The Eu^{2+} ion is a strong reducing agent in | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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+

Vedclass · Accepted Answer

$4f^{7}$

Vedclass · Answer

(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 $Eu=63$]

Similar Questions

Provide the electronic configurations of lanthanoids $(4f)$ series elements $(Ln)$ and their ions $Ln^{2+}$ and $Ln^{3+}$,and state their characteristics.

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

Which of the following is responsible for the lanthanoid contraction?

The Lanthanoid contraction is responsible for the fact that

What is the general formula of lanthanoid hydroxide? (Consider $Ln$ as any lanthanoid element)

Vedclass Test Series

Exam Paper Generator

Online Exam Module