Edexcel UK AS 8CH0 / A-Level 9CH0 · Topic 1 · Year 12 / AS

Part 3: Shells, orbitals and electron configurations

Reviewed 9 October 2026.

Replace the fixed-orbit picture with shells, subshells and orbitals, then use electron counts to build configurations through krypton.

Three levels of organisation

A principal quantum shell is labelled n = 1, 2, 3, … and contains subshells. A subshell contains one or more orbitals. An orbital is a region of space in an atom associated with a high probability of finding an electron; it can hold up to two electrons with opposite spins. It is not a circular path followed by an electron.

The first four shells can contain at most 2, 8, 18 and 32 electrons respectively, following 2n². Capacity is different from filling order: the third shell can hold 18, although potassium’s next electron enters 4s before 3d fills. Spin is a quantum property represented by an arrow; opposite arrows do not mean electrons travel around a circle in opposite directions.

Subshell capacities
SubshellNumber of orbitalsMaximum electronsShape required here
s12Spherical
p36Each orbital is dumbbell-shaped, with two lobes
d510Detailed shapes not required
f714Explains the capacity of shell 4; configurations beyond Z = 36 are not required

An s orbital is a sphere; a p orbital has two lobes

A sketch of an s orbital represents a three-dimensional spherical region, often drawn as a circle with a curved equatorial line. A p orbital has two equal lobes on opposite sides of the nucleus. Both lobes belong to one orbital, so together they can hold at most two electrons. The three p orbitals are oriented at right angles to one another.

In 8CH0/01 June 2023 Q1(b), the examiner report distinguishes a sphere from a circle and opposite spin from opposite directions of travel. These distinctions describe the model accurately, rather than supplying alternative names for the same thing.

Spherical s orbital, two-lobed p orbital and oxygen 2p boxes with one opposite-spin pair and two singly occupied orbitals.

Swipe horizontally to view the whole diagram.

Schematic orbital boundaries and occupancy. These are probability-region representations, not electron trajectories or measured images.

Build configurations in a controlled order

For neutral atoms through Z = 36, use the filling sequence 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. Put one electron into each equal-energy orbital of a subshell before pairing them, with parallel spins for those singly occupied orbitals. When a pair forms, its spins must be opposite. The total superscripts must equal the number of electrons.

Worked example: sulfur has Z = 16. Filling 1s² 2s² 2p⁶ uses ten electrons; 3s² uses two more, leaving four for 3p. Its configuration is 1s² 2s² 2p⁶ 3s² 3p⁴. The three 3p boxes contain one pair and two single electrons. Pairing two boxes while leaving the third empty would violate the singly-before-pairing rule.

For bromine, Z = 35, the configuration is [Ar] 3d¹⁰ 4s² 4p⁵. Writing occupied subshells in principal-shell order is common even though 4s was filled before 3d. Chromium and copper are exceptions to the simplest filling prediction: Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹. Their closely spaced subshell energies make these arrangements lower in energy overall; do not force 4s² for these two atoms.

Count electrons first, then change the outer occupancy

For the s- and p-block ions required in Topic 1, remove outer-shell electrons to form cations and add electrons to the outer subshell to form anions. Calcium, [Ar] 4s², becomes Ca²⁺, [Ar]. Sulfur, [Ne] 3s² 3p⁴, becomes S²⁻, [Ne] 3s² 3p⁶ = [Ar]. Both ions have 18 electrons, but different nuclear charges and chemical behaviour.

For aluminium, [Ne] 3s² 3p¹, remove the 3p electron and then the two 3s electrons to give Al³⁺ = [Ne]. A bromide ion has 35 + 1 = 36 electrons and configuration [Ar] 3d¹⁰ 4s² 4p⁶. Check the final electron count independently; changing the nuclear charge to obtain the ion would change the element.

Topic 1 requires neutral atom configurations through krypton and s-/p-block ion configurations. Transition-metal ion configurations and the loss of 4s electrons from them are taught in Topic 15.

Chlorine outer orbitals have paired 3s and 3p occupancy 2,2,1; chloride has paired 3s and three full 3p pairs.

Swipe horizontally to view the whole diagram.

Electrons-in-boxes comparison for an atom and its ion. One box represents one orbital; each paired set has opposite spins.

Electron configuration explains the Periodic Table

The block identifies the subshell being filled in the periodic pattern: Groups 1 and 2 are s block, Groups 13–18 are p block, and the central series is d block. Helium has 1s², so its configuration is s block although it is placed with the chemically unreactive noble gases.

Elements in the same main group have the same number and pattern of outer electrons. Group 2 atoms have an outer ns² configuration and tend to lose two electrons; Group 7 halogens have ns²np⁵ and can gain one. Chemical properties repeat because these outer configurations repeat. The highest occupied principal shell gives the period, but the subshell filled across a d-block row belongs to the shell below it.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Q1. How many electrons can one p orbital, a p subshell and the third shell each hold?Show answer

One p orbital holds two; three p orbitals make a p subshell holding six; the third shell contains 3s, 3p and 3d and holds 2 + 6 + 10 = 18. Do not use the subshell capacity for one orbital.

Q2. Write the full configuration of phosphorus (Z = 15) and describe its three 3p boxes.Show answer

1s² 2s² 2p⁶ 3s² 3p³. Each 3p orbital contains one electron, with the three arrows pointing the same way. Pairing begins only after all three are singly occupied.

Q3. Write the configurations of K⁺ (Z = 19) and Se²⁻ (Z = 34).Show answer

K⁺ has 18 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶. Se²⁻ has 36: [Ar] 3d¹⁰ 4s² 4p⁶. Potassium loses one electron; selenium gains two.

Q4. A student writes chromium as [Ar] 3d⁴ 4s². Correct it and check the total.Show answer

The ground-state configuration is [Ar] 3d⁵ 4s¹. The count is 18 + 5 + 1 = 24. Chromium is an exception to the simple filling prediction because the occupied subshell energies are close.

Q5. An atom ends 3s² 3p⁵. Identify its block and explain one likely ion charge.Show answer

It is a p-block atom, chlorine, with seven outer electrons. Gaining one electron completes 3p, producing Cl⁻. The gained electron changes charge but does not change the number of protons.

Sources

Sources and examiner guidance (reviewed 9 October 2026)

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