Actinide Organometallics

Question 1

Describe PUREX process to extract U, Pu and Am

Answer 1

1. Start with Pu, U and other fission products eg. Am
2. First chop up rods into small bits
3. Dissolve in Conc HNO3- doesn’t dissolve Zirconium so left behind
4. Oxidising conditions so move to preferred OS: UO2(NO3)2(H2O)2 (U6+), Pu(NO3)4 (Pu4+), Am(NO3)3 (Am3+) in H2O
5. Add hydrocarbon- kerosene and TBP- strongly polar bond so coordinates: Get separation of 2 phases
6. Actinide +3 and Ln 3+ elements and most of other waste products- aqueous phase
7. UO2(NO3)2(TBP)2 which extracts into the kerosene along with Pu(NO3)4.
8. Separate phases using long columns to remove aqueous phase leaving just U and Pu
9. The Pu(IV) is more readily reduced than U(VI) as stronger Zeff so a reducing agent is added along with distilled water.
10. The Pu(IV) is reduced to Pu(III) and extracts into the aqueous phase as Pu(NO3)3 using ion e.g. Fe
11. Left with pure U as nitrate in kerosene on top layer- separate layers
    Oxidise products- produce UO2 and PuO2

Question 2

How could you separate later actinides

Answer 2

1. As you would for the Lns

Question 3

How do you reduce plutonium selectively from U

Answer 3

1. To reduce the plutonium selectively, a reducing agent with a redox potential lower than +1.006 V but higher than that of uranyl must be added

Question 4

Describe which elements are generally used in actinide organometallics

Answer 4

1. Principally U and Th, although neptunium is now being investigated.

Question 5

What is the dominant OS for actinide organometallics

Answer 5

1. Dominated by Ac(IV) compounds although highly reactive +3 and some +2 compounds now known.

Question 6

What is some of the organometallic actinide behaviour dependent on

Answer 6

1. Some behaviour shown due to the symmetry and availability of the 5f orbitals
2. but also significant contribution from the 6d orbitals.

Question 7

Describe Uranium carbonyls

Answer 7

1. A few uranium carbonyls have been described.
2. Studies indicated that U(CO)6, whilst unstable, showed some level of U–>CO backbonding.
3. As lower V(CO) than free CO bond
4. Not seen in Lns
5. But not as stable as with Tungsten etc as larger so need more than 6 ligands to satisfy

Question 8

What does more back donation mean

Answer 8

1. Weaker bond
2. Lower stretching frequency
3. As occupying antibonding orbital

Question 9

How can you form a U carbonyl

Answer 9

1. First 3KCp* + UCl4 –> Cp*3UCl (salt metathesis)
2. Can’t start with U(III) Halides as very reactive and hard to handle so start with U(IV)
3. Reduce with Na –> NaCl + Cp*3U (not stable so can’t be isolated)
4. Add CO straight away in adduction reaction to produce Cp*3UCO

Question 10

Describe bond length of U-C in carbonyl and what that suggests

Answer 10

1. Shorter than single so has some double bond character- Indicates back bonding

Question 11

Describe cyclopentadienyl An compounds

Answer 11

1. Cp- and Cp*- are widely applied ligands.
2. Can be installed by salt metathesis with actinide halides and support a range of further reactions.

Question 12

What is the only all eta5 tetracyclopentadienyl

Answer 12

1. Cp4U

Question 13

Describe what happens when adding FeCl2 to Cp4U and how does this differ to Lns

Answer 13

1. does not transfer Cp- to FeCl2.
2. This reflects the size of the early actinides and their ability to engage in covalency.
3. Lanthanides produce LnCl3 + 3Cp3Fe as Fe-Cp is much more covalent and Ln-Cp is very ionic with no covalency
4. Similar seen with Th

Question 14

Why is U easier to work with than Ln in salt metathesis

Answer 14

1. Doesn’t undergo rapid equilibrium or form dimers and oligomers
2. Less polarity due to covalency

Question 15

How can you form Cp*2AnR2 species

Answer 15

1. Add LiR to Cp*2AnCl2 in salt metathesis

Question 16

Why is Th(IV) often a focus in this chemistry

Answer 16

1. It is paramagnetic and thus can be studied by NMR spectroscopy.

Question 17

What reactions can AnCp*2R2 species undergo

Answer 17

1. Sigma bond metathesis and insertion
2. Strong bonds but still polar and reactive

Question 18

Describe actinide sigma-alkyls and aryls

Answer 18

1. 5f elements similar to 4f elements tend to decompose via beta-hydride elimination of alkene

Question 19

What does decomposition beta hydride elimination mean for homoleptic σ-alkyls and aryls

Answer 19

1. Must be stabilised by:
   Sterically Bulky Ligands (without β-substituents)
2. ate complexes
3. Arene ligands

Question 20

Are organometallics good ligands for High Os

Answer 20

1. NO
2. C is readily oxidised

Question 21

What is a possible ligand for high oxidation state chemistry

Answer 21

1. COT 2-
2. Formed from COT by addition of 2K which adds to e- to form its stable 10 electron aromatic form
3. Sterically demanding

Question 22

How can you form uranocene

Answer 22

1. UCl4 + 2K2COT –> U(COT)2 + 4KCl

Question 23

Describe properties of uranocene

Answer 23

1. A green, pyrophoric, paramagnetic compound which is stable to hydrolysis.
2. COT rings are planar oriented in an eclipsed conformation
3. (eta8-C8H8)2U - 22 electron complex (6 + 8 + 8 or 2 + 10 + 10)
4. would lead to occupation of antibonding MOs in (18 electron) d metal compounds - would fall apart
5. so additional bonding MOs must be formed by interaction of ligand orbitals with 5f orbitals of appropriate symmetry