General electronic configuration (n−1)d¹⁻¹⁰ ns⁰⁻²: Worked examples

The scenario matrix

Before any example, here is the full landscape. Every question about a d-block configuration falls into exactly one of these cells. If we work one example per row, we have covered everything.

Cell	Scenario class	Degenerate / edge feature	Example
A	Ordinary neutral atom, no exception	$d$ fills 1–4 or 6–9, plain Madelung	Ex 1 (Ti)
B	Half-filled exception	$s \to d$ promotion to reach $d^{5}$	Ex 2 (Cr)
C	Fully-filled exception	$s \to d$ promotion to reach $d^{10}$	Ex 3 (Cu)
D	Positive ion (cation)	remove $n s$ before $(n - 1) d$	Ex 4 (Fe / Fe²⁺ / Fe³⁺)
E	Empty- $d$ boundary	$d^{0}$ — is it still "transition"?	Ex 5 (Sc³⁺, Ti⁴⁺)
F	Full- $d$ boundary	$d^{10}$ in atom and ion — not typical TM	Ex 6 (Zn / Zn²⁺)
G	Extreme ionisation	strip all valence electrons → noble-gas core	Ex 7 (Mn⁷⁺)
H	5th-period / heavier exception	$4 d /5 s$ swap; Pd is the odd one	Ex 8 (Pd)
I	Reverse / word problem	given a config or a property, name the species	Ex 9 (unknown ion)

Each example is tagged with its cell. Read the Forecast line and guess before scrolling.

The picture above is the mental model for the whole page: two "shelves", the outer $n s$ (higher once $d$ is occupied) and the inner $(n - 1) d$ . Electrons enter the lower shelf first but leave from the higher shelf first. Keep this red-highlighted $n s$ shelf in mind for every ion example.

Cell A — ordinary atom

Worked example Ex 1 · Titanium,

$Z = 22$ (Cell A) Write the ground-state configuration of the Ti atom.

Forecast: Guess the last two subshells before reading. (Hint: Ti is two steps past Ar.)

Fill the noble-gas core: Ar has $Z = 18$ , so we start from $[Ar]$ . Why this step? Aufbau fills the lowest 18 orbitals first; writing $[Ar]$ is just shorthand for $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$ .
Next lowest orbital is $4 s$ (its $n + ℓ = 4 + 0 = 4$ , beating $3 d$ 's $3 + 2 = 5$ ). Add both: $4 s^{2}$ . Count so far $= 20$ . Why this step? Smaller $(n + ℓ)$ = lower energy, from Aufbau Principle and Madelung Rule.
Two electrons remain ( $22 - 20 = 2$ ). The next available subshell is $3 d$ : $3 d^{2}$ . Why this step? $4 s$ is now full; Madelung sends the rest to $3 d$ .

Answer: $[Ar] 3 d^{2} 4 s^{2}$ .

Verify: electron count $= 18 + 2 + 2 = 22 = Z$ . ✓ And $d^{2}$ is in the plain $1$ – $4$ range, so no exception applies. Ti has a partially filled $d$ → it is a typical transition metal.

Cell B — half-filled exception

Worked example Ex 2 · Chromium,

$Z = 24$ (Cell B) Write the ground-state configuration and justify why the naïve answer is wrong.

Forecast: Naïve Madelung says $[Ar] 3 d^{4} 4 s^{2}$ . Do you trust it? Guess yes/no.

Naïve fill: $[Ar] 3 d^{4} 4 s^{2}$ (count $18 + 4 + 2 = 24$ ). This is a valid electron count but not the lowest-energy arrangement. Why this step? Always write the Madelung answer first so you can see what "cheats".
Move one $4 s$ electron into $3 d$ : $3 d^{5} 4 s^{1}$ (count $18 + 5 + 1 = 24$ , unchanged). Why this step? $3 d^{5}$ is half-filled: all five $d$ orbitals singly occupied, all spins parallel. From Exchange Energy and Hund's Rule, each extra pair of parallel-spin electrons adds an exchange stabilisation; five parallel spins maximise this bonus.
Compare costs: the promotion costs a little $s \to d$ energy but gains a large exchange + spherical-symmetry stabilisation. The gain wins. Why this step? Ground state = lowest total energy, not "textbook filling order".

