Question

What are the different oxidation states exhibited by the lanthanoids?

What are the oxidation states exhibited by lanthanoid elements?

What are the oxidation states exhibited by lanthanoids?

Answer 1

The members of group 3 of the periodic table are expected to have a uniform +3 oxidation state as a characteristic of the lanthanoids. The triply positive oxidation state is based on using the 6s² electron and the lone 5d-electron or one of the f-electrons if no 5d-electron is present. The sum of the first three ionization enthalpies is relatively low, so that these elements are highly electropositive and readily form +3 ions. However, in an aqueous solution and in the solid state, cerium (Ce⁴⁺) can give quadripositive ions and samarium, europium and ytterbium (Sm²⁺, Eu²⁺ and Yb²⁺) can give bipositive ions. Other elements can give a +4 state in the solid state. Reduction of MX₃ can give not only MX₂ but also complex reduced species under special conditions.

The concept of +3 oxidation state for lanthanoids has become relatively rigid and other oxidation states are called ‘almost anomalous’. Such anomalous oxidation states of various lanthanoids are shown as follows:

Different oxidation states are displayed by lanthanum and lanthanoids. Dotted lines show ambiguous or short-lasting valencies.

If we assume that special stability is associated with vacant, half-filled or full f-subshell, then the presence of +2 and +4 oxidation states can be reconciled with the electronic structures to a certain extent. Thus La, Gd and Lu form only tripositive ions because removing three electrons gives rise to the noble gas configuration in the La³⁺ ion. In Gd³ and Lu³⁺ ions, there is no removal of electrons from the stable configurations 4f⁷ and 4f¹⁴ respectively, because the lattice or hydration energies of M²⁺ or M⁺ ions will be smaller than those of M³⁺ ions or the additive lattice or hydration energies of salts of smaller M³⁺ ions.

The most stable bi- or quaternary ions are formed by elements that can do so by attaining f⁹, f⁷ and f¹⁴ configurations. Thus, cerium attains f⁰ configuration in the +4 oxidation state. Europium and ytterbium attain f⁷ and f¹⁴ configurations, respectively, in the +2 oxidation state. These facts seem to support the notion that the special stability of the f⁰, f⁷ and f¹⁴ configurations is important in determining the existence of oxidation states other than +3 for the lanthanoids, but this argument becomes less decisive when we see that samarium and thulium form M²⁺ ions, not M⁺ ions, having f⁶ and f¹³ configurations.

Also, praseodymium and neodymium form M⁴⁺ ions with f¹ and f² configurations but do not form penta- or hexavalent ions. There is no doubt that the Sm(II) and especially Tm(II), Pr(IV) and Nd(IV) states are very unstable, but the idea that merely approaching the f⁰, f⁷ or f¹⁴ configuration is conducive to stability even if no such configuration is attained is questionable. The existence of Nd²⁺(f⁴) is particularly conclusive evidence to believe that although the stability of the f⁰, f⁷ and f¹⁴ configuration may be one factor determining the stability of oxidation states, there are other thermodynamic and kinetic factors of equal or greater importance.

Answer 2

The lanthanoid elements (atomic numbers 57 to 71) primarily exhibit the +3 oxidation state, which is the most stable and common for all lanthanoids. However, some lanthanoids can also exhibit +2 and +4 oxidation states due to the variable involvement of 4f electrons.