Stability of half-filled and fully-filled subshells

WHAT is going on?

The two physical reasons behind this stability are:

Symmetrical distribution of electron density — a set of orbitals each with equal occupation (all singly or all doubly filled) gives a spherically symmetric, evenly-spread charge cloud that shields the nucleus uniformly and lowers repulsion.
Exchange energy — a purely quantum-mechanical stabilization that grows with the number of parallel-spin electron pairs. This is the dominant reason.

WHY: Exchange energy from first principles

HOW to count the stabilization (approximate model). Count the number of distinct pairs of parallel-spin electrons. If a subshell has $n$ electrons all with parallel spin, the number of exchange pairs is:

N_{\text{ex}} = \binom{n}{2} = \frac{n(n-1)}{2}

The $d^5$ vs $d^4$ argument (why $3d^5$ wins)

For $d^5$ all five electrons are parallel (Hund's rule), so among the $3d$ electrons: $N_{\text{ex}}(3d\text{–}3d) = \binom{5}{2} = 10 \text{ exchange pairs}$

Compare Chromium's two candidate configurations. Below we count within- $3d$ pairs (the biggest, most comparable contribution):

Config	$3d$ spins	$4s$ spins	$3d$ – $3d$ parallel pairs
$[Ar]3d^4 4s^2$	↑↑↑↑ (4 up)	↑↓ (paired)	$\binom{4}{2}=6$
$[Ar]3d^5 4s^1$	↑↑↑↑↑ (5 up)	↑ (1 up)	$\binom{5}{2}=10$

Going $3d^4 4s^2 \to 3d^5 4s^1$ gains 4 extra $3d$ – $3d$ exchange pairs ( $10-6=4$ ), i.e. $\approx -4K_{3d,3d}$ .

Similarly for Cu: $3d^9 4s^2 \to 3d^{10} 4s^1$ completes the $d$ subshell (fully-filled $d^{10}$ ), maximizing both symmetry and pairing stabilization.

Figure — Stability of half-filled and fully-filled subshells

Worked examples

Common mistake Steel-manning the wrong ideas

Wrong idea 1: "Half/full shells are stable because their energy orbitals are 'complete'." Why it feels right: symmetry and neatness suggest closure. The fix: the real driver is exchange energy (parallel-spin swaps) plus symmetric charge distribution — not mere "completeness."

Wrong idea 2: " $4s$ is always lower than $3d$ , so it always fills first and never empties." Why it feels right: Aufbau order lists $4s$ before $3d$ . The fix: $3d$ and $4s$ are very close in energy; once occupied, $3d$ can dip below $4s$ . So an atom can put an electron in $3d$ (Cr, Cu) or remove $4s$ first during ionization.

Wrong idea 3: "Every same-spin pair lowers energy by the SAME amount $K$ ." Why it feels right: the neat $\binom{n}{2}K$ formula treats all pairs identically. The fix: the exchange integral $K_{ij}$ depends on the spatial overlap of orbitals $i$ and $j$ , so a $3d$ – $3d$ pair, a $3d$ – $4s$ pair, and a $4s$ – $4s$ pair each have different $K$ . The single- $K$ rule is an approximation; the exact energy is $-\sum_{i<j}K_{ij}$ .

Wrong idea 4: "Cr is $3d^5 4s^1$ so it must have LESS repulsion than $3d^4 4s^2$ ." Why it feels right: fewer paired electrons = less repulsion. The fix: moving an electron into $3d$ actually adds some repulsion; the reason it still wins is the exchange stabilization ( $\approx -4K_{3d,3d}$ ) more than compensating.

Active recall

Recall Test yourself (hover/expand)

Which three $p/d/f$ occupations count as "half-filled"? → $p^3, d^5, f^7$ .
Formula for number of exchange pairs among $n$ parallel electrons? → $\binom{n}{2}=\frac{n(n-1)}{2}$ .
Is the exchange integral $K$ the same for every pair? → No — $K_{ij}$ depends on orbital overlap ( $3d$ – $3d \ne 3d$ – $4s$ ); single- $K$ is an approximation.
Why does Cr adopt $3d^5 4s^1$ ? → gains 4 extra $3d$ – $3d$ exchange pairs + symmetry, and $3d\approx4s$ energy.
Two physical reasons for extra stability? → symmetric charge distribution + exchange energy.

Recall Feynman: explain to a 12-year-old

Imagine electrons are kids on swings, and identical kids (same spin) love to swap swings. Every possible swap makes the whole playground a little happier (lower energy). But swaps between kids on nearby swings help more than swaps between kids far apart — so not every swap is worth the same. When each swing has exactly one kid facing the same way, there are LOTS of possible swaps, so the playground is super happy — that's a "half-filled" shell. Sometimes an atom will even move one kid from a nearby playground just to make more swaps possible. That's why Chromium and Copper "cheat" the normal seating order!

