# Why are s orbitals more “penetrating” than p orbitals?

In polyelectronic atoms, the reduction in the net central force due to electron-electron repulsion is accounted for through an effective nuclear charge that depends on a “shielding effect” of inner electrons.

The reason for the splitting of energies of orbitals in the same quantum shell (s<p<d<f) is usually given that s orbitals are more "penetrating" i.e have a greater probability density near the nucleus, so experience less shielding and thus a greater effective nuclear charge.

However, I can't understand why a comparatively larger p. density is used to measure the penetrating power. the mean orbital radius actually decreases from s to f orbitals. Sure, s orbitals have a relatively higher p.d near the nucleus but the rest of the p.density is also relatively more further away than that in p,d or f orbitals. Using the mean distance as the appropriate measure, shouldn't the energy be in the order f<d<p<s?

the only explanantion that comes to mind is that the decrease in energy from being unusually close some of the time outweighs the decrease in energy due to being more farther away for the rest of the time (and thus on avg), but this is just a guess and a rather non-intuituve explanation

• The point is, the mean distance is not an appropriate measure, due nonlinear distance dependence of electrostatic potential and also due the distance dependent effective nucleus charge. – Poutnik Jan 30 at 7:57
• – cngzz1 Jan 30 at 9:41
• Also do not see it as "most of the time". Is as it is. – Alchimista Jan 30 at 11:43

These are plots of the 2s and 2p orbitals for hydrogen (assuming 1 electron). The radial wave function is on the left and the probability density as a function of distance from the nucleus is on the right ($$4\pi r^{2}R(r)dr$$ if you understand the calculus).