 Solution

The unusually stable divalent configurations Eu(II) $4f^7$ and Yb(II) $4f^{14}$ make their third ionization particularly costly. Forming Eu(III), $4f^6$, or Yb(III), $4f^{13}$ removes an electron from a half-filled or filled f shell. The positive ionization-energy contribution therefore increases relative to the smoother background across the series.

Their trivalent ionic radii and hydration enthalpies do not supply a matching abrupt extra stabilization. Differences in atomization and the divalent character of elemental Eu and Yb partly compensate the ionization anomaly but do not erase the net less favourable trivalent aqueous formation enthalpy. This explanation concerns formation of the $+3$ aqueous ions, not a claim that $\mathrm{E}{\mathrm{u}}^{3+}$ or $\mathrm{Y}{\mathrm{b}}^{3+}$ is intrinsically the most stable isolated f configuration.