Hund's rule of maximum multiplicity: Visual walkthrough

Step 1 — What is an orbital, drawn as a box?

WHAT. Before any rule, we need a picture for "where an electron can live". We draw each orbital as a box, and an electron as an arrow inside it. An up-arrow ↑ means spin "up", a down-arrow ↓ means spin "down".

WHY. Rules about spin and pairing are impossible to state until we can see spin and pairing. The box-and-arrow picture is the smallest drawing that shows both.

PICTURE. Look at the figure: one empty box, one box with a single ↑, one box with a pair ↑↓. The word paired just means "two arrows in the same box, pointing opposite ways" — that opposite-pointing is forced by the Pauli principle.

Step 2 — Spin as an arrow, and the number $m_{s}$

WHAT. We attach a number to each arrow: $m_{s} = + \frac{1}{2}$ for ↑ and $m_{s} = - \frac{1}{2}$ for ↓.

$m_{s} = + \frac{1}{2} (↑), m_{s} = - \frac{1}{2} (↓)$

$m_{s}$ — the spin quantum number, the label that says which way the little arrow points.
$+ \frac{1}{2}$ / $- \frac{1}{2}$ — the only two allowed tilts; there is no third arrow.

WHY. To maximize spin later, we must be able to add spins up. Turning arrows into numbers $\pm \frac{1}{2}$ lets us literally sum them. (Origin of these numbers: Electron Spin Quantum Number.)

PICTURE. The figure shows two arrows: ↑ tagged $+ \frac{1}{2}$ , ↓ tagged $- \frac{1}{2}$ . When both sit in one box the tags cancel to $0$ ; when they sit in separate boxes and point the same way they add.

Step 3 — Two ways to place two electrons: the choice Hund settles

WHAT. Put 2 electrons into 2 degenerate boxes. Only two sensible layouts exist:

Option A (pair up): both in one box, ↑↓, other box empty.
Option B (spread out): one in each box, ↑ ↑ (parallel).

WHY. Hund's rule is the answer to "A or B?". To decide, we compare their energies — so first we must see the two options side by side.

PICTURE. The figure draws A and B. Notice in A the two arrows are crammed into one box (close together); in B they are in different boxes (farther apart). Distance is the first clue.

Step 4 — Reason 1: repulsion punishes crowding

WHAT. Same-box electrons occupy the same region of space, so they push on each other harder. Different-box electrons are farther apart → less repulsion.

WHY this comparison. Electrons are negative and repel. Energy goes up when like charges sit close. So Option A (crowded) sits at higher energy than Option B (spread).

PICTURE. The figure shows two clouds overlapping strongly in Option A (big red repulsion spring) versus two clouds far apart in Option B (weak spring). The tall spring = higher energy.

$E_{repulsion} (same box) > E_{repulsion} (different boxes)$

$E_{repulsion}$ — energy cost of two electrons pushing apart; bigger when they overlap.

This already favours Option B. But there is a second, deeper reason — the real star of Hund's rule.

Step 5 — Reason 2: exchange energy, counted by drawing swaps

WHAT. Electrons with the same spin in different boxes are indistinguishable — you can secretly swap any two of them and the atom looks unchanged. Each possible swap lowers the energy by a fixed amount $K$ (the exchange integral, always positive).

WHY a tool enters here — why counting pairs? Because the total discount is " $- K$ per swappable pair", the physics question becomes a counting question: how many same-spin pairs can I form? Counting pairs from a group is exactly the "choose 2" operation:

$swaps = (2 n) = \frac{n ( n - 1 )}{2}$

$n$ — number of parallel-spin (same-direction) electrons in the row of boxes.
$(2 n)$ — " $n$ choose 2", the number of distinct pairs you can pick from $n$ items.
$\frac{n ( n - 1 )}{2}$ — the plain-arithmetic value of that count.
Why $/2$ ? Picking electron $a$ then $b$ is the same pair as $b$ then $a$ ; dividing by $2$ removes the double-count.

WHY only parallel spins count. An ↑ and a ↓ are distinguishable (different arrows), so swapping them does change something → no discount. Only same-arrow pairs are secretly swappable.

$E_{exchange} = - K \cdot (2 n)$

The minus sign means exchange lowers (stabilizes) energy.
More parallel electrons → more swaps → deeper discount.

PICTURE. For $n = 3$ (three ↑ in three boxes) the figure draws the 3 curved swap arrows connecting each pair — you can literally count $(2 3) = 3$ lines. Each line is worth $- K$ . This is why Option B (parallel) beats even a "spread but antiparallel" layout: only parallel electrons earn swaps. (Deep dive: Exchange Energy and Half-filled Stability.)

Step 6 — Turning "most parallel electrons" into a number: multiplicity

WHAT. Add up the spin arrows. Each unpaired ↑ adds $+ \frac{1}{2}$ , so total spin is

$S = \frac{n}{2}$

$S$ — total spin quantum number, the sum of the individual $\pm \frac{1}{2}$ tags.
$n$ — number of unpaired parallel electrons; each contributes $+ \frac{1}{2}$ , hence divide by nothing extra — $n$ halves make $n /2$ .

Then define multiplicity:

$Multiplicity = 2 S + 1 = 2 (\frac{n}{2}) + 1 = n + 1$

$2 S + 1$ — the number of directions the total spin arrow can tilt (from $- S$ to $+ S$ ).
The algebra collapses it to the clean $n + 1$ : one more than the unpaired count.

