Coaching Centre
Electron configurations and orbitals
Energy levels, s, p, d and f subshells and orbitals, the Aufbau principle, Pauli exclusion principle and Hund's rule, writing full and noble-gas electron configurations and orbital box diagrams, exceptions such as chromium and copper, configurations of ions, and the blocks of the periodic table.
In Year 10 you wrote potassium's electron arrangement as 2, 8, 8, 1. But the third shell can hold 18 electrons, not 8. So why does potassium's 19th electron start a new, fourth shell when there is still room in the third? The answer is that shells are split into subshells with different energies, and the lowest available energy wins. Once you know the subshells, the shape of the periodic table stops being a convention and becomes a consequence.
Shells, subshells and orbitals
Electrons occupy energy levels or shells, numbered n = 1, 2, 3, 4 ... Shell n can hold up to 2n² electrons: 2, 8, 18, 32. Each shell is divided into subshells (also called sublevels), labelled s, p, d and f. Each subshell is made of orbitals: regions of space where an electron is likely to be found. Every orbital holds at most two electrons.
| Subshell | Orbitals | Maximum electrons | First appears in shell |
|---|---|---|---|
| s | 1 | 2 | n = 1 |
| p | 3 | 6 | n = 2 |
| d | 5 | 10 | n = 3 |
| f | 7 | 14 | n = 4 |
So shell 1 has only 1s; shell 2 has 2s and 2p (2 + 6 = 8); shell 3 has 3s, 3p and 3d (2 + 6 + 10 = 18); shell 4 has 4s, 4p, 4d and 4f (32).
Three rules for filling
- Aufbau principle: electrons occupy the lowest-energy subshell available first.
- Pauli exclusion principle: an orbital holds at most two electrons, and they must have opposite spins (shown as ↑↓).
- Hund's rule: within a subshell, electrons occupy separate orbitals with the same spin before any pair up, because this keeps them further apart.
The filling order is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p ... The diagonal diagram is an easy way to remember it.
Writing electron configurations
A configuration lists each occupied subshell with its number of electrons as a superscript: phosphorus (15 electrons) is 1s² 2s² 2p⁶ 3s² 3p³. To save space, the inner electrons can be replaced by the previous noble gas in square brackets: P is [Ne] 3s² 3p³. The electrons after the noble gas core are the outer or valence electrons.
Write the electron configuration of (a) sulfur, Z = 16, (b) titanium, Z = 22, and (c) bromine, Z = 35.
(a) 2 + 2 + 6 + 2 = 12 electrons fill to 3s; 4 more go into 3p: 1s² 2s² 2p⁶ 3s² 3p⁴, or [Ne] 3s² 3p⁴.
(b) [Ar] accounts for 18; 4s takes 2, and the last 2 enter 3d: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d², or [Ar] 4s² 3d².
(c) [Ar] 4s² 3d¹⁰ takes 30; 5 more go into 4p: [Ar] 4s² 3d¹⁰ 4p⁵. Bromine has 7 valence electrons (4s² 4p⁵), matching group 17.
Some books write 3d before 4s once the 3d is occupied ([Ar] 3d² 4s²), grouping by shell. Both orders are accepted in the HSC as long as the number of electrons in each subshell is right.
An orbital box diagram shows each orbital as a box and each electron as an arrow. It makes Hund's rule visible and shows how many electrons are unpaired.
Exceptions and ions
Chromium and copper break the simple filling order. A half-full (d⁵) or completely full (d¹⁰) d subshell is especially stable, and the 4s and 3d energies are so close that moving one electron is worthwhile:
- Cr (Z = 24): [Ar] 4s¹ 3d⁵, not [Ar] 4s² 3d⁴
- Cu (Z = 29): [Ar] 4s¹ 3d¹⁰, not [Ar] 4s² 3d⁹
Ions. For a negative ion, add electrons following the normal order: Cl⁻ is 1s² 2s² 2p⁶ 3s² 3p⁶, the same as argon. Ions and atoms with identical configurations are called isoelectronic. For positive ions of main-group elements, remove electrons from the outermost subshell. For transition metals, remove the 4s electrons first, even though 4s filled first. Once 3d is occupied, the 4s electrons are the outermost and highest in energy.
