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


by Chemist20
Tags: energy, orbitals
Chemist20
Chemist20 is offline
#1
Jan19-12, 06:32 AM
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!
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Ken G
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#2
Jan19-12, 09:01 AM
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Quote Quote by Chemist20 View Post
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.
Chemist20
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#3
Jan19-12, 12:57 PM
P: 87
Quote Quote by Ken G View Post
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! ;)

mjpam
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#4
Jan19-12, 07:07 PM
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.
Matterwave
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#5
Jan19-12, 08:06 PM
P: 2,041
Quote Quote by mjpam View Post
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.
You are right. The Hydrogen spectrum would be entirely dependent on n if relativistic effects are not taken into account.

Relativistic effects in the momentum-energy relation, the spin-orbit coupling, and the so called "Darwin term" account for the l (L, not I lol) dependence of the energy spectra of Hydrogen. The sum of these effects are usually called the "fine structure" of Hydrogen. They are called "fine structure" because these effects are alpha (~1/137) suppressed in comparison to non-relativistic effects (alpha being the fine structure constant). But, as Ken says, at the higher n's, the regular energy levels of Hydrogen get very close together, and these fine structure effects can become factors in which states have higher energy, etc.


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