Which of the following acts as a strong reducing agent? (Atomic number: Ce = 58, Eu = 63, Gd = 64, Lu = 71)

We need to identify which species acts as a strong reducing agent among the given lanthanide ions.

Analysing each option:

Lu$$^{3+}$$ (Z = 71): Lu has the configuration $$[Xe]\, 4f^{14}\, 5d^1\, 6s^2$$. Lu$$^{3+}$$ has a completely filled 4f shell ($$4f^{14}$$), which is very stable. It has no tendency to be further oxidized or reduced. It is not a reducing agent.

Gd$$^{3+}$$ (Z = 64): Gd has the configuration $$[Xe]\, 4f^7\, 5d^1\, 6s^2$$. Gd$$^{3+}$$ has a half-filled 4f shell ($$4f^7$$), which is very stable. It has no tendency to lose or gain electrons easily. It is not a strong reducing agent.

Eu$$^{2+}$$ (Z = 63): Eu has the configuration $$[Xe]\, 4f^7\, 6s^2$$. Eu$$^{2+}$$ has the configuration $$[Xe]\, 4f^7$$, which is a half-filled 4f shell. However, the common and stable oxidation state of lanthanides is $$+3$$. Eu$$^{2+}$$ readily loses one more electron to become Eu$$^{3+}$$ ($$4f^6$$), acting as a strong reducing agent.

Ce$$^{4+}$$ (Z = 58): Ce has the configuration $$[Xe]\, 4f^1\, 5d^1\, 6s^2$$. Ce$$^{4+}$$ has the noble gas configuration $$[Xe]$$, which is stable. Ce$$^{4+}$$ tends to gain an electron to become Ce$$^{3+}$$, making it an oxidizing agent, not a reducing agent.

Eu$$^{2+}$$ is a strong reducing agent because it is readily oxidized to the more stable Eu$$^{3+}$$ state.

The correct answer is Option 3: Eu$$^{2+}$$.