Electron configuration — filling shells and subshells — Chemistry

Before this

Protons, neutrons, electrons
The periodic table — groups and periods · coming soon

This leads to

Where the 2, 8, 8 rule breaks

In the last lecture, electrons filled shells as 2, 8, 8, and that rule worked for the first 20 elements. Potassium, element 19, is 2, 8, 8, 1. Calcium, element 20, is 2, 8, 8, 2. Notice something odd already: the third shell stopped at 8, and a fourth shell started.

Now go one step further, to scandium, element 21. You would expect 2, 8, 8, 3. But the real arrangement is 2, 8, 9, 2. The new electron goes back into the third shell. From scandium to zinc, the third shell keeps filling until it holds 18.

So the third shell could hold more than 8 all along. Then why did the fourth shell start before the third was full? The shell picture alone cannot answer this. To answer it, we need to look inside the shells.

Inside a shell: subshells and orbitals

Each shell is divided into subshells, labelled with the letters s, p, d and f. Each subshell is made of orbitals. An orbital is a region around the nucleus where an electron is likely to be found. It is not a fixed path like a planet's orbit. Each orbital can hold at most 2 electrons.

Subshell	Orbitals	Maximum electrons
s	1	2
p	3	6
d	5	10
f	7	14

Shell 1 has only an s subshell. Shell 2 has s and p. Shell 3 has s, p and d. Shell 4 has s, p, d and f. Now add up the capacities:

Shell 1: 1s holds 2.
Shell 2: 2s + 2p holds $2 + 6 = 8$ .
Shell 3: 3s + 3p + 3d holds $2 + 6 + 10 = 18$ .
Shell 4: 4s + 4p + 4d + 4f holds $2 + 6 + 10 + 14 = 32$ .

The pattern is $2 n^{2}$ for shell $n$ : 2, 8, 18, 32. So the third shell's true capacity is 18, as scandium showed. The "8" in the 2, 8, 8 rule was never the third shell's limit. It is the point where the 3s and 3p subshells are full.

The order of filling

Electrons fill the lowest-energy subshell available first. This is called the Aufbau principle, from the German for "building up". The key is that subshell energies do not simply go shell by shell:

Subshell energies from 1s to 4p, for the first 36 elements. Each box is one orbital. The 4s subshell lies a little below 3d, so it fills first.

1s \to 2s \to 2p \to 3s \to 3p \to 4s \to 3d \to 4p

Here is the answer to the opening puzzle. After 3p fills at argon, the next lowest subshell is 4s, not 3d. So potassium and calcium put their electrons into 4s. Only then does 3d begin to fill, from scandium onwards.

Why is 4s lower than 3d? The full answer needs quantum mechanics, but the main idea can be stated simply. An electron's energy depends on how strongly it is held by the nucleus. An electron in an s orbital spends some of its time very close to the nucleus, inside the inner shells. There, the inner electrons shield it less, and it feels the nucleus strongly. A d electron never gets that close. So even though 4s belongs to a higher shell, its electrons can end up slightly lower in energy than 3d electrons. The difference is small. That smallness explains the exceptions you will meet below.

Two more rules for the orbitals themselves

The Pauli exclusion principle. An orbital holds at most 2 electrons, and if it holds 2, they must have opposite spins. Spin is a property of the electron, drawn as an arrow pointing up or down. So a full orbital is drawn ↑↓.

Hund's rule. When electrons enter a subshell with several orbitals, they go into separate orbitals first, with parallel spins. Only when every orbital has one electron do they start to pair up. The reason is repulsion. Electrons are all negative and repel each other, so two electrons in separate orbitals are further apart and lower in energy than two squeezed into one.

For nitrogen, with three electrons in 2p, Hund's rule gives one electron in each p orbital, not a pair and a single:

Correct, 2p³: ↑ ↑ ↑ Not: ↑↓ ↑

Writing an electron configuration

Fill in energy order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. Each orbital holds 2 electrons with opposite spins. In a subshell, fill each orbital singly before pairing.

Write each subshell with its electrons as a superscript: sodium is 1s² 2s² 2p⁶ 3s¹.

Two writing conventions. First, a long configuration can be shortened by writing the previous noble gas in square brackets: sodium is [Ne] 3s¹, because neon is 1s² 2s² 2p⁶. Second, once 3d has electrons, we write the subshells in shell order, with 3d before 4s: iron is [Ar] 3d⁶ 4s². Some books write 4s first. Both mean the same atom, and examiners accept either.

Now build any of the first 36 elements below.

