To determine the electronic configurations of bivalent europium and trivalent cerium, we need to consider their respective atomic numbers and typical electron configurations:
Bivalent Europium (Eu2+): The atomic number of europium (Eu) is 63. In its neutral state, europium's electronic configuration is [Xe] 4f^7 6s^2. When europium becomes bivalent (Eu2+), it loses two electrons. The electrons are lost from the outermost shell first, which is the 6s orbital. Thus, the electronic configuration for Eu2+ becomes [Xe] 4f^7.
Trivalent Cerium (Ce3+): The atomic number of cerium (Ce) is 58. In its neutral state, cerium's electron configuration is [Xe] 4f^1 5d^1 6s^2. For trivalent cerium (Ce3+), three electrons are removed. The electrons are typically removed from the 6s orbital first, followed by the 4f orbital if needed. Therefore, the configuration for Ce3+ becomes [Xe] 4f^1.
The correct answer, therefore, is the electronic configurations [Xe] 4f^7 for bivalent europium and [Xe] 4f^1 for trivalent cerium. This matches the option:
$[Xe] 4f^7$ and $[Xe] 4f^1$
Given this reasoning, the other options do not accurately reflect the electron configurations of bivalent europium and trivalent cerium. They either contain incorrect occupancy in the 4f orbital or incorrect total electron count after ionization.
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