The Eu$$^{2+}$$ ion is a strong reducing agent in spite of its ground state electronic configuration (outermost):
[Atomic number of Eu = 63]

For europium the atomic number is given as $$Z = 63$$, which means a neutral Eu atom possesses 63 electrons.

First we recall the Aufbau order of filling: $$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p \dots$$

Up to xenon we have 54 electrons, so we may write the xenon core as $$[\,\text{Xe}\,]$$. The remaining electrons for Eu are $$63-54 = 9$$.

According to the order shown above, after $$[\,\text{Xe}\,]$$ the subshells are filled in the sequence $$6s \rightarrow 4f \rightarrow 5d$$. Thus

$$[\,\text{Xe}\,]\,6s^2$$ accounts for 2 electrons, leaving $$9-2=7$$ electrons still to be placed.

Those 7 electrons enter the $$4f$$ subshell:

$$[\,\text{Xe}\,]\,6s^2\,4f^7$$

(The next subshell, $$5d$$, remains empty for Eu.) So the outer-shell electronic configuration of the neutral atom is

$$4f^7\,6s^2$$

Now we pass to the divalent ion $$\text{Eu}^{2+}$$. For any metal ion the electrons are removed first from the subshell with the highest principal quantum number $$n$$. Here the highest $$n$$ is $$6$$ for the $$6s$$ electrons. Removing two electrons from $$6s^2$$ we obtain

$$\text{Eu}^{2+}: \; [\,\text{Xe}\,]\,4f^7$$

Hence the outermost (valence-shell) configuration of the $$\text{Eu}^{2+}$$ ion is simply

$$4f^7$$

This corresponds to a half-filled $$f$$ subshell, which is specially stable; nonetheless, $$\text{Eu}^{2+}$$ is a strong reducing agent because it can still be oxidised to the even more common $$\text{Eu}^{3+}$$ state in aqueous solution.

Comparing with the alternatives, only option B matches the configuration $$4f^7$$.

Hence, the correct answer is Option B.