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

Before anything, a reminder of the vocabulary we lean on, so no symbol appears un-earned:

Recall The four tools we reuse (click to open)

Aufbau principle — fill the cheapest (lowest-energy) orbital first. Order for us: $1 s < 2 s < 2 p < 3 s < 3 p < 4 s < 3 d < 4 p$ .
Pauli exclusion principle — one orbital = a box holding at most 2 electrons, opposite spins ( $↑↓$ ).
Hund's rule of maximum multiplicity — within a subshell, put one electron in each box (all spins $↑$ ) before you double any box up.
Quantum numbers (n, l, m, s) — $n$ = which shell (size), $l$ = subshell shape ( $s = 0, p = 1, d = 2$ ).

A subshell $n l$ is a set of equal-energy boxes: $s$ has 1 box (holds 2), $p$ has 3 boxes (holds 6), $d$ has 5 boxes (holds 10).

The scenario matrix

Every question this topic can throw is one of these cells. The examples below each carry a tag [Cell #] so you can see the whole space gets covered.

Cell	Case class	The specific trap it tests
1	Simple $s$ / $p$ block atom	Straight Aufbau, no $d$ , no exception
2	Normal $3 d$ transition metal	$4 s$ before $3 d$ ; count unpaired via Hund
3	Exception Cr ( $3 d^{5} 4 s^{1}$ )	Half-filled stability, exchange-pair count
4	Exception Cu ( $3 d^{10} 4 s^{1}$ )	Fully-filled stability
5	Cation (remove electrons)	Remove from highest- $n$ ( $4 s$ ) first, not $3 d$
6	Degenerate / zero case	$d^{0}$ , $d^{10}$ , closed shell → 0 unpaired, diamagnetic
7	Real-world word problem	Magnetism / paramagnetism prediction
8	Exam twist	"Which is more stable?" via exchange-energy arithmetic

Example 1 — Sulfur, a plain $p$ -block atom `[Cell 1]`

Worked example Statement

Write the full electronic configuration of sulfur ( $Z = 16$ ) and state how many unpaired electrons it has.

Forecast: guess the number of unpaired electrons before reading on — is it 0, 1, or 2?

Step 1 — Lay down the Aufbau order until 16 electrons are placed. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}$ . Count: $2 + 2 + 6 + 2 + 4 = 16$ . ✔ Why this step? Aufbau says fill cheapest first; $3 p$ ( $n + l = 4$ ) is the last open subshell reached before we run out of electrons.

Step 2 — Distribute the final $3 p^{4}$ among the three $p$ -boxes using Hund. Three boxes: put one $↑$ in each (3 electrons), then the 4th must pair up: $3 p : (↑↓) (↑) (↑)$ Why this step? Hund fills singly first; only the 4th electron is forced to share a box.

Step 3 — Count boxes with a lone electron. Two boxes hold a single $↑$ → 2 unpaired electrons.

Verify: total spins drawn $= 5 ↑ + 1 ↓$ inside $3 p$ gives 4 electrons — matches $3 p^{4}$ . The pairing pattern is forced, so 2 unpaired is unique. ✔

Example 2 — Titanium, a normal transition metal `[Cell 2]`

Example 3 — Chromium, the half-filled rebel `[Cell 3]`

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

Worked example Statement

Explain, by counting exchange pairs, why chromium ( $Z = 24$ ) is $[Ar] 3 d^{5} 4 s^{1}$ and not $[Ar] 3 d^{4} 4 s^{2}$ .

Forecast: which arrangement has MORE parallel-spin pairs — and does more pairs mean lower or higher energy?

Step 1 — Recall the exchange rule. Stability rises with the number of pairs of parallel-spin electrons in a subshell. For $k$ parallel electrons that number is $(2 k) = \frac{k ( k - 1 )}{2} .$ Why this step? Exchange energy and stability tells us electrons of the same spin can swap places, and each such swap-able pair lowers energy. So we literally count pairs.

