Electron Configurations -- Why is terbium's electron configuration [Xe] 6s2 4f9 ?

[i_love_science]

Why is terbium's electron configuration [Xe] 6s2 4f9, with no electrons in the d subshell and one extra in the f sub shell, while uranium's electron configuration is [Rn] 7s2 5f3 6d1, with one electron in the d subshell and none extra in the f subshell?

Are lanthanum and actinide d block elements or f block elements? Does this explain the difference between those electron configurations, and which is correct?

Thanks.

https://www.google.com/search?q=ura...LGc0KHVAHBXsQ_B16BAgBEAM#imgrc=Ljey_loSx3UArM

https://www.google.com/search?q=ter...pB50JHf2mDdQQ_B16BAgBEAM#imgrc=wiIp2saSuwaXbM

 

[DrDu]

There are many factors influencing the precise electron configuration of the lanthanides and actinides, but there are some trends. The f-electrons are bad in screening each other from the nuclear charge, hence, the more f-electrons in an atom, the more tightly they are bound. On the other hand, the f-electrons will screen the d-orbitals from nuclear charge. Both effects lead to an increasing energetic splitting of f and d orbitals. Therefore, both the heavier lanthanides and actinides hardly involve the d-orbitals. In actinides the energetic difference between f and d is lower than in lanthanides, hence the earlier actinides show higher oxidation states (and more often d1).

 

[HAYAO]

You shouldn't restrict that question to Tb and U.

In lanthanides, Pr, Nd, Pm, Sm, Eu, Tb, Dy, Ho, Er, Tm, Yb have [Xe]4fⁿ6s² type configuration. Meanwhile, La, Ce, Gd, Lu have [Xe]4fⁿ5d¹6s² type configuration.
In actinides, Pu, Am, Bk, Cf, Es, Fm, Md, No have [Rn]5fⁿ7s² type configuration. Meanwhile, Ac, Th, Pa, U, Np, Cm, Lr have [Rn]5fⁿ6d^m7s² type configuration.

I am not sure if you can consider these two series to have different d-orbital configuration solely as "lanthanide vs actinide" problem.

As @DrDu mentioned, there are many factors that contribute to the ultimate electron configuration of these elements, so it's hard to say what is exactly the reason for this trend. As a matter of fact, one reason don't explain all of the trends. I am going to answer the question with the best I know, and what I've heard in conferences. I'm not a professional in electronic energies of neutral lanthanides and actinides, so please take it with a grain of salt.

First, you should treat f⁰, f⁷, and f¹⁴ elements as exceptions. These have none, half, or fully occupied f-orbitals so they are relatively stable with that f-electron configuration. It is therefore expected that the remaining electrons will occupy other orbitals, namely d- and s-orbitals. This is why La, Gd, Lu, Ac, Cm, and Lr have d-orbitals occupied.

Second, lanthanides and actinides have different f- and d- orbital energy splitting. Lanthanides have larger energy splitting between f- and d-orbitals due to 4f-orbitals being relatively closer to the nucleus than 5d-orbitals (which is what @DrDu mentioned about nuclear charge screening effect, which changes the effective nuclear charge that orbitals feels). This causes actinides to be more susceptible to having d-orbitals occupied than lanthanides.

Third, you should treat lighter and heavier elements of the series differently. Relativistic effects cause larger energy splitting of f- and d-orbitals with larger atomic number. This means that lighter elements of series tend to have smaller f- and d-orbital splitting, hence the Ce in lanthanides and Th, Pa, U, Np in actinides, both being the lighter element of the series, having occupied d-orbitals. You can see that Th, the lightest actinide other than Ac (exception mentioned above) contains not one but two 6d electrons! You can also see that none of the heavier lanthanides or actinides have any electrons in the d-orbitals.

 

[DrDu]

I would not consider f0, f7 and f14 as exceptions. Binding energy increases and size of f orbitals decreases continuously with increasing nuclear charge. So evidently f0 and f14 are extremal. After f7, orbitals have to be occupied with two electrons, which costs more energy.
Also, while relativistic effects certainly are of high importance in actinides, I don't think they have huge impact on d and f electrons, as these don't get very near the nucleus (outer shells + high angular momentum).