quantum number-s- vs p- orbitals

[Nan_Guan_1L]

I know Dr.Lavelle went through this in lectures, but I don't think I really understand it: so for electrons in the same shell, why does s- electrons have a lower energy than p-electrons in the same shell? if the outer electrons feel a reduced electrostatic attraction, then why is the order of energy levels like s<p<d<f, instead of the opposite? Thank you!

[Sophia Hu 1A]

In the lecture, Dr. Lavelle described the reason the p-electrons have higher energy than the s-electrons in the same shell is a result of the different electrostatic interactions between the nucleus. p-, d-, and f-orbitals have nodal planes (zero electron density) which means there is a non-symmetric electron density distribution. On the other hand, s-orbitals have no nodal planes, so it has a symmetric electron density distribution.

Because of this different density distribution, s-electrons in an orbital have a higher nuclear charge than p-electrons in the same orbital which means it drops to a lower energy. This relates to reduced electrostatic attraction because since outer electrons feel a reduce electrostatic attraction, then the effective nuclear charge decreases which means the energy increases. Therefore, the order of the energy level would still be s<p<d<f.

[Pranav Kadiyala 1A]

To add on, Dr. Lavelle also mentioned how in Hydrogen (one electron one proton system), the s,p, and d orbitals are all degenerate (have the same energy). This is because there is no electron-electron shielding creating these energy differences.

Hope this helps :)

[Brittney Nguyen 2L]

Hi! I thought it would be valuable to add on the fact that the energies of the subshells are s<p<d<f because the s-orbital has an electron density that is closer to nucleus than the p-orbital, and the p-orbital has an electron density that is closer to the nucleus than the d-orbital, and so forth. If you remember from atomic spectra, electrons that are farther away from the nucleus have energy levels that approach 0 until the electron is completely detached, at which its energy is 0. Thus, subshells that are closer to the nucleus have lower energies.