Step 2 — Count for the naïve arrangement $3 d^{4} 4 s^{2}$ . $3 d^{4}$ = 4 parallel electrons → $(2 4) = 6$ pairs. The $4 s^{2}$ is one box with opposite spins → 0 parallel pairs. Total = 6. Why this step? Only same-spin electrons in the same subshell count; the paired $4 s$ contributes nothing.

Step 3 — Count for the real arrangement $3 d^{5} 4 s^{1}$ . $3 d^{5}$ = 5 parallel electrons → $(2 5) = 10$ pairs. $4 s^{1}$ is a lone electron → 0 pairs. Total = 10.

Step 4 — Compare. Gain of $10 - 6 = 4$ extra exchange pairs (look at the orange bar in the figure). That stabilization beats the tiny cost of promoting one electron from $4 s$ to the barely-higher $3 d$ . Why this step? Nature minimizes total energy, not "obey Aufbau literally."

Verify: $(2 4) = \frac{4 \cdot 3}{2} = 6$ , $(2 5) = \frac{5 \cdot 4}{2} = 10$ , difference $= 4$ . Electron total for $3 d^{5} 4 s^{1}$ : $18 + 5 + 1 = 24 = Z$ . ✔

Example 4 — Copper, the fully-filled rebel `[Cell 4]`

Worked example Statement

Show that copper ( $Z = 29$ ) is $[Ar] 3 d^{10} 4 s^{1}$ , and give its unpaired-electron count.

Forecast: a completely full $d$ subshell — how many unpaired electrons can it possibly have?

Step 1 — Naïve Aufbau prediction. $29 - 18 = 11$ valence electrons → $3 d^{9} 4 s^{2}$ . Why this step? Blind Aufbau puts 2 in $4 s$ , 9 in $3 d$ .

Step 2 — Promote one $4 s$ electron to complete the $d$ . $3 d^{9} 4 s^{2} \to 3 d^{10} 4 s^{1}$ . A full $3 d^{10}$ has maximal symmetry (every box $↑↓$ ) → extra stability. Why this step? A completely-filled subshell is spherically symmetric, so electrons shield each other most evenly → lowest repulsion. Same logic family as Cr.

Step 3 — Count unpaired. In $3 d^{10}$ every box is paired → 0. The lone $4 s^{1}$ → 1 unpaired electron. Final: $[Ar] 3 d^{10} 4 s^{1}$ .

Verify: $18 + 10 + 1 = 29 = Z$ . ✔ Only the single $4 s$ electron is unpaired, so Cu is weakly paramagnetic.

Example 5 — The Fe³⁺ ion: removal order `[Cell 5]`

Worked example Statement

Iron is $[Ar] 3 d^{6} 4 s^{2}$ . Write $Fe^{3 +}$ and count its unpaired electrons.

Forecast: we filled $4 s$ before $3 d$ — so when removing electrons, do the $3 d$ ones leave first?

Step 1 — Golden rule for ions: remove from the highest $n$ first. Highest $n$ present is $4 s$ ( $n = 4$ ). Remove both: $[Ar] 3 d^{6} 4 s^{0}$ . That's 2 electrons gone. Why this step? In the ion, the outer $4 s$ electrons are largest/most loosely held, so they leave first — even though they filled last. (Filling order ≠ removal order.)

Step 2 — Remove the 3rd electron from $3 d$ . $3 d^{6} \to 3 d^{5}$ . Final: $[Ar] 3 d^{5}$ . Why this step? After $4 s$ is empty, the next-highest available electrons are in $3 d$ .

Step 3 — Count unpaired in $3 d^{5}$ . Five boxes, one $↑$ each (Hund) → 5 unpaired.

Verify: electrons in ion $= 26 - 3 = 23$ . In $[Ar] 3 d^{5}$ : $18 + 5 = 23$ . ✔ Fe³⁺ is a classic $d^{5}$ high-spin ion — strongly paramagnetic. See Magnetic properties of transition metals.

