Organolanthanide Chemistry 2

Question 1

For which elements which are not expected to have 2+ OS have molecular precursors of divalent organometallics been found

Answer 1

1. Thulium 4f13
2. dysprosium 4f10
3. Neodymium 4f4

Question 2

Why are lanthanides not good at back donation

Answer 2

1. Don’t have appropriate symmetry orbitals

Question 3

How can you access +2 OS for Thulium, Dy, Neodynium

Answer 3

1. Add I2 1eq- careful with stoichiometry
2. Use DME solvent
3. Forms somewhat stable isoloble Dy2+ halide
4. Incredibly reducing
5. More than Sm

Question 4

How can you tell Sm-N is an adduct and not proper compound

Answer 4

1. In crystal structure N-N bond length is the same as if it were a gas - no reduction
2. Also if dissolve in donor solvent e.g. THF, THF displaces N and get bis-THF adduct of Sm2+ and lose Nitrogen

Question 5

Why would Tm show 2+ OS

Answer 5

1. 4f13
2. 6 unpaired electrons with same exchange energy as Sm

Question 6

Why do Sm and Dy show +2 OS

Answer 6

1.3rd Ionisation potentials are potentially high enough

Question 7

What happens when you react TmI2(DM3)3 + Cyclopentadienyl in N2

Answer 7

1. Synthesis of bulky cylopentadienyl derivatives of Tm(II) and Dy(II) results in coordination and reduction of an N2 unit.
2. This is accompanied by oxidation of the M2+ centre to M3+
3. Not an adduct
4. The [N2]2- is bound in a side-on fashion by the [Cp’2M] units
5. The ionic nature of the bonding is emphasised by the formation of a THF adduct without loss of N2.

Question 8

How can you get the rest of the Lns to the +2 OS

Answer 8

1. Careful reduction of tris –trimethylsilylcyclopentadienyl lanthanides with potassium graphite 2. allowed the isolation of the first Ln2+ derivatives as [Cp′3Ln]- anions of almost the entire remaining lanthanide series.

Question 9

Describe properties of [Cp’3Ln]- compounds

Answer 9

1. All very deeply coloured - no like normal Ln
2. [Cp’3La]- ueff= 1.73 BM
3. 1.73 BM spin only magnetic moment- not Lande- from where 1e- where s=1/2
4. Shows not reduction of f-orbitals but donation of electron into empty d-orbital - 4f05d1
5. Us= root(n(n+2)) = root(3)

Question 10

Describe electronic configuration of 2+ OS

Answer 10

1. All [Cp′3Ln]- except for Sm2+, Eu2+, Tm2+ and Yb2+ are rationalised as having a 4fn5d1 ground state.
2. i.e. magnetism shows orbital angular momentum of 5d electron is quenched by trigonal ligand field.
3. In contrast Sm2+, Eu2+, Tm2+ and Yb2+ are rationalised as having a ‘normal’ 4fn+1 configuration.
4. This reflects the variation in the 4f –> 5d promotion energies.

Question 11

Where does the extra electron reside in [Cp’3Ln]- for all elements except Sm, Eu, Yb, Tm

Answer 11

1. For [Cp′3Ln]- with 4fn5d1 ground state, the ‘extra’ electron resides in the 5dz2 orbital
2. which is stabilised by the trigonal ligand field. i.e. a classic crystal field effect.

Question 12

Describe crystal field effect that results in 4f-5d promotion energies

Answer 12

1. When you bring in carefully substituted ligands
2. Drive dz2 so far down in energy that it low enough to put an electron in
3. 4fn 5d1
4. Except for Yb, Eu, Sm, Tm

Question 13

What are the deep colours in [Cp’3Ln]- caused by

Answer 13

1. Arise from allowed 5d to LUMO MLCT
2. Metal ligand charge transfer
3. 5d to ligand transfer of electrons

Question 14

What element is involved in high oxidation state Ln organometallics

Answer 14

1. Ce4+

Question 15

Why is carbon bad at high oxidation state Ln organometallics

Answer 15

1. Not very electronegative
2. Easily oxidised

Question 16

What was first ligand used for high OS Ln organometallic chemistry

Answer 16

1. Cyclooctatetraene (COT)

Question 17

Describe COT

Answer 17

1. Anti aromatic system - 8 Pi electrons
2. one eta8-COT ligand will satisfy two metal valencies.
3. It is also sterically demanding and thus ideally suited to the formation of complexes with lanthanide ions.
4. Not planar

Question 18

How can you make COT stable and planar

Answer 18

1. Requires 2 electrons to achieve its stable 10 electron aromatic form
2. COT2-
3. COT + 2K –> COT2- (K+)2

Question 19

Describe property of COT2-

Answer 19

1. Decent donor- used for Transition metal chemistry
2. Sterically demanding

Question 20

What compounds do Lns form with COT

Answer 20

1. All the lanthanides form trivalent compounds of the type K[Ln(eta8-COT)2]
2. while only Ce forms a neutral sandwich molecule [Ce(eta8-COT)].

Question 21

How do you form K[Ln(eta8-COT)2]

Answer 21

1. React COT2-K+2 with LnCl3 in diglyme - salt metathesis
2. Produces K[Ln(eta8-COT)2] + 3KCl

Question 22

How do you form cerocene

Answer 22

1.[Ce(eta8-COT)]
2. Add AgI to K[Ce(eta8-COT)2] - oxidation
3. Produces cerocene + Ag + KI

Question 23

How do you go from cerocene to K[Ce(eta8-COT)2]

Answer 23

1. Add K/diglyme

Question 24

Describe properties of cerocene

Answer 24

1. The oxidation state of cerocene is apparently 4+.
2. However recent calculations and experiments have indicated that the Ce is best regarded as 3+ (ie [Xe]4f1 complexed by two COT1.5 ligands)- Ligands oxidised not Ce
3. As in all lanthanide organometallics the Ln-C bonding is best regarded as essentially ionic.
4. Contrasts quite markedly with the situation observed in actinide COT derivatives.

Question 25

What true Ce4+ compounds have been made

Answer 25

1. Cp3CeX

Question 26

What are the two ways you can make Cp3CeX

Answer 26

1. Salt metathesis - relies on a range of Ce(IV) Salts
2. Oxidation

Question 27

Describe salt metathesis to produce Cp3CeX

Answer 27

1. Ce(OtBu)2(Cl)2(THF)2 + 2 KCp
2. Produces Cerium bisalkoxide compound

Question 28

Describe oxidation to produce Cp3CeX

Answer 28

1. CeCp3 + 0.5 PhICl2 –> CeCp3Cl + PhI
2. Colour change from yellow to black
3. Very low yield

Question 29

What often happens when trying to produce Cp3CeX via oxidation

Answer 29

1. Often see side reactivity especially with perturbed sterics