Electronic Structure from Successive Ionisation Energies

UNIT 2 CHEMISTRY · PEARSON AAQ · A1.3.2

Electronic Structure from Successive Ionisation Energies

FREE CHEMISTRY REVISION · TOPIC 8

Find the jump. Count the electrons before it. Explain the shell change.

Use successive ionisation-energy tables and graphs as evidence for shell occupancy, and distinguish firm deductions from conclusions that need more information.

Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · A1.3.2

What you will learn

  • Explain how successive ionisation energies provide evidence for electronic shells.
  • Locate a large jump and count the electrons removed before it.
  • Deduce outer-shell electron numbers and, with suitable context, main-group membership.
  • Use several jumps in a complete sequence to reconstruct shell populations.
  • Read logarithmic axes and recognise what a short dataset cannot establish.

A1.3.2 concerns evidence from successive ionisation energies for the electronic structure of an atom. Definitions and equations were covered in A1.3.1. Trends between different elements are reserved for A1.3.3.

1. What is the experimental evidence?

Successive ionisation energies measure the energy needed for each further one-electron removal from an element’s increasingly positive gaseous species. The horizontal axis of a successive-energy graph counts electrons removed from one element; it does not run across the periodic table.

The values increase because the remaining electrons are more strongly bound as the species becomes more positive. However, the increase is not uniform. Some neighbouring values differ much more sharply than others. These large jumps provide evidence that electrons occupy distinct shells rather than a continuous, uniform distribution around the nucleus.

An outer-shell electron is generally farther from the nucleus and shielded by inner-shell electrons. Once that outer shell is emptied, the next electron must be removed from an inner shell. It is closer to the nucleus and experiences less shielding from inner electrons, so much more energy is required.

A large jump between IEₙ and IEₙ₊₁ indicates that n electrons were removed before reaching a more strongly bound inner shell.

Do not confuse the increase between every pair of successive values with a particularly large shell-boundary jump. Use the overall pattern, the scale of the change and any supplied electronic or periodic-table context.

2. Count before the jump, not after it

Suppose the first substantial shell-boundary jump is between the second and third ionisation energies. The first two electrons were removed relatively easily from the outer shell; the third is the first electron removed from an inner shell. Therefore, the neutral atom had two outer-shell electrons.

The jump starts after two removals, not after three. The higher value belongs to the first electron of the next, more tightly bound set.

Two outer electrons before an inner-shell removalFor magnesium, the first removal takes one 3s electron, the second takes the remaining 3s electron, and the third begins removal from the inner n equals 2 shell. The first two energies are 738 and 1451, then the third rises to 7733 kilojoules per mole.Magnesium: the first inner-shell removalIE₁ = 738remove 3sMg → Mg⁺IE₂ = 1451remove 3sMg⁺ → Mg²⁺IE₃ = 7733remove 2pMg²⁺ → Mg³⁺→→2 outer electrons removedInner shellValues rounded / kJ mol⁻¹ · all species gaseous
Original removal-sequence diagram using rounded magnesium values. The jump is after the second electron, so it identifies two outer-shell electrons.

For a main-group element, the first clear shell-boundary jump can identify its outer-shell population: one suggests Group 1; two suggests Group 2; three suggests Group 13; four suggests Group 14, and so on. Use the group-numbering convention requested; Groups 13–18 are sometimes described in school shorthand as Groups 3–0 or 3–8.

Limit this rule: do not automatically label every atom with two highest-shell electrons “Group 2”. Transition metals such as iron also have two 4s electrons in the neutral atom, while 3d electrons participate in further removals. Their sequences need subshell and configuration context. A brief dataset alone may not establish the element’s identity.

3. Read tables and graphs accurately

First confirm what the numbers represent: individual ionisation energies, not cumulative totals. Check units and read the ionisation stage numbers. Compare adjacent values and identify the unusually large change associated with entering an inner shell.

A ratio can help when later values are large. For magnesium, IE₂/IE₁ ≈ 1451/738 = 1.97, while IE₃/IE₂ ≈ 7733/1451 = 5.33. This supports the larger relative increase between the second and third removals. There is no universal numerical ratio threshold that defines every shell boundary.

