Step 1: Understanding the Concept:
In the Lanthanide series, the most stable and common oxidation state is $+3$.
An ion that exists in an oxidation state different from $+3$ will attempt to return to $+3$ by either gaining or losing electrons.
Oxidizing agents are species that undergo reduction (gain electrons). Therefore, ions in an oxidation state higher than $+3$ (like $+4$) will act as oxidizing agents.
Step 2: Detailed Explanation:
Cerium ($Ce^{4+}$):
Configuration: $[Xe] 4f^0$.
While $Ce^{4+}$ has a noble gas configuration, it is a powerful oxidizing agent in aqueous solution. Its reduction potential $E^\circ(Ce^{4+}/Ce^{3+})$ is +1.74 V. It gains one electron to reach the stable $+3$ state.
Terbium ($Tb^{4+}$):
Configuration: $[Xe] 4f^7$.
$Tb^{4+}$ has a half-filled $f$-subshell, which provides extra stability. However, the $+3$ state is still the most stable in general chemical environments for lanthanides. $Tb^{4+}$ acts as an oxidizing agent to reach $Tb^{3+}$.
Reducing Agents (Others):
$Eu^{2+}$ ($4f^7$) and $Yb^{2+}$ ($4f^{14}$) are in a state lower than $+3$. They lose one electron to become $+3$, acting as reducing agents.
$La^{3+}$ ($4f^0$) and $Lu^{3+}$ ($4f^{14}$) are already in their most stable $+3$ state and do not typically act as strong oxidizers or reducers.
Step 3: Final Answer:
The pair $Ce^{4+}$ and $Tb^{4+}$ consist of ions in the $+4$ state that act as oxidizing agents.