\varepsilon_i$ gaps the rule is built from.

Worked examples

Example 1 — Foundational: why does potassium's last electron go to 4s, not 3d? Potassium, $Z=19$ , has core $[\text{Ar}] = 1s^22s^22p^63s^23p^6$ (18 electrons) plus one more. Using the $(n+l)$ key from Step 1 above:

k(4,0) = (4+0)\times100 + 4 = 404 \qquad\text{vs.}\qquad k(3,2) = (3+2)\times100+3 = 503

Since $404 < 503$ , 4s fills first: $[\text{Ar}]4s^1$ , matching experiment exactly (and matching potassium's well-known alkali-metal chemistry — one loosely bound valence electron, first ionization energy 418.8 kJ/mol, far below any transition-metal value). Physically: with an $[\text{Ar}]$ core of 18 electrons already shielding, a 3d electron (no inner lobe) would sit almost entirely outside that core, feeling $Z_{\text{eff}} \approx 19-18=1$ — essentially a hydrogen-like $n=3$ electron. A 4s electron, by contrast, has small inner lobes that dip inside the Ar core and sample a substantially higher effective charge intermittently, which — despite the larger $n=4$ — pulls its average energy below the 3d level. Quantitatively, using Slater's rules (a crude but pedagogically standard estimate; see Effective Nuclear Charge), $Z_{\text{eff}}(4s) \approx 19 - 16.80 = 2.20$ while $Z_{\text{eff}}(3d) \approx 19-18 = 1.00$ (3d shielded almost completely by all 18 inner electrons, since none of them are themselves d electrons contributing less-than-full shielding as outer screeners would). Plugging into the hydrogenic estimate:

E_{4s} \approx -\frac{(2.20)^2}{4^2}\,\text{Ry} = -\frac{4.84}{16}\,\text{Ry} = -0.303\ \text{Ry} \approx -4.11\ \text{eV}

E_{3d} \approx -\frac{(1.00)^2}{3^2}\,\text{Ry} = -\frac{1.00}{9}\,\text{Ry} = -0.111\ \text{Ry} \approx -1.51\ \text{eV}

So $E_{4s} \ll E_{3d}$ (more negative, hence lower/more stable) — 4s wins decisively, consistent with experiment. (These Slater-rule numbers are approximate but the qualitative gap, several eV in K's favor of 4s, matches spectroscopic term values reasonably well.)

Limiting-case check: if we imagined switching off all $l$ -dependence of shielding (i.e., forced $S$ to depend on $n$ only, as if all $n=3$ and $n=4$ orbitals shielded identically), both 3d and 4s would revert to simple $-Z_{\text{eff}}^2/n^2$ with the same $Z_{\text{eff}}$ , and $n=3 < n=4$ would trivially put 3d below 4s — recovering the pure-hydrogen ( $l$ -independent) limit. The fact that real potassium reverses this ordering is direct proof that $l$ -dependent shielding, not just $n$ , controls the answer.

Example 2 — Applied: scandium and the ionization-order paradox. Scandium, $Z=21$ : Madelung ordering predicts, and experiment confirms, ground configuration

\text{Sc}: [\text{Ar}]\,4s^2 3d^1

Now ionize it once. Naively, since 4s filled "before" 3d, you might expect the electron removed to come from 3d (last-filled-electron intuition applied backwards — but actually the correct intuition, "last filled comes out first," would predict removing the 3d electron since it filled after 4s in the neutral-atom build-up; let's check this against data). Experimentally,

\text{Sc}^+ : [\text{Ar}]\,4s^13d^1, \qquad \text{NOT}\quad [\text{Ar}]4s^23d^0

Wait — the observed ground state of Sc⁺ is actually $[\text{Ar}]3d^14s^1$ (nearly degenerate with $4s^2$ , but this already signals something), and Sc²⁺ is unambiguously $[\text{Ar}]3d^1$ — the 4s electrons are gone entirely before 3d is touched. This is the textbook "ionization reverses the filling order" result: for Sc → Sc²⁺, both electrons removed are 4s electrons, even though 4s filled first in the neutral atom. The resolution is exactly the mechanism from Step 4 of the second derivation: once 3d starts to acquire real electron density (as it does the moment Sc has a $3d^1$ configuration), that d density is compact and close to the core, and it shields the more diffuse 4s electrons rather effectively, while itself being poorly shielded by them. Removing electrons progressively increases the effective nuclear charge felt by all remaining electrons, but it increases it more for 3d (whose energy was artificially close to 4s only because of the delicate penetration balance in the neutral, 19–21 electron regime). Once ionized, the orbital energy ordering flips: $E_{3d} < E_{4s}$ in the cation, so the atom preferentially retains the lower-energy 3d electrons and sheds the higher-energy 4s electrons first. This is a genuinely different balance from the neutral-atom filling calculation — it is not a violation of anything, but a reminder that " $(n+l)$ filling order" describes how a many-electron ground state is built up energetically as $Z$ increases (Aufbau bookkeeping), not a claim that the same orbital energy ordering persists once the electron count of that specific ion changes.

