For which pair is the magnetic moment mu = sq | 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.For $\mu = \sqrt{15} \

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$Co^{2+}, Cr^{3+}$

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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.For $\mu = \sqrt{15} \ BM$,we have $n(n+2) = 15$,which gives $n = 3$.Let us analyze the electronic configurations:$1$. $Co^{2+}$ $(Z=27)$: $[Ar] 3d^7$,unpaired electrons $n=3$.$2$. $Cr^{3+}$ $(Z=24)$: $[Ar] 3d^3$,unpaired electrons $n=3$.Both $Co^{2+}$ and $Cr^{3+}$ have $3$ unpaired electrons,so both have a magnetic moment of $\sqrt{3(3+2)} = \sqrt{15} \ BM$.

For which pair is the magnetic moment $\mu = \sqrt{15} \ BM$ correct?

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