Electron configuration and periodic trends: the filling order, the exceptions, and what the table tells you without a chart

An electron configuration lists the orbitals an atom's electrons fill, in the order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p and so on, with up to 2 electrons in each s subshell, 6 in each p and 10 in each d. Iron, with 26 electrons, is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶, or [Ar] 4s² 3d⁶. Below: the order built from the periodic table rather than memorised, iron worked line by line, the ion rule students get backwards, the two exceptions, and the four periodic trends that follow from the same picture.

The three rules, in plain words

Three rules decide every electron configuration.

The aufbau principle: fill the lowest-energy orbital available first. The Pauli exclusion principle: an orbital holds at most two electrons, and they must have opposite spins. Hund's rule: within a subshell, put one electron in each orbital before pairing any of them, all with the same spin.

The filling order is the part students memorise with a diagonal chart, and the chart is not needed. The periodic table is the chart. Read it left to right, row by row. The first two columns are the s block (two electrons per row). The six columns on the right are the p block (six electrons per row). The ten columns in the middle are the d block (ten electrons per row), and the d block of row 4 is the 3d subshell: its orbitals belong to shell 3, but their energy falls between 4s and 4p, so they fill in row 4. So a row reads: ns, then (n − 1)d if the row has a d block, then np. Row 4 reads 4s, 3d, 4p. That is the whole order, and the table carries it for you into the exam.

In Ontario this is SCH4U's Electron configurations and periodic trends, the idea after Quantum mechanical model of the atom; quantum numbers; the SCH4U course map shows the sequence, and every chemistry idea worked, in course order is the section this page belongs to. AP Chemistry's first unit asks the same, and the AP Chemistry page says how.

Worked example: iron, Z = 26

Write the subshells in order and keep a running count of electrons placed. Stop when the count reaches the atomic number.

1. 1s² → 2 electrons placed
2. 2s² → 4
3. 2p⁶ → 10
4. 3s² → 12
5. 3p⁶ → 18. This is argon's count, so the shorthand core is [Ar].
6. 4s² → 20. The 4s fills before the 3d: it is lower in energy for the neutral atom, and the table shows it, since potassium and calcium (4s) come before scandium (the first 3d element).
7. 3d⁶ → 26 = Z. Stop.

Check by adding the superscripts: 2 + 2 + 6 + 2 + 6 + 2 + 6 = 26. Full form: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶. Shorthand: [Ar] 4s² 3d⁶.

Now the orbital diagram for 3d⁶ by Hund's rule. Five boxes. One electron in each box first, all spin-up: 5 placed. The sixth pairs in the first box, spin-down. That leaves four boxes with a single electron, so iron has 4 unpaired electrons, which makes a single iron atom attracted to a magnetic field. (The strong magnetism of an iron nail also needs those spins to line up together across the metal.)

The wrong line. Filling 3d before 4s because "3 is less than 4", giving [Ar] 3d⁸ with no 4s. The count still reaches 26 (18 + 8), which is why this error survives a recount; only the order rule catches it. The check is the table: scandium comes after calcium, so 3d comes after 4s.

Ions: the rule that is backwards from what you expect

For a main-group ion the rule is the familiar one: add electrons to the next empty orbital for an anion, remove the outermost for a cation. Cl⁻ is [Ne] 3s² 3p⁶, the same configuration as argon (the two are isoelectronic). Na⁺ is [Ne].

For a transition-metal cation the rule surprises students. Remove the 4s electrons before the 3d. Although 4s filled first in the neutral atom, once the 3d orbitals hold electrons the 4s electrons sit highest in energy and leave first. Filling order and removal order are different lists.

• Fe²⁺: 26 − 2 = 24 electrons. Both 4s electrons go: [Ar] 3d⁶. Check: 18 + 6 = 24. Five boxes, one pair and four singles: 4 unpaired.
• Fe³⁺: 26 − 3 = 23 electrons. The 4s pair goes, then one 3d electron: [Ar] 3d⁵. Check: 18 + 5 = 23. Five boxes, one electron each: 5 unpaired electrons.

The wrong line. Fe³⁺ written as [Ar] 4s² 3d³, taking three electrons from the d subshell because it was written last. The count is right, 18 + 2 + 3 = 23, and the configuration is wrong. A marker sees this one on every test of this unit. The fix is one sentence to remember: a transition-metal ion has no 4s electrons.

The two exceptions in the first transition row

Two elements in the first row of the d block do not follow the filling order, and a Grade 12 course asks for both.

• Chromium, Z = 24: [Ar] 4s¹ 3d⁵, not 4s² 3d⁴. Check: 18 + 1 + 5 = 24. Six orbitals with one electron each: 6 unpaired electrons.
• Copper, Z = 29: [Ar] 4s¹ 3d¹⁰, not 4s² 3d⁹. Check: 18 + 1 + 10 = 29. One unpaired electron, the 4s.

The textbook reason is that a half-filled or completely filled d subshell is especially stable, so one 4s electron moves into 3d to make one. Be honest with yourself about what that sentence is: a description of the result, not a cause. The 4s and 3d energies are close in these atoms and the full accounting of electron repulsion is first-year physical chemistry. The course asks for the two configurations, and that is what to learn. The same pattern recurs lower down (molybdenum and silver, below chromium and copper), which a Grade 12 test rarely asks.

