The spin-only magnetic moment of [MnBr_4]^{x- | vedclass

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

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tetrahedral,$2$

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(D) The spin-only magnetic moment is given by the formula $\mu_{s} = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.Given $\mu_{s} = 5.9 \ BM$,we have $\sqrt{n(n+2)} = 5.9$.Squaring both sides,$n(n+2) \approx 34.81$,which gives $n = 5$.Since $Mn$ $(Z=25)$ has the electronic configuration $[Ar] 3d^5 4s^2$,to have $5$ unpaired electrons,$Mn$ must be in the $+2$ oxidation state $(Mn^{2+}: [Ar] 3d^5)$.$Br^-$ is a weak-field ligand,so it does not cause pairing of electrons,resulting in $sp^3$ hybridization.For the complex $[MnBr_4]^{x-}$,the oxidation state of $Mn$ is $x-4 = -2$,so $x = 2$.The geometry of $sp^3$ hybridized complexes is tetrahedral.

The spin-only magnetic moment of $[MnBr_4]^{x-}$ is $5.9 \ BM$. The geometry of the complex and $x$ respectively are

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