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(C) The electronic configuration of Lanthanoids is $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$. For $Eu$ $(Z=63)$,the configuration is $[Xe] 4f^7 6s^2$. When $Eu$

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$Eu^{2+}$

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(C) The electronic configuration of Lanthanoids is $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$. For $Eu$ $(Z=63)$,the configuration is $[Xe] 4f^7 6s^2$. When $Eu$ forms $Eu^{2+}$ ion,it loses two $6s$ electrons,resulting in the configuration $[Xe] 4f^7$. $Pr^{3+}$ $(Z=59)$ has configuration $[Xe] 4f^2$. $Lu^{3+}$ $(Z=71)$ has configuration $[Xe] 4f^{14}$. $Ce^{4+}$ $(Z=58)$ has configuration $[Xe] 4f^0$. Therefore,the ion with $4f^7$ configuration is $Eu^{2+}$.

The ion with $4f^7$ configuration is

Similar Questions

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Choose the most appropriate answer from the options given below:

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