Explore — build the configuration, one electron at a time

Move the slider to add electrons one at a time. Watch where each new electron goes.

Atomic number Z = 11

Some examples, built in energy order:

Element	Configuration	Shells
nitrogen (7)	1s² 2s² 2p³	2, 5
sodium (11)	[Ne] 3s¹	2, 8, 1
potassium (19)	[Ar] 4s¹	2, 8, 8, 1
scandium (21)	[Ar] 3d¹ 4s²	2, 8, 9, 2
chromium (24)	[Ar] 3d⁵ 4s¹ (exception)	2, 8, 13, 1
iron (26)	[Ar] 3d⁶ 4s²	2, 8, 14, 2
copper (29)	[Ar] 3d¹⁰ 4s¹ (exception)	2, 8, 18, 1

Two famous exceptions: chromium and copper

The rules predict chromium (24) as [Ar] 3d⁴ 4s². The real atom is [Ar] 3d⁵ 4s¹. Copper (29) is predicted as [Ar] 3d⁹ 4s², but is really [Ar] 3d¹⁰ 4s¹.

In both cases, one electron moves from 4s into 3d. The result is a d subshell that is exactly half full (chromium) or completely full (copper). These arrangements are especially stable. Normally that extra stability would be too small to matter. But 4s and 3d are so close in energy that this small effect decides the result. Do not learn these two as random facts. Learn the reason, and they follow: a half-full or full d subshell is worth one electron moving.

Stop and think

Iron is [Ar] 3d⁶ 4s². When iron forms the ion Fe²⁺, it loses two electrons. Which two?

Show the answer

The two 4s electrons. Fe²⁺ is [Ar] 3d⁶, not [Ar] 3d⁴ 4s². This surprises many students, because 4s filled first. But once the 3d orbitals contain electrons, they pull the 3d energy down, and 4s ends up as the outermost, highest-energy subshell. The electrons lost first are the outermost ones. A useful way to remember it: 4s fills first, and 4s also empties first.

Worked examples

Worked Example 1 — Phosphorus

Write the full configuration of phosphorus (Z = 15) and draw its 3p orbitals.

Fill in order, counting as you go: 1s² (2 used), 2s² (4), 2p⁶ (10), 3s² (12). That leaves 3 electrons for 3p.

Configuration: 1s² 2s² 2p⁶ 3s² 3p³, or [Ne] 3s² 3p³. By Hund's rule, the three 3p electrons go one in each orbital: ↑ | ↑ | ↑. Phosphorus has three unpaired electrons.

Worked Example 2 — Iron, counting carefully

Write the configuration of iron (Z = 26) and draw its 3d orbitals.

1s² 2s² 2p⁶ 3s² 3p⁶ uses 18 electrons. 4s² comes next: 20 used. That leaves $26 - 20 = 6$ for 3d. Subtract as you go. It is the easiest way to avoid placing one electron too many or too few.

Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s², or [Ar] 3d⁶ 4s².

3d boxes: five electrons go in singly, then the sixth pairs up in the first orbital: ↑↓ | ↑ | ↑ | ↑ | ↑. Iron has four unpaired electrons.

Worked Example 3 — Ions of main-group elements

Write the configurations of Ca²⁺ and Cl⁻.

Calcium is [Ar] 4s². It loses its two outermost electrons, the 4s pair, so Ca²⁺ is 1s² 2s² 2p⁶ 3s² 3p⁶, the same as argon.

Chlorine is [Ne] 3s² 3p⁵. It gains one electron into the next space, 3p, so Cl⁻ is [Ne] 3s² 3p⁶, also the same as argon.

Two different elements end with the same arrangement as a noble gas. That is not a coincidence. It is a clue to why these ions form at all, which you will meet in bonding.

Worked Example 4 — Why sodium and potassium behave alike

Compare the outer electrons of sodium and potassium.

Sodium is [Ne] 3s¹. Potassium is [Ar] 4s¹. Their inner electrons are different, but each has exactly one electron in an outer s subshell. These outer electrons are the valence electrons: the ones that meet other atoms in a reaction. Because the outer pattern is the same, the two elements react in very similar ways, and they sit in the same group of the periodic table.

One mistake is more common than any other in this topic. Find it in the student's work below.

Spot the error — a configuration for iron

A student writes the configuration of iron (Z = 26). One line is wrong. Tap the line you think it is.

