Lesson 4: Ionisation Energy Trends

The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms: X(g) → X⁺(g) + e⁻. It is always endothermic — you pay to pull an electron away from a nucleus — and its trends across the periodic table (Structure 1.3 and 3.1) are the cleanest evidence we have for shell and subshell structure. Three factors control it: nuclear charge (more protons pull harder), distance from the nucleus (further electrons are held more weakly), and shielding by inner electrons (which cancels part of the nuclear charge). Every trend question in this topic is an argument about which of these factors wins.

The trends — and the dips that prove the theory

Across a period, ionisation energy generally increases: protons are added while electrons enter the same shell, so shielding barely changes, the radius shrinks, and the outer electrons are held ever tighter. Down a group it decreases: despite the rising nuclear charge, the new shell's distance and shielding dominate. Period 3 shows two famous dips. Magnesium (738 kJ mol⁻¹) exceeds aluminium (578): aluminium's outer electron sits in a 3p orbital, higher in energy, further away and shielded by the filled 3s² — so it leaves more easily despite the extra proton. Phosphorus (1012) exceeds sulfur (1000): sulfur's 3p⁴ has one paired orbital, and the repulsion between the paired electrons makes one of them easier to remove than phosphorus's half-filled, singly-occupied 3p³ arrangement. Examiners adore both dips because they distinguish students who memorised "it increases" from students who understand subshells.

Successive ionisation energies: reading the group number

Remove electrons one after another and the energies climb steadily — then jump when you break into an inner shell. Sodium's values (kJ mol⁻¹) run 496, then 4562: the enormous jump after the first electron exposes the n = 2 core and confirms one valence electron. The rule is absolute: the number of electrons removed before the first big jump equals the group number. A log₁₀ plot of successive ionisation energies makes shells visible as clusters of points, and the pattern of small rises within a shell versus the cliff between shells is direct experimental evidence for the quantum model of Lesson 3 — data first, theory second.

Worked example 1: deducing a group from data

An element has successive ionisation energies (kJ mol⁻¹): 578, 1817, 2745, 11 578, 14 831. Deduce its group and suggest its identity. Ratios: 578 → 1817 → 2745 are steady rises (same shell, increasing charge on the ion). Then 2745 → 11 578 is a jump by a factor of about 4 — a new, inner shell has been breached. Three electrons were removed before the jump, so the element has three valence electrons: Group 13. The first value, 578 kJ mol⁻¹, matches aluminium in the data booklet — consistent with a Period 3 element whose 3p¹ electron is removed first. Cross-checking the first ionisation energy against the data booklet is a legitimate and encouraged confirmation step.

Worked example 2: the log plot — data handling with real numbers

Take magnesium's twelve successive ionisation energies from the data booklet and plot log₁₀(IE) against the number of the electron removed. The first two points sit low (log ≈ 2.87, 3.16), the next eight cluster higher (the n = 2 shell), and the last two leap again (the 1s core). The graph does three jobs at once: it confirms the 2, 8, 2 arrangement without any theory assumed in advance; it shows why the *second* ionisation energy of magnesium (1451 kJ mol⁻¹) is roughly double the first (the same shell, but the ion's increased charge contracts what remains); and it exposes a subtlety — the rise within the n = 2 cluster is not smooth, because the 2s and 2p subshells empty at slightly different energies. As practical data work, note the technique: plotting twelve values spanning 738 to 107 000 kJ mol⁻¹ is impossible on a linear scale, so the logarithm is a choice forced by the data range — exactly the kind of justified decision the IA analysis section wants. Uncertainty here is not instrumental but interpretive: deciding where the "jump" begins is a judgement, and two students may defend adjacent boundaries; state your criterion (a rise of more than ×3, say) explicitly.

Nature of Science — and the real world

Ionisation energies were among the first quantitative data sets precise enough to *constrain* atomic theory: the shell model's numbers came from spectroscopy and ionisation measurements, not from philosophy. The pattern-seeking is the science. Practically, ionisation energy explains why the noble gases are inert (their ionisation energies are the highest in each period), why sodium is violently reactive (496 kJ mol⁻¹ is cheap), and why photoelectron spectroscopy — which measures these energies directly with photons — is used to fingerprint materials in semiconductor manufacture.

Common mistakes

• Defining ionisation energy without "gaseous atoms" and "one mole" — the definition is worth a mark and is marked strictly.
• Explaining the Mg → Al dip with "aluminium has more shielding" alone; the full answer needs the 3p electron's higher energy/greater distance and its shielding by 3s².
• Reading successive ionisation energies backwards: the jump comes after the valence electrons are gone, so count the values *before* the jump.
• Claiming the second ionisation energy is smaller because "the atom wants a full shell" — teleology scores nothing; argue from charge, distance and shielding.

IB Exam Experience

Question 1. Which element has the highest first ionisation energy?

• (A) Na
• (B) Mg
• (C) P
• (D) Ar

Question 2. Why is the first ionisation energy of aluminium lower than that of magnesium?