Numerical sanity check using ionization energies: the first ionization energy of Sc is 633.1 kJ/mol (removing a 4s electron); the second ionization energy is 1235.0 kJ/mol (removing the second 4s electron, from Sc⁺ configuration $\sim 4s^13d^1$ ); the third ionization energy, 2388.6 kJ/mol, removes the 3d electron from Sc²⁺. The steep rise from IE₂ to IE₃ (nearly double) is consistent with switching from removing an s electron in a mixed configuration to removing a d electron from an already-doubly-charged, much more tightly bound ion — not simply a smooth trend, confirming the qualitative orbital-crossing picture.

Example 3 — Exam/competition level: rationalize why palladium is $[\text{Kr}]4d^{10}5s^0$ , the most extreme Madelung exception in the d-block. The Madelung rule predicts, marching along $(n+l)=6$ then $(n+l)=7$ diagonals for $Z=46$ :

[\text{Kr}]\,5s^2 4d^8

The observed ground configuration is

\text{Pd}: [\text{Kr}]\,4d^{10}5s^{0}

— a complete emptying of 5s, not merely a $5s^1$ half-promotion as in Cu or Cr. This is the largest single deviation from the naive rule anywhere in the periodic table, and it is a favorite "explain this" exam question.

Step-by-step rationale. Apply the same energetics balance as in the Cr/Cu derivation, but now push it to its logical extreme: 1. One-electron cost: promoting both 5s electrons into 4d costs $2(\varepsilon_{4d}-\varepsilon_{5s})$ in one-electron energy — the largest such cost of any first-to-transition-metal comparison, since $\varepsilon_{4d}$ and $\varepsilon_{5s}$ are, for 4d-series metals, unusually close together (closer than the 3d/4s gap, because 4d orbitals are more diffuse/penetrating than 3d, having one more radial node structure available to sneak inward). 2. Exchange gain: filling 4d completely, $4d^{10}$ , gives a filled-subshell configuration. While a filled shell has no net spin (so it isn't a "half-filled, all-parallel-spin" Hund's-rule maximization like Cr's $3d^5$ ), a completely filled d-subshell carries a large, distinct stabilization from having all ten spin-orbitals occupied — every possible pairwise exchange interaction among same-spin electrons within the shell is realized, and additionally filled-shell configurations have spherically symmetric charge density, minimizing electron-electron repulsion asymmetries. 3. Because Pd sits in the 4d row (where the s/d one-electron gap is anomalously small — smaller than in either the 3d or 5d rows, a real periodic effect from relativistic and radial-node systematics), the exchange/shell-closure stabilization from $4d^{10}$ comfortably outweighs the one-electron promotion cost for both 5s electrons simultaneously, something that never happens for any 3d-series element. 4. Confirmation: Pd is famous experimentally for having zero unpaired electrons and a closed-shell $^1S_0$ ground state, unlike every other element in its row — directly verifying $4d^{10}5s^0$ , since a $5s^24d^8$ configuration would necessarily have two unpaired d electrons (Hund's rule on 8 electrons in 5 d-orbitals) and would not be diamagnetic. Palladium's well-documented diamagnetism (in contrast to paramagnetic Ni or Pt) is the empirical fingerprint proving the exception.

Limiting-case check: contrast with platinum ( $5d$ row, $Z=78$ ), which is $[\text{Xe}]4f^{14}5d^96s^1$ — only a partial, single-electron promotion, much more Cu-like than Pd-like. This shows the "complete emptying of s" behavior is not generic to column 10 (Ni, Pd, Pt all column 10) but specific to the unusually small 4d/5s gap of period 5 — confirming that the exceptions really do trace back to quantitative differences in the same underlying penetration competition, not to some entirely separate rule.

