← Unit 1: How can the diversity of materials be explained?
How can knowledge of elements explain the properties of matter?
electron configurations of atoms up to atomic number 36 using the Schrödinger model (shells, subshells, orbitals; spdf notation), and the explanation of trends in the periodic table including atomic radius, first ionisation energy and electronegativity in terms of core charge, shielding and shell number
A focused VCE Chemistry Unit 1 answer on electron configuration. Covers the Schrödinger model (shells, subshells, orbitals), spdf notation up to Z=36, and the explanation of atomic radius, first ionisation energy and electronegativity trends across periods and down groups.
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What this dot point is asking
VCAA wants you to write electron configurations of atoms and ions up to Z = 36 using the Schrödinger model (shells, subshells, orbitals) with spdf notation, and to explain periodic trends (atomic radius, first ionisation energy, electronegativity) in terms of core charge, shielding and shell number.
The answer
Shells, subshells, orbitals
The Schrödinger model places electrons in regions called orbitals, grouped into subshells, grouped into shells.
- Shell (principal quantum number n = 1, 2, 3, ...): roughly an energy level / distance from the nucleus.
- Subshell (s, p, d, f): grouped orbitals of the same shape within a shell.
- Orbital: a 3D region with up to 2 electrons (opposite spins).
Subshell capacities:
| Subshell | Orbitals | Max electrons |
|---|---|---|
| s | 1 | 2 |
| p | 3 | 6 |
| d | 5 | 10 |
| f | 7 | 14 |
Filling order (aufbau)
Subshells fill from lowest energy upward. For neutral atoms up to Z = 36 the order is:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p
The 4s subshell fills before the 3d (because, when both are empty, 4s sits slightly lower in energy). Once 3d begins filling, the 4s rises slightly above 3d, which is why transition-metal cations lose 4s electrons before 3d.
spdf notation
Write the subshells in order with the number of electrons as a superscript. The exponents must sum to the total number of electrons. Examples up to Z = 36:
- Carbon (Z = 6): 1s^2 2s^2 2p^2
- Sodium (Z = 11): 1s^2 2s^2 2p^6 3s^1
- Argon (Z = 18): 1s^2 2s^2 2p^6 3s^2 3p^6
- Calcium (Z = 20): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2
- Iron (Z = 26): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6
- Bromine (Z = 35): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^5
Two anomalies you need to know: Cr is [Ar] 4s^1 3d^5 and Cu is [Ar] 4s^1 3d^10 (a half-filled or fully-filled 3d is slightly more stable than a paired 4s).
Ions
For anions, add electrons to the next available orbital. O (1s^2 2s^2 2p^4) gains two electrons to become O^2- (1s^2 2s^2 2p^6).
For cations, remove electrons from the highest-n shell first. For transition metals, that means removing 4s electrons before 3d:
- Fe (4s^2 3d^6) becomes Fe^2+ (3d^6) and Fe^3+ (3d^5).
- Cu (4s^1 3d^10) becomes Cu^+ (3d^10) and Cu^2+ (3d^9).
Periodic trends
Three factors explain almost every trend:
- Core charge (effective nuclear charge): the net positive charge felt by a valence electron after accounting for inner-shell shielding. Roughly equal to the group number for main-group elements.
- Shielding: electrons in inner shells reduce the pull of the nucleus on the valence electrons.
- Shell number (n): a larger n means valence electrons sit further from the nucleus.
Atomic radius
- Across a period (left to right): core charge increases, shielding is roughly constant, shell number is constant. Valence electrons are pulled closer. Radius decreases.
- Down a group: shell number increases (added shells), shielding increases. Valence electrons sit further out. Radius increases.
First ionisation energy
The energy to remove the most loosely held electron from a gaseous atom.
- Across a period: stronger pull on valence electrons (higher core charge, same n) means ionisation energy increases.
- Down a group: weaker pull on valence electrons (higher n, more shielding) means ionisation energy decreases.
Small dips exist within a period (Mg to Al, P to S) due to subshell effects, but the overall trend is upward.
