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

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$[CoF_3(H_2O)_3]$

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(A) The spin only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{24} \, BM$,we have $n(n+2) = 24$,which implies $n = 4$.Let us analyze the species:$A$: In $[CoF_3(H_2O)_3]$,$Co$ is in $+3$ oxidation state $(d^6)$. With weak field ligands,$n = 4$.$B$: In $[CoCl_4]^{2-}$,$Co$ is in $+2$ oxidation state $(d^7)$. With weak field ligands,$n = 3$.$C$: In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(d^8)$. With weak field ligands,$n = 2$.$D$: In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in $+2$ oxidation state $(d^8)$. With weak field ligands,$n = 2$.Thus,$[CoF_3(H_2O)_3]$ has $4$ unpaired electrons.

The species with spin only magnetic moment of $\sqrt{24} \, BM$ is

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