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(A) To determine the spin-only magnetic moment $(\mu = \sqrt{n(n+2)} \ B.M.)$,we find the number of unpaired electrons $(n)$ for each complex:$(i)$ $[

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$i > iii > iv > ii$

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(A) To determine the spin-only magnetic moment $(\mu = \sqrt{n(n+2)} \ B.M.)$,we find the number of unpaired electrons $(n)$ for each complex:$(i)$ $[FeF_{6}]^{3-}$: $Fe^{3+}$ is $3d^{5}$. $F^{-}$ is a weak field ligand $(WFL)$,so electrons remain unpaired. $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \ B.M.$$(ii)$ $[Co(NH_{3})_{6}]^{3+}$: $Co^{3+}$ is $3d^{6}$. $NH_{3}$ is a strong field ligand $(SFL)$,causing pairing. $n = 0$,$\mu = 0 \ B.M.$$(iii)$ $[NiCl_{4}]^{2-}$: $Ni^{2+}$ is $3d^{8}$. $Cl^{-}$ is a $WFL$. In tetrahedral geometry,$n = 2$,$\mu = \sqrt{2(2+2)} = \sqrt{8} \ B.M.$$(iv)$ $[Cu(NH_{3})_{4}]^{2+}$: $Cu^{2+}$ is $3d^{9}$. $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \ B.M.$Comparing the values: $\sqrt{35} > \sqrt{8} > \sqrt{3} > 0$,which corresponds to $i > iii > iv > ii$.

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

In which of the following orders are the given complex ions arranged correctly with respect to their decreasing spin-only magnetic moment?
$(i)$ $[FeF_{6}]^{3-}$
$(ii)$ $[Co(NH_{3})_{6}]^{3+}$
$(iii)$ $[NiCl_{4}]^{2-}$
$(iv)$ $[Cu(NH_{3})_{4}]^{2+}$

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In which of the following orders are the given complex ions arranged correctly with respect to their decreasing spin-only magnetic moment?$(i)$ $[FeF_{6}]^{3-}$$(ii)$ $[Co(NH_{3})_{6}]^{3+}$$(iii)$ $[NiCl_{4}]^{2-}$$(iv)$ $[Cu(NH_{3})_{4}]^{2+}$