Which of the following ions shows the highest | vedclass

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(B) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
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Vedclass · Accepted Answer

$Mn^{2+}$

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(B) The spin-only magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
$1$. For $Co^{2+}$ $(3d^7)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.
$2$. For $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.
$3$. For $Ti^{2+}$ $(3d^2)$: $n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$.
$4$. For $Fe^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.
Comparing the values,$Mn^{2+}$ has the highest number of unpaired electrons $(n=5)$,hence it shows the highest magnetic moment.

Which of the following ions shows the highest spin-only magnetic moment value?

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