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(B) The magnetic moment $(\mu)$ is related to the number of unpaired electrons '$n$' by the formula: $\mu_{eff} = \sqrt{n(n+2)} \ BM$. Given $\mu_{eff

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$Fe^{2+}$

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(B) The magnetic moment $(\mu)$ is related to the number of unpaired electrons '$n$' by the formula: $\mu_{eff} = \sqrt{n(n+2)} \ BM$. Given $\mu_{eff} = \sqrt{24} \ BM$, we equate $\sqrt{n(n+2)} = \sqrt{24}$, which gives $n(n+2) = 24$. Solving for $n$, we get $n^2 + 2n - 24 = 0$, which factors to $(n+6)(n-4) = 0$. Since $n$ must be positive, $n = 4$. Now, let us check the number of unpaired electrons for each ion: $Mn^{2+}$ $(Z=25)$: $[Ar] 3d^5$, $n = 5$. $Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$, $n = 4$. $Fe^{3+}$ $(Z=26)$: $[Ar] 3d^5$, $n = 5$. $Co^{2+}$ $(Z=27)$: $[Ar] 3d^7$, $n = 3$. Thus, $Fe^{2+}$ has $4$ unpaired electrons and corresponds to a magnetic moment of $\sqrt{24} \ BM$.

Which of the following metal ions has a calculated magnetic moment value of $\sqrt{24} \ BM$?

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