# Effective nuclear charge on an inner electron

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

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

• 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.
– Karl
Jul 20 '17 at 21:35

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.