Stability of half-filled and fully-filled subshells: Visual walkthrough

Prerequisites we lean on (build them first if shaky): Hund's Rule of Maximum Multiplicity, Aufbau Principle, (n+l) Rule and Orbital Energies, Pauli Exclusion Principle.

Step 1 — Draw a subshell as a row of boxes

WHAT. Before any energy talk, let us agree on a picture. A subshell is just a group of orbitals that share the same shape-family. The $d$ subshell has five orbitals. We draw each orbital as a box, and an electron as an arrow: up-arrow $↑$ = one spin direction, down-arrow $↓$ = the opposite.

WHY. Energy in this whole story depends only on how many arrows point the same way. If we can see the arrows, we can count the stability directly — no algebra needed yet.

PICTURE. Five boxes in a row. Look at the figure: the top row is $d^{4}$ (four up-arrows, one empty box), the bottom row is $d^{5}$ (five up-arrows). We will keep re-using exactly this drawing.

Step 2 — Two same-spin electrons can "swap" — and that swap lowers energy

WHAT. Take any two arrows that point the same way. Because electrons are truly identical, the universe cannot tell whether electron $a$ sits in box $i$ and electron $b$ in box $j$ , or the other way around. These two situations are the same physical state. Quantum mechanics blends them, and that blend sits at a lower energy than if the two were forbidden to mix.

WHY this tool — "exchange". We reach for the idea of swapping (called exchange) because it is the only new ingredient that distinguishes same-spin from opposite-spin electrons. Opposite-spin arrows can be told apart (their spins differ), so no such blend, no bonus. So exchange is precisely the knob that answers our question: why do parallel spins help?

PICTURE. Two boxes, two up-arrows, and a curved double-headed arrow between them labelled "swap allowed → energy drops by $K$ ." Beside it: two boxes, one up one down, with a red ✗ — "swap not allowed, no bonus."

Step 3 — Count ALL the swaps: the pair-counting picture

WHAT. One swap gives $- K$ . But with many same-spin arrows there are many possible pairs to swap. We must count every distinct unordered pair.

WHY "choose 2". The question "how many ways to pick 2 arrows out of $n$ ?" is answered by the combination $(2 n)$ — read " $n$ choose $2$ ". We use combinations, not permutations, because a swap of $(a, b)$ is the same swap as $(b, a)$ ; order does not matter.

PICTURE. Five dots (the five $d^{5}$ arrows) drawn as points, with a line drawn between every pair. Count the lines: exactly $10$ . The figure colours these $10$ links and writes $(2 5) = 10$ .

Total stabilization (using one average $K$ for every pair): $E_{exch} \approx - N_{ex} K = - \frac{n ( n - 1 )}{2} K$

Step 4 — The $d^{4}$ side: count its swaps

WHAT. Chromium's naive configuration is $[A r] 3 d^{4} 4 s^{2}$ . Inside the $3 d$ subshell there are four parallel up-arrows (Hund's rule fills them singly and parallel — see Hund's Rule of Maximum Multiplicity).

WHY. We count only the $3 d$ – $3 d$ pairs here because these are the largest and most comparable $K$ values; the $4 s$ electrons are paired ( $↑↓$ ), which cancels their same-spin bonus.

PICTURE. The four up-arrows drawn as four dots; every pair linked → $6$ lines. Figure writes $(2 4) = 6$ .

$N_{ex} (d^{4}) = (2 4) = \frac{4 \cdot 3}{2} = 6$

the four dots ::: the four parallel $3 d$ electrons.
$6$ ::: the number of $3 d$ – $3 d$ swaps available, i.e. stabilization $\approx - 6 K$ .

Step 5 — The $d^{5}$ side: promote one electron, gain swaps

WHAT. Move one $4 s$ electron up into the empty fifth $d$ box. Now the $3 d$ subshell holds five parallel up-arrows, and $4 s$ has just one.

WHY. We spend a little: promoting an electron and un-pairing the $4 s$ costs some energy, and $3 d$ is very slightly higher than $4 s$ ((see (n+l) Rule and Orbital Energies)). We earn something bigger: extra exchange swaps. The whole page is about showing the earnings beat the cost.

PICTURE. Five dots fully linked → $10$ lines, sitting beside the $d^{4}$ picture's $6$ lines, with the 4 new lines highlighted in pink.