Example 6 — Zinc: the fully-closed (degenerate) case `[Cell 6]`

Example 7 — Word problem: "the strongest magnet in the row" `[Cell 7]`

Worked example Statement

A lab has neutral atoms of Mn ( $Z = 25$ ), Fe ( $Z = 26$ ) and Cr ( $Z = 24$ ). Which single neutral atom has the most unpaired electrons (biggest spin magnetism)?

Forecast: guess the winner before the arithmetic.

Step 1 — Write each valence configuration.

Cr: $[Ar] 3 d^{5} 4 s^{1}$ (exception).
Mn: $[Ar] 3 d^{5} 4 s^{2}$ .
Fe: $[Ar] 3 d^{6} 4 s^{2}$ . Why this step? Unpaired count needs the actual boxes, so we write the real (not naïve) configs.

Step 2 — Count unpaired for each.

Cr: $3 d^{5}$ (5 singles) $+ 4 s^{1}$ (1 single) → 6.
Mn: $3 d^{5}$ (5 singles) $+ 4 s^{2}$ (paired) → 5.
Fe: $3 d^{6}$ = 5 boxes filled singly then 1 pairs → 4 unpaired in $d$ ; $4 s^{2}$ paired → 4. Why this step? Hund gives the single-vs-paired split in each subshell.

Step 3 — Pick the max. $6 > 5 > 4$ → Chromium wins with 6 unpaired electrons.

Verify: Cr's exception ( $4 s \to 3 d$ promotion) is exactly what gives it 6 rather than Mn's 5 — the half-filled trick pays off in magnetism too. Totals: Cr $18 + 5 + 1 = 24$ , Mn $18 + 5 + 2 = 25$ , Fe $18 + 6 + 2 = 26$ . ✔

Example 8 — Exam twist: prove the stability numerically `[Cell 8]`

Worked example Statement

Compare the exchange stabilization of $3 d^{5}$ versus $3 d^{4}$ and of $3 d^{10}$ (as two spin-sets of 5) versus $3 d^{9}$ , to justify both exceptions in one calculation.

Forecast: which jump gains more exchange pairs — the $d^{4} \to d^{5}$ (Cr) or the "top-up" toward $d^{10}$ (Cu)?

Step 1 — Cr side: $3 d^{4} \to 3 d^{5}$ . Parallel electrons go $4 \to 5$ . $Δ_{Cr} = (2 5) - (2 4) = 10 - 6 = 4 pairs.$ Why this step? All $d^{5}$ electrons share one spin, so we count within a single parallel set.

Step 2 — Cu side: count both spin-sets in $d^{10}$ . A full $d^{10}$ = 5 up + 5 down. Exchange pairs occur only within a spin set, so total $= (2 5) + (2 5) = 10 + 10 = 20$ . For $d^{9}$ = 5 up + 4 down: $(2 5) + (2 4) = 10 + 6 = 16$ . $Δ_{Cu} = 20 - 16 = 4 pairs.$ Why this step? Opposite-spin electrons don't exchange, so we sum each spin block separately (see the two coloured columns in the figure).

Step 3 — Read the result. Both promotions ( $4 s \to 3 d$ ) buy the atom +4 exchange pairs, plus the symmetry bonus of a half/full shell — enough to overcome the small promotion cost each time. That's why both Cr and Cu bend the rule.

Verify: $Δ_{Cr} = 10 - 6 = 4$ ; $Δ_{Cu} = 20 - 16 = 4$ ; equal, both positive. ✔ (See Periodic trends: the same logic reappears at Mo, Ag, Au further down.)

Recall Self-test (click)

How many unpaired electrons in $Fe^{3 +}$ ? ::: 5 (config $[Ar] 3 d^{5}$ ). Why does Cr adopt $3 d^{5} 4 s^{1}$ ? ::: Half-filled $d^{5}$ gains 4 extra exchange pairs + symmetry, beating the $4 s \to 3 d$ promotion cost. Is Zn paramagnetic? ::: No — $3 d^{10} 4 s^{2}$ has 0 unpaired electrons, so diamagnetic. When ionizing a transition metal, which electrons leave first? ::: The highest- $n$ ones ( $4 s$ ), not the last-filled $3 d$ .