The student's work, with the error marked:

1. Iron has 26 electrons. Correct: Z = 26.
2. 1s² 2s² 2p⁶ 3s² 3p⁶ uses 18 electrons. Correct.
3. The other 8 go into 3d, because shell 3 comes before shell 4: 3d⁸. Wrong. Electrons fill in energy order, and 4s is lower than 3d. The correct line is 4s² (20 used), then 3d⁶.
4. So iron is [Ar] 3d⁸. Wrong, because it follows from line 3. The correct answer is [Ar] 3d⁶ 4s².

Three classic mistakes

Trap 1 — Filling 3d before 4s

Fill by energy, not by shell number: 3p, then 4s, then 3d. Potassium is [Ar] 4s¹, not [Ar] 3d¹.

Trap 2 — Pairing electrons too early

In a p or d subshell, put one electron in each orbital before any pairing. Oxygen's 2p⁴ is ↑↓ | ↑ | ↑, with two unpaired electrons, not ↑↓ | ↑↓ | (empty).

Trap 3 — Removing the wrong electrons from a transition metal

Transition metals lose their 4s electrons first. Fe²⁺ is [Ar] 3d⁶, and Fe³⁺ is [Ar] 3d⁵. Writing [Ar] 3d⁴ 4s² for Fe²⁺ is wrong.

Try it yourself

Count the electrons as you place them. Open a hint only when you are stuck.

Write the full electron configurations of magnesium (Z = 12) and chlorine (Z = 17).

Hint 1
Follow the order 1s, 2s, 2p, 3s, 3p, subtracting as you go.

Hint 2
After 1s² 2s² 2p⁶, you have used 10 electrons.

Full answer
Magnesium: 1s² 2s² 2p⁶ 3s² (12 electrons). Chlorine: 1s² 2s² 2p⁶ 3s² 3p⁵ (17 electrons).
Draw the 2p orbital boxes for oxygen (Z = 8). How many unpaired electrons does oxygen have?

Hint 1
Oxygen is 1s² 2s² 2p⁴. You need to place 4 electrons in 3 boxes.

Hint 2
Put one in each box first. Where does the fourth go?

Full answer
↑↓ | ↑ | ↑. One electron goes into each of the three orbitals, and the fourth pairs up in the first. Oxygen has 2 unpaired electrons.
Write the short-form configurations of potassium (Z = 19) and scandium (Z = 21), and give the number of electrons in each shell.

Hint 1
Both start from [Ar], which has 18 electrons.

Hint 2
After argon, the order is 4s, then 3d.

Full answer
Potassium: [Ar] 4s¹, shells 2, 8, 8, 1. Scandium: [Ar] 3d¹ 4s², shells 2, 8, 9, 2. The 3d electron belongs to the third shell, which is why scandium's third shell has 9.
Write the configuration of copper (Z = 29), then of the ion Cu²⁺. How many unpaired electrons does Cu²⁺ have?

Hint 1
Copper is one of the two exceptions: one electron moves from 4s to make the d subshell full.

Hint 2
A transition metal loses its 4s electrons first. Cu²⁺ loses two electrons, but copper has only one in 4s.

Full answer
Copper is [Ar] 3d¹⁰ 4s¹. To form Cu²⁺, it loses the 4s electron first and then one 3d electron, giving [Ar] 3d⁹. Nine electrons in five d orbitals: four pairs and one single, so Cu²⁺ has 1 unpaired electron.

The bigger idea

Look at the shape of the periodic table. On the left are two columns, on the right six, in the middle ten, and below, a separate row of fourteen. Those numbers are 2, 6, 10 and 14: the capacities of the s, p, d and f subshells. Chemists call these regions the s-block, p-block, d-block and f-block. An element's position tells you which subshell its last electron entered.

Mendeleev arranged the table in 1869, by how the elements behaved, decades before anyone knew that electrons existed. The explanation of its shape only came with quantum theory in the 1920s. The pattern was discovered first. The reason was found later, and the two fit exactly.

Key concepts in this lesson

Shells contain subshells (s, p, d, f) holding 2, 6, 10 and 14 electrons, so shell $n$ holds $2 n^{2}$ : 2, 8, 18, 32.
Electrons fill the lowest-energy subshell first, in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, because 4s is slightly lower than 3d.
Each orbital holds at most two electrons with opposite spins, and in a subshell electrons occupy orbitals singly before pairing.
Chromium is [Ar] 3d⁵ 4s¹ and copper is [Ar] 3d¹⁰ 4s¹, because a half-full or full d subshell is especially stable.
Transition metals lose their 4s electrons first when they form ions, so Fe²⁺ is [Ar] 3d⁶.
The outer (valence) electrons decide how an element reacts, and the s, p, d and f blocks of the periodic table follow the subshell being filled.