The magnetic moment of an octahedral homolept | vedclass

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(C) The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. For an octahedral complex,the magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,whe

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$NCS^{-}$

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(C) The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. For an octahedral complex,the magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. $A$ magnetic moment of $5.9 \ BM$ corresponds to $n = 5$ unpaired electrons. This indicates that the complex is a high-spin complex,which occurs when the ligand is a weak field ligand. Among the given options,$NCS^{-}$ is a weak field ligand,while $ethylenediamine$,$CN^{-}$,and $CO$ are strong field ligands that would cause pairing of electrons. Therefore,the complex is $[Mn(NCS)_6]^{4-}$.

The magnetic moment of an octahedral homoleptic $Mn(II)$ complex is $5.9 \ BM$. The suitable ligand for this complex is:

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