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.

List-$I$ (Complex)	List-$II$ (Hybridisation of central metal ion)
$A$. $[CoF_6]^{3-}$	$I$. $d^2sp^3$
$B$. $[NiCl_4]^{2-}$	$II$. $sp^3$
$C$. $[Co(NH_3)_6]^{3+}$	$III$. $sp^3d^2$
$D$. $[Ni(CN)_4]^{2-}$	$IV$. $dsp^2$

$Mn^{2+}$ 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?

Similar Questions

Choose the correct statement for the $[Ni(CN)_4]^{2-}$ complex ion (Atomic number of $Ni=28$).

Match List-$I$ with List-$II$.
List-$I$ (Complex) List-$II$ (Hybridisation of central metal ion)
$A$. $[CoF_6]^{3-}$ $I$. $d^2sp^3$
$B$. $[NiCl_4]^{2-}$ $II$. $sp^3$
$C$. $[Co(NH_3)_6]^{3+}$ $III$. $sp^3d^2$
$D$. $[Ni(CN)_4]^{2-}$ $IV$. $dsp^2$
Choose the correct answer from the options given below:

The shape of the $[Fe(CN)_6]^{4-}$ ion is:

Which one of the following has a square planar geometry?

What is the color of the complex $[Co(H_2O)_6]Cl_3$?

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