The most common oxidation state of Lanthanoid | vedclass

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Question

(A) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$. After losing $3$ electrons,$Ce^{3+}$ has the configuration $[Xe] 4f^1$. Ho

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$Ce$ (At. No. $58$)

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(A) The electronic configuration of $Ce$ $(Z=58)$ is $[Xe] 4f^1 5d^1 6s^2$. After losing $3$ electrons,$Ce^{3+}$ has the configuration $[Xe] 4f^1$. However,$Ce$ can easily lose one more electron to form $Ce^{4+}$,which has the configuration $[Xe] 4f^0 5d^0$. This $Ce^{4+}$ ion achieves a stable noble gas configuration,making it easy for $Ce$ to deviate from the $+3$ oxidation state.

The most common oxidation state of Lanthanoid elements is $+3$. Which of the following is likely to deviate easily from $+3$ oxidation state?

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