Describe electron configuration in terms of shells, subshells (s, p, d) and orbitals using the (1s 2s 2p 3s 3p 4s 3d 4p) filling order, and explain the periodic trends in atomic radius, first ionisation energy and electronegativity using effective nuclear charge and shielding

What this dot point is asking

QCAA wants you to write electron configurations using s, p and d subshells with the standard filling order, predict the configuration of ions (including the 4s-removed-first rule for transition metals), and explain the three core periodic trends (atomic radius, first ionisation energy, electronegativity) in terms of effective nuclear charge and shielding. This is the foundation for every bonding argument in Topic 2.

The answer

Electrons occupy shells (energy levels labelled by principal quantum number n) made up of subshells (s, p, d, f) that contain orbitals. Orbitals fill in a specific order set by their energies, following the aufbau principle. Periodic trends arise because of how the effective nuclear charge experienced by the outer electrons changes across periods and down groups.

Shells, subshells and orbitals

The capacity of shell n is $2n^2$ electrons. Shell 1: 2 (1s). Shell 2: 8 (2s, 2p). Shell 3: 18 (3s, 3p, 3d). Shell 4: 32 (4s, 4p, 4d, 4f).

Each orbital holds up to two electrons of opposite spin (Pauli exclusion). Within a subshell, orbitals are filled singly before pairing (Hund's rule), which minimises electron-electron repulsion.

The filling order

QCE Chemistry expects you to use the standard aufbau order. For Z up to 36 (krypton), the order is:

The 4s subshell fills before 3d in neutral atoms because 4s is at lower energy than 3d at that stage. Beyond Kr the order becomes 5s, 4d, 5p, and so on.

Worked configurations:

• Na (Z = 11): $1s^2 2s^2 2p^6 3s^1$
• Cl (Z = 17): $1s^2 2s^2 2p^6 3s^2 3p^5$
• Ca (Z = 20): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$
• Fe (Z = 26): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$
• Br (Z = 35): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5$

Two QCAA-relevant exceptions (chromium and copper) gain extra stability by half-filled or fully filled d subshells:

• Cr (Z = 24): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$ (not $4s^2 3d^4$).
• Cu (Z = 29): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}$ (not $4s^2 3d^9$).

Electron configurations of ions

For main-group ions, electrons are added or removed from the outermost subshell.

• F (Z = 9): $1s^2 2s^2 2p^5$. F-: $1s^2 2s^2 2p^6$ (noble-gas Ne configuration).
• Mg (Z = 12): $1s^2 2s^2 2p^6 3s^2$. Mg^2+: $1s^2 2s^2 2p^6$.

For transition metals forming cations, remove the s electrons before the d electrons.

• Fe (Z = 26): $...4s^2 3d^6$. Fe^2+: $...3d^6$ (4s lost). Fe^3+: $...3d^5$ (4s and one 3d lost).
• Cu (Z = 29): $...4s^1 3d^{10}$. Cu^+: $...3d^{10}$. Cu^2+: $...3d^9$.

This is a frequent QCAA test point because it contradicts the naive "last in, first out" expectation.

Effective nuclear charge and shielding

Effective nuclear charge (Z_eff) is the net positive charge experienced by an outer electron, after accounting for shielding (also called screening) by inner electrons.

where S is the screening constant. Inner-shell electrons shield outer electrons effectively; electrons in the same subshell shield each other only weakly.

Two principles drive every periodic trend:

1. Across a period: Z rises by one each step; S stays roughly constant (electrons added to the same shell shield each other poorly). So Z_eff rises strongly.
2. Down a group: Z rises but S also rises sharply (a whole new inner shell each step). Z_eff stays roughly constant, but the principal quantum number n rises, so the outermost electrons are physically further from the nucleus.

Atomic radius

Atomic radius is the typical distance from the nucleus to the outermost occupied shell.

• Across a period: radius decreases. Z_eff rises, electrons in the same shell pulled in tighter.
• Down a group: radius increases. Outermost shell has higher n; the electron is intrinsically further out.

Ionic radii follow:

• Cations are smaller than their parent atoms (lose a whole outer shell, or at least reduce electron-electron repulsion).
• Anions are larger than their parent atoms (added electron increases repulsion in the outer shell).

First ionisation energy

First ionisation energy (IE_1) is the energy required to remove the most loosely held electron from a gaseous atom.

