Stability of half-filled and fully-filled subshells: Exercises

Before we start, one tool we use over and over — the exchange-pair counter:

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

Level 1 — Recognition

L1.1

State whether each subshell occupation is half-filled, fully-filled, or neither: $2 p^{6}$ , $3 d^{4}$ , $4 f^{7}$ , $3 d^{10}$ , $2 p^{3}$ , $3 d^{6}$ .

Recall Solution

A subshell is half-filled when every orbital holds exactly one electron; fully-filled when every orbital holds two.

$p$ has 3 orbitals ⇒ half = $p^{3}$ , full = $p^{6}$ .
$d$ has 5 orbitals ⇒ half = $d^{5}$ , full = $d^{10}$ .
$f$ has 7 orbitals ⇒ half = $f^{7}$ , full = $f^{14}$ .

Occupation	Verdict
$2 p^{6}$	fully-filled
$3 d^{4}$	neither
$4 f^{7}$	half-filled
$3 d^{10}$	fully-filled
$2 p^{3}$	half-filled
$3 d^{6}$	neither

L1.2

Write the actual ground-state configuration of Cr ( $Z = 24$ ) and Cu ( $Z = 29$ ), and name the type of extra-stable subshell each achieves.

Recall Solution

Cr $= [Ar] 3 d^{5} 4 s^{1}$ → half-filled $3 d^{5}$ .
Cu $= [Ar] 3 d^{10} 4 s^{1}$ → fully-filled $3 d^{10}$ .

Both promote one $4 s$ electron into $3 d$ to reach a specially stable subshell. See Electronic Configuration of Transition Elements.

Level 2 — Application

L2.1

Count the number of parallel-spin exchange pairs for $2 p^{3}$ (nitrogen's valence set), where all three $p$ electrons are parallel.

Recall Solution

All 3 are parallel (Hund's rule), so: $N_{ex} = (2 3) = \frac{3 \cdot 2}{2} = 3 pairs.$

L2.2

For a fully parallel $3 d^{5}$ set, compute $N_{ex}$ and the approximate exchange stabilization in units of $K$ .

Recall Solution

$n = 5$ parallel electrons: $N_{ex} = (2 5) = \frac{5 \cdot 4}{2} = 10 pairs, E_{exch} \approx - 10 K .$

L2.3

A subshell has 4 parallel-spin electrons. How many exchange pairs? What if it gains a 5th parallel electron — how many new pairs appear?

Recall Solution

$n = 4$ : $(2 4) = 6$ pairs.
$n = 5$ : $(2 5) = 10$ pairs.
New pairs = $10 - 6 = 4$ . Adding the 5th electron creates one pair with each of the 4 already present. This "+4" is exactly the Cr driving force.

Level 3 — Analysis

L3.1

Chromium's two candidate configs: $[Ar] 3 d^{4} 4 s^{2}$ vs $[Ar] 3 d^{5} 4 s^{1}$ . Counting only within- $3 d$ parallel pairs, find the exchange gain (in units of $K_{3 d, 3 d}$ ) for the flip.

Recall Solution

$3 d^{4}$ (4 parallel up): $(2 4) = 6$ pairs.
$3 d^{5}$ (5 parallel up): $(2 5) = 10$ pairs.
Gain $= 10 - 6 = 4$ pairs $\Rightarrow Δ E_{exch} \approx - 4 K_{3 d, 3 d}$ .

This favorable exchange (plus $3 d \approx 4 s$ energies and the symmetric $d^{5}$ cloud) outweighs the cost of moving one electron $4 s \to 3 d$ , so Cr adopts $3 d^{5} 4 s^{1}$ .

L3.2

Same analysis for Copper: $[Ar] 3 d^{9} 4 s^{2} \to [Ar] 3 d^{10} 4 s^{1}$ . Explain why counting parallel $3 d$ – $3 d$ pairs alone is misleading here, and what really stabilizes $d^{10}$ .

Recall Solution

In a full $d^{10}$ subshell, each orbital holds an ↑ and a ↓ electron. Parallel-spin pairs exist within the spin-up set (5 up) and within the spin-down set (5 down): $N_{ex} = (2 5) + (2 5) = 10 + 10 = 20 pairs (both spin sets).$ Compared to $3 d^{9}$ (5 up, 4 down): $(2 5) + (2 4) = 10 + 6 = 16$ pairs. Gain $= 20 - 16 = 4$ pairs. Why "parallel only" misleads: you must count parallel pairs separately inside each spin channel — ↑ electrons never exchange with ↓ electrons. The real stabilization of $d^{10}$ is this maximal, perfectly symmetric filling plus the spherically symmetric charge cloud.

L3.3

Nitrogen ( $Z = 7$ ) has an unusually high first ionization energy compared with oxygen ( $Z = 8$ ). Using exchange pairs, explain why.

Recall Solution

N: $2 p^{3}$ , all parallel ⇒ $(2 3) = 3$ exchange pairs. Half-filled, very stable.
O: $2 p^{4}$ = one paired orbital + two singly filled ⇒ 3 parallel-up electrons give $(2 3) = 3$ pairs, and the extra electron is paired (adds repulsion, no new parallel pair).

Removing an electron from N destroys a stable half-filled set (costs extra). Removing the paired electron from O relieves repulsion and returns O $^{+}$ to the stable $2 p^{3}$ . So N's first ionization energy is higher than O's — a well-known dip. See Ionization Energy Trends.

Level 4 — Synthesis

L4.1

Manganese is $[Ar] 3 d^{5} 4 s^{2}$ (it does not flip like Cr). Explain why Mn keeps $4 s^{2}$ while Cr flips to $4 s^{1}$ .

