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The electronic configuration of bivalent europium and trivalent cerium are:
[Atomic number: Xe = 54, Ce = 58, Eu = 63]
A. \[[Xe]4{f^7}6{s^2}\, and \,\,[Xe]4{f^2}6{s^2}\]
B. \[[Xe]4{f^4}\, and \,\,[Xe]4{f^9}\]
C. \[[Xe]4{f^2}\, and \,\,[Xe]4{f^7}\]
D. \[[Xe]4{f^7}\, and \,\,[Xe]4{f^1}\]

Last updated date: 14th Jul 2024
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Hint: To solve this question, first identify the electronic configuration of both the given elements. Then understand which ions of the given elements are required. After that, remove the corresponding number of electrons from the elements and write their final electronic configurations.

Complete Step-by-Step answer:
The atomic numbers of the given elements, Europium and Cerium are 63 and 58 respectively. From this, the electronic configuration of Europium and Cerium can be written as follows:
Electronic configuration of europium = Eu = \[[Xe]4{f^7}6{s^2}\]
Electronic configuration of cerium = Ce = \[[Xe]4{f^1}5{d^1}6{s^2}\]
Now the conditions that have been given to us are that the europium atom is made bivalent and the cerium atom is made trivalent. Making an atom trivalent means removing three electrons from the atom.
Hence, after removing 2 electrons from europium and 3 electrons from cerium, the electron configurations thus obtained are:
For Europium: \[[Xe]4{f^7}\]
For cerium: \[[Xe]4{f^1}\]

Hence, Option D is the correct option.

Note: While all lanthanides form relatively large trivalent (3+) ions, Eu and cerium (Ce) have additional valances, europium forms 2+ ions, and Ce forms 4+ ions, leading to chemical reaction differences in how these ions can partition versus the 3+ REEs.