A) half-filled d-orbitals
B) all paired electrons in d-orbitals
C) empty d-orbitals
D) fully filled d-orbitals
Correct Answer: A
Solution :
\[Ce(58)=[Xe]4{{f}^{1}}5{{d}^{1}}6{{s}^{2}}\]predicted \[[Xe]4{{f}^{2}}5{{d}^{0}}6{{s}^{2}}\]observed \[C{{e}^{4+}}=[Xe]4{{f}^{0}}5{{d}^{0}}6{{s}^{0}}\] Since, in + 4 oxidation state, all orbitals are empty and it gains the stable configuration of nearest inert gas, therefore \[\text{C}{{\text{e}}^{\text{4+}}}\]is most stable.
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