Skip to content

Chemistry · Ch 3 — Periodic Classification of Elements

Ionic radius

3.5.2

Ionic radius

Definition. The ionic radius of an ion is the distance from the centre of its nucleus out to which it exerts influence on the surrounding electron cloud.

Pauling's method. For a uni-univalent ionic crystal, Pauling calculated the individual cation and anion radii from the experimentally measured inter-ionic distance dd, starting from two assumptions: the ions are perfect touching spheres, so

d=rC++rA−...(1)d = r_{C^+} + r_{A^-} \qquad \text{...(1)}

and, for ions that have a noble-gas electronic configuration, the ionic radius is inversely proportional to the effective nuclear charge felt at the ion's periphery:

rC+∝1(Zeff)C+...(2)rA−∝1(Zeff)A−...(3)r_{C^+} \propto \frac{1}{(Z_{eff})_{C^+}} \qquad \text{...(2)} \qquad\qquad r_{A^-} \propto \frac{1}{(Z_{eff})_{A^-}} \qquad \text{...(3)}

Dividing (2) by (3) gives the ratio of the two radii purely in terms of their effective nuclear charges (found via Slater's rules, Section 3.5.1), and combining that ratio with equation (1) lets both individual radii be solved for.

Worked example: Na+ and F- in NaF, given the inter-ionic distance dNa−F=231d_{Na-F} = 231 pm. Both ions have the neon (Ne) electron configuration 1s2 2s2 2p61s^2\,2s^2\,2p^6, so Slater's rules give the same screening constant for both: for a 2p electron in this configuration, S=(7×0.35)+(2×0.85)=2.45+1.70=4.15S = (7 \times 0.35) + (2 \times 0.85) = 2.45 + 1.70 = 4.15. Then:

(Zeff)Na+=11−4.15=6.85(Zeff)F−=9−4.15=4.85(Z_{eff})_{Na^+} = 11 - 4.15 = 6.85 \qquad\qquad (Z_{eff})_{F^-} = 9 - 4.15 = 4.85 …