A) \[+3,\,\,+4\]
B) \[+2,\,\,+3\]
C) \[+2,\,\,+4\]
D) \[+3,\,\,+5\]
Correct Answer: A
Solution :
Cerium \[C{{e}_{58}}[Xe]4{{f}^{1}}5{{d}^{1}}6{{s}^{2}}\] Its most stable oxidation state is \[+3\] but \[+4\] is also existing. Electronic configuration of cerium is \[_{58}Ce\Rightarrow [{{X}_{e}}]4{{f}^{1}},5{{d}^{1}},6{{s}^{2}}\] The most common oxidation state of lanthanoids series is \[+3\]. So, cerium form stable compound in \[+3\] oxidation state. Apart from \[+3\] oxidation state, cerium is stable in \[+4\] oxidation state because cerium acquire noble gas configuration \[(4{{f}^{0}},\,\,5{{d}^{0}},\,6{{s}^{0}})\] by the loss of four electrons.
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