P-, d-, f- Orbitals

[Scot Widjaja Dis 1J]

Hey guys,

I just had a quick question from Lecture 9 regarding p-, d-, and f- orbitals. What does Dr. Lavelle mean when he says that p-, d-, and f- orbitals have a nodal plane? And how does having a nodal plane make it so that there is a 0 probability for an orbital to have an electron density?

[Jade_Tai_2L]

The wave function Ψ^2 represents the probability of finding electron density. So in p-, d-, and f- orbitals, there's places in the orbital where there's 0 probability of having electron density. If you watch his lecture again and look at the diagrams he showed, you can see areas on the orbitals where it comes to a "point" which is the nodal plane. Since the s-orbital is spherical, it does not have any risk of going through the actual nucleus of the atom (the Heisenberg principal says there's almost no chance of finding electrons here, hence 0 electron density). Hopefully this helps, and please correct me if I'm wrong!

[Kristina Krivenko 3I]

You are correct. A nodal plane is a region where there is zero probability of finding an electron in that area. Thus, the fact that p-, d- and f-orbitals have nodal planes means that there are areas in those orbitals (the area is the nodal plane) where the likelihood of finding electrons is zero.

[apurva-3E]

A nodal plane is where the probability of finding an electron is 0

[Leo Naylor 2F]

All these answers are correct. To provide some further information, the nodal planes of the p-, d-, and f-orbitals are the reason that these states are a higher energy level than the s-orbital when they are in the same shell, because the different arrangement means that the effective nuclear charge for p-, d-, and f-orbitals is greater than for s-orbitals.

[Scot Widjaja Dis 1J]

Ok, great! Thanks guys!