Question

$C{e}^{4+}$ is stable. This is because of

• A. half-filled $d$ -orbitals
• B. all paired electrons in $d$ -orbitals
• C. empty orbital
• D. fully filled $d$ -orbital

Answer 1

Answer: (C)

The electronic configuration of Ce is

$C{e}_{58}=[Xe]4f^{1}5d^{1}6s^2$ (predicted)

or $=[Xe]4f^{2}5d^{0}6s^2$ (observed)

$C{e}^{4+}=[Xe]4f^{0}5d^{0}6s^{\circ }$

Since, in +4 oxidation state, all (ie, $4f,5d$ and $6s$ ) orbitals are empty and $Ce$ gains the stable configuration of nearest inert gas, $C{e}^{4+}$ is most stable