Electronegativity
The tendency of an atom to attract a shared pair of electrons in a covalent bond.
- Across a period: electronegativity increases for the same reason as ionisation energy.
- Down a group: electronegativity decreases.
Fluorine sits in the top right (excluding noble gases) and is the most electronegative element on the Pauling scale (4.0). Caesium and francium are the least electronegative.
Worked example
Compare the first ionisation energies of Na, Mg, Al and write configurations.
Configurations:
- Na (Z = 11): 1s^2 2s^2 2p^6 3s^1
- Mg (Z = 12): 1s^2 2s^2 2p^6 3s^2
- Al (Z = 13): 1s^2 2s^2 2p^6 3s^2 3p^1
Predicted order Na < Mg < Al (core charge rising left to right). Actual order Na < Al < Mg.
The dip from Mg to Al happens because Al's outermost electron is in the 3p subshell, slightly higher in energy and more shielded by the filled 3s^2 below it, so it is easier to remove than Mg's 3s^2 electron.
Common traps
Writing 3d before 4s in a neutral atom. Filling order for neutral atoms is 4s before 3d. (Order in cations is the reverse.)
Removing 3d electrons first when ionising a transition metal. Always remove the 4s electrons first.
Forgetting Cr and Cu are anomalies. Both promote one 4s electron into 3d for extra stability.
Saying nuclear charge instead of core charge. Nuclear charge is the full +Z. Core charge is what's left after the inner shells shield it. Use the latter when explaining periodic trends.
Mixing radius and ionisation energy trends. They go in opposite directions (radius and ionisation energy are inversely related across both periods and groups).
In one sentence
Electrons occupy shells, subshells and orbitals in the Schrödinger model, the configuration is written in spdf notation with 4s filling before 3d for neutral atoms but emptying first for transition-metal cations, and atomic radius, ionisation energy and electronegativity trends across the periodic table follow directly from how core charge, shielding and shell number change.
Past exam questions, worked
Real questions from past VCAA papers on this dot point, with our answer explainer.
2024 VCE3 marks(a) Write the full electron configuration (spdf notation) of a neutral iron atom (Z = 26). (b) Write the configuration of the Fe^2+ ion.Show worked answer →
A 3-mark answer needs the correct order, the right number of electrons, and the right ion configuration.
(a) Neutral Fe (26 electrons), filling 1s, 2s, 2p, 3s, 3p, 4s, 3d (the 4s subshell fills before 3d in the neutral atom):
1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6
(b) For transition-metal cations, electrons are removed from the 4s subshell before the 3d (the 4s electrons sit furthest from the nucleus once the 3d is partly filled):
1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 (the 4s^2 is removed)
A common mistake is removing 3d electrons first because 3d "filled last". The 4s electrons are easiest to remove because, after the 3d begins filling, the 4s orbital is slightly higher in energy and more shielded.
2025 VCE4 marksExplain why first ionisation energy generally increases across Period 3 (from Na to Ar) but decreases down Group 1 (from Li to Cs). Use core charge, shielding and shell number in your explanation.Show worked answer →
A 4-mark answer needs both trends explained in terms of the same three factors.
Across Period 3 (Na to Ar):
Core charge (effective nuclear charge felt by valence electrons) increases from +1 to +8 across the period. Shielding is roughly constant (the same inner shells, 1s^2 2s^2 2p^6, shield each valence electron). Shell number is constant (all valence electrons are in shell n = 3). Stronger attraction to the nucleus pulls valence electrons closer, so more energy is required to remove one. First ionisation energy therefore increases.
Down Group 1 (Li to Cs):
Core charge felt by the valence electron is roughly constant (each new shell adds one proton but also adds a full inner shell that shields it). Shielding increases (more inner shells of electrons). Shell number increases, so the valence electron is further from the nucleus. The valence electron is held less tightly. First ionisation energy therefore decreases.
A 4-mark answer names all three factors and the direction each pushes. Markers also accept noting the small dips at Mg to Al (3s^2 vs 3p^1) and at P to S (paired-electron repulsion in the 3p orbitals), although that level of detail is not required.
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