Electron Configuration Calculator

Continue in the tool

More pages for this tool

electron configuration of ions cations anions

Electron configuration of ions: cations (Fe3+, Cu2+) and anions (O2-, Cl-)

The most common mistake with ions is removing electrons from the last subshell filled. The right rule is different: they leave first from the shell with the highest n. Type the ion, such as Fe3+ or O2-, and the tool shows the configuration and what is isoelectronic with it.

Cations lose electrons from the outer shell

To form a cation you start from the neutral atom and remove electrons from the shell with the highest n; if that shell has more than one subshell, the one with the highest ℓ goes first. In iron, [Ar] 4s2 3d6, the first two electrons to leave are the 4s ones (n = 4), and only then the 3d. So Fe2+ is [Ar] 3d6 and Fe3+ is [Ar] 3d5, with five unpaired electrons, a half-filled subshell. Anyone removing electrons from 3d, the last filled, would reach [Ar] 4s2 3d4 for Fe2+, and be wrong. The same holds for copper, [Ar] 3d10 4s1: Cu+ loses the 4s and ends up [Ar] 3d10, and Cu2+ also loses one from 3d and ends up [Ar] 3d9.
Cations of main-group metals empty the valence shell and end up with the configuration of the preceding noble gas: Na+, Mg2+ and Al3+ are [Ne], K+ and Ca2+ are [Ar]. Metals in the p block may lose only the p electrons and keep the s ones, the so-called inert pair effect: Pb2+ is [Xe] 4f14 5d10 6s2, with the 6s2 pair intact, and Sn4+ is [Kr] 4d10. Among transition metals Zn2+ is [Ar] 3d10, Cr3+ is [Ar] 3d3 and Mn2+ is [Ar] 3d5. In every case the number of electrons left is Z minus the charge: Fe3+ has 26 - 3 = 23 electrons.
Anions do the opposite. The extra electrons go into the first subshell not yet full: oxygen, 2p4, becomes O2- with 2p6, chlorine, 3p5, becomes Cl- with 3p6, and nitrogen, 2p3, becomes N3- with 2p6. All of them end up with a noble gas configuration and are isoelectronic with it: N3-, O2-, F-, Na+, Mg2+ and Al3+ all have the 10 electrons of neon, 1s2 2s2 2p6, and the tool lists this group in the Configuration → element tab. Within the group the radius shrinks as Z increases, because more protons hold the same 10 electrons. Anions with a charge below -4 are not handled, since they do not exist as stable ions in the gas phase.

Ions and their configurations

Iron(II)

Input: Fe2+
Expected output: [Ar] 3d6

Both 4s electrons leave; 26 - 2 = 24 electrons.

Iron(III)

Input: Fe3+
Expected output: [Ar] 3d5

One more leaves from 3d; half-filled subshell, 5 unpaired.

Copper(II)

Input: Cu2+
Expected output: [Ar] 3d9

Starts from the real Cu, [Ar] 3d10 4s1.

Chromium(III)

Input: Cr3+
Expected output: [Ar] 3d3

Starts from the real Cr, [Ar] 3d5 4s1: the 4s leaves and two from 3d.

Oxide

Input: O2-
Expected output: 1s2 2s2 2p6

Same as neon: 8 + 2 = 10 electrons.

Chloride

Input: Cl-
Expected output: 1s2 2s2 2p6 3s2 3p6

Same as argon: 17 + 1 = 18 electrons.

Full tool FAQ

How do you write the electron configuration of an element?

Take the atomic number (the number of electrons in the neutral atom) and distribute them over the subshells in Madelung order, filling each to its maximum (s 2, p 6, d 10, f 14). For chlorine (Z = 17): 1s2 2s2 2p6 3s2 3p5, or [Ne] 3s2 3p5, with 2, 8 and 7 electrons in shells K, L and M.

What is the filling order of the Madelung diagram?

What are the exceptions to the filling order?

How do you write the configuration of an ion?

What are the valence shell and the highest-energy subshell?

How do you find the quantum numbers of the last electron?

What is Hund's rule and what is paramagnetism?

What does [Ar] 4s2 3d6 mean?

How do you know how many electrons each shell has?

Where does the data come from, and does it cover 109 to 118?

Frequently asked questions

Why does 4s leave before 3d in iron?

