A very simple test for the presence of alcohols in the lab involves adding ceric(IV) ammonium nitrate solution which gives a pinkish red colour in the presence of alcohol.

When I looked up the structure of this compound it turns out to be quite interesting. So a couple of questions about lanthanides came up in my mind (a Google search only comes up with very complicated research papers):

  1. Wouldn't this compound be highly unstable since there 12 donor atoms (forming an icosahedral structure) surrounding it (but it is quite stable)?
  2. Is the formation of complexes in lanthanides and actinides similar to that of transition metals? Do $\ce{f}$ orbitals undergo crystal field splitting just like $\ce{d}$ orbitals?
  3. Are higher coordination numbers like 12 more common (than octahedral and tetrahedral structures) among lanthanides and actinides because they have more orbitals for bonding, or is this compound an exception?

3 Answers 3


After a lot of digging around, I finally found something about this in J.D. Lee's Concise Inorganic Chemistry. I thought I'd post whatever I found over here anyway since it's very interesting:

Answer to question 2

Complex formation by lanthanides is different from that of actinides. In lanthanides, the $\mathrm{4f}$ orbitals are well shielded by the larger $\mathrm{5d}$ and $\mathrm{6s}$ orbitals and are deep inside the atom. So $\mathrm{f}$ orbitals do not participate in any bonding and complex formation is similar to that of transition metals. However in the actinides, the $\mathrm{5f}$ orbitals extend outwards and participate in bonding much more easily making the interactions much more complex than in the transition metals.

Answer to question 3

Higher coordination numbers are apparently extremely common among f-block elements (strange right?). Octahedral (6) and tetrahedral (4) structures are very rare except when bulky ligands are present. Most common coordination numbers among lanthanides are

  • 7 (Capped trigonal prismatic) yttrium acetylacetonate hydrate (Yttrium seems to be grouped with the f-block elements due to similar properties)

  • 8 (Square antiprismatic and Dodecahedral) cerium acetylacetonate and holmium tropolonate

  • 9 (Tri-capped trigonal prismatic) nonaaquaneodymium(III) complex $\ce{[Nd(H2O)9]^3+}$

10 (very complex) and 12 (icosahedral) are seen only in the larger elements like cerium and thorium with small ligands like $\ce{NO3-}$ and $\ce{SO4^2-}$

The actinides also commonly form exotic structures like chained tricapped trigonal prismatic in $\ce{[ThF8]^4-}$ and $\ce{[PaF7]^3-}$ distorted cubic in $\ce{[PaF8]^3-}$

The text mentions that the nature of bonding in $\ce{[Ce(NO3)6]^2-}$ is still not understood because it would imply bond orders of less than 1 or participation of f orbitals.

Could anyone explain anything about the first question since there is nothing mentioned on the stability of these compounds.

  • $\begingroup$ Maybe cerium's $f$ orbitals are more exposed early in the lanthanide series? $\endgroup$ Oct 6, 2017 at 23:08

1) No, because it is not donor-acceptor relationship, but electrostatic one

2) No, f-orbitals of lanthanide and actinides are unavailable for valence interactions

3) No, they are common because amount of ligands in the case is determined by geometry, and not electronic structure. Lanthanide cations are big.

  • $\begingroup$ I don't understand what you meant by answers 1 and 3.. could you explain why having an electrostatic interaction makes it more stable. $\endgroup$
    – kaliaden
    Mar 9, 2013 at 9:49
  • $\begingroup$ The d-element complexes often have heavy dose of covalent bounding. This results in octahedral, tetrahedral or planar square coordination, as most closely resembling symmetry of d-orbitals. Complexes of lanthanides are purely electrostatic, and coordination number is determined by amount of ligands that can be in direct contact with central atom simulationely, the same way complexes of late alkali and alcali earths are built. Note: actinides are much trickier matter, as 5f- and 6d- orbitals in them have close energy. $\endgroup$
    – permeakra
    Mar 9, 2013 at 13:59
  • $\begingroup$ If lanthanide complexes don't have crystal field splitting of the f orbitals can you explain why most lanthanide complexes are coloured? Crystal field splitting is ultimately just the lowering of symmetry leading to non-degenrate orbitals in the coulomb field, and this as the bonding in lanthanide complexes is generally considered to be ionic this will be the case here. $\endgroup$
    – Ian Bush
    Dec 1, 2019 at 9:46
  • $\begingroup$ @IanBush Quick googling shows that crystal field splitting is minimal for f-orbitals in Ln complexes and optical $Ln^{3+}$ spectra have minnmal dependence on ligands. Apparently, their optical spectra are dominated by f-f transitions with energy difference arising from total angular momentum (including spin participation). I welcome any correction. $\endgroup$
    – permeakra
    Dec 1, 2019 at 10:59

They form complexes compound but their tendency to form complexes is less their transition metal ions .the lanthanide ions spite a high charge have low charge density due to their large size.they form stable complexes with chelating ligands such as EDAT ,oxime beta-ketones.

  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. For more information in general have a look at the help center. At the moment, while undoubtedly correct, this reads more like a comment than an actual answer, and only addresses a subset of the question. Please elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer. $\endgroup$ Jun 7, 2018 at 11:32

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