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Is the shielding effect in a multi-electronic species caused due to mutual repulsion of electrons?

If it were so, while calculation the $Z_\text{eff}$ on an electron in say the $\mathrm{2p}$ orbital of a $\ce{Na}$ $\mathrm{(1s^2,2s^2,2p^6,3s^1)}$ atom, why don't we also consider the repulsive force of outer electrons in $\mathrm{3s}$ and add it to $Z_\text{eff}$ instead of $Z_\text{eff} = Z - \text{repulsion due to inner electrons}$?

Why doesn't the equation read

$$Z_\text{eff} = Z - \text{repulsion due to inner electrons} + \text{attraction due to outer electrons}?$$

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  • $\begingroup$ 10th grade physics: The inside of a charged sphere is free of field lines. Same as with gravity: If you dig 1km deep, your weight gets lower as if the earth's diameter had shrunk by two kilometers. $\endgroup$
    – Karl
    Jul 20 '17 at 21:35
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Firstly, effective nuclear charge is just a model for predicting trends in atomic properties. The actual solution of the quantum mechanical equations does use the full nuclear charge and explicit calculation of the all electron-electron repulsions (and electron-nuclear attraction).

The point of $Z_\text{eff}$ is to approximate the value of the nuclear charge felt by electrons at a given distance. The answer to your question then is Gauss's Law, which states that, to determine the field at a given point, you consider all and only the charge inside a certain surface (on which your point sits). Moreover, it states that the field there is equivalent to if all of the charge inside that surface was a point charge at the center.

So if we apply Gauss's Law to our atom, we add up all the charge inside a sphere centered on the nucleus ($Z-\text{effective number of core electrons}$), and regard the system as an effective nucleus with that charge, no core electrons and the electron of interest being the "first" electron and so we ignore the outer electrons.

Here are some physics discussions for why, intuitively, Gauss's Law allows you to ignore outside electrons from the physics stackexchange.

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