Question

$ Ce^{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

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

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

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

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