Answer: $[Ar] 3 d^{5} 4 s^{1}$ .

Verify: count $= 18 + 5 + 1 = 24 = Z$ ✓. Unpaired electrons $= 5$ (five singly-filled $d$ ) $+ 1$ ( $4 s$ ) $= 6$ — the largest possible for period 4, matching Cr's famously high magnetic moment (cross-check via Magnetic Properties (spin-only formula)).

Cell C — fully-filled exception

Cell D — cation, the removal rule

Worked example Ex 4 · Iron and its ions,

$Z = 26$ (Cell D) Write Fe, Fe²⁺, Fe³⁺.

Forecast: When Fe loses electrons, which leave first — the $4 s$ or the $3 d$ ? Guess before step 3.

Fe atom: $[Ar] 3 d^{6} 4 s^{2}$ (count $18 + 6 + 2 = 26$ ). No exception ( $d^{6}$ is in the plain $6$ – $9$ range). Why this step? Straight Madelung; $d^{6}$ is not near a half/full boundary.
Fe²⁺: remove two $4 s$ electrons → $[Ar] 3 d^{6}$ . Why this step? Once $3 d$ is occupied, rising nuclear charge pulls $3 d$ below $4 s$ (see Effective Nuclear Charge and Shielding); so $4 s$ is now the outermost, highest-energy electron and leaves first. We do not touch $3 d$ yet.
Fe³⁺: the $4 s$ is already gone, so the third electron comes from $3 d$ → $[Ar] 3 d^{5}$ . Why this step? With $s$ empty, the next-highest occupied orbital is $3 d$ . And $3 d^{5}$ is half-filled → extra stable, part of why Fe³⁺ is so common.

Answers: $Fe = [Ar] 3 d^{6} 4 s^{2}$ ; $Fe^{2 +} = [Ar] 3 d^{6}$ ; $Fe^{3 +} = [Ar] 3 d^{5}$ .

Verify: Fe²⁺ has $18 + 6 = 24$ electrons; Fe³⁺ has $18 + 5 = 23$ . Charge $=$ protons $-$ electrons: $26 - 24 = + 2$ ✓, $26 - 23 = + 3$ ✓. Both ions have partially filled $d$ → Fe is unquestionably a transition metal, and it shows Variable Oxidation States of Transition Metals.

Cell E — empty- $d$ boundary ( $d^{0}$ )

Worked example Ex 5 · Sc³⁺ and Ti⁴⁺ (Cell E — degenerate low end)

Write both ions and decide if they are "transition ions".

Forecast: Sc atom is $[Ar] 3 d^{1} 4 s^{2}$ . What does Sc³⁺ have left?

Sc ( $Z = 21$ ): $[Ar] 3 d^{1} 4 s^{2}$ . Remove 3 electrons for the $+ 3$ ion: first the two $4 s$ , then the single $3 d$ → $[Ar]$ , i.e. $3 d^{0}$ . Why this step? Removal order $n s$ then $(n - 1) d$ ; only one $d$ electron existed, so it also goes.
Ti ( $Z = 22$ ): $[Ar] 3 d^{2} 4 s^{2}$ . For $+ 4$ : remove $4 s^{2}$ then $3 d^{2}$ → $[Ar]$ , $3 d^{0}$ . Why this step? Same order; both $d$ electrons leave to reach the very stable noble-gas core.

Answers: $Sc^{3 +} = [Ar] (3 d^{0})$ ; $Ti^{4 +} = [Ar] (3 d^{0})$ .

Verify: Sc³⁺ has $21 - 3 = 18$ electrons $=$ Ar ✓; Ti⁴⁺ has $22 - 4 = 18 =$ Ar ✓.

