The lanthanoid that does NOT show +4 oxidatio | vedclass

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Question

(B) The lanthanoids exhibit a common oxidation state of $+3$. However,some elements show $+2$ or $+4$ oxidation states due to stable electronic config

Vedclass · Accepted Answer

$Eu$

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(B) The lanthanoids exhibit a common oxidation state of $+3$. However,some elements show $+2$ or $+4$ oxidation states due to stable electronic configurations (like $f^0$,$f^7$,or $f^{14}$).$Ce$ $(Z=58)$ shows $+4$ to achieve a stable $f^0$ configuration.$Tb$ $(Z=65)$ shows $+4$ to achieve a stable $f^7$ configuration.$Dy$ $(Z=66)$ can show $+4$ in certain compounds.$Eu$ $(Z=63)$ shows $+2$ oxidation state to achieve a stable $f^7$ configuration,but it does not exhibit a $+4$ oxidation state.

The lanthanoid that does $NOT$ show $+4$ oxidation state is

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