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$\mathrm{Ce}^{4+}$ is stable. This is because of
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Verified Answer
The correct answer is:
empty orbital
The electronic configuration of Ce is
$$
\mathrm{Ce}_{58}=[\mathrm{Xe}] 4 f^{1} 5 d^{1} 6 s^{2} \text { (predicted) }
$$
or $=[\mathrm{Xe}] 4 f^{2} 5 d^{0} 6 s^{2}$ (observed)
$$
\mathrm{Ce}^{4+}=[\mathrm{Xe}] 4 f^{0} 5 d^{0} 6 s^{0}
$$
Since, in $+4$ oxidation state, all (ie, $4 f, 5 d$ and $6 s$ ) orbitals are empty and Ce gains the stable configuration of nearest inert gas, $\mathrm{Ce}^{4+}$ is most stable.
$$
\mathrm{Ce}_{58}=[\mathrm{Xe}] 4 f^{1} 5 d^{1} 6 s^{2} \text { (predicted) }
$$
or $=[\mathrm{Xe}] 4 f^{2} 5 d^{0} 6 s^{2}$ (observed)
$$
\mathrm{Ce}^{4+}=[\mathrm{Xe}] 4 f^{0} 5 d^{0} 6 s^{0}
$$
Since, in $+4$ oxidation state, all (ie, $4 f, 5 d$ and $6 s$ ) orbitals are empty and Ce gains the stable configuration of nearest inert gas, $\mathrm{Ce}^{4+}$ is most stable.
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