WHY this tool — why $2 S + 1$ and not just $n$ ? Multiplicity is the measurable fingerprint (it names states "singlet/doublet/triplet…" and links to Magnetic Moment of Atoms). "Maximum multiplicity" is just a fancy way of saying "maximum $n$ " — maximum unpaired parallel electrons — which Steps 4–5 proved is the low-energy choice.

PICTURE. The figure lines up three rows: $n = 1, 2, 3$ boxes filled ↑, with the ladder of $m_{S}$ values on the right growing taller ( $2$ , $3$ , $4$ rungs). Taller ladder = higher multiplicity.

Step 7 — The full recipe applied: N, then O (the degenerate & pairing cases)

WHAT. Now fill real atoms using the rule "all boxes singly & parallel first, then pair", which Aufbau hands us after deciding it's the $2 p$ subshell's turn.

Nitrogen ( $2 p^{3}$ ): three electrons, three boxes → ↑ ↑ ↑.

$n = 3$ , so multiplicity $= n + 1 = 4$ (a quartet), swaps $= (2 3) = 3$ .
This is the degenerate case done fully — every box gets exactly one electron.

Oxygen ( $2 p^{4}$ ): the 4th electron has no empty box left, so it must pair: ↑↓ ↑ ↑.

WHY it pairs only now: single-occupancy is exhausted; only then does Pauli-pairing begin.
$n = 2$ unpaired, multiplicity $= 3$ (a triplet).

PICTURE. The figure shows the $2 p$ row for N (three lonely ↑) and for O (one box now doubled ↑↓, two still single). The doubled box is highlighted — that's the forced pairing.

Step 8 — Edge & degenerate cases you must never trip on

WHAT. Four boundary situations, each drawn:

Empty subshell / zero electrons: $n = 0$ , multiplicity $= 0 + 1 = 1$ . Nothing to arrange.
Fully filled ( $2 p^{6}$ , $d^{10}$ ): every box ↑↓, so unpaired $n = 0$ , multiplicity $= 1$ (a singlet) — many electrons but zero net spin. This kills the myth "more electrons = more multiplicity".
Exactly half-filled ( $d^{5}$ , ↑ ↑ ↑ ↑ ↑): $n = 5$ , swaps $= (2 5) = 10$ — the peak of exchange discount before pairing must start. This is why Cr grabs $[Ar] 3 d^{5} 4 s^{1}$ .
One past half-filled ( $d^{6}$ ): first five ↑, sixth pairs → $n = 4$ , multiplicity $= 5$ . Pairing has begun, yet Hund still forces the maximum unpaired count of 4.

WHY these matter. They are the exact places students give the wrong multiplicity. Each obeys the same recipe — the drawings just show the recipe never breaks.

PICTURE. The figure stacks all four rows of boxes with their $n$ and multiplicity labelled, and marks the $d^{4} \to d^{5}$ jump in swaps ( $6 \to 10$ ) with a big arrow.

The one-picture summary

PICTURE. One figure compressing everything: a row of degenerate boxes; arrows entering singly & parallel first (green), the forced pair entering last (coral), a side ladder showing multiplicity $= n + 1$ rising, and the swap-count curve $(2 n)$ peaking at the half-filled shell.

Recall Feynman retelling of the whole walkthrough

Draw each orbital as an empty chair at the same comfy height (Step 1). Each new kid carries a little arrow — up or down (Step 2). When two kids arrive we can either cram them on one chair or give each their own (Step 3). Crammed kids elbow each other → costs energy (Step 4). Kids sitting alone facing the same way can secretly swap seats, and every possible swap gives the whole room a discount $- K$ ; count the swaps with "choose 2" (Step 5). We score the room by its multiplicity $n + 1$ — bigger means more happy same-facing singles (Step 6). So nitrogen seats three kids, each alone facing up (quartet); oxygen's fourth kid finds no empty chair and must double up (triplet) — but only after everyone tried a solo chair first (Step 7). Empty and totally full rooms have zero lonely kids (multiplicity 1); a half-filled room has the most swaps of all — that's the sweet spot chromium chases (Step 8). One rule, one picture: singles before doubles, all facing up.

Recall Forecast the answers

Multiplicity of $2 p^{6}$ ? ::: $1$ (all paired, $n = 0$ ). Swaps in $d^{5}$ vs $d^{4}$ ? ::: $10$ vs $6$ — the jump that stabilizes $d^{5}$ . Multiplicity of $d^{6}$ (Fe²⁺)? ::: $5$ (four unpaired). Why do antiparallel spread electrons earn no exchange discount? ::: Opposite arrows are distinguishable, so swapping them changes something → no $- K$ .

Connections

2.1.09 Hund's rule of maximum multiplicity (Hinglish) — the same story in Hinglish.
Aufbau Principle — chooses which subshell; hands us the row of boxes.
Pauli Exclusion Principle — forces paired arrows to point opposite ways.
Electron Spin Quantum Number — where $m_{s} = \pm \frac{1}{2}$ comes from.
Exchange Energy and Half-filled Stability — the deep dive on $- K (2 n)$ .
Paramagnetism and Diamagnetism — unpaired arrows ⇒ paramagnetic.
Magnetic Moment of Atoms — turns $n$ into $μ = n (n + 2) μ_{B}$ .