Electronic configuration of elements (Z = 1 to 30) — exceptions Cr, Cu: Visual walkthrough

Step 1 — What a "box" and an "arrow" mean

WHY this picture? The Pauli exclusion principle says one box holds at most two electrons, and only if their arrows point opposite ways. So a box is either empty, half-full ( $↑$ ), or full ( $↑↓$ ). This is the entire alphabet of the derivation — everything below is built from boxes and arrows.

PICTURE — the three legal states of a single box, and the one forbidden state:

Step 2 — The subshells as rows of boxes

WHY 5 boxes for $d$ ? The letter $d$ means the angular-momentum label $l = 2$ , and the number of boxes in any subshell is $2 l + 1$ . For $d$ : $2 (2) + 1 = 5$ . Those five boxes hold at most $5 \times 2 = 10$ electrons — that is where the " $d^{10}$ " ceiling comes from.

PICTURE — the two subshells that fight in this whole story, drawn to the same energy scale:

Step 3 — The naïve fill: obey Aufbau blindly

WHY $4 s$ before $3 d$ ? The Madelung $(n + l)$ score: $4 s$ has $4 + 0 = 4$ , while $3 d$ has $3 + 2 = 5$ . Lower score = lower energy = filled first (this comes from Shielding and penetration effect — the $4 s$ cloud dips close to the nucleus). So the naïve recipe puts $2$ arrows in $4 s$ , then the remaining $4$ into $3 d$ , spread out one-per-box by Hund's rule of maximum multiplicity.

PICTURE — the naïve Chromium, $3 d^{4} 4 s^{2}$ :

Notice the $d$ -row: four boxes filled, one empty. That lonely empty box is the whole problem. Hold that image.

Step 4 — The hidden currency: exchange energy

WHY does parallel spin help? Quantum mechanically, same-spin electrons avoid each other in space (they can never sit in the same box, and they dodge each other even between boxes). Less crowding → less repulsion → lower energy. The more parallel pairs you can make, the more coins you collect.

HOW MANY pairs? If a subshell has $k$ arrows all pointing the same way, count how many ways to pick $2$ of them:

PICTURE — the coins drawn as links between arrows, for $k = 4$ vs $k = 5$ :

Count the links: $4$ arrows make $(2 4) = 6$ links; $5$ arrows make $(2 5) = 10$ links. Going from $4$ to $5$ parallel arrows is worth $10 - 6 = 4$ extra coins.

Step 5 — The trade: promote one electron

WHY would nature pay to move an electron uphill? Moving $4 s \to 3 d$ costs a tiny bit of energy (because $3 d$ sits slightly higher). But look at what we buy: the $d$ -subshell jumps from $4$ parallel arrows to $5$ , gaining $4$ exchange coins (Step 4). Near $Z = 24$ the $4 s$ and $3 d$ energies are almost equal, so the tiny promotion cost is smaller than the $4$ -coin bonus. Net total energy drops. Nature minimises total energy, not "obedience to Aufbau".

PICTURE — the before/after with the moving arrow highlighted, and the coin tally underneath:

Step 6 — Copper: the same trick, taken to completion

WHY is $d^{10}$ special? A completely filled subshell is perfectly symmetric — the electron cloud is a smooth sphere, so electrons shield each other with maximum efficiency and repulsion is minimised. That symmetric-completion bonus, plus every possible exchange coin already collected, outweighs the promotion cost. Same logic as Chromium, pushed all the way to a full shell.

PICTURE — naïve $3 d^{9} 4 s^{2}$ vs real $3 d^{10} 4 s^{1}$ :

Step 7 — The degenerate & boundary cases (why only Cr, Cu ≤ 30)

WHY each neighbour stays honest:

V ( $3 d^{3} 4 s^{2}$ ): promoting gives $3 d^{4} 4 s^{1}$ → $P (4) - P (3) = 6 - 3 = 3$ coins, but you don't reach the magic half-full $d^{5}$ , so there is no symmetry bonus and the promotion cost wins. Stays $3 d^{3} 4 s^{2}$ .
Mn ( $3 d^{5} 4 s^{2}$ ): it is already at the magic $d^{5}$ and has a filled $4 s^{2}$ . Promoting would break $4 s$ for nothing — no empty $d$ -box left to gain a coin. Stays put.
Zn ( $3 d^{10} 4 s^{2}$ ): already $d^{10}$ and $4 s^{2}$ ; nothing left to promote. Stays put.

So the exception fires only when a single promotion lands exactly on $d^{5}$ or $d^{10}$ from one box short — that is precisely $Z = 24$ and $Z = 29$ in this range.

PICTURE — the neighbour map: who can reach a magic shell in one hop and who cannot:

Recall Check yourself

Why is Manganese ( $3 d^{5} 4 s^{2}$ ) NOT an exception even though it has the magic $d^{5}$ ? ::: It reaches $d^{5}$ without promoting — its $4 s$ is already full and there is no empty $d$ -box to move into, so no trade is possible or needed. Why does Vanadium not become $3 d^{4} 4 s^{1}$ ? ::: The promotion would gain only $3$ exchange coins and would not land on a magic half/full shell, so the promotion cost outweighs the bonus.

The one-picture summary

The whole derivation compressed: start from boxes+arrows → count exchange coins $(2 k)$ → when one promotion lands on the magic $d^{5}$ or $d^{10}$ , the coin+symmetry bonus beats the tiny $4 s \to 3 d$ cost → Cr and Cu (and only them, for $Z \leq 30$ ) rearrange.

Recall Feynman retelling — say it in plain words

Imagine electrons as kids picking seats, and every time two kids facing the same way sit in the same row they get a friendship coin. Nature loves coins. The rule-book (Aufbau) says fill the cheap $4 s$ chair first, then the five $d$ chairs. For most atoms that's the cheapest arrangement overall. But at Chromium, the $d$ -row is one kid short of being completely one-directional (all five facing the same way). If one kid hops out of the nearly-equal-cost $4 s$ chair into that last empty $d$ seat, the row goes from four to five same-facing kids — and the friendship coins jump from six to ten. Four bonus coins is worth more than the tiny cost of the hop, so nature makes the swap: Chromium becomes $3 d^{5} 4 s^{1}$ . Copper is the same story but the hop completes the whole $d$ -row into perfectly balanced pairs ( $3 d^{10}$ ), which is even better, so it swaps too. Every other atom nearby either can't reach a magic row in one hop, or is already sitting on one — so they don't bother. That's the entire secret: nature counts total coins, not rule-book obedience.

This connects onward to Magnetic properties of transition metals (unpaired arrows = magnetism) and Periodic trends.

Flashcards

How many exchange pairs do

k

parallel electrons make?

(2 k) = \frac{k ( k - 1 )}{2}

Exchange coins gained going

3 d^{4} \to 3 d^{5}

10 - 6 = 4

pairs.

Why does Cr promote a

4 s

electron?

The

4

extra exchange pairs from reaching half-filled

d^{5}

outweigh the tiny

4 s \to 3 d

promotion cost.

Why is Cu

3 d^{10} 4 s^{1}

A completely filled, spherically symmetric

d^{10}

is extra stable, beating the promotion cost.

Why is Mn (

3 d^{5} 4 s^{2}

) not an exception?

It reaches

d^{5}

without any promotion;

4 s

is already full and no empty

d

-box remains.