Chemistry of the Lanthanides Flashcards
Describe the radial distribution functions of the lanthanide elements.
> 4f orbitals penetrate closer to the nucleus than 5d or 6s, therefore, 4f electrons are held more tightly.
6s/5d electrons are lost first on ionisation before any 4f.
Why is the chemistry of the Lanthanides dominated by the +3 oxidation state?
> As positive charge increases (upon ionisation), 4f orbitals are stabilised by higher Zeff so much that beyond 3+, ionisation is very difficult. +3 oxidation state therefore is most stable/common oxidation state.
What exceptions are there to this? (+3 ox state)
> Eu2+ : [Xe}4f7 and Yb2+ : [Xe}4f14 - Stability of half filled or filled shell makes 3rd ionisation energy high in these cases so 2+ state is important.
> Ce4+ : [Xe]4f0 and Tb4+: [Xe]4f7 - the ionisation energy is low in these cases as 3+ ions have 1 electron surplus to empty or half-filled shell. So, 4+ state is important.
> i.e. 2+/4+ state is accessible if 4f0/4f7/4f14 configuration results.
What is the lanthanide contraction?
> Steadily increasing nuclear charge along the series is not compensated for by extra electron (poor shielding) giving higher Zeff - leads to reduction in ionic/atomic radii.
> i.e. The lanthanide contraction cancels out expected increase in atomic/ionic radii.
What are the exceptions to the lanthanide contraction? (anomalies)
> Eu/Yb have slightly larger atomic nuclei than expected as they only donate 2 electrons to the conduction band (stable 2+ ions).
Coordination chemistry of the Lanthanides(III): What important differences are there compared to the d block?
> No significant overlap between highly contracted 4f orbitals on metals with ligand orbitals - bonding is essentially ionic.
> High charge (+3) and lack of covalent boning means that Ln 3+ ions prefer ‘hard’ ligands (O,N - donors mostly, especially with negative charges).
> No specific directional nature to bonding; ligands pack around metal cation so as to minimise steric/electrostatic repulsions.
> No CFSE effects: 4f orbitals and ligand orbitals barely interact, therefore, no electronic preference for particular geometries (4f cant split). Also means no ‘kinetic inertness’ - kinetically labile complexes with fast ligand substitution reactions.
> Ions are large (r= 1Å) so coordination numbers are generally higher than for d block (around 8-10) because more ligands can pack around the metal.
All Ln3+ cations behave similarly with small differences due to steady decrease in ionic radius,. what does this cause?
(i) Increased bond strengths due to higher charge density on cation.
(ii) Reduced coordination numbers with smaller/heavier ions (huge contrast with d block)
Why are aqua complexes only stable in acidic solution (pH <6)?
> Hydrolysis occurs to give insoluble hydroxides.(Ln(OH)3 ppt).
Why are water stable Ln complexes difficult to achieve and how is it achieved?
> Ln3+ - OH2 interaction is sufficiently strong that other monodentate ligands (amines, halides) do not bind strongly in water.
> To get water stable complexes requires multidentate chelating ligands (High positive ∆H and high negative ∆S).
> Geometries = whatever results from minimising ligand/ligand repulsions- no specific requirements from the metal.
How can low coordination complexes be stabilised?
> Low coordination numbers can be stabilised with very bulky ligands.
What complications are there with the separation and purification of lanthanides?
(i) ores contain all lanthanides together: main ones are Ln(PO4) (monazite) and LnF(CO3) (basthaesite).
(ii) separation of lanthanide (+3) ions from a mixture is difficult due to similar sizes, changes and coordination behaviour of ions (indiscriminate mixture).
How are lanthanides separated and purified?
> First get crude mixture of lanthanide salts in solution:
(i) dissolve ore in hot conc. NaOH -> slurry of hydrated oxides Ln2O3.xH2O (sodium carbonate and sodium phosphate remain in aqueous solution). Loading: extraction of aqueous solution of Ln3+ salts with
HL in kerosene. Smaller, heavier ions bind more strongly to L- so are preferentially extracted.
Organic Phase = slightly enriched with heavier La
Aqueous phase = slightly enriched with lighter La
(ii) dissolve oxides in boiling HCl (strong acid with pH 3.5) -> solution of mixed [Ln(H20)n]3+ as chloride salts. Stripping: La ions ‘stripped’ from organic phase using aqueous acid which reverses above process (ligand is protonated and drops off). Smaller, heavier ions need stronger acid to separate them from L- and get them back into the aqueous phase. Biggest La ion= most weakly bound to ligand due to lower charge density (acid doesn’t have to be as strong to extract).
> The steps are combined in continuous flow system to give complete separation of Ln3+ ions. Pure Ln 3+ ions precipitate as hydroxides and then calcined (oxidised) to give pure Ln2O3.
> Separation method depends on scale: Large (industrial) scale: solvent extraction. Small (lab) scale: ion-exchange chromatography. In both cases, separation is based on small, steady increase in stability constants across series as ionic radius decreases.
Outline the process of ion exchange chromatography
> Ion exchange chromatography uses a resin based on polystyrene beads (column support) with pendant anionic sulfonate groups.
> Ln3+ ions have high affinity for sulfonates (they stick to the column) and displace Na+ ions.
