Successive Ionisation Energies

Spec Mapping — OCR H432 Module 3.1.1 — Periodicity, covering the definition of successive ionisation energies, the explanation of why successive IEs always increase, the interpretation of jumps in successive IE data as evidence for the shell model of the atom, the identification of group number from the position of the first large jump, and the use of log-IE-against-ionisation-number plots to deduce electron arrangement (refer to the official OCR H432 specification document for exact wording).

If first ionisation energies map the outer electron configurations of elements across the periodic table, then successive ionisation energies — IE₂, IE₃, IE₄, …, IE$_Z$Z​ — provide a "depth probe" of the inner structure of a single atom. By stripping electrons one at a time from a gaseous atom and measuring the energy required at each step, you generate a numerical fingerprint that reveals the electron's shell number, sub-shell type, and ultimately the total electron count. This is the most direct experimental evidence we have for the quantum-mechanical shell model — predating the more refined photoelectron-spectroscopy data of the modern era. OCR Paper 1 frequently asks you to interpret an unknown set of successive IEs: identify the group (from the position of the first large jump), the period (from the number of jumps), and often the element itself (from the magnitude of IE₁). This lesson builds that fluency through worked examples and a clear conceptual framework.

Key Definition: the n-th ionisation energy of an element is the energy required to remove one mole of electrons from one mole of gaseous $(n{-}1){+}$(n−1)+ ions to form one mole of gaseous $n{+}$n+ ions: $\text{X}^{(n-1)+}\text{(g)} \rightarrow \text{X}^{n+}\text{(g)} + \text{e}^- \qquad \Delta H = \text{IE}_n$X(n−1)+(g)→Xn+(g)+e−ΔH=IEn​

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Defining successive ionisation energies

The first few in sequence:

Ionisation	Equation	Energy
1st	$\text{X(g)} \rightarrow \text{X}^+\text{(g)} + \text{e}^-$X(g)→X+(g)+e−	IE₁
2nd	$\text{X}^+\text{(g)} \rightarrow \text{X}^{2+}\text{(g)} + \text{e}^-$X+(g)→X2+(g)+e−	IE₂
3rd	$\text{X}^{2+}\text{(g)} \rightarrow \text{X}^{3+}\text{(g)} + \text{e}^-$X2+(g)→X3+(g)+e−	IE₃
⋮	⋮	⋮
n-th	$\text{X}^{(n-1)+}\text{(g)} \rightarrow \text{X}^{n+}\text{(g)} + \text{e}^-$X(n−1)+(g)→Xn+(g)+e−	IE$_n$n​

Two universal facts:

• All successive IEs are endothermic (positive) — energy must always be supplied to detach an electron from a cation against the electrostatic attraction of the nucleus.
• Each successive IE is larger than the one before it: IE₁ < IE₂ < IE₃ < … < IE$_Z$Z​.

The reason each IE is larger than the last:

1. Effective nuclear charge per electron increases. After each electron is removed, the same number of protons (Z) attracts a smaller pool of remaining electrons, so each remaining electron experiences a larger $Z_{\text{eff}}$Zeff​.
2. The ion becomes smaller. Reducing the electron count contracts the cation's radius, so all remaining electrons sit closer to the nucleus and are held more tightly.
3. Shielding decreases. Each departing electron also stops contributing to the shielding of the others, so the inner electrons feel a more naked nuclear pull.

These three effects compound on every successive ionisation, so the energy required rises monotonically.

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Evidence for shell structure: jumps between shells

While successive IEs rise monotonically, they do not rise smoothly. There are characteristic large jumps — "step changes" — whenever the next electron to be removed must come from an inner shell (smaller $n$n, much closer to the nucleus, much less shielded). The size of these jumps is what provides the experimental evidence for discrete electron shells.

Worked example: sodium (Z = 11)

Sodium has configuration 1s² 2s² 2p⁶ 3s¹ — that is, 1 electron in shell 3, 8 in shell 2, 2 in shell 1.

