Stability of half-filled and fully-filled subshells: Worked examples

This page is a calculator you can trust. The parent note (parent) told you why half-filled and full subshells are special. Here we grind through every kind of counting problem the topic can throw at you, so no exam scenario is a surprise.

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The scenario matrix

Every problem this topic can ask lands in one of these cells. The worked examples below are tagged with the cell they cover.

Cell	Case class	What makes it tricky	Example
A	Small $p$ subshell, half-filled	tiny $n$ , easy check	Ex 1 (N, $2 p^{3}$ )
B	Two competing configs (the flip)	count the difference	Ex 2 (Cr)
C	Fully-filled $d^{10}$	both spin groups count	Ex 3 (Cu)
D	Ionization (remove an electron)	$Δ$ between before/after	Ex 4 (N vs O IE)
E	Zero / degenerate input	$n = 0$ or $n = 1$ : no pairs	Ex 5 (edge cases)
F	Mixed spins (partly filled beyond half)	up-group and down-group	Ex 6 (O, $2 p^{4}$ )
G	Cross-subshell terms	$3 d$ – $4 s$ small $K$ vs $3 d$ – $3 d$	Ex 7 (Cr, full count)
H	Real-world / exam twist	"why NOT flip?" balance	Ex 8 (large gap)

Prerequisites you may want open: Hund's Rule of Maximum Multiplicity, Pauli Exclusion Principle, Aufbau Principle, (n+l) Rule and Orbital Energies.

Worked examples

Worked example Ex 1 — Nitrogen

$2 p^{3}$ (Cell A: small half-filled $p$ ) Statement: Nitrogen is $1 s^{2} 2 s^{2} 2 p^{3}$ . All three $2 p$ electrons are parallel (Hund's rule). How many $2 p$ – $2 p$ exchange pairs, and what is the stabilization in units of $K$ ?

Forecast: Guess before reading — is it 2, 3, or 6 pairs?

List the parallel group. Three $2 p$ electrons, all ↑. So $n = 3$ . Why this step? Only same-spin electrons form exchange pairs; here all three qualify.
Apply the formula. $N_{ex} = (2 3) = \frac{3 \cdot 2}{2} = 3.$ Why this step? $(2 n)$ counts unordered same-spin pairs — that's the definition of an exchange.
Convert to energy. $E_{exch} = - 3 K .$ Why this step? Each pair contributes $- K$ in the single- $K$ model.

Verify: Name the 3 pairs directly: (1,2),(1,3),(2,3) — exactly 3, no repeats. ✓ Units: $K$ is an energy, so $- 3 K$ is an energy. ✓

Worked example Ex 2 — Chromium: which config wins? (Cell B: competing configs)

Statement: Compare the within- $3 d$ exchange pairs of $[A r] 3 d^{4} 4 s^{2}$ and $[A r] 3 d^{5} 4 s^{1}$ . How many extra pairs does the flip buy?

Forecast: More pairs for $d^{4}$ or $d^{5}$ ? By how many — 2, 4, or 10?

Count $d^{4}$ . Four $3 d$ electrons all ↑ ⇒ $n = 4$ ⇒ $(2 4) = \frac{4 \cdot 3}{2} = 6$ pairs. Why this step? In $3 d^{4}$ Hund's rule keeps all four unpaired and parallel.
Count $d^{5}$ . Five $3 d$ electrons all ↑ ⇒ $n = 5$ ⇒ $(2 5) = \frac{5 \cdot 4}{2} = 10$ pairs. Why this step? Promoting a $4 s$ electron into $3 d$ makes the fifth parallel arrow.
Take the difference. $Δ N_{ex} = 10 - 6 = 4$ extra $3 d$ – $3 d$ pairs, i.e. $\approx - 4 K_{3 d, 3 d}$ . Why this step? Only the change matters for deciding which config the atom picks.

Verify: Adding the 5th electron to a group of 4 creates exactly one new pair with each of the 4 existing ones $\Rightarrow + 4$ . ✓ Matches $10 - 6 = 4$ . ✓

Worked example Ex 3 — Copper

$3 d^{10}$ (Cell C: fully-filled, both spin groups) Statement: In $[A r] 3 d^{10} 4 s^{1}$ the $3 d$ subshell is full: 5 electrons ↑ and 5 electrons ↓. Count total $3 d$ – $3 d$ exchange pairs.

Forecast: Since arrows now split into two groups, will the total be more or less than $d^{5}$ 's 10?

Count the up-group. 5 ↑ electrons ⇒ $(2 5) = 10$ pairs. Why this step? Exchange only happens within a spin direction.
Count the down-group. 5 ↓ electrons ⇒ $(2 5) = 10$ pairs. Why this step? The ↓ group is an independent set of parallel spins.
Add (never cross). Total $= 10 + 10 = 20$ pairs. No ↑–↓ pairs are counted. Why this step? An ↑ and a ↓ electron are distinguishable by spin, so swapping them gives a genuinely new state — no exchange stabilization.

Verify: Two disjoint groups of 5 ⇒ $2 (2 5) = 20$ . ✓ This is why $d^{10}$ is so stable: it maximizes pairs in both directions.

