Why is effective nuclear charge used for explaining the periodicity of size of atoms and ions, even though radius depends upon many other factors?

The size of $\ce{Na+}$ is smaller than that of $\ce{F-}$ and $\ce{Ne}$, as $Z_\mathrm{eff}$ of $\ce{Na+}$ is higher than that of other ions. This is written at this site. But I don't understand how $Z_\mathrm{eff}$ can be used as the only deciding factor for determining the size of the ions (or atoms for that matter).

It is clear that the size of an atom is determined by the net potential energy of valance electrons, with respect to the nucleus: More energy means less attraction hence bigger radius, and less energy means more attraction hence smaller size.

Now the formula of net potential energy for an atom is: $$E_\mathrm{net} = \frac{KZ^2e^2}{r}-\text{potential energy due to electrons}$$ where $Z$ is atomic number $e$ is 1C charge and $r$ is radius of atom.

It is clear that the potential energy of two atoms (or ions) can't simply be compared on the basis of $Z_\mathrm{eff}=Z- e_\mathrm{insh}$ (the number of inner shell electrons is $e_\mathrm{insh}$). It also depends upon other factors as well.

But many textbooks explain the periodicity of the size of atoms on this basis, but why?

  • $\begingroup$ Na+, Ne and F- are isoelectric. While in general there is more factors affecting the radius, in this particular case are those factors constant. $\endgroup$
    – Poutnik
    Commented Aug 2, 2021 at 6:15
  • $\begingroup$ P.S.: I have meant independent factors. As the bigger electron repulsion, weakening the contraction trend, is induced as the negative feedback by the primary reason for the contraction - higher Z_eff. $\endgroup$
    – Poutnik
    Commented Aug 2, 2021 at 8:29
  • 1
    $\begingroup$ I think Poutnik's answer here would be helpful. $\endgroup$ Commented Aug 2, 2021 at 17:37


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