Ionisation energies
Ionisation energies
First and successive ionisation energies
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Ionisation energy measures how much energy is needed to remove electrons. First ionisation energy removes one electron from each gaseous atom. The precise definition below includes the amount, gaseous state and ions formed.
Removing a negative electron from the attraction of a positive nucleus needs energy, so ionisation is endothermic. The unit is kJ mol⁻¹.
The second ionisation energy starts with gaseous 1+ ions, not neutral atoms. Each successive ionisation removes another electron from an increasingly positive ion.
- First ionisation energy
- The energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.
X an atom of the element in the gas state; X+ a gaseous ion of the element with charge +1; e− an electron; (g) gas.
| Step | Equation |
|---|---|
| First | Mg(g) → Mg⁺(g) + e⁻ |
| Second | Mg⁺(g) → Mg²⁺(g) + e⁻ |
| Third | Mg²⁺(g) → Mg³⁺(g) + e⁻ |
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
- Define first ionisation energy. [3]
- Write the first-ionisation equation for magnesium, with state symbols. [1]
- State the units. [1]
Total: 5 marks
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- Energy required to remove one electron from each atom [1] in one mole of gaseous atoms [1] to form one mole of gaseous 1+ ions [1].
- Mg(g) → Mg⁺(g) + e⁻. [1]
- kJ mol⁻¹. [1]
- Write the equation for the second ionisation energy of magnesium. [1]
- Write the equation for the third ionisation energy of aluminium. [1]
- Is ionisation endothermic or exothermic? [1]
Total: 3 marks
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- Mg⁺(g) → Mg²⁺(g) + e⁻. [1]
- Al²⁺(g) → Al³⁺(g) + e⁻. [1]
- Endothermic. [1]
What controls ionisation energy?
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Ask one central question: how strongly is the electron being removed attracted to the nucleus? Stronger attraction means more energy is needed to remove it.
- More protons give a greater nuclear charge and tend to strengthen attraction.
- An electron further from the nucleus is less strongly attracted.
- Inner electrons shield an outer electron from some of the nuclear attraction.
- Subshell energy and repulsion between paired electrons can explain exceptions to a simple trend.
Compare the relevant factors together. Saying only that an atom has more protons is not enough if the electron is also further away and more shielded.
| Factor | Effect |
|---|---|
| More protons | Greater nuclear charge strengthens attraction. |
| Greater distance | Weaker attraction to the electron. |
| More inner-shell shielding | Weaker attraction to the electron being removed. |
| Higher-energy sub-shell | Less additional energy needed to remove the electron. |
| Electron pairing | Repulsion within an orbital can make one electron easier to remove. |
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
- State the three main factors controlling attraction between the nucleus and the electron being removed. [3]
- Why is an electron in a higher-energy sub-shell easier to remove? [2]
Total: 5 marks
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- Nuclear charge [1], distance from the nucleus [1], shielding by inner electrons [1].
- It is already at higher energy / less strongly attracted [1], so less additional energy is required to remove it [1].
Na has IE₁ = 496 and IE₂ = 4562 kJ mol⁻¹. Mg has IE₁ = 738 and IE₂ = 1451 kJ mol⁻¹. Explain why the second IE of sodium is much greater than that of magnesium. Atomic numbers: Na = 11, Mg = 12.
Total: 4 marks
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- Na⁺ is 1s² 2s² 2p⁶: the second electron is removed from n = 2. [1]
- Mg⁺ is 1s² 2s² 2p⁶ 3s¹: its second electron is removed from 3s. [1]
- The electron in Na⁺ is closer to the nucleus and less shielded. [1]
- It is more strongly attracted and requires much more energy to remove. [1]
First ionisation energy across Period 3
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Across Period 3, first ionisation energy generally rises. The nucleus gains protons, while the added electrons enter the same main shell. Inner-shell shielding stays broadly similar.
The outer electron is attracted more strongly and the atomic radius decreases. More energy is therefore needed to remove that electron.
From Mg to Al, the removed electron is in the higher-energy 3p subshell. From P to S, the removed 3p electron is paired, so electron-electron repulsion makes removal easier. Use these specific explanations for the exceptions.
- Mg → Al
- Mg: [Ne] 3s². Al: [Ne] 3s² 3p¹. The electron removed from Al is in a higher-energy, slightly more shielded 3p sub-shell, so is easier to remove.
- P → S
- P: [Ne] 3s² 3p³. S: [Ne] 3s² 3p⁴. Sulfur has a pair in one 3p orbital. Repulsion in that pair makes an electron easier to remove.
