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

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:

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:

1. 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.
2. Shielding: electrons in inner shells reduce the pull of the nucleus on the valence electrons.
3. 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.

Examples in context

Example 1. Lithium extraction at Greenbushes Mine, Western Australia. Talison's Greenbushes operation in WA is the world's largest source of lithium. Lithium has electron configuration $1s^2 \, 2s^1$. The single $2s$ electron sits far from a core charge of $+1$, so it is loosely held; the first ionisation energy is only $520 \, \text{kJ}$ per mole, the lowest among period-2 elements. This is why lithium reacts vigorously with water and is so easily oxidised to $\text{Li}^+$. During spodumene processing, lithium is leached as $\text{Li}_2 \text{SO}_4$, then precipitated as $\text{Li}_2 \text{CO}_3$. The same low ionisation energy makes lithium an excellent anode metal in batteries, since the electron can be released with little energy input.

Example 2. Selenium-doped solar cells at Monash University. Researchers at Monash develop CIGS thin-film solar cells doped with selenium. Selenium is below sulfur in Group 16 with configuration $[\text{Ar}] \, 3d^{10} \, 4s^2 \, 4p^4$. Its larger atomic radius (about $120 \, \text{pm}$) and lower electronegativity ($2.55$) compared to sulfur ($1.84$, $2.58$) mean selenium forms longer, weaker covalent bonds with metals like copper and indium. This narrows the semiconductor band gap to around $1.04 \, \text{eV}$, ideal for absorbing visible sunlight. The periodic trend (radius increases down a group, electronegativity decreases) directly determines the photovoltaic response of the material.

Try this

Q1. Write the full electron configuration of (a) iron, $\text{Fe}$, and (b) the $\text{Fe}^{3+}$ ion. State which orbitals lose electrons first when iron is ionised. [3 marks]

• Cue. (a) $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^6 \, 4s^2$. (b) $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^5$. Lose $4s$ before $3d$ when ionising.

Q2. First ionisation energies (kJ/mol) for period 3 are: Na 496, Mg 738, Al 577, Si 786. Explain the dip from Mg to Al. [2 marks]

• Cue. Al's outermost electron is in $3p$, higher in energy and partially shielded by full $3s^2$; easier to remove than Mg's $3s$ electron.

Q3. Compare and contrast trends in atomic radius across period 3 and down group 1. (a) State each trend. (b) Explain each using core charge and shell number. (c) Predict which is larger: $\text{Na}$ or $\text{K}$; $\text{Na}$ or $\text{Mg}$. [2+2+2 marks]

• Cue. (a) Across: decreases; down: increases. (b) Across: core charge increases, same shell. Down: new shell added each row. (c) K larger than Na; Na larger than Mg.