0
$\begingroup$

What I am taught-

Inner shell electrons shield the outermost electrons from the attraction of nucleus. More closer the inner shell is to the nucleus, more its shielding effect. thus $n_s > n_p > n_d > n_f$ and such (in terms of screening effect as "s" orbitals are closest to the nucleus). $Z_{\text{eff}} = Z - \sigma$; where $\sigma$ is the screening constant and can be calculated via Slater's rule.

Why I think this happens -

Nucleus attracts the outermost electron just as well according to coulomb's formulae ; but the inner shell electrons provide an repulsive force towards the outermost electrons and thus "decrease" the attractive forces. But I don't get how $n_s > n_p > n_d > n_f$ is working; like s is closer to the nucleus thus is more far away from the outermost electrons thus it should provide less repulsive forces, right ? But what I think this order is due to is the shapes of these orbitals; like the p orbitals are like dumbells due to which some repulsive forces may get cancelled out (like derivation of electric dipoles and stuff); but I don't think that shows the full picture.

Adding to it; I know about the penetrating power of the orbitals ; but doesn't it just plot a graph of $4 \pi^2 \psi_r^2$ as a function of $r$ (distance from nucleus) of different orbitals and seeing the graph we can predict about the distance of the orbitals from nucleus; and we can deduct that "s" orbital stays the closest and the distance increases from there on. But it again pulls the question why then s orbital has highest screening effect, because if its closest to the nucleus it should be farthest from outer shell electrons; thus should have least screening.

$\endgroup$
3
  • 1
    $\begingroup$ As the $1s$ electrons are nearly against a nucleus with $Z$ protons, the outer electrons feelsthe attraction of $Z$ protons minus the $2$ electronic charges due to these $1s$ electrons. If these $1$s electrons were more cumbersome, and distributed around a large domain around the atom, the outer electrons will "feel" the attraction of the $Z$ protons minus something smaller than $2$ $\endgroup$
    – Maurice
    Oct 10, 2021 at 14:55
  • $\begingroup$ Welcome to Chemistry Stack Exchange! I've made some formatting edits to your post and added some MathJax. Please feel free to edit further. $\endgroup$
    – uhoh
    Oct 10, 2021 at 23:51
  • $\begingroup$ Screening is direct application of classical electrostatics. Inner spherical charge distribution gives full screening like if the charge were in a nucleus. Deviations from spherical symmetry and reaching outer regions behind the screened electrons decrease the screening. Therefore, overlapping of 2 orbitals negatively affects screening by the more inner orbital. $\endgroup$
    – Poutnik
    Oct 11, 2021 at 5:29

0

Your Answer

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