Worked example Ex 4 — Nitrogen vs Oxygen ionization energy (Cell D: remove an electron)

Statement: First ionization removes one electron. Compare the loss of $2 p$ – $2 p$ exchange pairs for N ( $2 p^{3} \to 2 p^{2}$ ) and O ( $2 p^{4} \to 2 p^{3}$ ). Which loses more exchange stabilization?

Forecast: N's IE is famously higher than O's. Which one should therefore lose more pairs on ionization?

N before/after. $2 p^{3}$ (3 ↑) has $(2 3) = 3$ pairs. $2 p^{2}$ (2 ↑) has $(2 2) = 1$ . Loss $= 3 - 1 = 2$ pairs. Why this step? Removing an electron shrinks the parallel group by one.
O before/after. $2 p^{4}$ = 3 ↑ + 1 ↓ ⇒ up-group $(2 3) = 3$ , down-group $(2 1) = 0$ , total $3$ . After removing the lone ↓ we get $2 p^{3}$ (3 ↑) $= 3$ pairs. Loss $= 3 - 3 = 0$ pairs. Why this step? O's easiest electron to remove is the paired ↓ one, which was in no exchange pair.
Compare. N loses $2 K$ of stabilization; O loses $0 K$ . So N clings to its electron harder ⇒ higher IE for N. ✓ Why this step? Exchange loss is a real energetic cost of ionizing — it explains the N>O bump.

Verify: N loss $= 3 - 1 = 2$ ; O loss $= 3 - 3 = 0$ ; $2 > 0$ matches the observed IE dip at oxygen. ✓

Worked example Ex 7 — Chromium, full cross-subshell count (Cell G:

$3 d$ – $4 s$ terms) Statement: For $[A r] 3 d^{5} 4 s^{1}$ , all six outer electrons (5 in $3 d$ , 1 in $4 s$ ) are ↑. Count all same-spin pairs, then split into $3 d$ – $3 d$ and $3 d$ – $4 s$ .

Forecast: Will including the $4 s$ electron change the decision Cr makes?

Total pairs among 6 parallel electrons. $(2 6) = \frac{6 \cdot 5}{2} = 15.$ Why this step? Treat all six ↑ electrons as one parallel set first.
Within $3 d$ . $(2 5) = 10$ pairs (weight $K_{3 d, 3 d}$ ). Why this step? These are the big, tightly-overlapping pairs.
Cross $3 d$ – $4 s$ . $15 - 10 = 5$ pairs (one $4 s$ electron × 5 $3 d$ electrons), weight $K_{3 d, 4 s} ≪ K_{3 d, 3 d}$ . Why this step? The lone $4 s$ ↑ pairs with each of the five $3 d$ ↑, but the small spatial overlap makes these worth little.

Verify: $10 + 5 = 15 = (2 6)$ . ✓ Decomposition is exact; and since $K_{3 d, 4 s} ≪ K_{3 d, 3 d}$ , the $+ 4$ within- $3 d$ gain from Ex 2 remains the decision-maker for the transition-metal flip. ✓

Worked example Ex 8 — When the atom does

not flip (Cell H: exam twist, the balance) Statement: Suppose promoting an electron into a subshell would gain $Δ N_{ex} = 4$ exchange pairs (worth $\approx 4 K$ ), but costs a promotion energy $Δ E_{gap}$ to jump the orbital gap. Give the condition for the flip to happen, and decide two cases: (a) $Δ E_{gap} = 3 K$ , (b) $Δ E_{gap} = 6 K$ .

Forecast: Is the anomalous config always adopted? Guess: yes / no / depends.

Write the balance. Flip is favourable when the exchange gain exceeds the promotion cost: $4 K > Δ E_{gap} .$ Why this step? Nature picks the lower-energy config; the flip only wins if it lowers energy overall.
Case (a). $4 K > 3 K$ is true ⇒ atom flips (like $3 d \approx 4 s$ where the gap is small). Why this step? Small gap ⇒ exchange bonus dominates.
Case (b). $4 K > 6 K$ is false ⇒ atom keeps normal Aufbau filling. Why this step? A large orbital-energy gap swamps the exchange bonus — so the anomaly is not universal.

Verify: (a) $4 > 3$ ⇒ flip. (b) $4 < 6$ ⇒ no flip. ✓ Consistent with "flip only when $n s$ and $(n - 1) d$ are energetically close."

Recall

Recall Quick self-test

Exchange pairs among 5 parallel electrons? ::: $(2 5) = 10$ $N_{ex}$ for $d^{10}$ (both spins)? ::: $2 (2 5) = 20$ Extra $3 d$ – $3 d$ pairs on Cr's flip? ::: $10 - 6 = 4$ Exchange loss when ionizing N ( $2 p^{3} \to 2 p^{2}$ )? ::: $2 K$ Exchange loss when ionizing O ( $2 p^{4} \to 2 p^{3}$ )? ::: $0 K$ $(2 1)$ and $(2 0)$ ? ::: both $= 0$ (no pair from ≤1 electron) Flip condition in words? ::: exchange gain $>$ orbital-gap cost