# How is it possible that 3d orbitals are more contracted than 4f?

by Chemist20
Tags: energy, orbitals
 P: 87 just thinking... the energy of an orbital depends on: a)quantum number n b)quantum number l. an orbital will be of less energy when it's more contracted. So technically, 3d should have a lower energy than 4f and hence be contracted. BUT IT'S THE OTHER WAY AROUND!!! help? thank you!
PF Gold
P: 3,136
 Quote by Chemist20 an orbital will be of less energy when it's more contracted. So technically, 3d should have a lower energy than 4f and hence be contracted. BUT IT'S THE OTHER WAY AROUND!!!
I think you mean the 4s, not 4f, correct? Your reasoning works if you compare orbitals that are similar except for their n value, so if they have the same l value then the n will control how contracted they are and what their energy is. But when you also change the l value, you can no longer characterize the orbital by just its "size", there's also an issue about its "shape". It turns out that s orbitals have a probability of the electron being extremely close to the nucleus, so s orbitals have significantly lower energies then the other l states. As n rises, the energy differences between different n states becomes less than this dip in the s states.
P: 87
 Quote by Ken G I think you mean the 4s, not 4f, correct? Your reasoning works if you compare orbitals that are similar except for their n value, so if they have the same l value then the n will control how contracted they are and what their energy is. But when you also change the l value, you can no longer characterize the orbital by just its "size", there's also an issue about its "shape". It turns out that s orbitals have a probability of the electron being extremely close to the nucleus, so s orbitals have significantly lower energies then the other l states. As n rises, the energy differences between different n states becomes less than this dip in the s states.
Nope! I actually mean the 4f. the 4f are more contracted than 3d. And I don't see why! ;)

 P: 76 How is it possible that 3d orbitals are more contracted than 4f? If I remember correctly (and I'm vaguely remember something I read in an inorganic chemistry text book several years ago), it has something to with relativistic effects. I try to find the actual reference rather than just waving my hands.