Find the electronic configuration of europium, $Z=63$

Neutral europium has configuration: $[Xe]4f^7 6s^2$

Write the configuration of $Eu^{2+}$

On forming $Eu^{2+}$, the two outermost $6s$ electrons are removed first: $Eu^{2+} : [Xe]4f^7$

So the outermost configuration of $Eu^{2+}$ is: $4f^7$

Why is $Eu^{2+}$ a strong reducing agent?

A reducing agent gets oxidized easily. Thus $Eu^{2+}$ tends to lose one electron to form $Eu^{3+}$: $Eu^{2+} \rightarrow Eu^{3+} + e^-$

The given statement says it is a strong reducing agent in spite of its ground-state configuration. This means although $Eu^{2+}$ itself has the especially stable half-filled configuration $4f^7$, it still behaves as a reducing agent.

Match with the options

• A: $4f^7 6s^2$ — configuration of neutral Eu, not $Eu^{2+}$
• B: $4f^6$ — not $Eu^{2+}$
• C: $4f^7$ — correct
• D: $4f^6 6s^2$ — incorrect

Conclusion

The outermost electronic configuration of $Eu^{2+}$ is: $\boxed{4f^7}$

Hence the correct option is C.