To solve this problem, we need to determine which pair of lanthanoid ions is diamagnetic.Diamagnetism occurs when all the electrons in an ion are paired.$\newline$Let's analyze each option:$\newline$1. $Gd^{3+}$ and $Eu^{3+}: \newline$• $Gd^{3+}$ has 7 unpaired electrons $(4f^7 configuration)$, making it paramagnetic.$\newline$• $Eu^{3+}$ has 6 unpaired electrons $(4f^6 configuration)$, making it paramagnetic.$\newline$2. $Pm^{3+}$ and $Sm^{3+}: \newline$• $Pm^{3+}$ has 4 unpaired electrons $(4f^4 configuration)$, making it paramagnetic.$\newline$• $Sm^{3+}$ has 5 unpaired electrons $(4f^5 configuration)$, making it paramagnetic.$\newline$3. $Ce^{4+}$ and $Yb^{2+}: \newline$• $Ce^{4+}$ has no unpaired electrons $(4f^0 configuration)$, making it diamagnetic.$\newline$• $Yb^{2+}$ has no unpaired electrons $(4f^{14} configuration)$, making it diamagnetic.$\newline$4. $Ce^{3+}$ and $Eu^{2+}: \newline$• $Ce^{3+}$ has 1 unpaired electron $(4f^1 configuration)$, making it paramagnetic.$\newline$• $Ce^{3+}$ has 1 unpaired electron $(4f^1 configuration)$, making it paramagnetic.$\newline$• $Eu^{2+}$ has 7 unpaired electrons $(4f^7 configuration)$, making it paramagnetic.$\newline$Therefore, the correct pair of diamagnetic ions is $Ce^{4+}$ and $Yb^{2+}. \newline$This corresponds to Option 3.