Ionic Radius and Isoelectronic Series

Sodium loses its single 3s electron to become Na⁺, which empties the n = 3 shell entirely. The ion is left with a neon-like n = 2 outer shell, and the same 11 protons now pull on only 10 electrons. Na⁺ (about 102 pm) is far smaller than Na (about 186 pm).

The nucleus does not gain protons when the atom becomes a cation. Na and Na⁺ both have eleven protons; the dramatic shrinkage occurs because the occupied 3s shell disappears. A cation can also shrink without losing an entire principal shell: Fe²⁺ and Fe³⁺ both retain 3d electrons, but removing an additional electron reduces repulsion and allows the remaining cloud to contract. Do not use “one fewer shell” unless the configuration actually shows it.

1. For an atom and its own ion, hold nuclear charge fixed and examine electron gain or loss. Cations are generally smaller and anions larger than the parent atom.
2. For ions of the same element, the more positively charged ion is generally smaller under comparable conditions.
3. For an isoelectronic series, verify equal electron count first, then rank by proton count: more protons generally gives a smaller radius.
4. For unrelated ions in different shells, use configurations and provided radius data. Dividing proton count by electron count is not a universal radius formula.

Measured ionic radii also depend on coordination and, for some metals, spin state. The introductory rankings here assume comparable environments. This is why quoting a single radius without its context is less useful than explaining the electronic comparison.

The same ten-electron arrangement is drawn progressively inward as proton count increases. Decreasing radius is therefore O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺. If the question requests increasing radius, reverse the list. Charge is a useful consequence of the electron-count calculation here, but the nuclear-charge argument works because the species are isoelectronic. Na⁺ and Cl⁻ do not belong in the same ten-electron comparison: Cl⁻ has eighteen electrons.