In which of the following order the given complex ions are 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) $$[Cu(NH_3)_4]^{2+}$$

We need to determine the number of unpaired electrons in each complex ion to calculate the spin-only magnetic moment using $$\mu = \sqrt{n(n+2)}$$ BM, where $$n$$ is the number of unpaired electrons.

For $$[FeF_6]^{3-}$$: $$Fe^{3+}$$ has the configuration $$[Ar]3d^5$$. Since $$F^-$$ is a weak field ligand, the complex is high spin with all five d-electrons unpaired. So $$n = 5$$ and $$\mu = \sqrt{5 \times 7} = \sqrt{35} = 5.92$$ BM.

For $$[Co(NH_3)_6]^{3+}$$: $$Co^{3+}$$ has the configuration $$[Ar]3d^6$$. Since $$NH_3$$ is a strong field ligand, the complex is low spin. All six electrons pair up in the three $$t_{2g}$$ orbitals, giving $$n = 0$$ and $$\mu = 0$$ BM.

For $$[NiCl_4]^{2-}$$: $$Ni^{2+}$$ has the configuration $$[Ar]3d^8$$. This is a tetrahedral complex (since $$Cl^-$$ is a weak field ligand and $$Ni^{2+}$$ commonly forms tetrahedral complexes with halides). In a tetrahedral field with $$d^8$$, there are 2 unpaired electrons. So $$n = 2$$ and $$\mu = \sqrt{2 \times 4} = \sqrt{8} = 2.83$$ BM.

For $$[Cu(NH_3)_4]^{2+}$$: $$Cu^{2+}$$ has the configuration $$[Ar]3d^9$$. With 9 d-electrons, there is always 1 unpaired electron regardless of the field strength. So $$n = 1$$ and $$\mu = \sqrt{1 \times 3} = \sqrt{3} = 1.73$$ BM.

Arranging in decreasing order of magnetic moment: $$[FeF_6]^{3-}$$ (5.92) $$>$$ $$[NiCl_4]^{2-}$$ (2.83) $$>$$ $$[Cu(NH_3)_4]^{2+}$$ (1.73) $$>$$ $$[Co(NH_3)_6]^{3+}$$ (0), i.e., (i) $$>$$ (iii) $$>$$ (iv) $$>$$ (ii). This corresponds to option (3).