Lanthanides, Actinides and Organometallic Chemistry

18-electron rule and metal–metal bond order

The 18-electron rule states that many stable organometallic complexes, especially those of the d-block, have a valence-electron count of 18. This matches the nine valence orbitals (five d, one s, three p) available for bonding. To count electrons, start from the metal's group number, add the electrons donated by each ligand, and correct for the overall charge by subtracting one electron per unit of positive charge on the metal and adding one per unit of negative charge. Typical donations are 2 electrons for CO, phosphines and amines, 5 for a neutral \eta^5-cyclopentadienyl ring (6 when counted as an anion), 3 for an \eta^3-allyl and 2 for an \eta^2-alkene. A bridging carbonyl is shared between metals, so each metal it bridges receives one electron from it in the covalent count.

A metal–metal bond contributes one electron to each metal, so every M–M bond adds one electron to the count of both partners. If a metal is short of 18 electrons, the deficit tells you how many metal–metal bonds it must form, and the bond order between two identical metals equals the number of such bonds. For example, a \mathrm{CpMo(CO)_2} fragment has 6 + 5 + 2\times 2 = 15 electrons, so each Mo needs three more electrons. That means three Mo–Mo bonds and a bond order of 3.

Dinuclear cyclopentadienyl carbonyls of iron are handled the same way. The total electron count is distributed across the metal centres, and the Fe–Fe bond count is chosen to bring each metal to 18. Bridged dimers that stay at 15 or 17 electrons per metal do not satisfy the rule without additional bonding, so they are not 18-electron species.

Magnetic moments of lanthanide ions

Lanthanide(III) ions have partly filled 4f shells that are well shielded by the filled 5s and 5p subshells. Their magnetism therefore follows the Russell–Saunders scheme with spin–orbit coupling, and the orbital contribution is not quenched as it usually is for first-row d-block ions. The ground term ^{2S+1}L_J is found from Hund's rules: maximise S, then maximise L, then take J = |L-S| for a shell that is at most half filled and J = L+S for a shell more than half filled. The effective moment is \mu_{\text{eff}} = g_J\sqrt{J(J+1)} BM, where the Landé factor is g_J = 1 + \dfrac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)}. The spin-only formula \sqrt{4S(S+1)} underestimates lanthanide moments because it ignores the orbital contribution.

Example: for Yb(III), 4f^{13} gives L = 3, S = \tfrac12 and J = \tfrac72. Then g_J = \tfrac87 and \mu_{\text{eff}} = \tfrac87\sqrt{\tfrac{63}{4}} \approx 4.54 BM.

The same procedure applies to a complex anion. The nitrate ion is uninegative, so in [\mathrm{Ce(NO_3)_5}]^{2-} the five nitrates contribute -5, which requires Ce to be in the +3 state. The ion is then treated as a free Ce(III) ion with the ground term from its 4f^1 configuration.

Radioactive decay modes and nuclear reactions

Nuclear decay and nuclear reactions both conserve mass number A and atomic number Z, so the decay mode is identified from the change in (A, Z). An α-particle (^4_2He) lowers A by 4 and Z by 2. A β⁻ emission leaves A unchanged and raises Z by 1. β⁺ emission and electron capture lower Z by 1 with A unchanged. Gamma emission changes neither A nor Z. Neutron emission lowers A by 1 and keeps Z, while proton emission lowers both A and Z by 1. Products formed by neutron or proton emission from a nucleus that is already neutron-rich or proton-rich are usually radioactive.

In a compound nucleus formed by capturing a light particle, add the particle's A and Z to the target, then subtract the values for each emitted particle. Each step must balance.

Worked method: for ^{232}_{90}Th decaying to ^{208}_{82}Pb, the number of α-particles is (232-208)/4 = 6. The charge balance 90 - 2(6) + n_\beta = 82 gives n_\beta = 4, so the total number of α and β particles emitted is 10. This is the standard way to balance a decay series from the mass and charge changes.

Actinide properties and oxidation states

The actinides are the elements from actinium to lawrencium, in which the 5f subshell is filled. Compared with the 4f orbitals of the lanthanides, the 5f orbitals are more radially extended. The 5f electrons therefore experience a weaker effective attraction from the nucleus and are bound less tightly, which lets them take part in bonding and gives the actinides a wider range of oxidation states than the lanthanides.

All actinides are radioactive. None has a stable nuclide, although thorium and uranium have half-lives long enough to occur in nature. Elements beyond uranium, with Z > 92, do not occur in significant natural amounts and are prepared synthetically, usually by neutron capture and β-decay in reactors or by nuclear bombardment.

Actinide contraction, the steady decrease in ionic radii across the series, does occur. Poor shielding of the 5f electrons by one another raises the effective nuclear charge as Z increases. The series therefore differs from the lanthanides in having greater variability of oxidation states and in showing actinide contraction as a real trend.

Nickelocene reduction

Nickelocene, (\eta^5\text{-C}_5\text{H}_5)_2\mathrm{Ni}, is a 20-electron sandwich complex in which both cyclopentadienyl rings are bonded in the \eta^5 mode. Reducing it with sodium amalgam in an alcohol solvent adds two electrons to one ring. The reduced ring is then protonated by the solvent, and the protonation converts the aromatic \eta^5-C_5H_5 into the cyclopentenyl ligand \eta^3-C_5H_7.

The key point is the stoichiometry. A two-electron reduction and two protons add one \mathrm{C_5H_7} ligand, so only one ring is converted and the other ring keeps its \eta^5-bonding. The product (\eta^5\text{-C}_5\text{H}_5)(\eta^3\text{-C}_5\text{H}_7)\mathrm{Ni} is an 18-electron species, since Ni contributes 10, the \eta^5-ring 5 and the \eta^3-cyclopentenyl 3. A product with two \mathrm{C_5H_7} rings would need four electrons and four protons, so it is not the major product of this step.

Lanthanide electron configurations

Lanthanides have the general configuration [\mathrm{Xe}]4f^n 5d^{0\text{–}1} 6s^2 in the neutral atom. Most lanthanides form the +3 ion, which has [\mathrm{Xe}]4f^n with n equal to the atomic number minus 57 for the neutral metal's trivalent ion. The 4f electrons are shielded and do not change much with the chemical environment. To find the configuration of an ion in a compound, first fix the metal's oxidation state from the charge of the ligands and the overall charge, then subtract electrons from the neutral configuration.