Why does plutonium have more oxidation states than samarium?

Electron configuration of Pu: $\ce{[Rn] 5f^6 7s^2}$

Electron configuration of Sm: $\ce{[Xe] 4f^6 6s^2}$

I thought that only the valence electrons affected the oxidation states, so why does plutonium have more oxidation states (6,5,4,3) than samarium (2,3) whilst they both have the same valence electrons (including f-orbitals): ..$\ce{f^6}$..$\ce{s^2}$ ?

Elements such as scandium (Sc),Yttrium (Y) and Lanthanum (La) all have the same valence electrons (including d-orbitals): ..$\ce{d^1}$..$\ce{s^2}$ , but these elements have the same oxidation state : 3 and that doesn't change, even when going down that group and looking at actinium.

Does this have to do with the ionization energy, or the energy level or ....?


This is yet another interesting effect of the anomalous compactness of orbitals in the first appearance of each type of subshell ($1s$, $2p$, $3d$, $4f$, $5g$, etc). The solutions to the Schrödinger equation for electron wavefunctions in hydrogen-like atoms are such that these subshells are composed of orbitals with no radial nodes. This means the electrons in these subshells are closer to the nucleus than you might expect, and therefore they are more strongly bound.

The striking chemical similarity of the lanthanides comes from the fact that the $4f$ subshell is both anomalously compact and subjected to a high effective nuclear charge. The electrons end up being held so closely to the nucleus that they effective behave as core electrons, and participate very little in chemical interactions; lanthanides do not form coordination compounds using their $4f$ orbitals, and it is very difficult to go past an oxidation state of +3 in most because few conditions can compensate the large energy required to ionize a fourth electron coupled with the relatively weakly bound compounds they would form.

In the actinides, the $5f$ orbitals do not suffer from the same lack of radial node, and therefore are more chemically available. The first half of the actinides show many compounds with high oxidation numbers, and have significant coordination chemistry using $f$ orbitals. Curiously, as you go form curium to berkelium, there is a sudden increase in reluctance to display high oxidation numbers, and the actinides beyond curium tend to behave similarly (though of course exploration of their chemical properties is significantly hindered by their rarity and instability). This is likely because even though the $5f$ orbitals have a radial node, in the second half of the actinides they are subjected to such a high effective nuclear charge that become too strongly bound to display significant participation in chemistry.

This same effect explains why the metals in the first row of transition metals do not display as many stable compounds with high oxidation states compared to the second and third rows, as the $3d$ orbitals are much smaller than $4d$ or $5d$ orbitals.


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