Periodic Trends: Atomic Radius and Ionisation Energy

Spec Mapping — OCR H432 Module 3.1.1 — Periodicity, covering the definition and trends in atomic radius across Periods 2 and 3 and down Groups 1, 2 and 17, the definition of first ionisation energy, the trend in first ionisation energy across Periods 2 and 3 (with characteristic dips at Group 13 and Group 16 elements), the trend in first ionisation energy down a group, and the four factors used to explain ionisation-energy variation: nuclear charge, atomic radius, shielding and sub-shell occupancy (refer to the official OCR H432 specification document for exact wording).

Atomic radius and first ionisation energy (IE₁) are the two foundational periodic properties from which every reactivity argument in inorganic chemistry is derived. Together they explain why fluorine is the most aggressive oxidising agent, why caesium reacts with water explosively, why magnesium burns with a brilliant white flame but beryllium will not, and why the noble gases are chemically lazy. The trends are not monotonic — Period 3 shows two characteristic dips in IE₁ (at Al and at S) that turn out to be direct experimental evidence for the sub-shell structure of electrons predicted by quantum mechanics. The OCR specification expects you to quote and apply four factors when explaining any ionisation-energy variation: nuclear charge, atomic radius (distance from nucleus to outer electron), shielding by inner electrons, and sub-shell occupancy (spin pairing). This lesson builds that fluency and equips you to construct the kind of multi-step reasoning that OCR Paper 1 demands.

Key Definition: the first ionisation energy (IE₁) of an element is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions: $\text{X(g)} \rightarrow \text{X}^+ \text{(g)} + \text{e}^- \qquad \Delta H = \text{IE}_1$X(g)→X+(g)+e−ΔH=IE1​ The three non-negotiable elements of this definition — per mole, gaseous, 1+ ions — must each appear in every OCR answer.

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Atomic radius

Atomic radius is a measure of the size of an atom — conventionally half the distance between the nuclei of two atoms of the same element joined by a single covalent bond (the covalent radius). Because electron density falls off smoothly with distance from the nucleus and an atom has no sharp edge, the radius depends on the measurement context. OCR uses three conventions:

• Covalent radius — half the bond length in X–X (e.g. Cl–Cl in Cl₂); applies to non-metals and most metalloids.
• Metallic radius — half the shortest interatomic distance in a metal lattice.
• Van der Waals radius — half the closest non-bonded approach of two atoms in a solid; applies to noble gases.

The three measures give slightly different numerical values for the same element (van der Waals > metallic > covalent), so when you quote a periodic trend, be consistent about which convention you are using.

Trend across a period

Across a period, atomic radius decreases from left to right.

Period 3 element	Na	Mg	Al	Si	P	S	Cl	Ar (vdW)
Atomic radius / pm	186	160	143	117	110	104	99	71

Explanation (three steps):

1. As you move from Na (Z = 11) to Cl (Z = 17), the nuclear charge increases from +11 to +17 (six more protons).
2. Each successive electron is added to the same shell (the 3rd shell), so the shielding by inner electrons (1s² 2s² 2p⁶ = 10 inner electrons) stays roughly constant.
3. The outer electron therefore experiences an increasing effective nuclear charge ($Z_{\text{eff}} = Z - \sigma$Zeff​=Z−σ, where $\sigma$σ is the shielding constant), and is drawn closer to the nucleus.

The same logic applies in Period 2 (Li → Ne). Notice the radius does not fall smoothly — there are small bumps that reflect sub-shell structure, but the dominant trend is a monotonic decrease.

Trend down a group

Down any group, atomic radius increases.

Group 1	Li	Na	K	Rb	Cs
Radius / pm	152	186	227	248	265

Three contributing effects:

• More shells are occupied as you go down → outer electrons are in higher principal quantum levels ($n = 2, 3, 4, 5$n=2,3,4,5…), which are larger.
• More inner electrons shield the outer electron from the full nuclear charge ($\sigma$σ rises).
• Nuclear charge also rises, but the effects of distance and shielding dominate.

The net result is a steady increase in radius. Hence Group 7 atoms (F < Cl < Br < I) become progressively larger and progressively less aggressive at gaining an electron — directly relevant to the halogen reactivity lesson later in the course.

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First ionisation energy

The first ionisation energy (IE₁) is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions:

$\text{X(g)} \rightarrow \text{X}^+ \text{(g)} + \text{e}^- \qquad \Delta H = \text{IE}_1$X(g)→X+(g)+e−ΔH=IE1​

The three non-negotiable elements of the definition you must include in every OCR answer:

1. Energy required (it is the energy input, not the energy released).
2. Per mole of gaseous atoms — both starting material and product are specified as 1 mole each.
3. Gaseous atoms → gaseous 1+ ions — state symbols are essential.

The units are kJ mol⁻¹ and the value is always endothermic (positive) because energy must be supplied to overcome the electrostatic attraction between the outgoing electron and the nucleus.

