Which of the following transition metal ions | vedclass

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(A) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 3.87 \ B

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$Co^{2+}$

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(A) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu = 3.87 \ BM$,we solve for $n$: $3.87 = \sqrt{n(n+2)} \implies 15 \approx n(n+2) \implies n^2 + 2n - 15 = 0 \implies (n+5)(n-3) = 0$. Thus,$n = 3$.Now,check the electronic configurations:$Co^{2+}$ $(Z=27)$: $[Ar] 3d^7$,unpaired electrons $n=3$.$Co^{3+}$ $(Z=27)$: $[Ar] 3d^6$,unpaired electrons $n=4$.$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$,unpaired electrons $n=4$.$Fe^{3+}$ $(Z=26)$: $[Ar] 3d^5$,unpaired electrons $n=5$.Therefore,$Co^{2+}$ has $3$ unpaired electrons,corresponding to $3.87 \ BM$.

Which of the following transition metal ions has a magnetic moment of $3.87 \ BM$?

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