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
| Element | First IE / kJ mol⁻¹ |
|---|---|
| Na | 496 |
| Mg | 738 |
| Al | 578 |
| Si | 787 |
| P | 1012 |
| S | 1000 |
| Cl | 1251 |
| Ar | 1521 |
- Describe the general trend across Period 3. [1]
- Explain the trend. [3]
- Identify the two decreases. [1]
- What do the decreases show about electron arrangement? [1]
Total: 6 marks
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- First IE generally increases from Na to Ar. [1]
- Nuclear charge increases [1]. Electrons enter the same principal shell, so inner-shell shielding is broadly similar [1]. Stronger attraction means more energy is required [1].
- Mg → Al and P → S. Both needed. [1]
- Evidence for different sub-shell energies and electron pairing within orbitals. [1]
Mg is [Ne] 3s²; Al is [Ne] 3s² 3p¹.
- Which sub-shell loses the first electron in Al? [1]
- Explain why Al has a lower first IE than Mg. [3]
Total: 4 marks
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- 3p. [1]
- The 3p electron is higher in energy than 3s [1] and slightly more shielded [1], so it is less strongly attracted / easier to remove [1].
- State the final sub-shell occupancy of phosphorus and sulfur. Atomic numbers: P = 15, S = 16. [2]
- Explain why sulfur has a slightly lower first ionisation energy than phosphorus. [2]
Total: 4 marks
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- P: 3p³ [1]. S: 3p⁴ [1].
- Sulfur has an electron pair in one 3p orbital [1]. Repulsion within that pair makes an electron easier to remove [1].
| Species | Configuration | First IE / kJ mol⁻¹ |
|---|---|---|
| A | [Ne] 3s² | 738 |
| B | [Ne] 3s² 3p¹ | 578 |
| C | [Ne] 3s² 3p³ | 1012 |
| D | [Ne] 3s² 3p⁴ | 1000 |
All four are in Period 3: Na, Mg, Al, Si, P, S, Cl, Ar.
- Identify A, B, C and D. [4]
- Explain why the first ionisation energy of B is lower than that of A. [3]
- Explain why the first ionisation energy of D is lower than that of C. [2]
Total: 9 marks
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- A = Mg; B = Al; C = P; D = S. [4]
- B loses 3p rather than 3s [1]. The 3p electron is higher in energy / slightly more shielded [1], so it is easier to remove [1].
- D has a pair in a 3p orbital [1]. Pair repulsion makes one electron easier to remove [1].
First ionisation energy down Group 2
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Down Group 2, first ionisation energy falls even though the nucleus contains more protons. The electron being removed is now in a shell further from the nucleus.
- There are more occupied shells.
- The outer electron is further away and more shielded by inner electrons.
- These effects outweigh the larger nuclear charge, so the attraction is weaker.
- Less energy is needed to remove the outer electron.
The repeated drop when a new outer shell begins to fill is evidence for the shell model.
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
| Element | First IE / kJ mol⁻¹ |
|---|---|
| Be | 900 |
| Mg | 738 |
| Ca | 590 |
| Sr | 550 |
| Ba | 503 |
- Describe the trend in first ionisation energy down Group 2. [1]
- Explain this trend in first ionisation energy. [4]
Total: 5 marks
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- First IE decreases. [1]
- The outer electron occupies a higher principal shell [1], is further from the nucleus [1] and has more shielding [1]. These outweigh increased nuclear charge, so attraction is weaker and less energy is required [1].
Successive ionisation energies and shell structure
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Each successive ionisation removes an electron from an ion that is already positive, so the energy rises. A particularly large jump tells you that the next electron comes from an inner shell.
Inner-shell electrons are closer to the nucleus and less shielded. They are much more strongly attracted than the outer electrons just removed.
- Locate the first large jump.
- Count the electrons removed before it. That is the number in the outer shell.
- For a main-group element, connect this count to the group: three outer electrons means Group 3; six means Group 6.
- If more large jumps are shown, use them to work out further shell populations.
| Period 3 element | First major jump after |
|---|---|
| Na | IE₁ |
| Mg | IE₂ |
| Al | IE₃ |
| Si | IE₄ |
| P | IE₅ |
| S | IE₆ |
| Cl | IE₇ |
| Ar | IE₈ |
Identify a Period 3 element
Successive IEs: 578, 1817, 2745 and 11 578 kJ mol⁻¹.
These are the first four successive ionisation energies of a Period 3 element. Identify the element and explain your reasoning. Write its full electron configuration.
