Eigenvalue of La

[Vts]

In old periodic tables La was listed in 5d subshell, that corresponds to n=5 and l=2 quantum numbers. Modern IUPAC table lists it in 4f subshell, that corresponds to n=4, l=3. It is known that La has no 4f electron, but it did not stop IUPAC from listing it in 4f. What is correct eigenvalue of the last distinct electron to be used for the solving Schrodinger equation for example? Should n=4 and l=3 be used?

 

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[Klaynos]

wikipedia says

 

5d, 6s

 

So that l=2, n=5 An unfilled band

 

And n=6 l=0 A filled band... But probably the higher energy config.

[thedarkshade]

Lanthanides and Actinides are considered as "inner passing elements" (dunno if the spelling is correct), this is because these elements start to fill the f subshell. Nor La neither Ac have electrons in f, but the other lanthanides and actinides do have. If the name lanthanides is bothering you, that is only because the elements in this group have very very similar characteristics with La. Why would La be counted in f when when the config. is 5d 6s ...!

[Vts]

"Why would La be counted in f when when the config. is 5d 6s ...!"

 

Now I know why. The periodic's law basic rule is "n+l" rule. That is, periodicity follows "n+l", and not "n" or "l" separately.

 

Value of "n+l" for determining electron in La is 7 (either 4f or 5d orbitals), so as for Lanthanides.

 

If you want to see how it works, go to perfectperiodictable.com. It is demonstrated there clearly.