Limiting cases, scope & common traps

Regime	What Madelung $(n+l)$ predicts	What actually happens	Why
$Z\le 20$ (through Ca)	Strict $(n+l)$ , then- $n$ order	Matches exactly	One-electron gaps dominate; exchange/repulsion corrections are small perturbations
Cr, Mo ( $3d^54s^1$ , $4d^54s^1$ or $4d^5$ )	$d^4s^2$	$d^5s^1$ (half-filled d)	Exchange stabilization of half-filled subshell exceeds promotion cost
Cu, Ag, Au ( $3d^{10}4s^1$ etc.)	$d^9s^2$	$d^{10}s^1$	Filled-subshell stabilization exceeds promotion cost
Pd	$4d^85s^2$	$4d^{10}5s^0$	Anomalously small 4d/5s one-electron gap in period 5; both electrons promote
Ionized transition metals (any M $^{n+}$ )	(rule doesn't apply to ions directly)	s-electrons always removed before d	Once d subshell is populated, it depresses in energy below s; ionization order ≠ filling order
Lanthanides (e.g. La, Ce, Gd)	Regular $4f$ filling	Several irregular $4f^{n}5d^1$ vs $4f^{n+1}5d^0$ swaps	$4f$ and $5d$ become nearly degenerate; tiny energy differences flip the ground state
$n\to\infty$ , hydrogen-like limit (one electron only)	(rule is meaningless — no shielding)	Pure $n$ -only degeneracy, all $l$ equal-energy	Recovers hydrogenic Coulomb degeneracy exactly; confirms $l$ -splitting is a screening effect, not intrinsic

Common trapMisconception: "4s fills before 3d, so 4s is always higher in energy than 3d is drawn in the orbital diagram."

Many students draw 4s below 3d in energy-level diagrams for neutral atoms (correct, matching the fact that it fills first) but then assume this ordering is fixed and use it to justify which electron ionizes first — incorrectly predicting that 3d electrons are removed before 4s in cations. In reality, the energy ordering inverts upon ionization or upon significant d-orbital population, because the mutual shielding relationship between 4s and 3d is not symmetric (d electrons shield s electrons more effectively than s electrons shield d electrons, once both are populated). The correct rule for ions: always remove valence $s$ electrons before $d$ electrons in transition-metal cations, regardless of neutral-atom filling order.

The chemist behind it

Origins · 1 minErwin Madelung and how this was discovered

Erwin Madelung

1881–1972 · German theoretical physicist, University of Frankfurt

Madelung is best known to chemistry students for the empirical $(n+l)$ electron-filling rule that bears his name, but his broader career was in mathematical and theoretical physics — he made foundational contributions to lattice-energy calculations in ionic crystals (the Madelung constant, still the standard tool for computing electrostatic lattice energies of salts like NaCl) and to the hydrodynamic formulation of quantum mechanics (the Madelung equations, an alternative representation of the Schrödinger equation as fluid-like continuity and Euler equations for a probability density and phase). The electron-configuration rule was noted by Madelung around 1936 as a numerological regularity in observed atomic spectra and ground-state configurations; it was independently rediscovered and popularized by the Russian chemist Vsevolod Klechkowsky in the 1950s, which is why the rule is called the Madelung rule in much of the West and the Klechkowsky rule in much of the Russian-language literature. Neither derived it from first-principles quantum mechanics — both recognized it as an empirical pattern in existing spectroscopic and X-ray term-value data, later rationalized (though not exactly derived) by Hartree-Fock-level penetration/shielding arguments of the kind developed in this article.

History

- 1925 — Pauli exclusion principle established, giving a rigorous basis for shell/subshell electron capacity.

- 1928–1930 — Hartree (and later Hartree–Fock, with Fock's 1930 exchange corrections) self-consistent-field calculations begin quantitatively confirming

l

-dependent orbital energies in multielectron atoms.

- 1936 — Erwin Madelung notes the empirical

(n+l)

regularity in atomic ground-state configurations.

- 1951 — Vsevolod Klechkowsky independently formulates and publishes the same rule, widely adopted in Soviet/Russian pedagogy.

- 1960s–present — systematic X-ray photoelectron spectroscopy (XPS) and atomic beam spectroscopy catalog the full list of Madelung-rule exceptions across the d- and f-blocks, refining the modern understanding of exchange-driven anomalies.

Why does energy depend on l in atoms but not in hydrogen?