Edge insight: $d^{0}$ is the empty boundary. These ions have no partially filled $d$ in this state — that is exactly why Sc³⁺ and Ti⁴⁺ are colourless (no $d$ electron to make a $d$ – $d$ jump; see Colour and d-d Transitions). The elements Sc and Ti still count as transition metals because the atom (or another stable ion) does have a partial $d$ .

Cell F — full- $d$ boundary ( $d^{10}$ in atom AND ion)

Worked example Ex 6 · Zinc,

$Z = 30$ (Cell F — degenerate high end) Write Zn and Zn²⁺, then classify.

Forecast: Is Zn a typical transition metal? Guess yes/no, then check the definition.

Zn atom: $[Ar] 3 d^{10} 4 s^{2}$ (count $18 + 10 + 2 = 30$ ). Why this step? $3 d$ fills completely at $Z = 30$ ; no promotion needed — it's already full.
Zn²⁺: remove the two $4 s$ → $[Ar] 3 d^{10}$ . Why this step? $n s$ leaves first; $3 d^{10}$ is untouched and stays fully filled.

Answers: $Zn = [Ar] 3 d^{10} 4 s^{2}$ ; $Zn^{2 +} = [Ar] 3 d^{10}$ .

Verify: Zn²⁺ has $30 - 2 = 28$ electrons; charge $= 30 - 28 = + 2$ ✓.

Edge insight: The IUPAC definition requires a partially filled $d$ in the atom or a stable ion. Zn is $d^{10}$ as an atom and $d^{10}$ as its only common ion → never partially filled → not a typical transition metal, even though it sits in the d-block. This is the exact mirror of Cell E: E is stuck at $d^{0}$ , F is stuck at $d^{10}$ ; only the in-between is "true transition".

Cell G — extreme ionisation

Worked example Ex 7 · Manganese in permanganate, Mn⁷⁺,

$Z = 25$ (Cell G) Write Mn, Mn²⁺, and the Mn⁷⁺ found in $MnO_{4}^{-}$ .

Forecast: Mn has 7 valence electrons ( $3 d^{5} 4 s^{2}$ ). What is left after removing all seven?

Mn atom: $[Ar] 3 d^{5} 4 s^{2}$ (count $18 + 5 + 2 = 25$ ). Note $3 d^{5}$ is reached naturally here — no promotion, because Mn simply has 5 $d$ electrons. Why this step? $18 + 2 + 5 = 25$ ; the half-filled $d^{5}$ is just where the count lands.
Mn²⁺: remove $4 s^{2}$ → $[Ar] 3 d^{5}$ . Why this step? $n s$ first; leaves the stable half-filled $d^{5}$ .
Mn⁷⁺: remove all 7 valence electrons ( $4 s^{2} + 3 d^{5}$ ) → $[Ar]$ , a bare noble-gas core. Why this step? The $+ 7$ oxidation state in $MnO_{4}^{-}$ formally strips every valence electron.

Answers: $Mn = [Ar] 3 d^{5} 4 s^{2}$ ; $Mn^{2 +} = [Ar] 3 d^{5}$ ; $Mn^{7 +} = [Ar]$ .

Verify: Mn⁷⁺ has $25 - 7 = 18$ electrons $=$ Ar ✓.

Twist insight: $Mn^{7 +}$ has $d^{0}$ — no $d$ electron — yet $MnO_{4}^{-}$ is intensely purple. The colour therefore cannot be a $d$ – $d$ transition; it comes from ligand-to-metal charge transfer instead (contrast with the $d$ – $d$ mechanism in Colour and d-d Transitions). A common exam trap.

Cell H — heavier-period exception

Worked example Ex 8 · Palladium,

$Z = 46$ (Cell H) Write Pd's ground state. It is the most extreme d-block exception.

Forecast: Naïve Madelung gives $[Kr] 4 d^{8} 5 s^{2}$ . But Pd empties $5 s$ completely — guess the real config.