Worked answer
- Large jump between IE₃ and IE₄.
- Three outer-shell electrons, then an inner-shell electron is removed.
- The element is in Group 3. In Period 3, this is aluminium.
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
A Period 3 element has successive IEs of 786, 1577, 3232, 4356 and 16 091 kJ mol⁻¹. Period 3: Na, Mg, Al, Si, P, S, Cl, Ar.
- Where is the large jump? [1]
- How many outer-shell electrons? [1]
- Identify the element. [1]
- Write its full electron configuration. [1]
- Explain the jump. [1]
Total: 5 marks
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- Between IE₄ and IE₅. [1]
- Four. [1]
- Silicon, Si. [1]
- 1s² 2s² 2p⁶ 3s² 3p². [1]
- The fifth electron comes from an inner shell and is more strongly attracted to the nucleus. [1]
A Period 3 element has successive IEs of 1000, 2250, 3360, 4560, 7000, 8500, 27 100 and 31 800 kJ mol⁻¹. Period 3: Na, Mg, Al, Si, P, S, Cl, Ar.
- Identify the largest jump. [1]
- State the number of outer-shell electrons. [1]
- Identify the element. [1]
- Write its full configuration. [1]
- Explain the sharp rise. [3]
Total: 7 marks
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- Between IE₆ and IE₇. [1]
- Six. [1]
- Sulfur, S. [1]
- 1s² 2s² 2p⁶ 3s² 3p⁴. [1]
- The seventh electron comes from n = 2 [1]. It is closer and less shielded [1], so is much more strongly attracted and requires far more energy [1].
Al has successive IEs of 578, 1817, 2745 and 11 578 kJ mol⁻¹.
- Write the fourth-ionisation equation, including state symbols. [1]
- Explain why IE₄ is much greater than IE₃. [3]
- Calculate IE₄ / IE₃ to three significant figures. [1]
- What does this show about the number of outer-shell electrons? [1]
Total: 6 marks
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- Al³⁺(g) → Al⁴⁺(g) + e⁻. [1]
- Three outer electrons have already been removed [1]. The fourth comes from an inner shell, closer / less shielded [1], so is much more strongly attracted [1].
- 11 578 / 2745 = 4.22. [1]
- Al has three outer-shell electrons. [1]
The graph shows all thirteen successive IEs of an unknown Period 3 element. Period 3: Na, Mg, Al, Si, P, S, Cl, Ar.
- Identify both large jumps. [2]
- State the number of electrons in the outer shell and second shell. [2]
- Deduce the shell arrangement and identify the element. [2]
- Write its full configuration. [1]
- Explain the first large jump. [2]
- Why do successive IEs increase even before a new shell is reached? [1]
Total: 10 marks
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- After IE₃ and IE₁₁. [2]
- Outer shell: 3. Second shell: 8. [2]
- 2,8,3; aluminium. [2]
- 1s² 2s² 2p⁶ 3s² 3p¹. [1]
- The next electron comes from n = 2 [1], closer / less shielded and more strongly attracted [1].
- Each electron is removed from an increasingly positive ion, so remaining electrons are more strongly attracted. [1]
M has isotopes at m/z 24, 25 and 26 with abundances 78.99%, 10.00% and 11.01%. Relevant atomic numbers: Na = 11, Mg = 12, Al = 13, Ca = 20.
- Calculate Aᵣ to two decimal places. [2]
- Identify M. [1]
- Write the full configuration of M. [1]
- Write the configuration of M²⁺. [1]
- Write the third-ionisation equation. [1]
- The first three IEs are 738, 1451 and 7733 kJ mol⁻¹. Explain what these show about the outer electrons and the next electron removed. [4]
- Calcium has first IE 590 kJ mol⁻¹. Why is this lower than M’s? [2]
Total: 12 marks
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- Aᵣ = Σ(isotopic mass × abundance)/Σ(abundance)
Aᵣ = [24(78.99) + 25(10.00) + 26(11.01)]/100 [1] = 24.32 [1]. - Magnesium, Mg. [1]
- 1s² 2s² 2p⁶ 3s². [1]
- 1s² 2s² 2p⁶ or [Ne]. [1]
- Mg²⁺(g) → Mg³⁺(g) + e⁻. [1]
- Large jump between IE₂ and IE₃ [1], hence two outer electrons [1]. The third comes from an inner shell [1], closer / less shielded and much more strongly attracted [1].
- Ca has another occupied shell, so the outer electron is further away and more shielded [1]. Attraction is weaker, requiring less energy [1].