f-block elements Flashcards
What is the abundance trend of the lanthanides?
Every other element has a higher abundance, followed by an element with a lower abundance. The abundance of all elements decreases as the atomic number increases.
Why are lanthanides so difficult to seperate and purify? How are they seperated?
They all stay in the +3 oxidation state. Mixed solutions are of the +3 states are slowly seperated in ion exchange columns.
Briefly describe the stages in the purification of uranium.
The ore is oxidised to UO22+(aq), which is complexed with NH3 to give yellow cake.
Yellow cake is heated to give UO3, which is reduced to UO2. This is reacted with HF to give UF4 which gives U and then UF6.
UF6 can have the isomers seperated to give 235U.
Compare the f and d-orbitals.
F-orbitals are similar to d-orbitals but are more lumpy as they have more angular nodes.
The f-orbitals are more shielded by other electrons but they oxidise quickly, especially the 4f orbitals.
Describe the lanthanide contraction and the relativistic effect.
Progressing from left to right along the row, the ionic radius decreases like the other rows in the table.
This size reduction is mostly due to increasing Zeff like the other rows, however around 30% is due to the relativistic effect.
As the electrons approach the nucleus, their speed increase to approximately 0.7c. Because of this the mass increases to 1.35me, meaning the electrons spend more time near the nucleus (radius has a 1/mass dependancy). This affects the s electrons most strongly, which defines the ionic radius.
Describe the f-orbitals and why lanthanides have weak interactions with ligands.
The f-orbitals are all ungerade symmetry and have a high angular nodality. 4f orbitals are very core-like as the electrons spend more time near the nucleus as they have no nodes in the radial distribution function.
They also are hidden below the 6s and 5d, p and s orbitals. This makes them very core like and poor at interacting with orbitals.
Using lanthanide contraction, explain why the actinides can interact with ligands and lanthanides cannot.
The radius of the ions decreases as the electrons increase in speed (1/mass dependance), contracting the s orbitals - the lanthanide contraction.
This greater screening of the nucleus causes expansion of the f-orbitals. For the lanthanides, 4f only expands a small amount meaning ligand interactions are ionic. For the actinides, the expansion is large, meaning they can interact covalently with ligands.
Describe any unsual features of neutral lanthanides and actinides, as well as the nature of the orbitals as the atoms are ionised.
Like the d-orbitals, the half-filled and full orbitals are favoured. However in this case, one electron is added to the 5/6d orbitals, rather than pairing into the f-orbitals or when the f orbitals are filled (behaviour shown by Gd and Lu).
The f orbitals are the highest energy before ionisation, however as the atom is ionised, it drops rapidly in energy as fewer nodes in an orbital means faster energy decrease on ionisation. At +3 atomic charge, the f orbitals are lower energy than the d and s orbitals filled above it.
Why does the +3 ox. state dominate lanthanide chemistry?
The 4f electrons drop so rapidally in energy that once the +3 state is reached, the atom cannot be ionised any more. (One electron from 4f, two electrons from 6s)
What is the trend in metallic radii of lanthanides and actinides?
Lanthanides generally decrease across the row, however Eu and Yb form +2 states, not +3 as they are half and fully filled so they have higher radii.
The actinides have covalent character so there is no obvious trend. At first the radii is smaller and decreases to U and Pu but then increases above the lanthanides.
What determines if the +2 or +3 oxidation states of the lanthanides is taken? How is this related to other thermodynamic values?
Eu and Yb have higher I3 values than the other lanthanides due to their half and fully filled 4f orbitals which have unusual stability, making the +2 ox. state more stable. This trend is also seen in the atomisation enthalpies.
Generally describe the stability of the actinides.
A much wider range of oxidation states are avaliable, at least 2 per element, as the 5f orbital is spacially extended.
The most common oxidation states are +3 and +4 early in the row and peaking at +7 for Pu. At Cf, the +2 and +3 states are the only states that can be taken.
Describe what controls the energy levels of the lanthanides.
Electron-electron repulsion is the largest term followed by the spin-orbit coupling term. The ligand field interaction is very weak for the lanthanides.
How can the spin-orbit coupling of the lanthanides be estimated?
How is this used to work out the term symbol?
Using Russell-Saunders coupling (summing s1+s2+s3…= S and the same for L) is a good estimate for the experimental coupling.
Therefore to calculate the term symbol, find S and L, then combine them to give 2S+1LJ where J = L+S, L+S-1,…|L-S|.
Describe why the lanthanides are inherently massive magnets compared to the transition metals.
The ligand-field energy is very significant for the transition metals, meaning electrons are tied up in ligand bonding and cannot circulate. This quenches the orbital angular momentum meaning the magnetism comes only from electron spin.
In lanthanides, the ligand interactions are ionic meaning the orbital angular momentum is not quenched. This means the spin-orbit couping of the electrons is significant, leading to massive magnetism.
What equation determines the effective magnetic moment of the lanthanides? Why can’t this be applied to the actinides?
The Landé formula.
