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Ionisation energies provide some of the most powerful evidence for the arrangement of electrons in atoms. By measuring how much energy is needed to remove electrons, we can map out the shell and subshell structure of atoms -- and explain the trends we see across the periodic table.
The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions:
X(g) -> X+(g) + e-
Key conditions:
The second ionisation energy is the energy required to remove a second electron:
X+(g) -> X2+(g) + e-
Each successive ionisation energy is higher than the last because you are removing an electron from an increasingly positive ion -- the remaining electrons are held more tightly.
If the substance were in the solid or liquid state, additional energy would be needed to overcome intermolecular forces (or metallic/ionic bonding) before the electron could be removed. Using the gaseous state ensures we measure only the electron-nucleus interaction.
flowchart TD
A[Factors Affecting Ionisation Energy] --> B[Nuclear Charge]
A --> C[Atomic Radius / Distance]
A --> D[Electron Shielding]
A --> E[Subshell Type and Electron Pairing]
B --> F[More protons = stronger attraction = higher IE]
C --> G[Greater distance = weaker attraction = lower IE]
D --> H[More inner shells = more shielding = lower IE]
E --> I[Paired electrons experience repulsion = lower IE]
E --> J[s electrons penetrate more than p = higher IE for s]
More protons in the nucleus means a greater attractive force on the electrons. Higher nuclear charge -> higher ionisation energy.
The further the outer electron is from the nucleus, the weaker the attraction. Greater distance -> lower ionisation energy. The force follows Coulomb's law (inverse-square relationship): F is proportional to 1/r-squared.
Inner electrons repel outer electrons, reducing the effective nuclear charge felt by the outermost electron. More inner shells -> more shielding -> lower ionisation energy.
Effective nuclear charge (Zeff) = Nuclear charge (Z) - Shielding (S)
For example, sodium (Z = 11) has the configuration 1s2 2s2 2p6 3s1. The 3s electron is shielded by 10 inner electrons, so Zeff is approximately 11 - 10 = +1. This is why sodium's outer electron is so easy to remove.
An electron in a paired orbital experiences extra repulsion from the other electron in that orbital. Also, electrons in p orbitals are slightly easier to remove than electrons in s orbitals of the same shell because s orbitals penetrate closer to the nucleus.
The general trend across a period is increasing first ionisation energy because:
| Element | Li | Be | B | C | N | O | F | Ne |
|---|---|---|---|---|---|---|---|---|
| Z | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Config (outer) | 2s1 | 2s2 | 2p1 | 2p2 | 2p3 | 2p4 | 2p5 | 2p6 |
| 1st IE (kJ/mol) | 520 | 900 | 801 | 1086 | 1402 | 1314 | 1681 | 2081 |
Beryllium (1s2 2s2) -> Boron (1s2 2s2 2p1)
Boron's outer electron is in a 2p orbital, which is higher in energy and further from the nucleus than beryllium's 2s orbital. The 2p electron is also shielded by the 2s2 electrons. Despite boron having one more proton, these effects outweigh the increased nuclear charge.
Exam tip: When explaining the Be -> B drop, always mention three points: (1) boron's electron is in a 2p orbital, not 2s; (2) 2p is higher in energy; (3) the 2p electron is shielded by the 2s2 pair.
Nitrogen (1s2 2s2 2p3) -> Oxygen (1s2 2s2 2p4)
Nitrogen has three unpaired 2p electrons (one in each orbital). In oxygen, the fourth 2p electron must pair up in one of the orbitals. The paired electron experiences extra electron-electron repulsion, making it easier to remove despite oxygen having a higher nuclear charge.
Exam tip: When explaining the N -> O drop, always mention: (1) nitrogen has three unpaired 2p electrons; (2) oxygen has one paired 2p orbital; (3) the paired electron experiences extra repulsion.
The same general trend and anomalies appear:
| Element | Na | Mg | Al | Si | P | S | Cl | Ar |
|---|---|---|---|---|---|---|---|---|
| Z | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| Config (outer) | 3s1 | 3s2 | 3p1 | 3p2 | 3p3 | 3p4 | 3p5 | 3p6 |
| 1st IE (kJ/mol) | 496 | 738 | 578 | 789 | 1012 | 1000 | 1251 | 1521 |
Magnesium (1s2 2s2 2p6 3s2) -> Aluminium (1s2 2s2 2p6 3s2 3p1)
Same reason as Be -> B. Aluminium's outer electron is in a 3p orbital, which is higher in energy and more shielded than magnesium's 3s.
Phosphorus (... 3p3) -> Sulfur (... 3p4)
Same reason as N -> O. Sulfur has a paired electron in one of its 3p orbitals, increasing repulsion.
When we remove all the electrons from an atom one by one, the ionisation energies increase but not uniformly. There are large jumps where we break into a new (inner) shell.
