The spin-only magnetic moment of [MnBr_4]^{2- | vedclass

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(A) The oxidation state of $Mn$ in $[MnBr_4]^{2-}$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. Since the coordination number is

Vedclass · Accepted Answer

Tetrahedral

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(A) The oxidation state of $Mn$ in $[MnBr_4]^{2-}$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. Since the coordination number is $4$,the geometry can be either tetrahedral $(sp^3)$ or square planar $(dsp^2)$. $A$ magnetic moment of $5.9 \ BM$ corresponds to $n = 5$ unpaired electrons,calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$. For $Mn^{2+}$ $(3d^5)$,having $5$ unpaired electrons indicates a high-spin complex. Therefore,the geometry must be tetrahedral,as square planar geometry would involve pairing of electrons,resulting in a lower magnetic moment.

List-$I$ (Tetrahedral Complex)	List-$II$ (Electronic configuration)
$A$. $TiCl_4$	$I$. $e^2, t_2^0$
$B$. $[FeO_4]^{2-}$	$II$. $e^4, t_2^3$
$C$. $[FeCl_4]^{-}$	$III$. $e^0, t_2^0$
$D$. $[CoCl_4]^{2-}$	$IV$. $e^2, t_2^3$

The spin-only magnetic moment of $[MnBr_4]^{2-}$ is $5.9 \ BM$. Predict the geometry of the complex ion.

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