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The electronic configuration of divalent cation of Eu is

$(a)\;4f^6\qquad(b)\;4f^7\qquad(c)\;5f^7\qquad(d)\;5f^8$

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Answer: $[Xe]4f^7$
The electronic confuguration of Europium is $[Xe]4f^76s^2 \rightarrow$ for its divalent cation, you lose the $6s^2 \rightarrow [Xe]4f^7$
Although usually trivalent, europium readily forms divalent compounds. This behavior is unusual to most lanthanides, which almost exclusively form compounds with an oxidation state of +3.
The +2 state has an electron configuration $4f^7$ because the half-filled f-shell gives more stability.
answered Jan 29, 2014 by sreemathi.v
edited Aug 4, 2014 by balaji.thirumalai
 

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