Magnesium first three successive ionisation energiesA linear vertical axis from zero to eight thousand kilojoules per mole plots magnesium energies of 738, 1451 and 7733 at stages one, two and three. The large rise occurs from the second to the third point.Ionisation energy / kJ mol⁻¹0200040006000800073814517733123Ionisation stage
Original quantitative plot of the rounded data used above. The vertical axis is linear and starts at zero. Connecting lines guide the eye between discrete removal stages.

Logarithmic axes: a long sequence can span several orders of magnitude. On a log₁₀ scale, equal vertical distances represent equal multiplicative factors: values 10³, 10⁴ and 10⁵ are equally spaced even though their differences are not equal. Read the tick labels rather than treating the axis as linear. If the graph plots log₁₀(IE / kJ mol⁻¹), a vertical value of 4 corresponds to IE = 10⁴ = 10,000 kJ mol⁻¹.

When asked to explain a jump, write the electronic reason. “The line gets steeper” describes a graph; it does not explain why the next electron is harder to remove.

4. Use several jumps to reconstruct shells

A complete sequence contains one ionisation stage for each electron originally present in the neutral atom. For magnesium, twelve stages correspond to twelve electrons. The major principal-shell boundaries occur between IE₂ and IE₃, and between IE₁₀ and IE₁₁.

  1. Stages 1–2 remove two electrons from n = 3.
  2. Stages 3–10 remove eight electrons from n = 2: 10 − 2 = 8.
  3. Stages 11–12 remove two electrons from n = 1: 12 − 10 = 2.

Removal order: 2 → 8 → 2 (outer to inner).
Conventional shell arrangement: 2, 8, 2 (inner to outer).

The numbers happen to read the same way for magnesium. That is not true for all atoms: a 13-electron main-group atom with major boundaries after stages 3 and 11 has removal groups 3 → 8 → 2, but shell arrangement 2, 8, 3.

Magnesium removal groups reveal shell populationsA schematic sequence groups magnesium stages 1 to 2 in shell n equals 3, stages 3 to 10 in shell n equals 2, and stages 11 to 12 in shell n equals 1. These group sizes are 2, 8 and 2. This is not an energy graph.Count removals between shell boundariesStages 1–2Stages 3–10Stages 11–122 electrons8 electrons2 electronsn = 3n = 2n = 1→→Outer → inner removal; not an energy scaleWrite the shell arrangement from inner to outer
Original grouping diagram. The widths of the panels do not encode energy or electron number; the labelled stage ranges provide the counts.

Within a principal shell, electrons occupy different subshells and interact with one another. Smaller structure in a sequence can reflect subshell changes and electron interactions. Do not interpret every increase as the start of a new principal shell, or claim that one large jump alone specifies every orbital occupancy.

5. Worked examples: make and justify deductions

Example 1 · Magnesium’s first three values

IE₁ = 738; IE₂ = 1451; IE₃ = 7733 kJ mol⁻¹ (rounded).

  1. Changes: 1451 − 738 = 713; 7733 − 1451 = 6282 kJ mol⁻¹.
  2. The marked rise is between IE₂ and IE₃.
  3. Two outer electrons have been removed before the inner-shell electron is reached.
  4. For Mg, this supports [Ne] 3s²: after two removals, Mg²⁺ has the neon core, and the third removal takes a 2p electron.

Model explanation: “The large increase after the second ionisation shows that the first two electrons occupied an outer shell. The third is removed from an inner shell, closer to the nucleus and less shielded, so it is more strongly attracted.”

Example 2 · An unknown main-group atom

An original illustrative exam dataset gives IE₁–IE₄ = 600, 1800, 2800, 12,000 kJ mol⁻¹. These are teaching values, not measured values for a named element. The question states that X is a main-group element in Period 3.

  1. The clearest shell-boundary jump is from the third to the fourth value.
  2. Three electrons are removed before reaching the inner shell: three outer electrons.
  3. A Period 3 main-group element with three outer electrons is in Group 13, so the context identifies aluminium.
  4. Its configuration is [Ne] 3s² 3p¹, or shell arrangement 2, 8, 3.

The energy pattern gives the outer-shell count; the supplied period and main-group context permit identification. Do not present the illustrative numbers as experimental aluminium measurements.

Example 3 · Recover all shell populations

A complete sequence for an unknown main-group atom has 13 removals and major shell-boundary jumps between IE₃/IE₄ and IE₁₁/IE₁₂.

