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

Vedclass · Accepted Answer

$NCS^{-}$

Vedclass · Answer

(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 $[Mn(NCS)_6]^{4-}$.

List-$I$ (Complex)	List-$II$ (Spin only magnetic moment in $B$.$M$.)
$A$. $[Cr(NH_3)_6]^{3+}$	$I$. $4.90$
$B$. $[NiCl_4]^{2-}$	$II$. $3.87$
$C$. $[CoF_6]^{3-}$	$III$. $0.0$
$D$. $[Ni(CN)_4]^{2-}$	$IV$. $2.83$

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

Similar Questions

Which of the following outer octahedral complexes have the same number of unpaired electrons?
$1$. $[MnCl_6]^{3-}$
$2$. $[FeF_6]^{3-}$
$3$. $[CoF_6]^{3-}$
$4$. $[Ni(NH_3)_6]^{2+}$

The pair not correctly matched is:

Which complex ion has the ability to absorb visible light?

The spin-only magnetic moment value of an octahedral complex among $CoCl_{3} \cdot 4NH_{3}$,$NiCl_{2} \cdot 6H_{2}O$ and $PtCl_{4} \cdot 2HCl$,which upon reaction with excess of $AgNO_{3}$ gives $2$ moles of $AgCl$ is $....$ $B.M.$ (Nearest Integer)

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