Identify the lanthanoid that exhibits zero ef | vedclass

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(B) The effective magnetic moment $(\mu_{eff})$ is calculated using the formula $\mu_{eff} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired e

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$Lu$

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(B) The effective magnetic moment $(\mu_{eff})$ is calculated using the formula $\mu_{eff} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. For a lanthanoid to exhibit zero effective magnetic moment in the $+3$ state,it must have zero unpaired electrons $(n=0)$. $Lu$ (Lutetium) has the atomic number $71$ and its electronic configuration is $[Xe] \ 4f^{14} \ 5d^1 \ 6s^2$. In the $+3$ oxidation state,$Lu^{3+}$ loses three electrons to form $[Xe] \ 4f^{14}$. Since the $4f$ subshell is completely filled,there are no unpaired electrons $(n=0)$,resulting in a magnetic moment of $0 \ BM$.

Identify the lanthanoid that exhibits zero effective magnetic moment in $+3$ state.

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