Energy Levels in the same Shell

[Margia Adriano 2A]

Hi! In Lecture #10, it was concluded that s-electrons have a lower energy than p-electrons in the same shell. I am not sure about how we got to this conclusion. P-orbitals and d-orbitals have nodal planes, so that means that they have a nonsymmetric distribution of electron density, while s-orbitals have a symmetric one. I am not sure how the different electrostatic interactions between those electrons in the different orbitals and the nucleus leads to the conclusion that s-electrons are lower energy. Can someone please explain it to me? Thank you!

[Lucy Wang 2J]

Hey! I'm not 100% sure why it is that they have lower energy but I don't think we need to know why for now, just know that the order of increasing orbitals is s, p, d, f.

[Sami Siddiqui 1J]

To answer your question Margia,
We need to address two specific things. The first thing to consider is that the p, d, and f orbitals have a nodal point at the center of their wavefunctions. What this means is that the electron density is 0 at a very close distance to the nucleus for these types of orbitals, meaning that their electrons will be inherently further away from the nucleus. In other words, electrons of these orbitals have less penetration than say s-orbitals, which does not contain any nodal points and does have an electron density right by the nucleus. Because electrons in the s-orbitals can penetrate closer to the nucleus, they will tend to be of a lower energy given that an electron of basically any other orbital type because these electrons will tend to be more stable. Also, we must consider the second thing: shielding. Because the s-orbital electrons penetrate further, they do somewhat shield the electrons of p-,d-, and f-orbitals and limit the attraction of their electrons to the nucleus. This in it of itself is another reason for the energy state classification.