Four factors affecting ionisation energy

OCR expects you to quote and apply four factors when explaining any ionisation-energy variation:

Factor	Direction of effect
Nuclear charge	Higher $Z$Z → stronger attraction on outer e⁻ → higher IE
Atomic radius (distance)	Larger radius → weaker attraction ($\propto 1/r^2$∝1/r2) → lower IE
Shielding	More inner electrons shield outer e⁻ from full nuclear charge → lower IE
Sub-shell occupancy (spin pairing)	Paired electrons in the same orbital repel → electron easier to remove → lower IE; new sub-shell (e.g. 3s² → 3p¹) → lower IE

Different combinations of these factors dominate in different parts of the periodic table, and the skill OCR wants you to develop is identifying which factor is decisive for any given comparison.

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Trend across Period 3

Element	Configuration	IE₁ / kJ mol⁻¹
Na	[Ne] 3s¹	496
Mg	[Ne] 3s²	738
Al	[Ne] 3s² 3p¹	578 (dip)
Si	[Ne] 3s² 3p²	789
P	[Ne] 3s² 3p³	1012
S	[Ne] 3s² 3p⁴	1000 (dip)
Cl	[Ne] 3s² 3p⁵	1251
Ar	[Ne] 3s² 3p⁶	1521

flowchart TD
  A[IE1 across Period 3] --> B[Overall rise Na to Ar]
  A --> C[Dip at Al]
  A --> D[Dip at S]
  B --> B1[Z increases, r decreases, sigma constant]
  C --> C1[Outer e- in 3p higher than 3s, slightly shielded by 3s2]
  D --> D1[3p4 pairs in one 3p orbital, pair repulsion eases removal]

Explaining the Al dip (Mg → Al)

Mg has [Ne] 3s²; Al has [Ne] 3s² 3p¹. The electron removed from Al is in 3p, not 3s. Three contributions:

• Higher sub-shell energy: 3p is at higher energy and on average further from the nucleus than 3s.
• Slight extra shielding: the filled 3s² pair partially shields the 3p¹ electron from the nucleus.
• These two effects outweigh the small increase in nuclear charge from Mg (+12) to Al (+13).

The 3p electron is therefore easier to remove than a 3s electron, and IE₁(Al) < IE₁(Mg). The Mg/Al dip is direct experimental evidence that sub-shells exist within the third shell — the s and p sub-shells are not at the same energy.

Explaining the S dip (P → S)

In P (3p³), Hund's rule places one electron in each of the three 3p orbitals with parallel spins — no pairing, no in-orbital repulsion. In S (3p⁴), the fourth 3p electron is forced to pair up in one of the 3p orbitals with the existing electron. The two electrons now occupy the same region of space and experience strong electron-electron repulsion. That repulsion destabilises the paired electron, making it easier to remove than if it were unpaired.

The result is IE₁(S) < IE₁(P), even though S has one more proton in its nucleus. The P/S dip is direct experimental evidence for Hund's rule and the three-orbital nature of the 3p sub-shell.

The same dips appear in Period 2 between Be/B (the 2s/2p sub-shell change) and between N/O (the spin-pair dip in 2p). The fact that the pattern repeats across periods is itself evidence of periodicity: the same quantum-mechanical structure is reasserting itself one shell out.

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Trend down a group

Ionisation energy decreases down a group:

Element	IE₁ / kJ mol⁻¹
Be	899
Mg	738
Ca	590
Sr	549
Ba	503

Halogen	IE₁ / kJ mol⁻¹
F	1681
Cl	1251
Br	1140
I	1008

Explanation: down a group, the outer electron sits in a shell further from the nucleus (larger atomic radius) and is shielded by more complete inner shells. Although nuclear charge increases, the distance and shielding effects dominate, so the outer electron is held less tightly and less energy is required to remove it.

This is why Group 1 and Group 2 reactivity increases down the group (the metal loses electrons more easily), while Group 7 reactivity decreases (the halogen gains electrons less easily). The two opposite trends both follow from the same atomic-radius logic.

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Worked examples

Example 1 — the Mg/Al dip

Q: The first ionisation energies of Mg, Al and Si are 738, 578 and 789 kJ mol⁻¹. Explain why Al has a lower IE₁ than Mg, but Si has a higher IE₁ than Al.

Answer:

Mg → Al (dip): Mg has configuration [Ne] 3s²; Al has [Ne] 3s² 3p¹. The electron removed from Al is in the 3p sub-shell, which is at higher energy and (on average) slightly further from the nucleus than the 3s sub-shell. The 3p¹ electron is also slightly shielded by the filled 3s² pair. These two effects outweigh the small increase in nuclear charge (Z 12 → 13), so less energy is required to remove the outer electron from Al than from Mg.

Al → Si (rise): Si (3s² 3p²) has one more proton than Al (3s² 3p¹). The electron removed is still from the 3p sub-shell, so there is no sub-shell change. Shielding is essentially unchanged (the extra electron sits in the same sub-shell as the previous one and shields the nucleus only weakly). The increased nuclear charge therefore dominates and IE₁ rises.

Example 2 — the P/S dip

Q: Explain, using electron configurations, why the first ionisation energy of sulfur is lower than that of phosphorus.