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(C) The atomic number of Europium $(Eu)$ is $63$.The ground state electronic configuration of $Eu$ is $[Xe] 4f^{7} 6s^{2}$.When $Eu$ forms the $Eu^{2+

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$4f^{7}$

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(C) The atomic number of Europium $(Eu)$ is $63$.The ground state electronic configuration of $Eu$ is $[Xe] 4f^{7} 6s^{2}$.When $Eu$ forms the $Eu^{2+}$ ion,it loses two electrons from the $6s$ orbital.Therefore,the electronic configuration of $Eu^{2+}$ is $[Xe] 4f^{7}$.This configuration is highly stable due to the half-filled $f$-orbital $(f^{7})$,which makes $Eu^{2+}$ a strong reducing agent as it tends to lose an electron to reach the even more stable $Eu^{3+}$ state $(4f^{6})$.

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

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