Ionisation Energies

Ionisation energies provide direct experimental evidence for the existence of electron shells and sub-shells. Understanding trends in ionisation energy is essential for explaining periodicity and the properties of elements.

Spec mapping (AQA 7405): This lesson maps to §3.1.1.3 (electron configuration — continued: ionisation energies as evidence for sub-shell structure) and §3.2.1 (periodicity — period 3 trends in first ionisation energy). The two "drops" in the trend across a period (Be → B / Mg → Al, and N → O / P → S) are explicitly named in the spec. Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives: Defining first ionisation energy and stating the equation is AO1. Explaining trends across periods and down groups in terms of nuclear charge, atomic radius and shielding is AO2 — and is heavily examined on every Paper 1 and Paper 2. Using successive ionisation energy data to identify an element's group is AO3 — typically a 3-4 mark item.

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Definitions

First ionisation energy: The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.

Equation: X(g) → X⁺(g) + e⁻

Second ionisation energy: The energy required to remove one electron from each ion in one mole of gaseous 1+ ions to form one mole of gaseous 2+ ions.

Equation: X⁺(g) → X²⁺(g) + e⁻

In general, the nth ionisation energy is always defined for the removal of one electron from the (n−1)+ ion.

Key Point: Ionisation energies are always positive (endothermic) because energy must be supplied to overcome the attraction between the electron and the nucleus. They are measured in kJ mol⁻¹.

Conditions

The definition specifies gaseous atoms because all electrons must be in their normal energy levels (not involved in bonding or intermolecular forces). This ensures a fair comparison between elements.

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Factors Affecting Ionisation Energy

Three main factors determine the size of an ionisation energy:

1. Nuclear Charge

A greater number of protons in the nucleus means a stronger attraction for the outer electrons, so more energy is needed to remove an electron. Increasing nuclear charge increases ionisation energy.

2. Atomic Radius (Distance from Nucleus)

The further an electron is from the nucleus, the weaker the attraction (the force of attraction follows an inverse square law: F ∝ 1/r²). Increasing distance decreases ionisation energy.

3. Electron Shielding

Inner electrons repel outer electrons and reduce the effective nuclear charge experienced by the outer electron. More inner shells means more shielding. Increasing shielding decreases ionisation energy.

Key Definition: The effective nuclear charge is the net positive charge experienced by an outer electron, after accounting for the shielding effect of inner electrons.

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Trend Across a Period (e.g. Period 2: Li to Ne)

The general trend across a period is increasing first ionisation energy. This is because:

• The nuclear charge increases (more protons) as you go across the period.
• Electrons are being added to the same shell, so the shielding remains approximately constant.
• The atomic radius decreases slightly.

Therefore, the effective nuclear charge increases, and the outer electron is held more tightly.

Exceptions Across Period 2

There are two drops in the otherwise increasing trend:

Drop 1: Be (900 kJ mol⁻¹) to B (801 kJ mol⁻¹)

Beryllium: 1s² 2s² Boron: 1s² 2s² 2p¹

The outer electron of boron is in a 2p sub-shell, which is higher in energy and further from the nucleus than the 2s sub-shell of beryllium. The 2p electron is also partially shielded by the 2s² electrons. Therefore, less energy is required to remove it.

Drop 2: N (1402 kJ mol⁻¹) to O (1314 kJ mol⁻¹)

Nitrogen: 1s² 2s² 2p³ — three singly occupied 2p orbitals Oxygen: 1s² 2s² 2p⁴ — one 2p orbital contains a pair of electrons

In oxygen, the paired electrons in the 2p orbital experience electron-electron repulsion within the same orbital. This additional repulsion makes it easier to remove one of the paired electrons, so the ionisation energy of oxygen is lower than that of nitrogen.

