My text material says:

$\mathrm{Z_{eff}= Z-S}$ where $\mathrm{Z_{eff}}$ is the effective nuclear change on that specified electron; $\mathrm{Z}$ is the atomic number and $\mathrm{S}$ is the shielding constant or screening constant which is the measure of net repulsion on that electron caused by other electrons in the same shell and inner shells.

So I infer that the Effective Nuclear Force on the electron of our interest (supposing it lies in the outermost orbit) is the resultant of the Force of the Nucleus minus the repulsion of the other electrons. Am I correct in assuming that? Can "charge" and "force" be analogous in here? If so, by Coulomb's law, shouldn't the closer orbitals repulse the electron more than the ones that are further away? Or in other words, shouldn't the Shielding constant be getting higher as we approach the outermost orbit?

But the values in Slater's rule shows the opposite:

the shielding experienced by an $\mathrm{s-}$ or $\mathrm{p-}$ electron, electrons within same group shield $\mathrm{0.35}$, except the $\mathrm{1s}$ which shield $\mathrm{0.30}$ electrons within the $\mathrm{n-1}$ group shield $\mathrm{0.85}$ electrons within the $\mathrm{n-2}$ or lower groups shield $\mathrm{1.00}$ the shielding experienced by $\mathrm{nd}$ or $\mathrm{nf}$ valence electrons electrons within same group shield $\mathrm{0.35}$ electrons within the lower groups shield $\mathrm{1.00}$

Implying that the further we move from the outermost electron, the Shielding or the Repulsion increases, which sounds the opposite of Coulomb's law to me.

What am I missing here?


  1. Chemistry LibreText - Slater's rules
  2. Concise Inorganic Chemistry by J.D. Lee

1 Answer 1


There is important knowledge from classical electrostatics (going as far as to Newton classical gravity and his invention of integral calculus, as it is common for quadratic forces):

  • Two distributed non-overlapping charges with spherical distribution symmetries have on each other the same force effect as if they were point charges, located at the distributed charge center.

If we consider inner s orbitals, or superposition of all 3 filled p orbitals, they have spherically symmetric electron probability density.

Therefore, their shielding coefficients would be the same (1.00) as if located in the nucleus, if not decreased by overlapping of the shielding and shielded orbital. As the overlapping brings partial negative shielding.

Because, when a shielding electron is outside the shielded electron(wrt the nucleus), repulsion acts toward the nucleus, effectively being like an additional nucleus charge (= "negative shielding" ). If a point charge is overlapped by spherically symmetric charge distribution, it is attracted/repulsed only by the inner part of this distribution.

The more inner shielding orbital is, the smaller is overlapping and the better is shielding of the shielded orbital. Breaking spherical symmetry decreases the shielding even more.

See also Slater's rules topic on Wikipedia and Libretexts


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.