Hydrogen has exactly one electron orbiting a pure $-Ze^2/4\pi\epsilon_0 r$ Coulomb potential, which possesses an extra hidden symmetry (conservation of the Laplace–Runge–Lenz vector) that forces all orbitals of a given $n$ to share exactly the same energy, regardless of $l$ . Add a second electron, and each electron now moves in the nucleus's field plus the non-uniform, partially-screening charge cloud of the other electrons. That combined potential is no longer pure $1/r$ , so the special symmetry is broken, and the energy becomes sensitive to the orbital's shape — specifically to how much of its density penetrates the inner, poorly-screened region, which depends on $l$ as well as $n$ .

Is the Madelung (n+l) rule ever taught as exact, and if so is that a mistake?

It is often presented as if it were exact because it correctly reproduces ground-state configurations for the overwhelming majority of elements, especially all of periods 1–3 and most main-group elements throughout the table. However, it is fundamentally an empirical regularity, not a derived law, and it is known to fail for at least fifteen elements (Cr, Cu, Nb, Mo, Ru, Rh, Pd, Ag, La, Ce, Gd, Pt, Au, and several actinides), because it ignores exchange and electron-correlation energy, which can outweigh the one-electron energy differences the rule is built from. Teaching it as a useful, highly reliable heuristic — rather than an exact law — is the accurate framing.

Why do 4s electrons get removed before 3d electrons during ionization, if 4s filled first?

Filling order in the neutral atom and removal order in the cation are governed by different energy balances, because the presence of 3d electrons changes the shielding environment. Once even one 3d electron is present, it shields the more diffuse 4s electrons quite effectively (poor mutual shielding runs the other way: s electrons barely shield d electrons), so as more electrons are added or removed, the 3d level drops below 4s in the effective energy ordering. In transition-metal cations, 3d is unambiguously the lower-energy, more tightly bound orbital, so the higher-energy 4s electrons ionize first — a real inversion, not a paradox.

Does the (n+l) rule apply to predicting which orbitals are used in bonding, not just which fill first?

No — filling order (Aufbau/Madelung bookkeeping) tells you the ground-state electron configuration of a free, gas-phase atom, but chemical bonding is governed by the orbitals that are energetically accessible and symmetry-appropriate in the bonded environment, which is a separate question addressed by molecular orbital theory, crystal field theory, and ligand field theory. For transition metals specifically, once you know the free-ion configuration (with s electrons removed first, as established above), the relevant valence orbitals for bonding are overwhelmingly the d orbitals (plus accessible s/p for hybridization), which is why d-orbital-based crystal field/ligand field arguments work even though s filled "first" in the neutral atom.

Why does palladium have zero unpaired electrons, unlike nickel and platinum in the same group?

Palladium's 4d and 5s orbitals happen to be unusually close in one-electron energy — closer than the analogous 3d/4s gap in nickel or the 5d/6s gap in platinum — because of period-specific systematics in radial node structure and relativistic effects. This small gap means the large stabilization from having a completely filled, spherically symmetric $4d^{10}$ subshell (maximizing favorable exchange interactions and minimizing electron-electron repulsion asymmetry) is enough to promote both 5s electrons into 4d, something that does not happen for Ni ( $3d^84s^2$ , two unpaired electrons) or Pt (only one electron promotes, $5d^96s^1$ ). The resulting $[\text{Kr}]4d^{10}5s^0$ configuration has no unpaired electrons at all, making elemental palladium diamagnetic — an experimentally verifiable fingerprint of this configuration.

How does penetration relate to effective nuclear charge and Slater's rules?

Penetration is the physical mechanism; effective nuclear charge $Z_{\text{eff}}$ and Slater's rules are the bookkeeping tools used to quantify it. An orbital that penetrates deeply into the region occupied by inner-shell electrons spends part of its time exposed to a higher nuclear charge than the naive "outer electron sees $Z$ minus number of inner electrons" picture would suggest, and Slater's empirical shielding constants are constructed (via different shielding contributions for $ns/np$ groups versus $nd$ or $nf$ groups) specifically to capture this $l$ -dependence numerically, even though they are a simplified fit rather than a first-principles calculation.

Why doesn't the (n+l) rule ever put an f orbital below the corresponding d or p orbital of the same shell?

Because $f$ orbitals ( $l=3$ ) have the largest centrifugal barrier of any subshell in a given principal shell, they are pushed furthest from the nucleus and penetrate least effectively through inner shielding electrons, so they are consistently the highest-energy subshell for any given $n$ . This is exactly the same penetration hierarchy (s < p < d < f in energy, for fixed $n$ ) that produces the s-before-d crossovers discussed throughout this article, just carried one step further along the angular momentum ladder — which is also why 4f only becomes energetically competitive with 5d and 6s many shells later than one might naively guess from $n$ alone.