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wise otter.Chemistry
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Atomic Structure

Topic 2 of 3

Electrons in orbitals

Build configurations from shell and subshell rules.

A-Level 8873, revised syllabus (2026-2027)

Replace circular paths with orbitals

An orbital describes where an electron may be found, not a tiny planetary orbit.

A shell is labelled by principal quantum number n. Within a shell, subshells contain orbitals: one s orbital, three p orbitals or five d orbitals. Each orbital holds at most two electrons with opposite spins. Orbitals in the same subshell have the same energy in an isolated atom unless an external influence splits them.

Orbitals required in this course
Shell/subshellNumber of orbitalsMaximum electrons
n = 1: 1s12
n = 2: 2s; 2p1; 32; 6
n = 3: 3s; 3p; 3d1; 3; 52; 6; 10
4s; 4p1; 32; 6

s and p orbital shapes

A sphere represents an s orbital. Two lobes on opposite sides of the nucleus represent one p orbital. The two colours distinguish lobes, not electric charges.

Boundary sketches show shape, not a hard surface or electron trajectory. The three p orbitals point along perpendicular axes; no wave functions are required.

Within a multi-electron shell, s is lower in energy than p, which is lower than d where present. The useful filling sequence for the atoms studied is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. Energies also depend on occupancy: this filling order is not a claim that 4s is always lower than 3d in every ion.

Build a configuration, then make the ion

Fill the available orbitals and count the electrons again after every change.

Writing an electron configuration
  1. Count electrons

    Use Z for a neutral atom; adjust for the signed ion charge.

  2. Fill lower levels first

    Follow the filling sequence, respecting the capacity of each subshell.

  3. Distribute within a subshell

    Place electrons singly with parallel spins in equal-energy orbitals before pairing them.

  4. Check the ion

    For transition-metal cations, remove 4s electrons before 3d electrons.

Configurations worth understanding
SpeciesElectronsConfiguration
N (Z = 7)71s2 2s2 2p3; one electron in each p orbital
O (Z = 8)81s2 2s2 2p4; one p orbital has a pair
Cl- (Z = 17)181s2 2s2 2p6 3s2 3p6
Fe (Z = 26)26[Ar] 3d6 4s2
Fe2+24[Ar] 3d6
Fe3+23[Ar] 3d5

The symbol [Ar] replaces the filled first 18 electrons. For Cr (Z = 24), use [Ar] 3d5 4s1; for Cu (Z = 29), use [Ar] 3d10 4s1. These are exceptions to simply assigning two electrons to 4s before 3d. When forming their positive ions, remove the 4s electron first.

Worked example

An ion does not fill like its neutral parent

Write the configuration of Cu2+ (Z = 29).

  1. Start with Cu: [Ar] 3d10 4s1.
  2. Remove the 4s electron, then one 3d electron.
  3. Count: 18 + 9 = 27 electrons, two fewer than 29 protons.
Answer

Cu2+ is [Ar] 3d9, not [Ar] 3d7 4s2.

Check your understandingWrite the configuration of S2- (Z = 16).Think it through, then reveal the answer
There are 18 electrons: 1s2 2s2 2p6 3s2 3p6. The two gained electrons complete 3p; the sulfur nucleus still has 16 protons.