JEE Main · 2025 · 24 Jan 2025 · Shift 1 · d and f Block Elements

Among lanthanide ions, the +3 state is the thermodynamically preferred, characteristic oxidation state for the series; ions trapped in unusual +2 or +4 states tend to revert to +3, and this tendency is what makes them good reducing or oxidising agents respectively.

$L{u}^{3+}$ ([Xe]$4f^{14}$) is already in the stable, characteristic +3 state with a completely filled f-subshell — it has no drive to change oxidation state and is a poor oxidiser.

$E{u}^{2+}$ ([Xe]$4f^7$) sits in an unstable +2 state relative to the preferred +3, so it readily loses an electron to become $E{u}^{3+}$ — it acts as a reducing agent, not an oxidiser.

$C{e}^{3+}$ is already in the normal, stable lanthanide oxidation state and shows no special oxidising or reducing tendency.

$T{b}^{4+}$ ([Xe]$4f^7$), by contrast, sits in an unusually high +4 state; since +3 is strongly preferred for lanthanides, $T{b}^{4+}$ readily accepts an electron to fall back to $T{b}^{3+}$, making it a strong oxidising agent.

$T{b}^{4+}$ is the strongest oxidising agent among the given ions.