$N_{ex} (d^{5}) = (2 5) = 10, Δ N_{ex} = 10 - 6 = 4$

$10$ ::: total $3 d$ – $3 d$ swaps in $d^{5}$ .
$Δ N_{ex} = 4$ ::: the four brand-new pink links — pure profit.
energy gained ::: $\approx - 4 K_{3 d, 3 d}$ , a genuine drop.

Step 6 — The cost side, and why exchange still wins

WHAT. List the costs of the promotion honestly: (a) a small orbital-energy cost because $3 d$ sits a hair above $4 s$ ; (b) a little extra electron–electron repulsion from crowding the $d$ set.

WHY show costs. A stability rule is a balance, not magic. We must weigh both pans of the scale, or we are just asserting the answer.

PICTURE. A balance/see-saw: left pan holds a small block "promotion + repulsion cost", right pan holds a bigger block " $- 4 K_{3 d, 3 d}$ exchange gain". The right pan tips down.

Step 7 — Edge case: when the atom does NOT flip

WHAT. The anomaly is not a universal law. If the $n s$ and $(n - 1) d$ energies are far apart, the promotion cost becomes large and the four-swap bonus can no longer pay for it.

WHY include this. A reader who thinks "half-filled always wins" will mispredict real atoms. We must cover the degenerate case where the scale tips the other way.

PICTURE. The same balance as Step 6, but now the left "cost" block is huge (large energy gap) and outweighs the fixed exchange block on the right — the scale tips left, "normal Aufbau filling stays."

Step 8 — Bonus case: the same picture explains nitrogen's ionization energy

WHAT. Nitrogen is $1 s^{2} 2 s^{2} 2 p^{3}$ — a half-filled $p^{3}$ with three parallel arrows.

WHY relevant. The identical swap-counting explains why nitrogen resists losing an electron more than oxygen does — a fact you meet in Ionization Energy Trends.

PICTURE. Three $p$ -dots fully linked → $3$ lines ( $(2 3) = 3$ ). An arrow shows "remove one electron → drops to $2$ dots → only $1$ link left → lose $2$ swaps," so removal is costly.

$N_{ex} (p^{3}) = (2 3) = 3 ⟶ N_{ex} (p^{2}) = (2 2) = 1$

$3 \to 1$ ::: pulling an electron out of the tidy half-filled set destroys two exchange swaps, so it costs extra energy — the reason N's first ionization energy pokes above the smooth trend.

The one-picture summary

Everything above collapses to one image: dots → lines → count. More parallel arrows ⇒ more lines ⇒ more $- K$ ⇒ more stable. Chromium climbs from $6$ lines to $10$ lines by promoting one electron; that $+ 4$ -line profit, cheap because $3 d \approx 4 s$ , is the whole reason for $[A r] 3 d^{5} 4 s^{1}$ .

Recall Feynman retelling — the whole walkthrough in plain words

Picture each orbital as a box and each electron as a little arrow. Arrows pointing the same way are like identical twins who love swapping seats — and every possible swap makes the room a tiny bit calmer (lower energy). Draw a dot for every same-way arrow and a line for every pair that could swap: the number of lines is the amount of "calm." Chromium could keep four twins in the $d$ -room and hide two paired electrons in the $4 s$ -room — that's $6$ swap-lines. But if it sends one $4 s$ electron up to join the $d$ -room, now there are five twins and ten swap-lines. Those four extra lines make the room much calmer, and because moving that electron barely costs anything (the two rooms are almost the same height), chromium happily does it — landing on $3 d^{5} 4 s^{1}$ . When the two rooms are far apart in height, the move costs too much and the atom stays put. The exact same dot-and-line trick explains why nitrogen ( $p^{3}$ , three twins, three lines) hates giving up an electron: pulling one out would erase two swap-lines.

Recall Quick self-check

How many $3 d$ – $3 d$ swaps in $d^{5}$ ? ::: $(2 5) = 10$ .
How many does $d^{4}$ have, and what is the profit of flipping? ::: $6$ , profit $= 10 - 6 = 4$ .
Does exchange always beat the promotion cost? ::: No — only when $n s$ and $(n - 1) d$ are energetically close.
Why is $K$ not the same for every pair? ::: it depends on how much the two orbitals overlap in space.