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

Before you start, keep these three anchors in mind — every trap below leans on one of them:

Aufbau is an approximate ordering (Aufbau principle), not a law of nature. The real law is "minimum total energy."
$4 s$ fills before $3 d$ because of $(n + l)$ (Quantum numbers (n, l, m, s)), but $3 d$ sits lower once occupied — that reversal is the source of half the confusion.
Half-filled ( $d^{5}$ ) and full ( $d^{10}$ ) subshells get an exchange + symmetry bonus that can beat the small promotion cost.

True or false — justify

Every

(n + l)

tie is broken by choosing the higher

n

first.

False — the tie-breaker in the Madelung rule picks lower

n

first (e.g.

3 p

before

4 s

, both

n + l = 4

), because a smaller shell keeps the electron closer to the nucleus and thus lower in energy.

Chromium is an exception because Aufbau is wrong for it.

False — Aufbau isn't "wrong," it's just approximate; near

Z = 24

the

3 d

and

4 s

energies nearly coincide, so the exchange stabilization of

d^{5}

tips the total energy in favour of

3 d^{5} 4 s^{1}

A completely filled

d^{10}

subshell is more stable than a half-filled

d^{5}

one.

True in the sense that

d^{10}

is a closed subshell with maximal symmetry, but they are stabilized by different effects —

d^{5}

wins on maximum parallel-spin exchange,

d^{10}

wins on complete symmetry and every orbital paired.

In the neutral atom,

4 s

is always lower in energy than

3 d

False — it is lower only while empty/filling; once electrons occupy

3 d

, the extra nuclear pull and reduced shielding push

3 d

below

4 s

, which is exactly why

4 s

leaves first on ionization (Shielding and penetration effect).

Hund's rule tells you which subshell to fill next.

False — Aufbau chooses the subshell; Hund's rule only governs how electrons spread within one degenerate subshell (singly first, parallel spins).

The Pauli principle is what forbids three electrons in one orbital.

True — Pauli says no two electrons share all four quantum numbers, so an orbital (fixed

n, l, m

) allows only two, distinguished by opposite spin

\pm \frac{1}{2}

Manganese (

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

) is an "exception" just like chromium.

False — Mn follows Aufbau perfectly; it happens to reach

d^{5}

without promoting any electron, so no rule was broken. Only Cr and Cu (in

Z \leq 30

) actually move an electron from

4 s

3 d

Exchange energy stabilization requires electrons to be in different orbitals.

True — exchange pairs are counted only between electrons of parallel spin in different orbitals of the same subshell; two paired electrons in one orbital contribute no exchange bonus.

Spot the error

"Cr is

[Ar] 3 d^{4} 4 s^{2}

because we fill

4 s

before

3 d

The filling order is right but the outcome is wrong — the atom promotes one

4 s

electron to reach the extra-stable

3 d^{5} 4 s^{1}

; nature minimizes total energy, not "obeys Aufbau step by step."

"Cu is

[Ar] 3 d^{9} 4 s^{2}

— nine into

d

, two into

s

Wrong — Cu is

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

; one

4 s

electron drops into

3 d

to complete the fully filled, maximally symmetric

d^{10}

shell.

"Sc is

[Ar] 4 s^{2} 3 d^{1}

, so we write

4 s

before

3 d

in the configuration."

The filling order is

4 s

then

3 d

, but the conventional written form groups by principal quantum number:

[Ar] 3 d^{1} 4 s^{2}

. Be consistent — the parent note flags this mixing as a classic slip.

"Fe³⁺ is

[Ar] 3 d^{3} 4 s^{2}

because

3 d

filled last, so

3 d

electrons leave first."

Wrong — electrons are removed from the highest $n$ first, so

4 s^{2}

goes before any

3 d

: Fe³⁺ is

[Ar] 3 d^{5}

, not

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

4 s^{2}

contributes an exchange bonus of one pair, helping stability."

No —

4 s

is a single orbital, so its two electrons must be paired (opposite spins); paired electrons give zero parallel-spin exchange pairs.

"Zn (

3 d^{10} 4 s^{2}

) is an exception because it has a full

d

subshell."

Not an exception — Zn fills

4 s^{2}

then

3 d^{10}

exactly as Aufbau predicts; no electron was promoted, so no rule was violated.

"Chromium's stability comes only from symmetry of

d^{5}

Incomplete — the dominant factor is maximum exchange energy (all five

d

-electrons parallel gives

(2 5) = 10

exchange pairs); symmetric charge distribution is a secondary contributor.

Edge cases

What is the configuration of a hydrogen atom, and does Hund's rule apply?

H is

1 s^{1}

; Hund's rule is irrelevant with a single electron since there's no second electron to pair or parallel with.

Does the Pauli principle limit how many electrons a subshell holds, or an orbital?

Directly it limits an orbital to 2; the subshell caps (

s = 2, p = 6, d = 10

) follow from multiplying by the number of orbitals (

1, 3, 5

For a

d^{5}

high-spin arrangement, how many electron pairs are there?

Zero paired electrons — all five occupy separate

d

orbitals with parallel spins (Hund), which is precisely what maximizes exchange stabilization.

Is Ca (

[Ar] 4 s^{2}

) affected by

3 d

stability at all?

No — at

Z = 20

the

3 d

orbitals are empty and higher in energy; filling

4 s^{2}

is simply the lowest-energy choice, with no promotion incentive yet.

What happens to the "half-filled bonus" argument for a paired subshell like

s^{2}

p^{6}

The exchange bonus counts only parallel-spin pairs; a filled

s^{2}

p^{6}

is stable through closed-shell symmetry, not through exchange among parallel electrons.

Recall One-line summary to keep

Aufbau orders the subshells, Pauli caps them, Hund arranges within them — and when $4 s$ – $3 d$ are near-degenerate, the exchange + symmetry bonus of $d^{5}$ / $d^{10}$ overrides the naïve order for Cr and Cu.

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

True or false — justify

Spot the error

Why questions

Edge cases