Atomic Radii and Ionisation Enthalpy: Trends, Anomalies and Reasoning
Updated 17 October 2026 · Fact-checked
Atomic radius decreases across a period (rising effective nuclear charge) and increases down a group (extra shells). Ionisation enthalpy generally rises across a period and falls down a group, with dips at Be → B and N → O. A large jump in successive ionisation enthalpies marks the removal of an inner-shell electron, which reveals the group of an element.
Notice that first ionisation enthalpy climbs across Period 2 while radius shrinks, except the dips at B (after Be) and at O (after N).
Illustration drawn for QuestionBuddy.
Understand Atomic Radii and Ionisation Enthalpy Trends
Shielding and effective nuclear charge. Inner electrons shield outer electrons from the nucleus. The effective nuclear charge is Z_eff = Z − S, where S is the screening by the other electrons. Across a period Z_eff rises because electrons are added to the same shell, which shields poorly; down a group the number of shells grows, which outweighs the rise in nuclear charge.
Atomic radii. Across a period atomic radius decreases (Z_eff pulls the shell in). Down a group it increases. A cation is smaller than its parent atom (fewer electrons, same nucleus, or an emptied shell) and an anion is larger (more electron-electron repulsion). In isoelectronic species (same number of electrons), the radius decreases as the nuclear charge increases: O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺.
Ionisation enthalpy. The first ionisation enthalpy (ΔᵢH₁) is the energy needed to remove the outermost electron from an isolated gaseous atom, X(g) → X⁺(g) + e⁻. It rises across a period and falls down a group. Successive values always increase: ΔᵢH₃ > ΔᵢH₂ > ΔᵢH₁, because the electron is pulled from a more positive ion.
Anomalies. B has lower ΔᵢH₁ than Be because the electron is removed from a 2p orbital, which is shielded by the 2s electrons and is farther out than 2s. O has a lower ΔᵢH₁ than N because the p⁴ configuration has a paired electron whose repulsion makes removal easier, whereas N has a stable half-filled p³.
Reading successive values. A sudden large jump after the removal of n electrons shows that n electrons are valence electrons and the next comes from a stable, inner shell. The element then belongs to the group with n valence electrons (for example, a big jump after the second ionisation implies group 2).
Units. ΔᵢH is in kJ mol⁻¹. To convert to a per-atom value, divide by N_A.
Atomic Radii and Ionisation Enthalpy Trends: formulas and key facts
- Effective nuclear charge
- S = screening by the other electrons.
- Trends
- Across a period: radius decreases, ΔᵢH increases | Down a group: radius increases, ΔᵢH decreases
- General trends with the exceptions noted.
- Successive ionisation
- A large jump after removing the valence electrons.
- Energy per atom
- ΔᵢH in kJ mol⁻¹.
Atomic Radii and Ionisation Enthalpy Trends in real life
Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.
Which element is it?
Suppose a lab report lists the successive ionisation enthalpies (kJ mol⁻¹) of an element X as 738, 1451, 7733 and 10 540.
Deduce its group and the formula of its chloride.
Try it first, then show the solutionSolution
- The big jump (1451 → 7733) comes after the second electron, so two electrons are valence electrons.
- The element is in group 2 (an alkaline earth metal).
- It forms X²⁺, so its chloride is XCl₂.
Answer: Group 2; chloride XCl₂.
Memory hook: Look for the largest ratio between successive values.
Tricks to remember Atomic Radii and Ionisation Enthalpy Trends
Short memory aids. Say each one aloud once, then use it on a question the same day.
Same shell, more charge
Across a period the electron shell stays the same but the nuclear charge grows, so atoms shrink.
Use it when: Explaining the period trend of radius.
The jump tells the group
Find where the successive ionisation enthalpy jumps; the number of electrons removed before it is the number of valence electrons.
Use it when: Identifying the group from data.
Atomic Radii and Ionisation Enthalpy Trends: solved examples
Ordered from board level to HOTS. Try each one before you open the solution.
Example 1: Radius trend in a period · 1 mark · MCQBoard level
Which of the following has the largest atomic radius?
