In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?
i) [FeF₆]^3– 

ii) [Co(NH₃)₆]³⁺

iii) [NiCl₄]^2– 

iv) [Cu(NH₃)₄]²⁺

• ${\left[{\mathrm{FeF}}_6\right]}^{3-}$ In this Flourine  act as weak field ligand thus electronic configuration is ${\mathrm{t}}_{2\mathrm{g}}^3{\mathrm{e}}_{\mathrm{g}}^2$; number of unpaired electron is 5.[Co(NH₃)₆]³⁺  ${\mathrm{t}}_{2\mathrm{g}}^6{\mathrm{e}}_{\mathrm{g}}^0$has No unpaired electron that shows NH₃ is strong field light which causes pairing, number of unpaired electron is zero.[NiCl₄]^2- ;Ni²⁺ 3d⁸ has 2 unpaired electrons.

[Cu(NH₃)₄]²⁺: Cu²⁺ 3d⁹ has 1 unpaired electrons. Higher the number unpaired electron higher is the spin only magnetic moment.

Hence, correct option is: (B) i > iii > iv > ii