Sodium has 11 electrons. Its successive ionisation energies are:
| IE | Value (kJ/mol) | Shell Being Ionised | Notes |
|---|---|---|---|
| 1st | 496 | 3rd shell (3s1) | Easy to remove -- one outer electron |
| 2nd | 4,562 | 2nd shell (2p6) | Huge jump -- breaking into inner shell |
| 3rd | 6,912 | 2nd shell (2p5) | Gradual increase within shell |
| 4th | 9,544 | 2nd shell (2p4) | |
| 5th | 13,354 | 2nd shell (2p3) | |
| 6th | 16,613 | 2nd shell (2p2) | |
| 7th | 20,117 | 2nd shell (2p1) | |
| 8th | 25,496 | 2nd shell (2s2) | Small jump -- 2s to 2p subshell boundary |
| 9th | 28,934 | 2nd shell (2s1) | |
| 10th | 141,362 | 1st shell (1s2) | Enormous jump -- breaking into 1s |
| 11th | 159,076 | 1st shell (1s1) |
The enormous jump between the 1st and 2nd ionisation energies confirms that sodium has one electron in its outer shell. The massive jump between the 9th and 10th confirms that the second shell holds 8 electrons, and the innermost shell holds 2.
The pattern of successive ionisation energies reveals the electronic structure. A big jump between the nth and (n+1)th ionisation energies tells you the element has n electrons in its outer shell:
Successive IEs (kJ/mol): 578, 1817, 2745, 11578, 14831, 18378
First ionisation energy decreases down a group because:
| Element | Z | Shells | 1st IE (kJ/mol) |
|---|---|---|---|
| Li | 3 | 2 | 520 |
| Na | 11 | 3 | 496 |
| K | 19 | 4 | 419 |
| Rb | 37 | 5 | 403 |
| Cs | 55 | 6 | 376 |
The decrease is not linear -- the largest drop is from Li to Na (adding the third shell for the first time), and the drops become smaller further down as additional shells have diminishing marginal effects on shielding and distance.
Misconception: "Ionisation energy decreases across a period because the atoms get bigger."
Correction: Atomic radius decreases across a period (electrons are added to the same shell while nuclear charge increases). This is one of the reasons ionisation energy increases across a period. The anomalies at Be->B and N->O (or Mg->Al and P->S) are due to subshell changes and electron pairing, not atomic radius.
Edexcel 9CH0 specification Topic 1, sub-topic 1.3 covers the definition of first ionisation energy (the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous unipositive ions); successive ionisation energies and the evidence they provide for shell structure; trends in first ionisation energy across periods 2 and 3 and down groups; the explanation of these trends in terms of nuclear charge, atomic radius, shielding and subshell type; and the explanation of anomalies — specifically the dips in first IE between Group 2 and Group 3 (Be→B, Mg→Al) and between Group 5 and Group 6 (N→O, P→S) (refer to the official Pearson Edexcel specification document for exact wording). Examined chiefly in Paper 1 (Advanced Inorganic and Physical Chemistry), with synoptic data-interpretation appearances in Paper 3 (General and Practical).
Question (8 marks):
The successive ionisation energies (kJ mol⁻¹) of an unknown element X are: 738, 1451, 7733, 10541, 13630, 17995, 21704, 25656.
(a) Identify element X and justify your answer using the data. (4) (b) Write the equation, including state symbols, for the second ionisation energy of X. (2) (c) Predict, with reasoning, whether the first ionisation energy of element (X+1) (one place to the right of X in the periodic table) is higher or lower than that of X. (2)
Solution with mark scheme:
(a) M1 — identify the large jump between IE2 (1451) and IE3 (7733): a factor of ~5.3 increase, far larger than the gradual rise between successive same-shell ionisations. M1 — interpret: two electrons are easy to remove (outer shell), then a sudden much harder removal indicates the next electron must come from an inner shell. M1 — therefore X has 2 outermost electrons → Group 2. A1 — given the magnitudes (IE1 = 738, comparable to magnesium's IE1 ≈ 738 kJ mol⁻¹), X is magnesium (Mg). Common error: candidates count "the third electron is hardest" without recognising that the jump (ratio between successive IEs) is the diagnostic feature.
(b) M1 — equation: Mg⁺(g) → Mg²⁺(g) + e⁻. A1 — state symbols all (g), as required by the IE definition; and the species removed is one electron from the gaseous unipositive ion to form the gaseous dipositive ion. Common error: writing Mg(g) → Mg²⁺(g) + 2e⁻ — that is a combined IE1+IE2, not the IE2 process.
(c) M1 — element (X+1) is aluminium (Al), in Group 3 with configuration [Ne] 3s² 3p¹. A1 — its outermost electron is in 3p, which is higher in energy than the 3s electrons of Mg, and is shielded slightly more effectively. Therefore IE1(Al) is lower than IE1(Mg) — the classic Group 2 → Group 3 dip. Common error: candidates predict IE1(Al) > IE1(Mg) on the basis of higher nuclear charge alone, missing the subshell argument.
Total: 8 marks (M3 A1 + M1 A1 + M1 A1).
Question (6 marks): The graph of first ionisation energies across period 2 (Li to Ne) shows a general increase but with two anomalies — a dip from Be to B and a dip from N to O.
(a) Explain the dip from Be to B. (3) (b) Explain the dip from N to O. (3)
Mark scheme decomposition by AO:
| Part | AO1 | AO2 | AO3 | Marks |
|---|---|---|---|---|
| (a) | 1 | 2 | 0 | 3 |
| (b) | 1 | 2 | 0 | 3 |
| Total | 2 | 4 | 0 | 6 |
(a) B1 (AO1.1) — Be is 1s² 2s²; B is 1s² 2s² 2p¹. M1 (AO2.1) — the outermost electron in B is in a 2p orbital, which is at higher energy and slightly further from the nucleus than the 2s. A1 (AO2.1) — therefore less energy is needed to remove the 2p electron in B than the 2s electron in Be, despite B's higher nuclear charge.
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