• Across a period: IE_1 increases. Higher Z_eff, smaller radius, electron more tightly held.
• Down a group: IE_1 decreases. Outer electron is in a higher shell, further from nucleus, more shielded.

Two subtle "dips" you should be ready to explain:

• Between Group 2 and Group 13 (e.g. Be to B): the outermost electron in B is in a 2p orbital, which is at slightly higher energy than the 2s; easier to remove.
• Between Group 15 and Group 16 (e.g. N to O): N has half-filled 2p (extra stability), so the next electron in O experiences pair-repulsion in the 2p orbital; easier to remove.

These dips are popular QCAA EA short-response prompts.

Electronegativity

Electronegativity is the tendency of an atom in a bond to attract bonding electrons toward itself. Pauling scale; fluorine is set at 4.0, the most electronegative element. Caesium is around 0.7.

• Across a period: electronegativity increases. Same Z_eff reasoning.
• Down a group: electronegativity decreases. Larger atom, bonding electrons further from the nucleus.

The most electronegative elements are concentrated in the upper-right of the periodic table (F, O, N, Cl). The least are bottom-left (Cs, Fr).

Electronegativity governs bond polarity (Topic 2 dot point on covalent bonding) and predicts whether a bond is ionic (large electronegativity difference), polar covalent (moderate difference), or non-polar covalent (small difference).

Putting the three trends together

A single line of reasoning (Z_eff plus shell structure) drives all three. QCAA EA short-response questions often ask you to apply that reasoning to a specific pair, e.g. "Why is the first ionisation energy of Mg higher than that of Na?" or "Why is F more electronegative than Cl?"

Examples in context

Example 1. Sodium reactivity at Townsville sugar refinery. Wilmar's Victoria Mill uses sodium hydroxide to clarify cane juice; the $\text{NaOH}$ originates from chlor-alkali electrolysis of brine. Sodium's electron configuration is $1s^2 2s^2 2p^6 3s^1$. The single $3s$ electron sits far from the nucleus with full inner-shell shielding, giving a low first ionisation energy of $496 \, \text{kJ mol}^{-1}$. This is why sodium readily forms $\text{Na}^+$ in any reaction with water or chlorine. Compare with magnesium ($738 \, \text{kJ mol}^{-1}$): magnesium has one more proton pulling the same shell tighter, illustrating the period-3 increase in ionisation energy across the row.

Example 2. Period-3 oxide acidity in QCAA data-test scenarios. A typical data-set tabulates $\text{Na}_2\text{O}$, $\text{MgO}$, $\text{Al}_2\text{O}_3$, $\text{SiO}_2$, $\text{P}_4\text{O}_{10}$, $\text{SO}_3$, $\text{Cl}_2\text{O}_7$ with their pH in water. Across period 3, electronegativity rises from $0.93$ ($\text{Na}$) to $3.16$ ($\text{Cl}$) because nuclear charge increases while shielding stays constant. Bond character in the oxides shifts from ionic (basic $\text{Na}_2\text{O}$, $\text{pH} \approx 14$) through amphoteric $\text{Al}_2\text{O}_3$ to covalent acidic ($\text{SO}_3$ giving $\text{pH} \approx 1$ when dissolved as sulfuric acid). Students must link the bonding trend to the periodic trend in electronegativity.

Try this

Q1. Write the full electron configuration of $\text{Cl}^-$ and identify the noble gas it is isoelectronic with. [2 marks]

• Cue. $1s^2 2s^2 2p^6 3s^2 3p^6$, isoelectronic with argon.

Q2. First ionisation energies (kJ mol$^{-1}$) across period 3 are: Na $496$, Mg $738$, Al $577$, Si $786$. Explain the dip from Mg to Al. [3 marks]

• Cue. Al loses a $3p$ electron higher in energy than the paired $3s$ electron of Mg; less tightly held despite more protons.

Q3. Consider $\text{F}$ and $\text{I}$. (a) Compare atomic radius. (b) Compare electronegativity. (c) Predict which forms a stronger ionic bond with sodium and justify. [2+2+3 marks]

• Cue. (a) $\text{F}$ smaller, fewer shells. (b) $\text{F}$ higher, $3.98$ vs $2.66$. (c) $\text{NaF}$ stronger: smaller anion, higher lattice energy. Application + judgement.