Which of the following complexes has a zero m | vedclass

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(D) The magnetic moment (spin only) is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. Zero magnetic moment implies

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$K_4[Fe(CN)_6]$

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(D) The magnetic moment (spin only) is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons. Zero magnetic moment implies $n = 0$.$1$. $[Ni(NH_3)_6]Cl_2$: $Ni^{2+}$ is $3d^8$. $NH_3$ is a weak field ligand,resulting in $sp^3d^2$ hybridization with $2$ unpaired electrons.$2$. $Na_3[FeF_6]$: $Fe^{3+}$ is $3d^5$. $F^-$ is a weak field ligand,resulting in $sp^3d^2$ hybridization with $5$ unpaired electrons.$3$. $[Cr(H_2O)_6]SO_4$: $Cr^{2+}$ is $3d^4$. $H_2O$ is a weak field ligand,resulting in $sp^3d^2$ hybridization with $4$ unpaired electrons.$4$. $K_4[Fe(CN)_6]$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing of electrons. It undergoes $d^2sp^3$ hybridization with $0$ unpaired electrons.Thus,$K_4[Fe(CN)_6]$ has a zero magnetic moment.

List-$I$ (Complex)	List-$II$ (Spin only Magnetic moment)
$A$. $[CoF_6]^{3-}$	$I$. $0$
$B$. $[Co(C_2O_4)_3]^{3-}$	$II$. $\sqrt{24}$
$C$. $[FeF_6]^{3-}$	$III$. $\sqrt{8}$
$D$. $[Mn(CN)_6]^{3-}$	$IV$. $\sqrt{35}$
	$V$. $\sqrt{15}$

Which of the following complexes has a zero magnetic moment (spin only)?

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Match the following:

List-$I$ (Complex) List-$II$ (Spin only Magnetic moment)

$A$. $[CoF_6]^{3-}$ $I$. $0$

$B$. $[Co(C_2O_4)_3]^{3-}$ $II$. $\sqrt{24}$

$C$. $[FeF_6]^{3-}$ $III$. $\sqrt{8}$

$D$. $[Mn(CN)_6]^{3-}$ $IV$. $\sqrt{35}$

$V$. $\sqrt{15}$

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