A-Level Chemistry May/June 2025 Q1(b)(iii): Arrange the three species F¯, Ne and Na⁺ in order of increasing radius. Explain your an…

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  Order of increasing radius:

  Na⁺ < Ne < F⁻

  Explanation:

  All three species, F⁻, Ne, and Na⁺, are isoelectronic, meaning they have the same number of electrons (10 electrons) and the same electronic configuration ($1s^22s^22p^6$).

  However, they have different numbers of protons in their nuclei:

  • Na⁺ has 11 protons
  • Ne has 10 protons
  • F⁻ has 9 protons

  The number of protons determines the nuclear charge. Na⁺ has the largest nuclear charge (11+). This results in the strongest nuclear attraction for the 10 electrons, pulling the electron shells in most tightly and giving it the smallest radius.

  Conversely, F⁻ has the smallest nuclear charge (9+). This results in the weakest nuclear attraction for the same 10 electrons, so its electron shells are held less tightly, giving it the largest radius.

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  How the marks are awarded

  • M1 — Stating the correct order of increasing radius: Na⁺ < Ne < F⁻.
  • M2 — Identifying that all three species have the same number of electrons or are isoelectronic.
  • M3 — Correctly linking the size of the species to its nuclear charge or number of protons (e.g., the largest species, F⁻, has the fewest protons).
  • M4 — Explaining that a smaller nuclear charge results in a weaker nuclear attraction for the outer electrons, causing a larger radius.

  Common mistakes

  • Reversing the order (F⁻ < Ne < Na⁺), incorrectly assuming that the species with the lowest atomic number is smallest.
  • Failing to state that the species are isoelectronic, which is the fundamental reason a direct comparison of nuclear charge is valid.
  • Providing a vague explanation like 'it depends on nuclear charge' without specifying the number of protons for each species and explaining how this affects the attraction on the electrons.
  • Confusing the number of protons with the overall charge of the ion when explaining the trend.

  Examiner tip: For isoelectronic species, the radius is determined by the nuclear charge; more protons mean a stronger attraction to the same number of electrons, resulting in a smaller radius.

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