Question 6

Diamagnetic species have no unpaired electrons. We check the electronic configurations of lanthanide ions: (A) $La$ ($Z=57$) is $[Xe] 5d^1 6s^2$. $La^{3+}$ is $[Xe] 4f^0$, which is diamagnetic. $Ce$ ($Z=58$) is $[Xe] 4f^1 5d^1 6s^2$. $Ce^{4+}$ is $[Xe] 4f^0$, which is diamagnetic. Both are diamagnetic. (B) $Yb$ ($Z=70$) is $[Xe] 4f^{14} 6s^2$. $Yb^{2+}$ is $[Xe] 4f^{14}$, which is diamagnetic (fully filled f-shell). $Lu$ ($Z=71$) is $[Xe] 4f^{14} 5d^1 6s^2$. $Lu^{3+}$ is $[Xe] 4f^{14}$, which is diamagnetic. Both are diamagnetic. (C) $La^{2+}$ is $[Xe] 5d^1$ (1 unpaired electron). $Ce^{3+}$ is $[Xe] 4f^1$ (1 unpaired electron). Both are paramagnetic. (D) $Yb^{3+}$ is $[Xe] 4f^{13}$ (1 unpaired electron). $Lu^{2+}$ is $[Xe] 4f^{14} 5d^1$ (1 unpaired electron). Both are paramagnetic. Correct pairs are (A) and (B).