Connections

Hund's Rule of Maximum Multiplicity — parallel spins → exchange energy.
Aufbau Principle — the order the anomalies break.
(n+l) Rule and Orbital Energies — why $3d$ and $4s$ are close.
Ionization Energy Trends — N vs O anomaly from half-filled $2p^3$ .
Electronic Configuration of Transition Elements — where Cr, Cu, Mo, Ag, Au appear.
Pauli Exclusion Principle — indistinguishability underlying exchange.

Which subshell occupations are "half-filled" and extra stable?

p^3

d^5

f^7

(one electron per orbital, all parallel spins).

Which subshell occupations are "fully-filled"?

p^6

d^{10}

f^{14}

(two electrons per orbital).

What are the TWO physical reasons for half/full-filled stability?

Symmetrical charge distribution + exchange energy (the dominant one).

Formula for number of exchange pairs among

n

parallel-spin electrons?

\binom{n}{2} = \dfrac{n(n-1)}{2}

Is the exchange integral the same for every same-spin pair?

No —

K_{ij}

depends on spatial overlap of orbitals

i,j

; the single-

K

\binom{n}{2}K

form is an approximation. Exact:

E=-\sum_{i<j}K_{ij}

Electronic configuration of Cr (

Z=24

[Ar]3d^5 4s^1

(not

3d^4 4s^2

Electronic configuration of Cu (

Z=29

[Ar]3d^{10} 4s^1

(not

3d^9 4s^2

How many extra 3d–3d exchange pairs does Cr gain going

3d^4 4s^2 \to 3d^5 4s^1

10 - 6 = 4

pairs, i.e.

\approx -4K_{3d,3d}

(ignoring smaller cross

3d

–

4s

terms).

Why is exchange energy negative (stabilizing)?

Indistinguishable same-spin electrons can swap positions; each swap lowers energy by its

K_{ij}

Why can an atom promote an electron from

4s

3d

3d

and

4s

energies are very close, so exchange + symmetry gains can outweigh the small cost.

Why is nitrogen's first ionization energy higher than oxygen's?

N has a stable half-filled

2p^3

; removing an electron disrupts it, costing extra energy.

Does the half/full-filled anomaly happen for ALL elements?

No — only when

ns

and

(n-1)d

energies are close enough for exchange gain to win.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, electrons ek subshell mein ek doosre ko repel karte hain, lekin ek chhupa hua quantum bonus bhi milta hai — jise exchange energy kehte hain. Jab do electrons ka spin same hota hai (dono up ya dono down), toh woh aapas mein jagah swap kar sakte hain, aur har possible swap se poore atom ki energy thodi si kam ho jaati hai. Jitne zyada parallel-spin electrons, utne zyada swaps, utni zyada stability. Isiliye half-filled ( $d^5$ ) aur fully-filled ( $d^{10}$ ) configurations extra stable hote hain.

Ek important baat: har swap ki energy ek jaisi nahi hoti. Exchange integral $K_{ij}$ do orbitals ke overlap par depend karta hai — $3d$ – $3d$ pair ka $K$ alag, $3d$ – $4s$ pair ka alag, aur $4s$ – $4s$ ka alag. Textbook wala neat formula $\binom{n}{2}K$ ek average $K$ maan leta hai — intuition aur exam ke liye badhiya, par yaad rakhna asli energy $-\sum_{i<j}K_{ij}$ hoti hai.

Number of swaps count karna easy hai: $n$ parallel electrons mein $\binom{n}{2}=\frac{n(n-1)}{2}$ pairs. Chromium mein dekho — $3d^4 4s^2$ mein 4 up $3d$ electrons (6 pairs), aur $3d^5 4s^1$ mein 5 up $3d$ electrons (10 pairs). Yaani 4 extra $3d$ – $3d$ exchange pairs $\approx -4K_{3d,3d}$ ! ( $3d$ – $4s$ cross terms bhi hote hain par chote hote hain kyunki overlap kam hai.) Isiliye Cr apni normal filling tod ke $3d^5 4s^1$ chunta hai. Same Copper ke saath — $3d^{10} 4s^1$ , fully-filled $d^{10}$ ke liye. Aur yeh cheating tabhi hoti hai jab $3d$ aur $4s$ ki energy close ho. Exam mein Cr, Cu ki config + exchange energy reason — high-yield 80/20 point hai!

Stability of half-filled and fully-filled subshells

WHAT is going on?

WHY: Exchange energy from first principles

The $d^5$ vs $d^4$ argument (why $3d^5$ wins)

Worked examples

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Quantum Atomic Structure

Connections