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(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. For $\mu = 4.91 \ BM$,we fin

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$Fe^{2+}$

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(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. For $\mu = 4.91 \ BM$,we find $n = 4$. For $\mu = 0 \ BM$,we find $n = 0$. $Fe^{2+}$ has a $d^6$ electronic configuration. In a high-spin complex (with weak field ligands),$Fe^{2+}$ has $4$ unpaired electrons $(n=4)$,resulting in $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$. In a low-spin complex (with strong field ligands),the electrons pair up,resulting in $0$ unpaired electrons $(n=0)$ and $\mu = 0 \ BM$.

$A$ complex of a certain metal has a magnetic moment of $4.91 \ BM$,whereas another complex of the same metal with the same oxidation state has zero magnetic moment. The metal ion could be:

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