Electron Configurations of Ions: Cations, Anions and Transition Metals

Writing the configuration of a neutral atom is only half the job. In chemistry, most elements are found as ions: Na⁺ in salt, Ca²⁺ in bones, Fe³⁺ in rust, Cl⁻ in seawater. Their electron configurations explain why these ions form, why some are more stable than others, and why many transition metal ions are coloured and magnetic. This article works through the rules with examples, from simple main-group ions to the trickier transition metals.

The basic rule

• Cations (positive ions): remove electrons from the neutral atom’s configuration.
• Anions (negative ions): add electrons to the neutral atom’s configuration.

The nucleus doesn’t change, only the number of electrons. See what is an ion?

Which electrons are removed first? Always those in the highest principal quantum number (n) shell, i.e. the outermost electrons. For main-group elements, this is also the last subshell filled. For transition metals, it isn’t, as we’ll see.

Main-group cations

Worked example 1: Na⁺

• Na (11 electrons): 1s² 2s² 2p⁶ 3s¹
• Remove 1 electron from 3s.
• Na⁺ (10 electrons): 1s² 2s² 2p⁶ = [Ne]

Worked example 2: Mg²⁺

• Mg: [Ne] 3s²
• Remove 2 electrons from 3s.
• Mg²⁺: [Ne]

Worked example 3: Al³⁺

• Al: [Ne] 3s² 3p¹
• Remove 3 electrons: first from 3p, then from 3s.
• Al³⁺: [Ne]

Main-group metals typically lose all their outer electrons to reach a noble gas configuration. This is why sodium forms Na⁺ and not Na²⁺: removing a second electron would mean breaking into the stable, tightly held 2p⁶ core, which needs far more energy. See ionization energy trend.

Main-group anions

Worked example 4: Cl⁻

• Cl (17): [Ne] 3s² 3p⁵
• Add 1 electron to 3p.
• Cl⁻ (18): [Ne] 3s² 3p⁶ = [Ar]

Worked example 5: O²⁻

• O (8): 1s² 2s² 2p⁴
• Add 2 electrons to 2p.
• O²⁻ (10): 1s² 2s² 2p⁶ = [Ne]

Worked example 6: N³⁻

• N (7): 1s² 2s² 2p³
• Add 3 electrons.
• N³⁻ (10): 1s² 2s² 2p⁶ = [Ne]

Isoelectronic species

Species with the same number of electrons (and the same configuration) are isoelectronic. All of these have the neon configuration, 1s² 2s² 2p⁶:

Species	Protons	Electrons	Ionic radius (pm, approx.)
N³⁻	7	10	146
O²⁻	8	10	140
F⁻	9	10	133
Ne	10	10	—
Na⁺	11	10	102
Mg²⁺	12	10	72
Al³⁺	13	10	54

With the same number of electrons, the species with more protons pulls the electrons in more strongly, so it’s smaller. This is a favourite exam question. See atomic radius trend.

Transition metal ions: 4s out first

Here’s the rule that catches almost everyone out:

When transition metals form ions, the 4s electrons are removed before the 3d electrons.

Even though 4s fills before 3d in the neutral atoms (see the Aufbau principle), once the 3d subshell is occupied, 4s electrons are the outermost (highest n) and are lost first.

Worked example 7: Fe²⁺ and Fe³⁺

• Fe (26): [Ar] 3d⁶ 4s²
• Fe²⁺ (24): remove both 4s electrons → [Ar] 3d⁶
• Fe³⁺ (23): then remove one 3d electron → [Ar] 3d⁵

Common error: writing Fe²⁺ as [Ar] 3d⁴ 4s². Wrong: the 4s electrons go first.

Fe³⁺ has a half-filled 3d subshell (five unpaired electrons), a relatively stable arrangement. This helps explain why iron(II) compounds are readily oxidised to iron(III) in air.

Worked example 8: Cu⁺ and Cu²⁺

• Cu (29): [Ar] 3d¹⁰ 4s¹ (an exception; see electron configuration exceptions)
• Cu⁺ (28): remove the 4s electron → [Ar] 3d¹⁰
• Cu²⁺ (27): remove one 3d electron → [Ar] 3d⁹

Cu⁺ has a full 3d subshell, so its compounds are usually colourless (no partly filled d subshell for d–d transitions). Cu²⁺, with 3d⁹, gives the familiar blue colour of copper sulfate solution.

