Which of the following ions is the strongest oxidizing agent? (Atomic Number of $$Ce = 58,\; Eu = 63,\; Tb = 65,\; Lu =  71$$)

We need to identify the strongest oxidising agent among the given lanthanide/actinide ions.

Key Concept: An oxidising agent gains electrons (gets reduced). The stronger the tendency to gain electrons, the stronger the oxidising agent. For lanthanide ions, unusual oxidation states are strongly driven towards the most stable +3 state, and ions that strongly want to reach a stable electronic configuration make the best oxidising agents.

Analysing each ion:

Option 1: $$Eu^{2+}$$ (Z = 63)

$$Eu^{2+}$$: [Xe] 4f$$^7$$. This has a half-filled 4f shell, which is very stable. $$Eu^{2+}$$ is actually relatively stable and is a reducing agent (it wants to lose an electron to form $$Eu^{3+}$$ [Xe]4f$$^6$$, but the half-filled stability resists this). It is not a strong oxidising agent.

Option 2: $$Tb^{4+}$$ (Z = 65)

$$Tb^{4+}$$: [Xe] 4f$$^7$$. Terbium in the +4 state also has a half-filled 4f$$^7$$ configuration. However, $$Tb^{4+}$$ strongly tends to gain an electron to become $$Tb^{3+}$$ [Xe]4f$$^8$$, because the +3 state is the most common and stable oxidation state for lanthanides. The high charge (+4) combined with the strong drive to reach +3 makes $$Tb^{4+}$$ a very strong oxidising agent.

Option 3: $$Lu^{3+}$$ (Z = 71)

$$Lu^{3+}$$: [Xe] 4f$$^{14}$$. This has a completely filled 4f shell and is already in the most stable +3 oxidation state. It has no tendency to gain or lose electrons easily. It is not an oxidising agent.

Option 4: $$Ce^{3+}$$ (Z = 58)

$$Ce^{3+}$$: [Xe] 4f$$^1$$. Cerium in the +3 state has one 4f electron. While $$Ce^{4+}$$ (4f$$^0$$, noble gas configuration) is a good oxidising agent, $$Ce^{3+}$$ itself is stable and not an oxidising agent.

Conclusion: $$Tb^{4+}$$ is the strongest oxidising agent because it has the strongest tendency to gain an electron to achieve the stable +3 state.

The correct answer is Option 2: $$Tb^{4+}$$.