Spin only magnetic moment of [MnBr_{6}]^{4-} | vedclass

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(B) In $[MnBr_{6}]^{4-}$,the oxidation state of $Mn$ is $+2$.Electronic configuration of $Mn^{2+}$ is $[Ar] 3d^{5}$.Since $Br^{-}$ is a weak field lig

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) In $[MnBr_{6}]^{4-}$,the oxidation state of $Mn$ is $+2$.Electronic configuration of $Mn^{2+}$ is $[Ar] 3d^{5}$.Since $Br^{-}$ is a weak field ligand,the electrons remain unpaired in the $d$-orbitals.Number of unpaired electrons $(n)$ = $5$.The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.91 \ B.M.$Rounding off to the closest integer,we get $6 \ B.M.$

Spin only magnetic moment of $[MnBr_{6}]^{4-}$ is $..... \ B.M.$ (round off to the closest integer)

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