How can I make cerium(IV) chloride? [closed]

Question

How can I make CeCl4?

I have some ideas but am worried I'm going to mess it up since this is the first time I've worked with lanthanides.

Answer 1

To the best of my knowledge, there are no known $\ce{LnCl4}$ species. Although $\ce{CeCl4}$ has been mentioned in older literature, more recent publications tend to refute its existence attributing the observed reactions to $\ce{CeCl3}$ and $\ce{[CeCl6]^2-}$ species, which could be fairly easily isolated and characterized.

Among cerium(IV) tetrahalides, only $\ce{CeF4}$ is known, and solely thanks to the fluorine's ability to support high oxidation states [1, pp. 17, 21]:

Fluorine is well known to promote high oxidation states. Factors associated with this are the high lattice energies associated with the small fluoride ion, along with the very small $\ce{F–F}$ bond energy (due to non-bonding electron pair repulsions) as well as high bond energies involving fluorine (not relevant in this case). Because of the larger size of the $\ce{Cl-}$ ion, there is going to be much less difference between the lattice energies of $\ce{CeCl3}$ and $\ce{CeCl4},$ and this is probably the determining factor. The higher $\Delta_\mathrm{at}H(\ce{Cl})$ of $\pu{121.5 kJ/mol}$ also mitigates against the formation of $\ce{CeCl4}.$

You can, however, stabilize cerium(IV) tetrachloride with neutral organic ligands [2, p. 310] such as dimethylsulfoxide, triphenylphosphine oxide, 2,2′-bipyridine-N,N′-dioxide and tris(dimethylaminido)phosphine oxide allowing to produce adducts such as $\ce{[CeCl4(dmso)3]}$ [3], $\ce{[CeCl4(tppo)2]}$ [3], $\ce{[CeCl4(bpyO2)2]}$ [4] and $\ce{[CeCl4(tdpo)2]}$ [5], respectively.

References

1. Cotton, S. A. Lanthanide and Actinide Chemistry; Inorganic Chemistry: A Textbook Series; Wiley: Chichester, England; Hoboken, NJ, 2006. ISBN 978-0-470-01005-1.
2. Handbook on the Physics and Chemistry of Rare Earths; Gschneidner, K. A., Ed.; Elsevier North Holland: Amsterdam, 2006; Vol. 36. ISBN 978-0-444-52142-2.
3. Březina, F. Verbindungen Des Vierwertigen Cers Mit Triphenylphosphin, Fallweise Mit Triphenylphosphinoxid. Collect. Czech. Chem. Commun. 1971, 36 (8), 2889–2894. DOI: 10.1135/cccc19712889.
4. Březina, F. Koordinationsverbindungen Des α,α’-Dipyridyl-N,N’-Dioxids Mit Salzen Des Vierwertigen Cers. Collect. Czech. Chem. Commun. 1974, 39 (8), 2162–2167. DOI: 10.1135/cccc19742162.
5. Du Preez, J. G. H.; Rohwer, H. E.; De Wet, J. F.; Caira, M. R. Preparation and Structure of Tetrachlorobis[Tris(Dimethylamido)Phosphine] Oxidecerium(IV). Inorganica Chimica Acta 1978, 26, L59–L60. DOI: 10.1016/S0020-1693(00)87177-7.

Answer 2

As noted by @ansdelisk, $\ce{CeCl4}$ hasn't been isolated yet as a pure compound. It exist in a solution although not for too long as it as hydrolyze/decomposes to $\ce{Ce(III)}$ and chlorine. As noted here:

Ceric chloride, $\ce{CeCl4}$, has not been isolated. It can be obtained in solution, but the solution cannot be kept without decomposing. Ceric hydroxide dissolves in cold concentrated hydrochloric acid with the production of a dark red solution; chlorine, however, is slowly evolved, and ultimately a solution of cerous chloride remains. The decomposition proceeds rapidly in hot solutions.

An old literature points out that $\ce{CeCl4}$ has been made using $\ce{CeO2}$ and a solution of $\ce{(POCl3 + SnCl4)}$:

$\ce{(POCl3 + SnCl4)}$ solutions were prepared by volume in a dry box. The $\ce{SnCl4}$ content was verified by chemical analysis for $\ce{Sn}$. This solution and $\ce{CeO2}$ were placed in sealed ampoules, heated to $\pu{120 °C}$ for 2 hours to increase the rate of solution, and then rotated in an air thermostat at $\pu{20 °C}$ for 2 hours. Without preheating, equilibrium was established after 200 hours. Preheating to $\pu{120 °C}$ lowered the equilibration time at $\pu{20 °C}$ to 2 hours.

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In general, Ce(IV) salts are metastable in aqueous solution and is a strong oxidizing agent that oxidizes hydrochloric acid to give chlorine gas. The reason is somewhat written in the Wikipedia article:

Despite the common name of cerium(IV) compounds, the Japanese spectroscopist Akio Kotani wrote "there is no genuine example of cerium(IV)". The reason for this can be seen in the structure of ceria itself, which always contains some octahedral vacancies where oxygen atoms would be expected to go and could be better considered a non-stoichiometric compound with chemical formula $\ce{CeO_{2−x}}$. Furthermore, each cerium atom in ceria does not lose all four of its valence electrons, but retains a partial hold on the last one, resulting in an oxidation state between +3 and +4. Even supposedly purely tetravalent compounds such as $\ce{CeRh3}$, $\ce{CeCo5}$, or ceria itself have X-ray photoemission and X-ray absorption spectra more characteristic of intermediate-valence compounds. The 4f electron in cerocene, $\ce{Ce(C8H8)2}$, is poised ambiguously between being localized and delocalized and this compound is also considered intermediate-valent.