Gen Chem_Ch 6_15th_Radial Distribution (3)
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Overview
Dr. Reza Latifi of Oklahoma State University explains how quantum numbers determine the nodes and radial distribution of the 3d orbital, then uses penetration to compare orbital energies. For orbitals with the same principal quantum number, he gives the order 3s < 3p < 3d; he also explains why penetration places 4s below 3d for electron filling.
Key takeaways
- A 3d orbital has l=2, so it has two angular nodes; because n-l-1=0, it has no radial nodes.
- For orbitals in the n=3 shell, the energy order is 3s < 3p < 3d, because increasing l corresponds to less penetration and higher energy.
- Penetration lowers orbital energy by allowing electron density closer to the nucleus, where electrostatic attraction is stronger.
- The 4s orbital penetrates more than 3d and is presented as lower in energy, so electrons fill 4s before 3d in Latifi's Aufbau ordering.
- A p subshell has three orbitals and a d subshell has five.
Chapters
- For n=3, the allowed angular momentum quantum numbers are l=0, 1, and 2, corresponding to 3s, 3p, and 3d.
- A 3d orbital has l=2 angular nodes and n-l-1=0 radial nodes, for two total nodes.
- Latifi depicts the angular nodes as planes through the d-orbital lobes and the radial distribution as one bump with no radial node.
- Comparing n=3 orbitals, Latifi uses angular momentum l to order energy: 3s (l=0) < 3p (l=1) < 3d (l=2).
- The 3s orbital penetrates closest to the nucleus, 3p penetrates less, and 3d shows no penetration in his comparison.
- Greater penetration exposes an electron to stronger nuclear attraction, lowering orbital energy; therefore an electron occupies 3s before 3p or 3d.
- A p subshell contains three orbitals, while a d subshell contains five.
- Although 4s has n=4 and 3d has n=3, Latifi explains that the subshells are close enough in energy for penetration to affect their ordering.
- The 4s orbital penetrates closer to the nucleus than 3d, stabilizing 4s below 3d in the filling order he presents.
- The resulting sequence is 4s before 3d; for orbitals with the same n, use increasing l, giving s before p, then d and f.
- Latifi flags 4s versus 3d as an important exception to relying on n alone and defers further electron-configuration discussion to the next lecture.
Transcript excerpt
0:11 [music] >> Hello and welcome back and thank you very much for watching another lecture from general chemistry. Last couple of lectures I mentioned that when n is a three or higher then you will have s p and d orbital. And I already covered s and p orbital but this lecture I'm going to show you how to get the radio distribution graph for d orbital and then we'll compare 3s, 3p and 3d orbital and see which one is going to have lower energy. So let's go with this that's we're talking about n equal three. n equal three and then we said possible answer for l it will be zero to n minus one so we get l equal to zero to three minus one. We get l zero, one and two. And that means the first orbital it will be the 3s orbital because l equal zero
1:00 means this is a p or s orbital. This one is going to be a p orbital belong to the third layer and this is a d orbital belong to the third layer. Okay, if we are dealing with a 3d orbital in that case we say l is going to be equal to two. And so angular node is going to be equal to l and equal to two. And totally remember that you have n number of nodes equal to n minus one so you have three minus one you have two nodes totally. Okay. And two of them already used for angular node. That means we are not going to have any radial node. Okay, and remember the radial node we show it by n minus one minus l so you say three minus one and l is equal to two so there will be no angular no radial nodes. We have only
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Related notes
Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Dr. Reza Latifi at Oklahoma State University, OSU.