• (A) Aluminium has a greater nuclear charge
• (B) The outer electron of aluminium is in a 3p orbital, which is higher in energy and shielded by the 3s electrons
• (C) Aluminium has fewer electrons than magnesium
• (D) The 3s electrons in magnesium are paired and repel each other

Question 3. An element has successive ionisation energies (kJ mol⁻¹) of 738, 1451, 7733 and 10 543. To which group does it belong?

• (A) Group 1
• (B) Group 2
• (C) Group 3
• (D) Group 13

Answer key. 1. (D) — argon completes Period 3 with the highest nuclear charge acting on the same shell; its first ionisation energy (1521 kJ mol⁻¹) is the period maximum. 2. (B) — the 3p electron is further from the nucleus, at higher energy and shielded by 3s², outweighing the extra proton. 3. (B) — two modest values, then the jump: two valence electrons (the data are magnesium's).

Structured question (Paper 2 style). (a) Define the term first ionisation energy and write the equation for the first ionisation of magnesium. [2] (b) Explain why the first ionisation energy increases across Period 3 from sodium to chlorine. [3] (c) Explain why the first ionisation energy of sulfur (1000 kJ mol⁻¹) is lower than that of phosphorus (1012 kJ mol⁻¹), even though sulfur has the greater nuclear charge. [2] Marking notes. (a) The energy required to remove one mole of electrons [1] from one mole of gaseous atoms [1]; equation Mg(g) → Mg⁺(g) + e⁻ (state symbols required for the equation mark). (b) Nuclear charge increases [1]; electrons are added to the same shell so shielding is approximately constant [1]; atomic radius decreases so the outer electrons experience stronger attraction [1]. (c) Phosphorus has a half-filled 3p³ arrangement with all electrons unpaired, while sulfur's 3p⁴ contains a paired orbital [1]; the repulsion between the paired electrons raises their energy, so one is removed more easily, outweighing the extra proton [1].

IA idea: There is no wet-lab route to ionisation energies — instead, build an IA around data: model the successive ionisation energies of Period 3 elements from the data booklet and test how well a simple Z_eff model predicts the observed dips at aluminium and sulfur.

Ionisation energy decides which ions exist

Why does sodium form Na⁺ and never Na²⁺ in its compounds, magnesium stop at Mg²⁺, and aluminium at Al³⁺? Ionisation energies answer it — read as an energy ledger. Forming Na²⁺ would cost 4562 kJ mol⁻¹ for the second ionisation alone, because that electron comes from the stable n = 2 core; no lattice or hydration energy a sodium compound could offer repays that, so Na²⁺ chemistry does not exist. Magnesium's first two ionisations cost 738 + 1451 = 2189 kJ mol⁻¹ in total — expensive, but the small, doubly-charged Mg²⁺ ion earns it back through a lattice enthalpy roughly four times that of a singly-charged salt of similar size. Aluminium pays 578 + 1817 + 2745 = 5140 kJ mol⁻¹ for Al³⁺ and recovers it only in strong lattices and concentrated hydration shells — which is why Al³⁺ compounds are markedly covalent in character (Lesson 5's continuum) and why AlCl₃ dimerises rather than forming a simple ionic lattice. The pattern generalises into a rule you can apply to any element: ionic charge stops where the next ionisation energy jumps, because the jump marks the core, and core electrons are never worth their price in chemical currency.

There is a mirror image. The electron affinity — the energy change when a gaseous atom gains an electron — is exothermic for the halogens (Cl: −349 kJ mol⁻¹) but the *second* electron affinity is always endothermic, because the incoming electron is repelled by an already negative ion (O⁻ → O²⁻ costs +844 kJ mol⁻¹). Oxide lattices nevertheless exist because the enormous lattice enthalpy of O²⁻ with small cations pays that bill too — the same ledger logic, run in the opposite direction, and the working principle behind the Born–Haber cycles of Lesson 22. Ion formation, in short, is never about atoms "wanting full shells"; it is about whether the total energy account balances, and ionisation energies are the largest entries in the account.

First ionisation energies down Group 1: reactivity priced in kJ

The first ionisation energies of the alkali metals fall steadily — Li 520, Na 496, K 419, Rb 403, Cs 376 kJ mol⁻¹ — and their reactivity with water climbs in the same order: lithium fizzes steadily, sodium melts and darts, potassium ignites its hydrogen with a lilac flame, and the trend continues violently below. The model explains both columns with one argument: down the group the valence electron occupies a shell further from the nucleus and shielded by more complete inner shells; nuclear charge rises too, but distance and shielding win, so the electron is cheaper to remove and the metal is a stronger reducing agent. Group 2 repeats the pattern at higher prices (Be 900, Mg 738, Ca 590, Sr 549, Ba 503) — the two-electron removal to reach M²⁺ still climbs down the group as the ions grow. Note the subtlety examiners probe: reactivity of the *metals* rises down Group 1 even though their melting points fall — different properties, different controlling factors (ionisation energy versus metallic bond strength, Lesson 8), and conflating them is a reliable way to lose an explanation mark.