Write configurations for Fe²⁺ and Fe³⁺. Explain why Fe³⁺ is more stable than you might expect.
Fe is [Ar] 4s² 3d⁶. Fe²⁺ loses the two 4s electrons: [Ar] 3d⁶.
Fe³⁺ loses one more, from 3d: [Ar] 3d⁵.
3d⁵ is a half-full subshell, with five unpaired electrons, one in each d orbital. This extra stability is one reason iron(II) compounds are slowly oxidised to iron(III) in air.
- Set Z = 26 (iron). Read off [Ar] 4s² 3d⁶ and 4 unpaired electrons. Now choose 3+: which electrons leave first?
- Step from Z = 23 to 24 to 25. What happens to the 4s subshell at chromium?
- Find Z = 29 (copper) and then make Cu⁺. How many unpaired electrons does Cu⁺ have?
- Make S²⁻, Cl⁻, K⁺ and Ca²⁺. What do they all have in common?
- Compare Z = 19 with the Bohr shells readout. Explain why the fourth shell starts before the third is full.
Configurations and the periodic table
The periodic table is laid out by the subshell being filled. The s-block (groups 1 and 2, plus helium) is filling an s subshell; the p-block (groups 13 to 18) a p subshell; the d-block (transition metals, groups 3 to 12) a d subshell; and the f-block (lanthanoids and actinoids) an f subshell. The width of each block, 2, 6, 10 and 14 columns, is the capacity of that subshell.
- The period number equals the highest shell number occupied (n of the outermost s electrons).
- For main-group elements, the group tells you the valence electrons: groups 1 and 2 have 1 and 2; groups 13 to 18 have 3 to 8 (group number − 10).
- Elements in the same group have the same outer configuration (all halogens end in ns² np⁵), which is why they react similarly.
An element has the configuration [Kr] 5s² 4d¹⁰ 5p². State its period, group and block, and identify it.
Highest shell is n = 5, so period 5. The last electron enters 5p, so it is in the p-block. Outer electrons 5s² 5p² = 4 valence electrons, so group 14.
Period 5, group 14 is tin, Sn (Z = 36 + 2 + 10 + 2 = 50).
Test yourself
Which is the ground-state configuration of arsenic, Z = 33?
18 + 2 + 10 + 3 = 33.
How many unpaired electrons does a nitrogen atom have in its ground state?
2p³: by Hund's rule one electron in each of the three 2p orbitals.
Which species has the configuration [Ar] 3d⁹?
Cu is [Ar] 4s¹ 3d¹⁰; removing 4s¹ then one 3d electron gives 3d⁹. Ni²⁺ is 3d⁸.
Which pair of species is isoelectronic?
Both have 10 electrons: 1s² 2s² 2p⁶.
An element's outermost electrons are 4s² 4p⁴. Which statement is correct?
n = 4 gives period 4; 2 + 4 = 6 valence electrons gives group 16 (selenium).
Label the blocks of the periodic table.
Order these subshells from first filled to last filled.
Each orbital can hold two electrons of opposite spin, according to the principle. Electrons occupy orbitals singly before pairing, according to rule. A p subshell contains orbitals and holds up to electrons. A d subshell holds up to electrons.
(a) Write the electron configurations of manganese (Z = 25) and the Mn²⁺ ion. (b) Draw an orbital box diagram for the 4s and 3d subshells of Mn²⁺ and state the number of unpaired electrons. (c) Explain why the 4s electrons are removed before the 3d electrons when Mn forms an ion.
(a) Mn: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵ = [Ar] 4s² 3d⁵. Mn²⁺: [Ar] 3d⁵.
(b) 4s: empty box. 3d: five boxes each with one up arrow. Five unpaired electrons.
(c) The 4s subshell fills before 3d because in K and Ca it is lower in energy. But once electrons occupy 3d, the 4s electrons are further from the nucleus and higher in energy, so they are the first to be removed when the atom is ionised.