The universally dominant and most stable oxidation state for all lanthanoids is . However, Cerium () — d and f Block Elements Chemistry Question
Question
The universally dominant and most stable oxidation state for all lanthanoids is $+3$. However, Cerium ($Z=58$) readily forms a $+4$ ion. How does $Ce^{4+}$ characteristically behave in aqueous reactions?
Answer: D
💡 Solution & Explanation
$Ce^{4+}$ achieves stability with an empty $4f^0$ core, but the thermodynamic hydration stability strongly favors $+3$. Therefore, $Ce^{4+}$ has an immense drive to capture an electron to become $Ce^{3+}$, acting as an aggressive oxidant.
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