(a) Na
(b) Mg
(c) Al
(d) Si
Try it first, then show the solutionSolution
- All are in period 3; the radius decreases from left to right, so Na is the largest.
Answer: (a) Na
Example 2: Isoelectronic ions · 3 marks · Short answerBoard level
Each of the ions N³⁻, O²⁻, F⁻, Na⁺ and Mg²⁺ has the same number of electrons. Rank them from the smallest to the largest radius, with the reason.
Try it first, then show the solutionSolution
- All have 10 electrons (isoelectronic). (1 mark)
- Nuclear charge: N 7, O 8, F 9, Na 11, Mg 12; a larger nuclear charge pulls the same electron cloud in more strongly. (1 mark)
- Order: Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻. (1 mark)
Answer: Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
Example 3: Anomalies in ionisation enthalpy · 4 marks · Short answerHarder board level
The first ionisation enthalpies (kJ mol⁻¹) of some second-period elements are Li 520, Be 899, B 801, C 1086, N 1402, O 1314, F 1681, Ne 2081 (as listed in standard tables). Explain the general trend and the two exceptions.
Try it first, then show the solutionSolution
- General trend: ΔᵢH₁ rises from Li to Ne as the nuclear charge rises and the radius falls. (1 mark)
- Be → B: the B electron is removed from the 2p orbital, which is higher in energy and shielded by 2s electrons than the 2s electron of Be, so B < Be. (1.5 marks)
- N → O: N has a stable half-filled 2p³; O (2p⁴) has a paired electron whose repulsion eases removal, so O < N. (1.5 marks)
Answer: Rising trend with dips at B (801 < 899) and O (1314 < 1402).
Example 4: Ionisation energy per atom · 4 marks · Short answerHarder board level
The first ionisation enthalpy of sodium is 496 kJ mol⁻¹. Find the energy to ionise one atom in joule and electronvolt, and the longest wavelength of light that can ionise a sodium atom. (N_A = 6.022 × 10²³, 1 eV = 1.602 × 10⁻¹⁹ J, hc = 1240 eV nm)
Try it first, then show the solutionSolution
- E = 496 × 10³/6.022 × 10²³ = 8.24 × 10⁻¹⁹ J. (1 mark)
- In eV: 8.24 × 10⁻¹⁹/1.602 × 10⁻¹⁹ = 5.14 eV. (1 mark)
- λ_max = hc/E = 1240/5.14. (1 mark)
- = 241 nm (ultraviolet). (1 mark)
Answer: 8.24 × 10⁻¹⁹ J = 5.14 eV; λ_max ≈ 241 nm.
Example 5: Identifying elements from data · 5 marks · Long answerHOTS / competency
The successive ionisation enthalpies (kJ mol⁻¹) of three elements are P: 496, 4562, 6910; Q: 738, 1451, 7733; R: 577, 1817, 2745, 11 577. Identify the group of each, the formula of the most likely oxide of each, and say why ΔᵢH₂ for P is so much larger than ΔᵢH₁.
Try it first, then show the solutionSolution
- P: the jump after the first electron (496 → 4562), so group 1, oxide P₂O. (1.5 marks)
- Q: the jump after the second (1451 → 7733), group 2, oxide QO. (1.5 marks)
- R: the jump after the third (2745 → 11 577), group 13, oxide R₂O₃. (1.5 marks)
- For P the second electron comes from a complete inner shell with a noble-gas configuration, closer to the nucleus and unshielded. (0.5 mark)
Answer: P: group 1, P₂O. Q: group 2, QO. R: group 13, R₂O₃.
Where students lose marks in Atomic Radii and Ionisation Enthalpy Trends
The slips examiners see most often, and what to write instead.
Mistake: Saying radius increases across a period
Reversing the trend.
Do this: Radius decreases across a period because Z_eff increases in the same shell.
Usually costs: 2 marks.
Mistake: Giving the nuclear charge as the only reason
Ignoring shielding and half-filled stability.
Do this: Mention Z_eff, shielding, and the stability of half-filled or filled subshells.
Usually costs: 1 to 2 marks.
Mistake: Comparing isoelectronic ions wrongly
Ranking by atomic number in the wrong direction.
