Explain why the successive ionisation energies of aluminium show two large increases: one between the third and fourth ionisation energies, and one between the tenth and eleventh ionisation energies.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
Model answer (4 marks)
Aluminium has the configuration [Ne] 3s² 3p¹, i.e. three valence electrons in the n=3 shell.
1. After the third ionisation the next electron removed is from the n=2 shell. Electrons in n=2 are closer to the nucleus and are less shielded, so the effective nuclear charge felt by this electron is much higher – a large increase in ionisation energy.
2. The tenth ionisation removes the last electron in the n=2 shell. The eleventh ionisation must remove an electron from the n=1 shell, which is even closer to the nucleus and has virtually no shielding. The effective nuclear charge is therefore much larger, giving another large jump in ionisation energy.
In both cases the large increase is due to removing an electron from a lower‑n shell, which experiences a higher effective nuclear charge and less shielding than the preceding electron.
1. After the third ionisation the next electron removed is from the n=2 shell. Electrons in n=2 are closer to the nucleus and are less shielded, so the effective nuclear charge felt by this electron is much higher – a large increase in ionisation energy.
2. The tenth ionisation removes the last electron in the n=2 shell. The eleventh ionisation must remove an electron from the n=1 shell, which is even closer to the nucleus and has virtually no shielding. The effective nuclear charge is therefore much larger, giving another large jump in ionisation energy.
In both cases the large increase is due to removing an electron from a lower‑n shell, which experiences a higher effective nuclear charge and less shielding than the preceding electron.
Examiner tips
- Use the electron configuration to identify the shell of the electron removed at each step.
- Explain that lower‑n shells are closer to the nucleus and less shielded, giving higher effective nuclear charge.
- Show the two specific jumps (3→4 and 10→11) and why they are large.
- Use the terms "effective nuclear charge" and "shielding" as expected by the mark scheme.
Common mistakes
- Confusing the jump between the 3rd and 4th ionisation with the 4th and 5th; the 4th ionisation is the key jump.
- Failing to mention that the 11th ionisation removes an electron from the n=1 shell, not just a higher‑energy n=2 electron.
- Using vague phrases like "more energy needed" without linking to effective nuclear charge and shielding.
Mark scheme (4 marks)
- Aluminium has the electron configuration [Ne] 3s² 3p¹ / 2,8,3, so there are three electrons in the third shell (valence electrons).
- After the third ionisation, the fourth electron is removed from a shell closer to the nucleus (the second shell / n=2), which is more strongly attracted to the nucleus / less shielded, requiring significantly more energy.
- The large increase between the tenth and eleventh ionisation energies occurs because the tenth electron is the last electron in the second shell (n=2), and the eleventh electron must be removed from the first shell (n=1), which is even closer to the nucleus and has virtually no shielding.
- In both cases, the large increase arises because the electron being removed is in a shell of lower principal quantum number (closer to the nucleus), experiencing greater effective nuclear charge / less shielding, compared to the preceding electron.
Key terms in this question
successive ionisation energies
Related
- All IB DP Chemistry Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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