The species that has a (spin only) magnetic m | vedclass

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(B) The spin-only magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.For $\mu = 5

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$[MnBr_4]^{2-}$ $(T_d)$

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(B) The spin-only magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.For $\mu = 5.9 \, BM$,we have $\sqrt{n(n+2)} \approx 5.9$,which implies $n = 5$.In $[MnBr_4]^{2-}$,the oxidation state of $Mn$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.Since $Br^-$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.Thus,$Mn^{2+}$ has $5$ unpaired electrons $(n=5)$.Therefore,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9 \, BM$.

The species that has a (spin only) magnetic moment of $5.9 \, BM$ is:

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