Quantum numbers for one electron

Each electron has an address of four quantum numbers. The principal number n is the shell (1, 2, 3 …). The angular number l is the subshell: l = 0 is s, 1 is p, 2 is d, 3 is f, and l runs from 0 to n − 1. The magnetic number mₗ is the orbital within the subshell and runs from −l to +l, so a d subshell (l = 2) has mₗ = −2, −1, 0, 1, 2: five orbitals. The spin number mₛ is +½ or −½.

Take the last electron placed in iron, the sixth 3d electron. It is in shell 3, so n = 3. It is in a d subshell, so l = 2. By the convention of filling the boxes from the left, starting at mₗ = −2, it pairs in the first box, so mₗ = −2. (Some courses start the boxes at +2; use your teacher's convention, since the choice is arbitrary.) It pairs, so its spin is opposite to the first electron's: mₛ = −½. The set is (3, 2, −2, −½).

Some sets are impossible, and tests ask you to spot them. n = 2 with l = 2 is impossible, because l can be at most n − 1 = 1: there is no 2d subshell, which is why the table's row 2 has no d block. mₗ = 2 with l = 1 is impossible, because a p subshell's mₗ stops at ±1. Any mₛ other than ±½ is impossible.

Periodic trends from the same picture

Every trend on this unit comes from two questions. How many shells does the atom have? How much of the nuclear charge reaches the outer electrons, after the inner shells shield it?

• Atomic radius increases down a group (more shells) and decreases across a period (more protons pulling on electrons in the same shell).
• First ionization energy, the energy to remove the outermost electron, increases across a period and decreases down a group. Down a group the outer electron is farther away and more shielded, so it is easier to remove.
• Electronegativity, the pull an atom exerts on shared electrons, increases across and decreases down; fluorine is the highest. What the difference between two atoms' values does to a bond is the electronegativity page, from the trend to the bond.
• Ion size: a cation is smaller than its atom (often a whole shell is gone, and the remaining electrons repel one another less), an anion larger (more repulsion in the same shell). Na⁺ is smaller than Na; Cl⁻ is larger than Cl.

One worked comparison a test likes. Across row 3, the first ionization energies of sodium, magnesium and aluminium are about 496, 738 and 578 kJ/mol. The rise from Na to Mg is the trend. The dip from Mg to Al is the exception: aluminium's outermost electron is alone in a 3p orbital, slightly higher in energy and a little better shielded than magnesium's paired 3s electrons, so it comes off more easily. The same kind of dip appears again between phosphorus and sulfur, where sulfur's fourth 3p electron is the first to pair and the pairing repulsion makes it easier to remove. Expect the trend to hold across a row with those two dips, and expect a test to ask about one of them.

The wrong line. "A bigger atom holds more electrons, so it is harder to ionize." It is the reverse: in a bigger atom the outer electron is farther from the nucleus and more shielded, so it is easier to remove. Ionization energy falls as radius grows.

Electron configuration on one screen

• The order is the table: each row reads ns, then (n − 1)d where the row has a d block, then np. No chart.
• Three rules: lowest energy first; two per orbital, opposite spins; one per orbital before pairing.
• Check the count: the superscripts add to the atomic number. A right count does not prove a right order.
• Ions: a transition-metal cation loses its 4s electrons first. Fe²⁺ [Ar] 3d⁶; Fe³⁺ [Ar] 3d⁵.
• Two exceptions: Cr [Ar] 4s¹ 3d⁵; Cu [Ar] 4s¹ 3d¹⁰.
• Four trends as arrows: radius ↓ across, ↑ down; ionization energy ↑ across, ↓ down; electronegativity ↑ across, ↓ down; cation smaller than atom, anion larger.

The order of this page, the rules, then one atom worked in full, then a problem, is chapter 3 of the free playbook.

Try one cold: cobalt and its ion

Cover everything above. Three questions.

1. Write the shorthand configuration of cobalt, Z = 27, and of Co²⁺, and say how many unpaired electrons Co²⁺ has.
2. Which is larger, K or K⁺?
3. Which has the higher first ionization energy, Cl or Br?

Check

• Co: [Ar] 4s² 3d⁷. Check: 18 + 2 + 7 = 27.
• Co²⁺: 27 − 2 = 25 electrons; the 4s pair goes: [Ar] 3d⁷. Check: 18 + 7 = 25. Five boxes: five singles, then two of them pair, leaving 3 unpaired electrons.
• K is larger. K⁺ has lost its only 4s electron and the whole fourth shell with it.
• Cl has the higher ionization energy. It has fewer shells, so its outer electron is closer to the nucleus and less shielded.

If you wrote Co²⁺ as [Ar] 4s² 3d⁵, you took from the d subshell; reread the ion rule.

Where this sits in the course

In the SCH4U pack this is idea 7 of 24, Electron configurations and periodic trends, in strand C. It follows Quantum mechanical model of the atom; quantum numbers, which supplies the orbitals and the four numbers this page uses. It is followed by Ionic and covalent bonding; Lewis structures; electronegativity, where the valence count this page gives is what a Lewis structure starts from, and then by Molecular shape (VSEPR) and polarity, where the shape and polarity of the molecule follow from that structure.

How to practise it

Configure ten elements from the table alone, no chart, mixed with five ions and three trend comparisons in one sitting. Mixing matters: Rohrer and Taylor (2007) found that students who practised problems shuffled across types did worse during practice and better on a test a week later, because they practised deciding which rule applies. Check yourself by counting electrons and by the ion rule, not by looking at an answer, and mark honestly: "I would have got it" is a miss. How to test yourself honestly has the marking rules. Roediger and Karpicke (2006) found that students who were tested on material remembered more of it a week later than students who reread it, so the ten configurations are a test you set yourself, not a review.