Worked example 9: Cr³⁺

• Cr (24): [Ar] 3d⁵ 4s¹
• Remove 4s¹ then two 3d electrons.
• Cr³⁺ (21): [Ar] 3d³

Worked example 10: Zn²⁺

• Zn (30): [Ar] 3d¹⁰ 4s²
• Zn²⁺: [Ar] 3d¹⁰

A full d subshell means zinc compounds are typically white or colourless, and zinc forms only one common ion. That’s why zinc is often not classed as a true transition metal: a transition metal is usually defined as forming at least one ion with a partly filled d subshell. See transition metals.

Worked example 11: Mn²⁺

• Mn (25): [Ar] 3d⁵ 4s²
• Mn²⁺: [Ar] 3d⁵, a half-filled d subshell. Mn²⁺ compounds are very pale pink.

Counting unpaired electrons in ions

Using Hund’s rule, the number of unpaired d electrons follows from the configuration (for free ions, in the simplest picture):

Ion	Configuration	Unpaired electrons	Magnetic behaviour
Sc³⁺	[Ar]	0	diamagnetic
Ti³⁺	[Ar] 3d¹	1	paramagnetic
Cr³⁺	[Ar] 3d³	3	paramagnetic
Mn²⁺	[Ar] 3d⁵	5	strongly paramagnetic
Fe²⁺	[Ar] 3d⁶	4	paramagnetic
Ni²⁺	[Ar] 3d⁸	2	paramagnetic
Cu²⁺	[Ar] 3d⁹	1	paramagnetic
Zn²⁺	[Ar] 3d¹⁰	0	diamagnetic

(In complexes, strong-field ligands can force electrons to pair, changing these numbers.)

Why this matters: stability, colour and reactivity

Ion configurations explain a surprising amount of everyday chemistry:

• Which ions form. Main-group elements form ions with noble gas configurations: Na⁺, Mg²⁺, Cl⁻, O²⁻. Aluminium forms Al³⁺, but never Al⁴⁺, because a fourth electron would have to come from the stable neon-like core.
• Relative stability. Fe³⁺ (d⁵, half-filled) is more stable than Fe²⁺ (d⁶), which is why iron(II) solutions slowly turn yellow-brown in air as Fe²⁺ is oxidised. Mn²⁺ (also d⁵) is unusually resistant to further oxidation.
• Colour. Ions with partly filled d subshells, such as Cu²⁺ (d⁹, blue), Ni²⁺ (d⁸, green) and Co²⁺ (d⁷, pink in water), are usually coloured. Ions with empty (Sc³⁺, Ti⁴⁺) or full (Zn²⁺, Cu⁺) d subshells are usually colourless.
• Biology. Fe²⁺ at the centre of haemoglobin binds oxygen; oxidation to Fe³⁺ stops it working. See elements in the human body.

Method summary

1. Write the neutral atom’s configuration (use noble gas shorthand).
2. For cations, remove electrons from the highest n first (4s before 3d; 5s before 4d; 6s and 6p before 5d and 4f).
3. For anions, add electrons to the next available orbitals.
4. Check that the total electrons = atomic number − charge.
5. Draw an orbital box diagram if you need to count unpaired electrons.

Practice questions

1. K⁺ (Z = 19)
2. S²⁻ (Z = 16)
3. Co²⁺ (Z = 27)
4. V³⁺ (Z = 23)
5. Which is larger, S²⁻ or Cl⁻? Explain.

Answers:

1. [Ne] 3s² 3p⁶ = [Ar]
2. [Ne] 3s² 3p⁶ = [Ar]
3. Co: [Ar] 3d⁷ 4s² → [Ar] 3d⁷ (3 unpaired)
4. V: [Ar] 3d³ 4s² → [Ar] 3d² (2 unpaired)
5. S²⁻ is larger. Both have 18 electrons (isoelectronic), but S²⁻ has only 16 protons compared with 17 for Cl⁻, so its electrons are held less tightly.

Key takeaways

• Cations lose electrons from the highest-n shell; anions gain electrons in the next available orbitals.
• Main-group ions usually reach a noble gas configuration.
• Isoelectronic species shrink as the number of protons increases.
• Transition metals lose 4s electrons before 3d: Fe²⁺ is [Ar] 3d⁶, not [Ar] 3d⁴ 4s².
• Partly filled d subshells explain the colour and magnetism of many transition metal ions.