Because the shell with the highest n loses electrons first, and 4s (n = 4) is more external than 3d (n = 3). In transition metals, an occupied 3d sits below 4s in energy, and 4s is the first to be ionized.

What are isoelectronic species?

Atoms or ions with the same number of electrons and the same configuration. Ne, O2-, F-, Na+ and Mg2+ have 10 electrons each. The Configuration → element tab lists them all for any configuration you type.

How do I type an ion in the tool?

Write the symbol with the charge, like Fe3+, Fe^3+, O2- or Cl-, or pick the charge in the selector next to the element. Typing the charge in the field already updates the selector.

Next steps

Cations lose electrons from the outer shell

To form a cation you start from the neutral atom and remove electrons from the shell with the highest n; if that shell has more than one subshell, the one with the highest ℓ goes first. In iron, [Ar] 4s2 3d6, the first two electrons to leave are the 4s ones (n = 4), and only then the 3d. So Fe2+ is [Ar] 3d6 and Fe3+ is [Ar] 3d5, with five unpaired electrons, a half-filled subshell. Anyone removing electrons from 3d, the last filled, would reach [Ar] 4s2 3d4 for Fe2+, and be wrong. The same holds for copper, [Ar] 3d10 4s1: Cu+ loses the 4s and ends up [Ar] 3d10, and Cu2+ also loses one from 3d and ends up [Ar] 3d9.

Cations of main-group metals empty the valence shell and end up with the configuration of the preceding noble gas: Na+, Mg2+ and Al3+ are [Ne], K+ and Ca2+ are [Ar]. Metals in the p block may lose only the p electrons and keep the s ones, the so-called inert pair effect: Pb2+ is [Xe] 4f14 5d10 6s2, with the 6s2 pair intact, and Sn4+ is [Kr] 4d10. Among transition metals Zn2+ is [Ar] 3d10, Cr3+ is [Ar] 3d3 and Mn2+ is [Ar] 3d5. In every case the number of electrons left is Z minus the charge: Fe3+ has 26 - 3 = 23 electrons.

Anions do the opposite. The extra electrons go into the first subshell not yet full: oxygen, 2p4, becomes O2- with 2p6, chlorine, 3p5, becomes Cl- with 3p6, and nitrogen, 2p3, becomes N3- with 2p6. All of them end up with a noble gas configuration and are isoelectronic with it: N3-, O2-, F-, Na+, Mg2+ and Al3+ all have the 10 electrons of neon, 1s2 2s2 2p6, and the tool lists this group in the Configuration → element tab. Within the group the radius shrinks as Z increases, because more protons hold the same 10 electrons. Anions with a charge below -4 are not handled, since they do not exist as stable ions in the gas phase.

Ions and their configurations

Iron(II)

Input: Fe2+
Expected output: [Ar] 3d6

Both 4s electrons leave; 26 - 2 = 24 electrons.

Iron(III)

Input: Fe3+
Expected output: [Ar] 3d5

One more leaves from 3d; half-filled subshell, 5 unpaired.

Copper(II)

Input: Cu2+
Expected output: [Ar] 3d9

Starts from the real Cu, [Ar] 3d10 4s1.

Chromium(III)

Input: Cr3+
Expected output: [Ar] 3d3

Starts from the real Cr, [Ar] 3d5 4s1: the 4s leaves and two from 3d.

Oxide

Input: O2-
Expected output: 1s2 2s2 2p6

Same as neon: 8 + 2 = 10 electrons.

Chloride

Input: Cl-
Expected output: 1s2 2s2 2p6 3s2 3p6

Same as argon: 17 + 1 = 18 electrons.

Full tool FAQ

How do you write the electron configuration of an element?

What is the filling order of the Madelung diagram?

What are the exceptions to the filling order?

How do you write the configuration of an ion?

What are the valence shell and the highest-energy subshell?

How do you find the quantum numbers of the last electron?

What is Hund's rule and what is paramagnetism?

What does [Ar] 4s2 3d6 mean?

How do you know how many electrons each shell has?

Where does the data come from, and does it cover 109 to 118?

Frequently asked questions

Why does 4s leave before 3d in iron?

What are isoelectronic species?

How do I type an ion in the tool?

Write the symbol with the charge, like Fe3+, Fe^3+, O2- or Cl-, or pick the charge in the selector next to the element. Typing the charge in the field already updates the selector.