Question

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

• A. \( 4f^7 \)
• B. \( 4f^6 \)
• C. \( 4f^76s^2 \)
• D. \( 4f^66s^2 \)

Hint:

Remember: For Lanthanoids, the stability of the \( +3 \) state often overrides the stability gained from half-filled or full-filled subshells in other states.

Answer 1

The Correct Option is A

The question asks about the ground state electronic configuration of the Eu²⁺ ion and why it is considered a strong reducing agent. The atomic number of Europium (Eu) is 63, which means it has 63 electrons.

1. Electronic Configuration of Neutral Europium (Eu):

   The electronic configuration of a neutral Europium atom (Eu) is:

   • [Xe] 4f^7 6s^2

   This configuration shows that Europium has 7 electrons in the 4f subshell and 2 electrons in the 6s subshell.

2. Formation of Eu²⁺ Ion:

   When europium forms a Eu²⁺ ion, it loses 2 electrons. These electrons are typically removed from the outermost shell, which is the 6s orbital in this case.

   • Electronic configuration for Eu²⁺ is: [Xe] 4f^7
3. Explanation for Strong Reducing Ability:

   The half-filled 4f^7 level is particularly stable due to exchange energy and symmetry reasons. Therefore, as a Eu²⁺ ion with a 4f⁷ configuration, it achieves significant stability. It readily loses an additional electron to revert to this stable 4f^7 configuration, hence making it a strong reducing agent.

Therefore, the correct answer is \( 4f^7 \). This configuration is the reason behind the strong reducing nature of the Eu²⁺ ion.