Question bank — Stability of half-filled and fully-filled subshells

False — completeness is a symptom, not the cause; the real driver is exchange energy (parallel-spin swaps) plus a symmetric charge cloud.

False — exchange only acts between electrons of the same spin; opposite-spin electrons cannot be swapped without producing a physically new state, so they contribute no exchange term.

False — the exchange integral $K_{ij}$ depends on the spatial overlap of orbitals $i$ and $j$, so $3d$–$3d$, $3d$–$4s$ and $4s$–$4s$ pairs each have a different $K$; the single-$K$ formula is only an approximation.

True — an exchange is an unordered choice of 2 electrons out of $n$, which is exactly the combination $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$.

False — promoting to $3d$ actually adds some repulsion; it wins because the exchange gain ($\approx -4K_{3d,3d}$, from 6→10 pairs) more than compensates.

False — it only happens when the $ns$ and $(n-1)d$ subshells are energetically close ($\Delta E\lesssim$ exchange gain); when the gap is large the exchange bonus can't overcome it and normal Aufbau filling holds.

False — $3d$ and $4s$ sit very close; once occupied $3d$ can dip below $4s$, so on ionization the ==$4s$ electron leaves first== (see Electronic Configuration of Transition Elements).

True — removing an electron from the stable, symmetric $2p^3$ (3 exchange pairs) costs extra, so N's ${\text{IE}}_1$ exceeds O's despite oxygen's higher nuclear charge.

True (as a decision) — cross $3d$–$4s$ terms exist but $K_{3d,4s}\ll K_{3d,3d}$, so the +4 within-$3d$ pairs is the dominant, deciding effect.

False — the five spin-up electrons still exchange among themselves ($\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ pairs) and the five spin-down do too; pairing does not switch exchange off within each spin set.

The error is calling less pairing the cause; the actual reason is the exchange stabilization from 4 extra parallel $3d$–$3d$ pairs, not merely reduced pairing.

The count is $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{n(n-1)}{2}$, which grows like $n^2/2$, and it counts only same-spin electrons — not all $n$ with all $n$.

Symmetry is one of two reasons; the exchange/pairing stabilization of the completed $d^{10}$ plus the near-equal $3d/4s$ energies together drive it, not symmetry alone.

Cr⁺ is $[Ar]3d^5$ (the $4s^1$ electron leaves), so it actually keeps the half-filled $d^5$ — the anomaly's stability persists through the first ionization.

The exchange integral is always positive ($K_{ij}>0$); it merely gets small for distant/low-overlap orbitals, so each pair only stabilizes (never destabilizes) — it just does so weakly.

They don't — the exchange bonus is maximal at the exactly half-filled $2p^3$ (nitrogen); with only 1 or 2 $p$ electrons there are 0 or 1 exchange pairs, far from the peak, so no such anomaly appears.

If we could tell them apart, swapping them would create a genuinely different arrangement; only because they are identical does the swap leave the physical state unchanged, producing the extra stabilizing quantum term.

The comparison hinges on same-type ($3d$–$3d$) pairs whose $K$ values are nearly equal, so approximating them by one $K$ preserves the ordering even though it distorts the absolute energy.

Hund's rule maximizes the number of parallel spins, and each extra parallel-spin electron adds more exchange pairs; maximum multiplicity is literally the arrangement that maximizes exchange stabilization.

Pauli demands the total wavefunction be antisymmetric, i.e. it flips sign when two electrons are swapped — exchange is exactly this built-in swap symmetry, not a violation; for same-spin electrons that antisymmetric spatial part keeps them apart, and the associated overlap term $-K_{ij}$ is the stabilizing energy Pauli's own structure produces.

Cramming a fifth electron into the compact $3d$ set raises inter-electron repulsion and the bare orbital energy slightly; the anomaly occurs only because exchange gain outweighs this cost.

$d^{10}$ combines maximal exchange within both spin sets with a perfectly spherical, symmetric charge cloud, so both stabilizing effects peak together, making the flip robust.

The $(n+l)$ rule only predicts the naive filling order $4s$ before $3d$; it has no term for exchange energy, so it cannot capture the anomalous flip that overrides that order.

It does — any exactly half-filled subshell qualifies; $2p^3$ has $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ parallel-spin exchange pairs, which is why nitrogen shows an IE anomaly.

$N_{\text{ex}}=\left(\genfrac{}{}{0ex}{}{0}{2}\right)=0$, so an empty subshell has no exchange stabilization — but this simply means the effect is absent, not that the atom is destabilized.

$\left(\genfrac{}{}{0ex}{}{1}{2}\right)=0$ — you need at least two same-spin electrons before any swap (and thus any exchange stabilization) is possible.

Each spin set is complete — 3 spin-up ($p$) exchange as $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ pairs and 3 spin-down likewise, giving $3+3=6$ pairs total (for $f^{14}$: $\left(\genfrac{}{}{0ex}{}{7}{2}\right)+\left(\genfrac{}{}{0ex}{}{7}{2}\right)=42$); combined with a perfectly spherical cloud this is the maximal, most stable full-shell case — exactly why noble-gas configurations are so inert.

No — it's an even stronger version of the same balance: the fully-filled $4d^{10}$ exchange/symmetry bonus is large enough that both $5s$ electrons drop into $4d$, again driven by exchange stabilization, not by Aufbau.

Yes in effect — with zero energy cost to move an electron, even a small exchange gain toward $d^5$ or $d^{10}$ tips the balance, so the half/full arrangement is always preferred.

Yes — any two parallel-spin electrons contribute a $K_{ij}$ term regardless of subshell, but the cross-subshell $K$ is small because their spatial overlap is low.