First Ionisation Energies Across Period 2

Element	Z	Configuration	1st IE (kJ mol⁻¹)
Li	3	1s² 2s¹	520
Be	4	1s² 2s²	900
B	5	1s² 2s² 2p¹	801
C	6	1s² 2s² 2p²	1086
N	7	1s² 2s² 2p³	1402
O	8	1s² 2s² 2p⁴	1314
F	9	1s² 2s² 2p⁵	1681
Ne	10	1s² 2s² 2p⁶	2081

The Same Pattern Repeats in Period 3

Element	Z	Configuration	1st IE (kJ mol⁻¹)
Na	11	[Ne] 3s¹	496
Mg	12	[Ne] 3s²	738
Al	13	[Ne] 3s² 3p¹	577
Si	14	[Ne] 3s² 3p²	786
P	15	[Ne] 3s² 3p³	1012
S	16	[Ne] 3s² 3p⁴	1000
Cl	17	[Ne] 3s² 3p⁵	1251
Ar	18	[Ne] 3s² 3p⁶	1521

The drop from Mg to Al is analogous to Be → B (s to p), and the drop from P to S is analogous to N → O (half-filled to paired p).

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Trend Down a Group

Down a group, the first ionisation energy decreases. This is because:

• The atomic radius increases (electrons are in higher shells, further from the nucleus).
• The shielding increases (more inner shells of electrons).
• Although nuclear charge also increases, the effects of distance and shielding outweigh it.

First Ionisation Energies Down Group 1

Element	1st IE (kJ mol⁻¹)
Li	520
Na	496
K	419
Rb	403
Cs	376

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Successive Ionisation Energies

Successive ionisation energies always increase because:

1. Each successive electron is removed from an increasingly positive ion, so the nuclear attraction on remaining electrons is greater.
2. There are fewer electrons providing shielding.

Evidence for Electron Shells

A graph of successive ionisation energies shows large jumps when an electron is removed from a shell closer to the nucleus (an inner shell). These jumps provide evidence for the arrangement of electrons in shells.

Worked Example 1: Successive Ionisation Energies of Sodium

Sodium (Z = 11) has the configuration 1s² 2s² 2p⁶ 3s¹.

IE	Value (kJ mol⁻¹)	Electron removed from
1st	496	3s
2nd	4562	2p
3rd	6912	2p
...	...	...
9th	16610	2s
10th	35 020	1s
11th	43 960	1s

There is a huge jump between the 1st and 2nd ionisation energies because the 1st electron is removed from the outer 3s shell, but the 2nd electron is removed from the inner 2p shell (much closer to the nucleus, less shielding).

Worked Example 2: Identifying an Element from Successive IEs

An element has the following successive ionisation energies (kJ mol⁻¹):

786, 1577, 3232, 4356, 16 091, 19 785, ...

There is a large jump between the 4th and 5th ionisation energies. This means 4 electrons are relatively easy to remove (outer shell), and the 5th is much harder (inner shell).

The element has 4 outer electrons → Group 14.

The element is in Period 3 (based on the magnitude of the values), so it is silicon (Si).

Worked Example 3: Evidence for Sub-shells from Successive IEs

The successive ionisation energies of aluminium (Z = 13, configuration [Ne] 3s² 3p¹) are:

577, 1817, 2745, 11 578, ...

• The 1st IE is relatively low (3p electron, shielded, higher energy).
• A jump between the 1st and 2nd (577 → 1817) is larger than expected from just increasing ion charge. This is partly because the 2nd electron comes from the 3s sub-shell, which is lower in energy than 3p.
• The very large jump between 3rd and 4th (2745 → 11 578) shows that the 4th electron is removed from a new inner shell (the 2p shell).

Exam Tip: When analysing successive ionisation energy data, look for large jumps. These indicate that the next electron is being removed from a shell closer to the nucleus. Count the number of electrons removed before the big jump to determine the group number.

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Linking Ionisation Energy to Reactivity

Group 1 Metals

Down Group 1, ionisation energy decreases, so it becomes easier to lose the outer electron. This explains why reactivity of Group 1 metals increases down the group.

