Which pair of ions have the same magnetic mom | vedclass

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(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. $Mn^{2+}

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$V^{2+}, Co^{2+}$

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(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. $Mn^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \ BM$.$2$. $Fe^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.$3$. $Ti^{2+}$ $(3d^2)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ BM$.$4$. $V^{2+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \ BM$.$5$. $Cr^{2+}$ $(3d^4)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.$6$. $Co^{2+}$ $(3d^7)$: $n = 3$,$\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87 \ BM$.Comparing the values,$V^{2+}$ and $Co^{2+}$ both have $n = 3$ and thus the same magnetic moment.

Which pair of ions have the same magnetic moment?

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