Diamagnetic Lanthanoid ions are:

We need to identify which pair of lanthanoid ions are diamagnetic. A species is diamagnetic if it has no unpaired electrons, i.e., all electrons are paired. For lanthanoid ions, this means the 4f subshell must be either completely empty ($$4f^0$$) or completely filled ($$4f^{14}$$).

Lanthanoids have the general configuration $$[Xe] 4f^n$$. When they form $$3+$$ or $$4+$$ ions, electrons are removed first from the 6s and 5d shells, and then from the 4f subshell.

For $$La^{3+}$$, lanthanum (Z = 57) has the configuration $$[Xe] 5d^1 6s^2$$. Removing three electrons yields $$[Xe] = 4f^0$$, which has no unpaired electrons and is therefore diamagnetic.

Cerium (Z = 58) in $$Ce^{4+}$$ has the configuration $$[Xe] 4f^1 5d^1 6s^2$$. Removing four electrons gives $$[Xe] = 4f^0$$, again with no unpaired electrons, so this ion is also diamagnetic.

Neodymium (Z = 60) in $$Nd^{3+}$$ starts from $$[Xe] 4f^4 6s^2$$. After removing three electrons, the configuration becomes $$[Xe] 4f^3$$, which has three unpaired electrons and is paramagnetic. Europium (Z = 63) in $$Eu^{3+}$$ has $$[Xe] 4f^7 6s^2$$ to start, and removing three electrons gives $$[Xe] 4f^6$$. By Hund’s rule, the six electrons occupy six of the seven 4f orbitals singly, resulting in six unpaired electrons and a paramagnetic species.

Lutetium (Z = 71) in $$Lu^{3+}$$ is described by $$[Xe] 4f^{14} 5d^1 6s^2$$ in the neutral atom. Removing three electrons leaves $$[Xe] 4f^{14}$$, which is completely filled and therefore diamagnetic.

Only $$La^{3+}$$, $$Ce^{4+}$$, and $$Lu^{3+}$$ are diamagnetic, but the pair consisting of $$La^{3+}$$ and $$Ce^{4+}$$ (both $$4f^0$$) matches one of the given options.

Answer: Option 2 — $$La^{3+}$$ and $$Ce^{4+}$$