Ln3+ (aq) + 3Na+ (resin) ↔ Ln3+ (resin) + 3Na+ (aq)
> Elute using solution of good ligand such as (edta)4-. Ln3+ is partitioned between sulfonate (stationary phase) and (edta)4- (mobile phase).
Ln3+ (resin) + (edta)4- (aq) + 3Na+ (aq) ↔ [Ln(edta)]- (aq) + 3Na+ (resin)
> Smaller, heavier 3+ ions have higher affinity for (edta)4- and elute first (logK is higher).
> Separation occurs on the basis of size/charge density of cation.
d-d transitions for TM ions arise principally from d-orbital splittings due to CFSE effects, does this occur for Ln3+ ions? Why?
> NO. This does not occur for Ln3+ ions as f orbitals are ‘buried’ close to the nucleus and are barely affected by ligands, therefore, the 7 4f orbitals remain degenerate.
> Instead, different arrangements of electrons within 4f orbitals have different energies because of different values of: S,L and J.
How do you arrange electrons in the ground state in f orbitals? Why?
> Electrons are added to the f orbital set (7 boxes) using Hund’s rules.
(i) start at LHS and
(ii) fill orbitals singly with spins aligned as far as possible..
… This gives the maximum number of unpaired spins as 1st priority (HUND 1) and maximum values of orbital angular momentum (L) as 2nd priority (HUND 2) - so immediately we have the ground state.
How do you determine the order of the ground state levels?
> Order of the ground state levels depends on whether the f shell is more or less half full.
> More than half full (≥f8) : highest J is lowest in energy.
> Less than half full (≤f8): lowest J is lowest in energy.
> Note: components of ground state term are not equally spaced. Spacing is proportional to J so spacing decreases at higher energies as J value decreases.
What are the 3 main difference between f-f absorption peaks and d-d absorption peaks?
(i) They are very low intensity (1-10 dm3mol-1cm-1). Both d-d and f-f transitions are laporte forbidden (no change in parity) - ‘g’-‘g’ vs ‘u’-‘u’.
Note: laporte rule= transitions must involve a change in parity (g ↔ u).
For d-d transitions, effects of low symmetry or molecular vibrations (which lower symmetry) relax the laporte rule and result in d/p mixing so ‘forbidden’ d-d transitions gain some d-p character which is parity allowed.
This does not happen with f-orbitals as they are so ‘buried’ that they do not feel changes in geometry or molecular vibrations, and remain pure f-orbitals, so f-f transition remain very weak and La(III) slats have very pale colours (or are colourless).
(ii) f-f transitions are very narrow (sharp) compared to broad d-d transitions. d-d transitions are broad because vibrations of ligands affect d-orbital energies (∆oct oscillates). Promotion of an electron is fast compared to molecular vibrations (Franck-Condon principle) so energy of transition depends when it occurs during the vibrational cycle and absorption peak is broad.
F orbital energies are unaffected by ligand vibrations so f-f absorptions are fixed in energy and peaks are very narrow.
(iii) Spin-forbidden transitions occur easily.
Note: spin rule = ∆S=0
High spin orbit coupling for heavy atoms means that pure ‘spin’ quantum number is no longer valid (S and L mixed up in J) so spin selection rule is circumvented.
Luminescence: In TM ions, d-d excited states usually lose their energy to molecular vibrations so light absorbed is converted to heat. Does this happen for Ln(III)?
> NO. Weak interactions between f orbitals and vibrating ligands mean that the likelihood of f-f excited state energy being converted to molecular vibrations is low so energy of excited state emitted as a photon: LUMINESCENCE.
Outline the process of luminescence in Ln(III) with some examples.
> All emission originates from single excited state, down to difference components of ground state ‘ladder’ giving multiple closely-spaced emission lines.
> Luminescence is particularly strong when there is a big energy gap between the excited state and the ground state manifold because the energy gap is too large for any one vibration to take the energy.
> Eu3+ (red) and Tb3+ (green) ions show strong luminescence in the vis region of the spectrum. (basis for a large number of useful applications e.g. lighting, imaging).
If absorption of light by Ln3+ complexes is so weak, how do they get into this excited state in the first place?
> Direct excitation would require intense light source(e.g. laser) perfectly matched in energy to that of the narrow absorption bands.
> Sensitisation of La luminescence occurs via strongly absorbing aromatic ligand (fully allowed π→ π* of aromatic ligand by absorption of UV) which transfers its energy to Ln3+ ion to generate the luminescent f-f excited state.
Outline the process by which a Ln(III) aromatic complex is luminescent.
(i) Absorption of UV light by the aromatic ligand occurs (fully allowed π→ π* transition) to give singet excited state of ligand (S0 → S1).
(ii) ‘Intersystem crossing’ (fast) occurs to give the ligand triplet excited state (T1): spin forbidden but possible due to spin orbit coupling of heavy metal ion.
(iii) Energy transfer from ligand to f-f excited state of La (fully allowed) - CRUCIAL STEP - T1 must be higher in energy than luminescent f-f level of Ln3+ ion for this transfer of energy to be possible.
note: NOT REDOX (no transfer of electrons).
(iv) Emission from La excited state to various components of ground state.
