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You can write the configuration of iron. Then a question asks for Fe³⁺, and the instinct is to take three electrons off the end of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶, leaving 3d³. That is wrong, and the reason applies across the periodic table, not just to transition metals. This post gives a three-step method, checks it against measured data, and uses it on magnetic moment questions.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list electronic configuration of elements in atomic structure, and electronic configuration again for the d- and f-block, lanthanoids and actinoids. The NCERT Class 12 chapter The d- and f-Block Elements (2026-27 reprint) states the key rule: "when d-block elements form ions, ns electrons are lost before (n – 1) d electrons". It explains that removing one electron "alters the relative energies of 4s and 3d orbitals", and gives the spin-only formula μ = √(n(n + 2)) BM. The Class 11 chapter Structure of Atom covers the filling order for neutral atoms. Check the syllabus for your year.
What is the three-step method?
Write the neutral atom using the filling order.
Regroup by shell, all n = 3 together before n = 4: iron becomes [Ar] 3d⁶ 4s².
Remove from the right: the highest n first, and within the same n, p before s. For anions, add electrons to the next empty spaces instead.
The filling order tells you how a neutral atom is built. It does not tell you which electrons are easiest to remove from an ion.
Worked examples across the blocks
Each configuration below matches the ground state listed in the NIST Atomic Spectra Database ionisation-energy tables.
Fe³⁺ (Z = 26): [Ar]3d⁶4s² loses both 4s and one 3d electron, giving [Ar]3d⁵. Five unpaired electrons.
Cu⁺ (Z = 29): copper is [Ar]3d¹⁰4s¹, so Cu⁺ is [Ar]3d¹⁰ and Cu²⁺ is [Ar]3d⁹.
Co²⁺ (Z = 27): [Ar]3d⁷4s² becomes [Ar]3d⁷, with three unpaired electrons.
Ga³⁺ (Z = 31): [Ar]3d¹⁰4s²4p¹ loses 4p then 4s, leaving [Ar]3d¹⁰. The 3d set stays full.
Sn²⁺ and Sn⁴⁺ (Z = 50): tin is [Kr]4d¹⁰5s²5p². Sn²⁺ loses the two 5p electrons: [Kr]4d¹⁰5s². Sn⁴⁺ also loses 5s: [Kr]4d¹⁰.
Gd³⁺ (Z = 64): gadolinium is [Xe]4f⁷5d¹6s². Remove two 6s and the 5d: [Xe]4f⁷, a half-filled f set with seven unpaired electrons.
S²⁻ (Z = 16): [Ne]3s²3p⁴ gains two electrons: [Ne]3s²3p⁶, the argon configuration.
How do magnetic moments test this?
The spin-only formula μ = √(n(n + 2)) BM turns a configuration into a number, so examiners use it to check that you removed the right electrons.
Example 1. Spin-only moment of Fe³⁺. It is 3d⁵ with n = 5, so μ = √35 ≈ 5.92 BM. If you wrongly wrote 3d³4s², you would get n = 3 and 3.87 BM, which is often an option.
Example 2. A divalent first-row ion has μ ≈ 2.84 BM. Which could it be? √(n(n + 2)) = 2.84 gives n = 2. Two unpaired d electrons come from d² or d⁸, so the ion is Ti²⁺ or Ni²⁺. NCERT's Table 4.7 lists both with a calculated 2.84 BM. Questions that ask for "the" answer usually rule one out with another clue.
Example 3. Cu⁺ and Zn²⁺ are both 3d¹⁰, so both have μ = 0. If your configuration for Cu⁺ gave an unpaired electron, the 4s electron was not removed first.
Does real data always follow the rule?
For the ions you meet in chemistry, yes. The NIST tables do show a curiosity: the isolated gas-phase Co⁺ ion has the ground state 3d⁸, not the 3d⁷4s¹ the rule predicts, because in free singly charged ions the 4s and 3d energies are close enough to compete. In compounds and solutions, and in every exam, use the rule. For why chromium and copper atoms break the filling order, see our post on Aufbau exceptions for chromium and copper.
Which mistakes cost marks?
Removing 3d before 4s. The most common error, for every first-row transition ion.
Not regrouping by shell before removing, so 4s looks like it is "in the middle".
Removing 5s before 5p for tin, lead or antimony ions.
Forgetting that anions add electrons to the same shell, not a new one.
Losing count. Total electrons must equal Z minus the charge: Fe³⁺ has 23.
When the counting is fine but the reason is hazy, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing. For which oxidation states are stable once you have the configuration, see our post on transition metal oxidation states.
A practice routine
Day 1: Write configurations for every first-row 2+ ion from Sc²⁺ to Zn²⁺, then compare with NCERT Table 4.7.
Day 2: Ten p-block and s-block ions, including Pb²⁺, Sb³⁺, Al³⁺, N³⁻ and Se²⁻.
Day 3: Five magnetic moment questions, in both directions.
Day 4: Three lanthanoid 3+ ions, then check your configurations against NCERT's lanthanoid table.
Tonight, write the configurations of Mn²⁺ and Mn⁴⁺ and calculate the spin-only moment of each.
