Mn^{2+} forms a complex with Br^{-} ion. The | vedclass

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(B) The magnetic moment of $5.92 \ B.M.$ corresponds to $n = 5$ unpaired electrons,indicating a high-spin $d^5$ configuration for $Mn^{2+}$.Since $Br^

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$[MnBr_4]^{2-}$,square planar

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(B) The magnetic moment of $5.92 \ B.M.$ corresponds to $n = 5$ unpaired electrons,indicating a high-spin $d^5$ configuration for $Mn^{2+}$.Since $Br^{-}$ is a weak field ligand,it does not cause pairing of electrons.For a coordination number of $4$,the $sp^3$ hybridization leads to a tetrahedral geometry.Square planar geometry for a $d^5$ system would require $dsp^2$ hybridization,which is not possible for a high-spin $Mn^{2+}$ complex as it would require pairing of electrons.Therefore,$[MnBr_4]^{2-}$ with square planar geometry is not possible.

$Mn^{2+}$ forms a complex with $Br^{-}$ ion. The magnetic moment of the complex is $5.92 \ B.M.$ What could not be the probable formula and geometry of the complex?

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