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(D) To determine the spin only magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:$1$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ i

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$[NiCl_4]^{2-}$

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(D) To determine the spin only magnetic moment,we calculate the number of unpaired electrons $(n)$ in each complex:$1$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. Configuration: $t_{2g}^5, e_g^0$. Unpaired electrons $(n)$ = $1$.$2$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand. Configuration: $t_{2g}^6, e_g^0$. Unpaired electrons $(n)$ = $0$.$3$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,forcing pairing to form $dsp^2$ hybridisation. Unpaired electrons $(n)$ = $0$.$4$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,no pairing occurs. Configuration: $3d^8$ with $sp^3$ hybridisation. Unpaired electrons $(n)$ = $2$.Since magnetic moment $\mu = \sqrt{n(n+2)} \ BM$,the complex with the highest number of unpaired electrons $(n=2)$ has the highest magnetic moment.Therefore,$[NiCl_4]^{2-}$ has the highest value.

List-$I$ (Complexes)	List-$II$ (Types)
$a$. $[Co(NH_3)_5NO_2]Cl_2$ and $[Co(NH_3)_5ONO]Cl_2$	$i$. Linkage isomerism
$b$. $[Cr(NH_3)_6][Co(CN)_6]$ and $[Cr(CN)_6][Co(NH_3)_6]$	$ii$. Coordination isomerism
$c$. $[Co(NH_3)_5(SO_4)]Br$ and $[Co(NH_3)_5Br]SO_4$	$iii$. Ionisation isomerism
$d$. $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$	$iv$. Solvate isomerism

Which complex among the following has the highest value of spin only magnetic moment?
$[Fe(CN)_6]^{3-}, [Fe(CN)_6]^{4-}, [Ni(CN)_4]^{2-}, [NiCl_4]^{2-}$

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Which complex among the following has the highest value of spin only magnetic moment?$[Fe(CN)_6]^{3-}, [Fe(CN)_6]^{4-}, [Ni(CN)_4]^{2-}, [NiCl_4]^{2-}$