When calculating the effective nuclear charge for an electron, why do we only consider the repulsive effect of the inner electrons on the outer electrons? The outer electrons also repel the inner ones (towards the nucleus) and so, when determining the effective nuclear charge for inner electrons, don't we need to account for this effect as well?

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    $\begingroup$ The outer electrons just don't penetrate the orbitals of the inner shells enough to have a "significant" effect. That isn't to say that there isn't any effect. This is chemistry where a few significant figures is generally good enough. $\endgroup$ – MaxW Jul 22 at 6:52
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    $\begingroup$ The effect of outer electrons may be also within the error ranges of the model, as there are already multiple simplifications in place to be ever able to compute anything. Be aware that even a classical 3-body gravitationally bound system cannot be solved exactly. $\endgroup$ – Poutnik Jul 22 at 6:57
  • $\begingroup$ You´ve got a bit of a wrong assumption here. If the electrons were in actual spherical shells, the outer electrons would indeed have zero influence. $\endgroup$ – Karl Jul 22 at 7:39

As chemists, we are typically more directly concerned with the effective nuclear charge on outer electrons than inner electrons, because it's only the outer electrons that interact with other atoms and thus determine an atom's chemical properties. [Though the effective nuclear charge on inner electrons will affect their radial probability distributions and thus create a second-order effect on the effective nuclear charge of outer electrons.]

Nevertheless, this is an interesting question.

When we are calculating the effective nuclear charge seen by electrons in an atom (say, for outer electrons) (which directly impacts, for instance, their ionization energies), we start by doing so for isolated atoms, i.e., atoms in the absence of an external field. Adding an external field (from the presence of other atoms, especially bonding atoms) is an added complication we won't consider here.

And in isolated atoms, the probability distributions of all electrons are spherically symmetrical.

So let's consider an inner electron in an isolated atom. According to Gauss's Law, any charge that is spherically distributed about the nucleus, and is farther from the nucleus than that electron, "cancels itself out" (it integrates to zero), i.e. it has no net effect on the effective charge felt by the electron.

So an inner electron is only affected by outer electrons to the extent that the outer electrons spend some time closer to the nucleus than the inner electron. [Again, when they are further from the nucleus than the inner electron, they have no effect on it, because of the spherical symmetry of their probability distributions.]

Based on the relative probability distributions of inner and outer electrons, the percent of the time this happens (outer electrons being closer to the nucleus than inner ones) is low. Having said that, when it does happen, it will cause the inner electrons to feel a reduced nuclear charge. I.e., to the extent that outer electrons affect inner electrons (in isolated atoms), they don't push them towards the nucleus, they push them away!

Hence there is no "reverse shielding" effect by outer electrons on inner electrons (again, in isolated atoms). There is only a (small) shielding effect.

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