Ionisation	IE / kJ mol⁻¹	Electron removed from
1st	496	3s (shell 3)
2nd	4563	2p (shell 2)
3rd	6913	2p
4th	9544	2p
5th	13352	2p
6th	16611	2p
7th	20115	2s
8th	25491	2s
9th	28934	2s? (final 2s)
10th	141367	1s
11th	159076	1s (very last e⁻)

Two dramatic features stand out:

• A huge jump between IE₁ (496) and IE₂ (4563) — a factor of ~9× — because IE₁ removes the lone 3s electron from shell 3, while IE₂ must come from the much closer, less shielded shell 2.
• A second huge jump between IE₉ (28934) and IE₁₀ (141367) — a factor of ~5× — because IE₁₀ comes from the 1s (innermost) shell.

Counting the shells from the log plot

If you plot $\log_{10}(\text{IE})$log10​(IE) against ionisation number, you see flat-ish "plateaus" separated by sudden rises (the shell boundaries). The number of electrons in each plateau tells you the number of electrons in that shell:

• Shell 3: 1 electron (IE₁ only) → matches 3s¹.
• Shell 2: 8 electrons (IE₂ – IE₉) → matches 2s² 2p⁶.
• Shell 1: 2 electrons (IE₁₀ – IE₁₁) → matches 1s².

Total: 1 + 8 + 2 = 11 electrons → Z = 11, sodium. The shell pattern (2, 8, 1) is reconstructable from the IE data alone, with no other input — a stunning experimental verification of the Bohr/quantum-mechanical shell model.

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Using successive IEs to identify an unknown's group

If you are not told the identity of the element, you can still work out its group number by looking at where the first big jump occurs:

flowchart TD
  A[Successive IEs given] --> B[Look for first large jump in ratio]
  B --> C{Jump after IE n}
  C -->|n electrons in outer shell| D[Group number = n]
  D --> E[n=1 → Group 1]
  D --> F[n=2 → Group 2]
  D --> G[n=3 → Group 13]
  D --> H[..]
  D --> I[n=7 → Group 17]

The logic: the first $n$n ionisations strip electrons from the outermost shell (relatively easy because those electrons are furthest from the nucleus and most shielded). The $(n{+}1)$(n+1)-th ionisation must attack the next-inner shell, which is much closer to the nucleus and less shielded → massive jump. Therefore $n$n = number of outer-shell electrons = group number (for s- and p-block elements; the d-block is more complex because 4s electrons leave before 3d).

Identifying the period

The number of large jumps + 1 equals the number of electron shells, which (for s- and p-block) equals the period number. For sodium above, there are two large jumps (after IE₁ and after IE₉) → 3 shells → Period 3. ✓

For an element in Period 4, you would expect three large jumps (between shells 4/3, 3/2, and 2/1) in the full successive-IE list.

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Worked examples

Example 1 — Group 3 (aluminium-like)

Q: The first seven ionisation energies of element Z (in kJ mol⁻¹) are: 577, 1817, 2745, 11578, 14831, 18378, 23296. Deduce the group of Z and suggest its identity.

Answer:

Look for the first big jump in the ratio $\text{IE}_{n+1}/\text{IE}_n$IEn+1​/IEn​:

• IE₁ → IE₂: 577 → 1817 (ratio 3.15, normal increase)
• IE₂ → IE₃: 1817 → 2745 (ratio 1.51, small)
• IE₃ → IE₄: 2745 → 11578 (ratio 4.22, BIG JUMP)
• IE₄ → IE₅: 11578 → 14831 (ratio 1.28, normal again)

The first large jump occurs after the third ionisation. This means Z has 3 electrons in its outermost shell and therefore belongs to Group 13 (Group 3 in older notation). Looking at the size of IE₁ (~580 kJ mol⁻¹, modest, just below Mg's 738), a reasonable candidate is aluminium (Z = 13, actual IE₁ = 578 kJ mol⁻¹). ✓

Example 2 — Group 2 (calcium-like)

Q: Element X has successive ionisation energies / kJ mol⁻¹: 590, 1145, 4912, 6491, 8153. Which group is X in?

