Cerium (Z = 58) is an important member of the | vedclass

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(C) Cerium $(Z = 58)$ has the electronic configuration $[Xe] 4f^1 5d^1 6s^2$. It exhibits $+3$ and $+4$ oxidation states. The $+4$ oxidation state is

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The $+4$ oxidation state of Cerium is not known in solution.

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(C) Cerium $(Z = 58)$ has the electronic configuration $[Xe] 4f^1 5d^1 6s^2$. It exhibits $+3$ and $+4$ oxidation states. The $+4$ oxidation state is achieved by losing all four valence electrons,resulting in a stable $[Xe]$ configuration. However,in aqueous solution,$Ce^{4+}$ is a strong oxidizing agent because it tends to revert to the more stable $Ce^{3+}$ state. Therefore,the statement that the $+4$ oxidation state is not known in solution is incorrect,as $Ce^{4+}$ compounds are well-known in analytical chemistry.

Cerium $(Z = 58)$ is an important member of the lanthanoids. Which of the following statements is $NOT$ correct for Cerium?

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