Which of the following compounds has a magnet | vedclass

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(B) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{15

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$[Cr(CN)_6]^{-3}$

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(B) The magnetic moment $\mu$ is given by the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{15} \ B.M.$,we have $n(n+2) = 15$,which implies $n^2 + 2n - 15 = 0$. Solving this quadratic equation gives $(n+5)(n-3) = 0$,so $n = 3$.Now,let us analyze the complexes:$A$: In $[Fe(H_2O)_6]^{+2}$,$Fe$ is in $+2$ oxidation state $(3d^6)$. $H_2O$ is a weak field ligand,so $Fe^{+2}$ has $4$ unpaired electrons $(n=4)$.$B$: In $[Cr(CN)_6]^{-3}$,$Cr$ is in $+3$ oxidation state $(3d^3)$. $CN^-$ is a strong field ligand,so $Cr^{+3}$ has $3$ unpaired electrons $(n=3)$.$C$: In $[Ni(NH_3)_6]^{+2}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $NH_3$ is a strong field ligand,so $Ni^{+2}$ has $2$ unpaired electrons $(n=2)$.$D$: In $[PtCl_4]^{-2}$,$Pt$ is in $+2$ oxidation state $(5d^8)$. $Pt$ is a $5d$ series metal,so it forms square planar complexes with $0$ unpaired electrons $(n=0)$.Thus,$[Cr(CN)_6]^{-3}$ has $3$ unpaired electrons,corresponding to $\sqrt{15} \ B.M.$

List $I$ (Complex)	List $II$ (Type of isomerism)
$A$. $[Pt(NH_3)_2Cl_2]$	$I$. Optical
$B$. $[Co(en)_3]^{3+}$	$II$. Solvate
$C$. $[Co(NH_3)_5NO_2]Cl_2$	$III$. Geometrical
$D$. $[Cr(H_2O)_6]Cl_3$	$IV$. Linkage

Which of the following compounds has a magnetic moment of $\sqrt{15} \ B.M.$?

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Match List $I$ with List $II$ :
List $I$ (Complex) List $II$ (Type of isomerism)
$A$. $[Pt(NH_3)_2Cl_2]$ $I$. Optical
$B$. $[Co(en)_3]^{3+}$ $II$. Solvate
$C$. $[Co(NH_3)_5NO_2]Cl_2$ $III$. Geometrical
$D$. $[Cr(H_2O)_6]Cl_3$ $IV$. Linkage

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