Do this: Higher nuclear charge, smaller radius, for isoelectronic species.
Usually costs: 1 to 2 marks.
Exam tips
- Quote the trend, then give the reason in terms of Z_eff and the number of shells.
- For exceptions, name the orbital and the configuration (2p, half-filled).
- In data questions, compute the ratios of successive values to find the jump.
- Use kJ mol⁻¹ for enthalpies; convert only when asked.
Practice questions on Atomic Radii and Ionisation Enthalpy Trends
Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.
Q1 · 1 mark · MCQBoard level
Which of the following has the highest first ionisation enthalpy?
(a) Li
(b) Na
(c) K
(d) RbHint
Ionisation enthalpy falls down a group.
Answer
(a) Li
Q2 · 2 marks · Short answerBoard level
Why is the radius of Na⁺ smaller than that of the Na atom?
Hint
Think about the shells and the number of electrons.
Answer
Na⁺ has lost its 3s electron, leaving only two shells; the same nuclear charge (+11) now holds fewer electrons, so it pulls them in.
Q3 · 3 marks · Short answerHarder board level
Explain why the first ionisation enthalpy of nitrogen is higher than that of oxygen even though oxygen has the higher nuclear charge.
Hint
Compare 2p³ and 2p⁴.
Answer
N has the stable, half-filled 2p³ configuration, which is hard to disturb. O (2p⁴) has one paired p electron; repulsion between the pair makes it easier to remove, which outweighs the greater nuclear charge.
Q4 · 3 marks · Short answerHarder board level
Arrange the following in increasing order of radius: Cl⁻, S²⁻, K⁺, Ca²⁺, Ar. Explain.
Hint
They are isoelectronic with 18 electrons.
Answer
Ca²⁺ (Z = 20) < K⁺ (19) < Ar (18) < Cl⁻ (17) < S²⁻ (16): for the same electron count, the greater the nuclear charge, the smaller the radius.
Q5 · 3 marks · Short answerHarder board level
The first and second ionisation enthalpies of an element are 899 and 1757 kJ mol⁻¹, and the third is 14 848 kJ mol⁻¹. Find its group and the energy in kJ needed to form 1 mol of X²⁺ ions from the gaseous atoms.
Hint
Add the first two values.
Answer
The huge jump after the second electron shows group 2. Energy = 899 + 1757 = 2656 kJ mol⁻¹.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the first ionisation enthalpies (kJ mol⁻¹) of the third-period elements are Na 496, Mg 738, Al 577, Si 787, P 1012, S 1000, Cl 1251, Ar 1521.
(i) State the general trend. (1 mark)
(ii) Identify the two exceptions. (1 mark)
(iii) Explain the exception at Al. (1 mark)
(iv) Explain the exception at S. (1 mark)Hint
Compare the exceptions with the second-period pattern.
Answer
(i) A general rise from Na to Ar. (ii) Al (577 < 738) and S (1000 < 1012). (iii) The Al electron comes from the 3p orbital, which is farther out and shielded by 3s. (iv) S (3p⁴) has a paired electron whose repulsion makes removal easier than for the half-filled 3p³ of P.
Test your recall
Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.
1.Define the first ionisation enthalpy.Show answer
The energy needed to remove the outermost electron from an isolated gaseous atom.
2.Why is the second ionisation enthalpy larger than the first?Show answer
The electron is removed from a positive ion, held more tightly.
3.Which is smaller: Na or Na⁺?Show answer
Na⁺.
4.What are isoelectronic species?Show answer
Atoms or ions having the same number of electrons.
Atomic Radii and Ionisation Enthalpy Trends: frequently asked questions
Why do noble gases have the highest ionisation enthalpy in each period?
They have complete, stable shells and the largest effective nuclear charge in the period, so removing an electron is very difficult.
Is the radius of an atom well defined?
Not exactly, because the electron cloud has no sharp edge. Covalent, metallic and van der Waals radii are defined in different ways.
Do ionisation enthalpies always decrease down a group?
Mostly yes, because the outer electron is farther away and shielded; small irregularities occur where d and f electrons shield poorly.