  1. Before the first jump: 3 outer-shell electrons.
  2. Between boundaries: 11 − 3 = 8 electrons in the next shell inward.
  3. After the second boundary: 13 − 11 = 2 innermost electrons.
  4. Reverse the removal groups 3 → 8 → 2 to write 2, 8, 3 from the nucleus outward.

Thirteen total removals mean the neutral atom had thirteen electrons, so Z = 13 and the atom is aluminium. The full sequence gives stronger identification evidence than only the first four energies.

Example 4 · Find the off-by-one error

A student says: “The jump is between the fourth and fifth energies, so the atom has five outer electrons.” The question specifies a Period 3 main-group atom.

  1. The fifth electron is already from an inner shell.
  2. Only the first four electrons were removed before the jump.
  3. The atom therefore has four outer-shell electrons: Group 14.
  4. Period 3 identifies silicon, [Ne] 3s² 3p², with shell arrangement 2, 8, 4.

Use the lower stage number at the boundary. “After the fourth removal” and “before the fifth removal” describe the same boundary.

Example 5 · What can a short table prove?

A table contains just IE₁ = 500 and IE₂ = 1000 kJ mol⁻¹ (illustrative). A student concludes that the atom has two outer electrons and must be magnesium.

  1. The data show only the first two removals.
  2. No third energy is provided, so a jump after IE₂ has not been demonstrated.
  3. The table cannot establish that the outer shell is emptied after two removals.
  4. It also lacks the period, full sequence or other evidence needed to identify magnesium.

A justified answer is: “More successive values are needed to locate the first shell-boundary jump; these two numbers alone do not identify the element.”

6. Common misconceptions and exam tips

  • “Jump from IE₃ to IE₄ means four outer electrons.” It means three were removed before entering the inner shell.
  • “Electrons enter a new shell at the jump.” Ionisation removes electrons; the next removal comes from a different shell.
  • “The nuclear charge increases each time.” Proton number stays fixed. The ion becomes more positive as electrons are removed.
  • “A large jump means a larger electron.” Electrons are identical; their environments and binding energies differ.
  • “A short sequence reveals every shell and subshell.” It may only reveal an outer-shell boundary. State what extra information is needed.
  • “Equal spacing on a log axis means equal energy differences.” It means equal multiplicative factors.
  • “Two outer electrons always means Group 2.” Apply the simple group link with appropriate main-group context.

Exam answer sequence: quote the boundary → count electrons before it → state the shell deduction → explain stronger attraction using smaller distance and reduced shielding → use period/group context only when supplied or justified.

Use “evidence supports” rather than claiming that a single jump proves a complete orbital diagram. If identifying an element, explain both the electron-count evidence and the additional information that fixes its position.

Your revision summary

  • Successive energies refer to repeated removal from one element’s gaseous species.
  • A particularly large jump indicates removal has reached a more tightly bound inner shell.
  • A jump between IEₙ and IEₙ₊₁ comes after n electron removals.
  • Differences between boundary positions give electron populations in successive shells.
  • Conventional shell arrangements are written inner to outer, opposite the removal sequence.
  • Read axis scales and avoid claiming more than the dataset establishes.

Magnesium values are rounded to whole kJ mol⁻¹ for teaching, consistent with NIST Atomic Spectra Database values 7.646236, 15.035271 and 80.1436 eV for the first three removals (1 eV per particle ≈ 96.485 kJ mol⁻¹). NIST magnesium ionisation-energy data. Datasets explicitly labelled illustrative are original teaching examples.

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Question 1 of 5 · Unit 2 A1.3.2

A main-group atom shows a large shell-boundary jump between IE₃ and IE₄. How many outer-shell electrons did it have?

Question 2 of 5 · Unit 2 A1.3.2

Why is an inner-shell electron much harder to remove after the outer shell has been emptied?

Question 3 of 5 · Unit 2 A1.3.2

A complete 13-electron main-group sequence has major boundaries after removals 3 and 11. What is the shell arrangement, inner to outer?

Question 4 of 5 · Unit 2 A1.3.2

A Period 3 main-group atom has a major jump between IE₄ and IE₅. Which deduction is justified?

Question 5 of 5 · Unit 2 A1.3.2

On an axis labelled log₁₀(IE / kJ mol⁻¹), a plotted value is 4. Which ionisation energy does it represent?

Scope verified against Pearson BTEC National AAQ Applied Science (Extended Certificate), Issue 7, May 2026, Unit 2 A1.3.2, printed page 36. Official Pearson specification (PDF).

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