Answer:
The element cerium (Ce) with atomic
number (Z = 58) has the electronic configuration \[[Xe]4{{f}^{1}}5{{d}^{1}}6{{s}^{2}}.Ce(III)\]
ion formed by losing three electrons has the configuration \[[Xe]4{{f}^{1}}\].
The ion can lose another electron readily to form Ce(IV)ion with configuration
of noble gas xenon (Xe).
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