Q3(c). Heavy-actinide separation; thorocene and uranocene
(i) Suggest a method for the separation of heavy actinides like Am, Cm, Bk, Cf, Es, Fm and Md with appropriate sketch.
(ii) Write the structure of Th(C₈H₈)₂ and that of U(C₈H₈)₂.
Answer in brief
(i) Cation-exchange chromatography: the An³+ ions are held on a sulfonate resin and eluted with ammonium 2-hydroxyisobutyrate; they come off in the order Md, Fm, Es, Cf, Bk, Cm, Am, the smallest ion first. (ii) Both are sandwich compounds [M(η⁸-C₈H₈)₂].
Key points
- Am to Md exist mainly as An³+, with almost the same chemistry; only microgram amounts (or a few atoms) are available, so a method that separates by small differences in ionic radius is needed.
- the mixed An³+ solution is passed onto a column of cation-exchange resin; the ions replace H⁺ or NH₄⁺ at the sulfonate sites.
- a solution of the complexing ligand 2-hydroxyisobutyrate is passed through. Across the series the actinide contraction makes the ions smaller; the smaller ion forms the more stable complex and so moves faster into the…
- each fraction is identified by its radioactivity (α energies). The elution curve is the same in form as that of the lanthanides (Lu first, La last), which helped to place the heavy actinides below the lanthanides.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
RESINsulfonated polystyrene, H⁺/NH₄⁺ formELUENT2-hydroxyisobutyrateORDERhighest Z first
(i) Ion-exchange separation [5 marks]
- Why ion exchange: Am to Md exist mainly as An³+, with almost the same chemistry; only microgram amounts (or a few atoms) are available, so a method that separates by small differences in ionic radius is needed.
- Loading: the mixed An³+ solution is passed onto a column of cation-exchange resin; the ions replace H⁺ or NH₄⁺ at the sulfonate sites.
- Elution: a solution of the complexing ligand 2-hydroxyisobutyrate is passed through. Across the series the actinide contraction makes the ions smaller; the smaller ion forms the more stable complex and so moves faster into the solution: Md elutes first, Am last.
- Detection: each fraction is identified by its radioactivity (α energies). The elution curve is the same in form as that of the lanthanides (Lu first, La last), which helped to place the heavy actinides below the lanthanides.
(ii) Structures of Th(C₈H₈)₂ and U(C₈H₈)₂ [5 marks]
FORMULA[M(η⁸-C₈H₈)₂]SYMMETRYsandwich, D₈ₕMETALTh⁴+ 5f⁰, U⁴+ 5f²
- Sandwich: the metal lies between two planar, parallel, eclipsed C₈H₈ rings, all 16 carbon atoms bonded equally (η⁸). Th and U adopt the same structure.
- Each ring is the aromatic cyclooctatetraenide dianion C₈H₈²- (10 π electrons, 4n+2 with n=2); the metal is M(IV).
- Bonding: the filled ring π combinations overlap the large 5f and 6d orbitals of the metal: δ (and π) interactions, with a possible φ contribution from 5f. This f-orbital participation is why the large C₈H₈²- ring suits the actinides.
- Preparation and properties: MCl₄ + 2K₂C₈H₈ →M(C₈H₈)₂ + 4KCl. Thorocene (5f⁰) is yellow and diamagnetic; uranocene (5f²) is green and paramagnetic. Both are stable towards water.
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All parts of Question 3
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- 3(b) Active site of oxyhaemerythrin10 marks
- 3(c) Heavy-actinide separation; thorocene and uranocene10 marks
- 3(d) Electrolysis of aqueous CuSO₄ between Pt electrodes20 marks