Core: Kr has $Z = 36$ , so start $[Kr]$ . Remaining electrons $= 46 - 36 = 10$ . Why this step? $[Kr]$ absorbs the first 36 electrons.
Naïve fill: $5 s^{2} 4 d^{8}$ ( $n + ℓ$ : $5 s = 5$ , $4 d = 6$ , so $5 s$ nominally first). Why this step? Standard Madelung ordering for period 5.
Actual: both $5 s$ electrons drop into $4 d$ , giving $[Kr] 4 d^{10} 5 s^{0}$ . Why this step? Reaching the fully filled, spherically symmetric $4 d^{10}$ (with $5 s$ empty) is lower in energy than $4 d^{8} 5 s^{2}$ . Pd is unique: it is the only element that empties its outer $s$ shell entirely in the ground state.

Answer: $[Kr] 4 d^{10}$ (i.e. $5 s^{0}$ ).

Verify: count $= 36 + 10 = 46 = Z$ ✓. This is the $n s^{0}$ end of the parent's $n s^{0 - 2}$ range — proof the " $0$ " in the recipe is real and not decorative.

Cell I — reverse / word problem

Worked example Ex 9 · Identify the species (Cell I)

A stable +3 ion of a period-4 d-block element has the configuration $[Ar] 3 d^{3}$ . Name the element and its atom's configuration.

Forecast: Work backwards. If the $+ 3$ ion has 3 $d$ electrons, how many electrons — and protons — does the neutral atom have?

Electrons in the ion $= 18 (Ar) + 3 = 21$ . Why this step? Count what's written.
A $+ 3$ ion has lost 3 electrons, so the neutral atom has $21 + 3 = 24$ electrons ⇒ $Z = 24$ . Why this step? Neutral atom: electrons $=$ protons $= Z$ .
$Z = 24$ is Chromium. Its atom (Cell B result) is $[Ar] 3 d^{5} 4 s^{1}$ . Why this step? We already derived Cr in Ex 2; cross-check that removing 3 electrons ( $4 s^{1}$ then two $3 d$ ) leaves $3 d^{3}$ — matches the given ion. ✓

Answer: Element = Cr ( $Z = 24$ ); atom $= [Ar] 3 d^{5} 4 s^{1}$ ; the ion $Cr^{3 +} = [Ar] 3 d^{3}$ .

Verify: $Cr^{3 +}$ electrons $= 24 - 3 = 21 = 18 + 3$ ✓, and removing $4 s^{1} + 2 \times 3 d$ from $3 d^{5} 4 s^{1}$ gives $3 d^{3}$ ✓.

Wrap-up: the matrix, filled

Recall Every cell in one glance

Which cell would strip all valence electrons to a noble-gas core? ::: Cell G (extreme ionisation, e.g. Mn⁷⁺). Which two cells sit at the "boundaries" of the $d^{1 - 10}$ range? ::: Cell E ( $d^{0}$ , empty) and Cell F ( $d^{10}$ , full in both atom and ion). In every cation example, which electrons leave first? ::: The $n s$ electrons, before any $(n - 1) d$ . Why is Zn (Cell F) not a typical transition metal but Cu (Cell C) is? ::: Cu forms $Cu^{2 +} = 3 d^{9}$ (partial $d$ ); Zn stays $d^{10}$ in atom and $Zn^{2 +}$ .

Connections

Parent topic
Aufbau Principle and Madelung Rule
Exchange Energy and Hund's Rule
Variable Oxidation States of Transition Metals
Colour and d-d Transitions
Magnetic Properties (spin-only formula)
f-Block (n-2)f orbitals filling
Effective Nuclear Charge and Shielding

Worked examples — General electronic configuration (n−1)d¹⁻¹⁰ ns⁰⁻²

The scenario matrix

Cell A — ordinary atom

Cell B — half-filled exception

Cell C — fully-filled exception

Cell D — cation, the removal rule

Cell E — empty- $d$ boundary ( $d^{0}$ )

Cell F — full- $d$ boundary ( $d^{10}$ in atom AND ion)

Cell G — extreme ionisation

Cell H — heavier-period exception

Cell I — reverse / word problem

Wrap-up: the matrix, filled

Connections

Concept Map