Group 17 (Halogens)

The trend in halogen reactivity (decreasing down the group) is NOT explained by ionisation energy directly, but rather by the decreasing ability to attract an electron into the outer shell (electron affinity). However, the atomic radius and shielding arguments are analogous.

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Synoptic Links

Ionisation energy connects to multiple other A-Level Chemistry topics:

• electron-configuration (previous lesson, §3.1.1.3): the existence of shells and sub-shells — and the s/p/d/f sub-shell ordering — is established empirically through successive ionisation energies. The "evidence for shells" interpretation here is the conceptual mirror of the "shell model" introduced in the previous lesson.
• periodicity (§3.2.1, A2): the period-3 trend in first ionisation energy is a key periodicity item, examined alongside trends in atomic radius, melting point, and electrical conductivity. The Mg → Al and P → S drops form the most-tested anomaly questions.
• bonding (§3.1.3): ionic bonding is driven by the trade-off between ionisation energy (energy cost to form a cation), electron affinity (energy gain to form an anion), and lattice enthalpy. Born-Haber cycles in lesson 8 of the bonding course use ionisation energies as one of their inputs.
• transition-metals (§3.2.5, A2): variable oxidation states reflect successive ionisation energies. Why iron forms Fe²⁺ and Fe³⁺ but rarely Fe⁴⁺ is explained by the relatively flat successive-IE profile of the 3d series until the d shell is empty.
• redox-electrochemistry (§3.1.7 + §3.1.11): standard electrode potentials reflect a thermodynamic cycle that includes ionisation energy as one of its terms. Trends in reactivity of Group 1 metals down the group (decreasing IE) explain the trend in standard reduction potential.

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Specimen Exam Question — modelled on the AQA A-Level Chemistry paper format

Question 1. [12 marks total]

(a) Define the term first ionisation energy. Include an equation in your answer. [2 marks]

(b) Explain the general trend in first ionisation energy across Period 3 from sodium to argon. [3 marks]

(c) Explain why the first ionisation energy of sulfur (1000 kJ mol⁻¹) is lower than that of phosphorus (1012 kJ mol⁻¹). [3 marks]

(d) The successive ionisation energies of an element X (in kJ mol⁻¹) are:

590, 1145, 4912, 6491, 8153, 10 496, ...

State which group X is in, and explain how you arrived at your answer. [2 marks]

(e) Predict whether the first ionisation energy of magnesium is larger or smaller than that of calcium. Explain your reasoning. [2 marks]

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Mark Scheme — Specimen Question 1 (modelled, not a real paper)

(a) Definition of first ionisation energy [2 marks, AO1]

• 1 mark: energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.
• 1 mark: equation X(g) → X⁺(g) + e⁻ (accept M(g) or any general symbol; state symbols are required).

Common error: forgetting to specify (g) state symbol. The definition is meaningless without gaseous atoms.

(b) Trend across Period 3 [3 marks, AO2]

• 1 mark: the general trend is an increase in first ionisation energy from Na to Ar.
• 1 mark: nuclear charge increases across the period (Na has 11 protons, Ar has 18).
• 1 mark: shielding is approximately constant (electrons added to the same shell); atomic radius decreases slightly; effective nuclear charge therefore increases.

Mention all three factors (nuclear charge, shielding, atomic radius). A response that only mentions one or two factors loses marks.

(c) Sulfur vs phosphorus [3 marks, AO3]

• 1 mark: phosphorus has electron configuration [Ne] 3s² 3p³ — three singly occupied 3p orbitals.
• 1 mark: sulfur has [Ne] 3s² 3p⁴ — one 3p orbital contains a pair of electrons.
• 1 mark: the paired electrons experience mutual repulsion within the same orbital; this destabilises one of the paired electrons, so it requires less energy to remove than a singly-occupied 3p electron in phosphorus.