Recall Solution

Mn already has a half-filled $3 d^{5}$ with its full $4 s^{2}$ . It has nothing to gain by promoting: moving a $4 s$ electron into $3 d$ would make $3 d^{6}$ (breaking the symmetric half-filled set and forcing a paired $d$ orbital) — a loss, not a gain. Cr, by contrast, is one electron short of $d^{5}$ ; promoting creates the half-filled set (+4 exchange pairs). The rule: the flip only happens when it completes a magic occupation.

L4.2

Predict the ground-state configuration of Molybdenum ( $Z = 42$ , same group as Cr) and explain using the exchange argument and (n+l) Rule and Orbital Energies.

Recall Solution

Mo lies below Cr, so its valence set is $4 d$ and $5 s$ . By analogy with Cr: $Mo = [Kr] 4 d^{5} 5 s^{1} .$ The half-filled $4 d^{5}$ gives $(2 5) = 10$ exchange pairs (vs 6 for $4 d^{4}$ ), a +4 gain, and $4 d \approx 5 s$ in energy. The flip is favorable for the same reason as Cr.

L4.3

Silver ( $Z = 47$ ) and Gold ( $Z = 79$ ) both end in $d^{10} s^{1}$ . Write both configs and state which magic occupation drives each.

Recall Solution

Ag $= [Kr] 4 d^{10} 5 s^{1}$ — fully-filled $4 d^{10}$ .
Au $= [Xe] 4 f^{14} 5 d^{10} 6 s^{1}$ — fully-filled $5 d^{10}$ (and the $4 f^{14}$ is also full).

Both complete a $d^{10}$ subshell (Cu-family behavior): maximum symmetry + maximum paired-set exchange stabilization outweighs the small $s \to d$ promotion cost because $d$ and $s$ energies are close in these rows.

Level 5 — Mastery

L5.1

For a hypothetical fully-parallel $f^{7}$ subshell (all 7 $f$ electrons up), compute (a) $N_{ex}$ , and (b) the number of extra exchange pairs gained relative to $f^{6}$ (6 parallel). Then state why $f^{7}$ (e.g. Eu, Gd) is famously stable.

Recall Solution

(a) $n = 7$ : $(2 7) = \frac{7 \cdot 6}{2} = 21$ pairs. (b) $f^{6}$ : $(2 6) = 15$ pairs. Extra $= 21 - 15 = 6$ pairs. The half-filled $f^{7}$ maximizes parallel-spin exchange (21 pairs, the most any single-spin $f$ set can have) and gives a spherically symmetric charge cloud — that is why Gd is $[Xe] 4 f^{7} 5 d^{1} 6 s^{2}$ and Eu $^{2 +}$ /Gd $^{3 +}$ ( $f^{7}$ ) are exceptionally stable.

L5.2

Suppose a $3 d$ – $3 d$ exchange integral is $K_{dd}$ and a $3 d$ – $4 s$ one is $K_{d s}$ , with $K_{d s} = 0.2 K_{dd}$ . For Cr, the flip gains $+ 4$ within- $3 d$ pairs and additionally makes the lone $4 s ↑$ parallel to all five $3 d ↑$ (5 new cross pairs), while the $3 d^{4} 4 s^{2}$ side has the $4 s$ paired (0 cross parallel pairs). Compute the total exchange gain in units of $K_{dd}$ .

Recall Solution

Within- $3 d$ gain: $+ 4 K_{dd}$ .
Cross $3 d$ – $4 s$ gain: the $4 s ↑$ is parallel to 5 $3 d ↑$ electrons ⇒ 5 new cross pairs $= 5 K_{d s} = 5 (0.2 K_{dd}) = 1.0 K_{dd}$ .
Total stabilization $\approx 4 K_{dd} + 1.0 K_{dd} = 5.0 K_{dd}$ .

So including the (smaller) cross terms increases the total exchange gain from $4 K_{dd}$ to $5 K_{dd}$ — but the decision-making part is still the dominant within- $3 d$ $+ 4 K_{dd}$ , consistent with the topic note.

L5.3

Using the single- $K$ model, show that the exchange stabilization of a fully parallel subshell grows quadratically with $n$ , and explain physically why "each added electron helps more than the last."

Recall Solution

$E_{exch} \approx - (2 n) K = - \frac{n ( n - 1 )}{2} K = - \frac{K}{2} (n^{2} - n) .$ The leading term is $- \frac{K}{2} n^{2}$ — quadratic in $n$ . Physically: adding the $n$ -th parallel electron creates $(n - 1)$ new pairs (one with every existing electron). Since $(n - 1)$ grows with $n$ , each successive electron adds more new pairs than the previous one — a running total that curves upward. This is why maximizing parallel spins (Hund's rule) and reaching the half-filled peak pays off so strongly.

Recall Master checklist (expand)

Magic occupations? ::: $p^{3}, d^{5}, f^{7}$ (half); $p^{6}, d^{10}, f^{14}$ (full).
Exchange pairs among $n$ parallel electrons? ::: $(2 n) = \frac{n ( n - 1 )}{2}$ .
Cr flip gain (within- $3 d$ )? ::: $10 - 6 = 4$ pairs $\approx - 4 K_{3 d, 3 d}$ .
Why Mn does NOT flip? ::: it already has $3 d^{5}$ ; promoting would break it.
Is single- $K$ exact? ::: No — $K_{ij}$ depends on orbital overlap.
Why does the flip need $n s \approx (n - 1) d$ ? ::: the exchange bonus must beat the promotion cost, only possible when the levels are close.