Lanthanoids: Electronic Configuration and Oxidation States

The fourteen lanthanoids are the elements cerium (Ce, $Z=58$) through lutetium (Lu, $Z=71$), which follow lanthanum (La, $Z=57$) in the periodic table. Across this series, electrons are progressively added to the 4f subshell, giving the general outer configuration

Irregular filling. Because the 4f and 5d orbitals are extremely close in energy across this part of the periodic table, the filling pattern is not perfectly regular: most lanthanoids have the configuration $[\text{Xe}]4f^n6s^2$ with no 5d electron, but a small number of members -- notably cerium (often written $[\text{Xe}]4f^15d^16s^2$ rather than the "expected" $[\text{Xe}]4f^26s^2$, reflecting the closeness in energy of promoting an electron between the two subshells), gadolinium ($[\text{Xe}]4f^75d^16s^2$, gaining the extra stability of a half-filled 4f subshell by keeping one electron in 5d), and lutetium ($[\text{Xe}]4f^{14}5d^16s^2$, with the 4f subshell already completely filled) -- retain a single 5d electron rather than following the simple $4f^n6s^2$ pattern. This is directly analogous to the chromium/copper exceptions seen earlier in the first-row transition series, driven by the same underlying preference for a half-filled or completely filled subshell.

Oxidation states. The overwhelmingly dominant oxidation state across the whole lanthanoid series is $\mathbf{+3}$, formed by the loss of the two outer $6s$ electrons together with either the single $5d$ electron (for La, Ce, Gd, Lu) or one $4f$ electron (for the remaining members). This uniformity in oxidation state -- in sharp contrast to the wide spread of oxidation states seen across the 3d transition series -- is because the 4f orbitals, lying deep within the atom (shielded by the outer $5s$ and $5p$ electrons), are essentially uninvolved in bonding for most lanthanoids; only the outer $5d^{0-1}6s^2$ electrons are readily lost.

A small number of lanthanoids also show $+2$ or $+4$ states, in each case because the resulting ion happens to achieve an especially stable 4f configuration: cerium readily forms $\text{Ce}^{4+}$ (losing one 4f electron in addition to the usual three, to reach the empty, noble-gas-like $4f^0$ configuration); europium forms $\text{Eu}^{2+}$ (retaining only $4f^7$, a half-filled and hence extra-stable subshell, rather than losing a third electron); and ytterbium forms $\text{Yb}^{2+}$ (retaining $4f^{14}$, a completely filled subshell). These exceptions are developed further, with the underlying reasoning, in a dedicated exercise later in this chapter. …