The \( \text{Eu}^{2+} \) ion is a strong reducing agent in spite of its ground state electronic configuration (outermost) : [Atomic number of Eu = 63]
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Remember: For Lanthanoids, the stability of the \( +3 \) state often overrides the stability gained from half-filled or full-filled subshells in other states.
Step 1: Understanding the Concept:
Lanthanoids generally show a stable oxidation state of \( +3 \). Some elements show \( +2 \) or \( +4 \) states if they result in empty (\( f^0 \)), half-filled (\( f^7 \)), or fully filled (\( f^{14} \)) subshells. Step 2: Key Formula or Approach:
Europium (\( \text{Eu} \)) Atomic number = 63.
Ground state configuration: \( [\text{Xe}] 4f^7 6s^2 \). Step 3: Detailed Explanation:
To form \( \text{Eu}^{2+} \), the two electrons from the \( 6s \) orbital are removed.
Configuration of \( \text{Eu}^{2+} = [\text{Xe}] 4f^7 \).
Even though \( 4f^7 \) is a stable half-filled configuration, \( \text{Eu}^{2+} \) acts as a reducing agent. This is because it has a strong tendency to lose one more electron to reach the \( +3 \) oxidation state (\( \text{Eu}^{3+} \)), which is the most stable and common oxidation state for all lanthanoids in aqueous solution. Step 4: Final Answer:
The ground state electronic configuration of \( \text{Eu}^{2+} \) is \( 4f^7 \).