Answer:

• IE₁ → IE₂: 590 → 1145 (ratio 1.94, normal)
• IE₂ → IE₃: 1145 → 4912 (ratio 4.29, BIG JUMP)
• IE₃ → IE₄: 4912 → 6491 (ratio 1.32)

First big jump after IE₂ → 2 outer electrons → Group 2. IE₁ of 590 kJ mol⁻¹ matches calcium (actual IE₁ = 590 kJ mol⁻¹). ✓

Example 3 — Group 1 (potassium-like)

Q: The successive IEs of unknown element Y are: 419, 3051, 4411, 5877, 7975, 9590, 11343, 14941, 16963, 48610, 54490 kJ mol⁻¹. State the group, period, and likely identity of Y.

Answer:

• IE₁ → IE₂: 419 → 3051 (ratio 7.28, BIG JUMP after IE₁)
• So Y has 1 outer-shell electron → Group 1.

Continue scanning for a second jump:

• IE₂ … IE₉: 3051 → 16963 (steady increase, factors ~1.2–1.4)
• IE₉ → IE₁₀: 16963 → 48610 (ratio 2.87, second BIG JUMP)

So there is a second shell boundary after the 9th ionisation. That means shell 2 holds 8 electrons (IE₂ to IE₉). Counting: 1 (outermost) + 8 (next) + 2 (innermost) = 11 — but Y has IEs going past that, so check: actually 1 + 8 + 2 = 11. Hmm — for K (Z = 19), shell pattern should be 2, 8, 8, 1: 1 outer + 8 + 8 + 2. The data above does not extend that far, but the visible pattern (jump after IE₁ — Group 1; jump after IE₉ — shell-3 to shell-2 boundary, since shell 3 has 8 e⁻) is consistent with K. Z = 19, potassium (IE₁ literature 419 kJ mol⁻¹). ✓

Example 4 — Group 7 (halogen pattern)

Q: Element W has successive IEs (kJ mol⁻¹): 1251, 2298, 3822, 5158, 6542, 9362, 11018, 33604, 38600, 43961. Identify the group and likely identity.

Answer:

Compute ratios:

• IE₇ → IE₈: 11018 → 33604 (ratio 3.05, BIG JUMP)
• All earlier ratios: 1.7, 1.7, 1.3, 1.3, 1.4, 1.2 — all "normal"

First big jump after IE₇ → 7 outer-shell electrons → Group 17 (halogen). The size of IE₁ (1251 kJ mol⁻¹) matches chlorine (Z = 17, literature IE₁ = 1251 kJ mol⁻¹). ✓

Example 5 — distinguishing Mg from Ca by IE pattern

Q: Two elements P and Q have the following first three successive IEs (kJ mol⁻¹):

• P: 738, 1451, 7733
• Q: 590, 1145, 4912

Identify the group of each, and explain how the IE values differentiate them.

Answer: For P: ratio IE₂/IE₁ = 1.97 (normal); IE₃/IE₂ = 5.33 (BIG JUMP) — first jump after IE₂ → Group 2. For Q: same pattern (ratio 4.29 after IE₂) → also Group 2. Both elements have two outer electrons.

The differences are quantitative, not pattern-based: P has larger IE₁ (738 vs 590), which means P's outer electron is more strongly bound — P is higher in Group 2 (smaller atom, closer outer e⁻, less shielding). So P is Mg; Q (with smaller IEs) is Ca. The same successive-IE pattern (Group 2) is shared, but the magnitudes locate each element on the down-group axis.

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Graphing successive IEs

When plotted as $\log_{10}(\text{IE})$log10​(IE) against ionisation number, the graph shows clear steps corresponding to shells. On a linear plot the inner-shell IEs are so large they squash the outer ones flat, so the log scale is preferred for any plot involving more than two shells.