Correct order of spin only magnetic moment of the following complex ions is:
(Given At. No. Fe : 26, Co : 27)

Find the correct order of spin-only magnetic moment for the given complex ions.

Determine the electronic configuration and unpaired electrons.

$$[\text{FeF}_6]^{3-}$$: Fe$$^{3+}$$ has configuration [Ar] 3d$$^5$$. F$$^-$$ is a weak field ligand (high spin). The 5 d-electrons occupy all five orbitals singly: 5 unpaired electrons.

$$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$$ BM

$$[\text{CoF}_6]^{3-}$$: Co$$^{3+}$$ has configuration [Ar] 3d$$^6$$. F$$^-$$ is a weak field ligand (high spin). In octahedral high spin d$$^6$$: t$$_{2g}^4$$ e$$_g^2$$, giving 4 unpaired electrons.

$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$$ BM

$$[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$$: Co$$^{3+}$$ has configuration [Ar] 3d$$^6$$. C$$_2$$O$$_4^{2-}$$ (oxalate) is higher than F$$^-$$ in the spectrochemical series, and Co$$^{3+}$$ has a high charge density. This results in a low spin configuration: t$$_{2g}^6$$ e$$_g^0$$, giving 0 unpaired electrons.

$$\mu = 0$$ BM

Order the magnetic moments.

$$[\text{FeF}_6]^{3-}$$ (5.92 BM) $$>$$ $$[\text{CoF}_6]^{3-}$$ (4.90 BM) $$>$$ $$[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$$ (0 BM)

The correct answer is $$[\text{FeF}_6]^{3-} > [\text{CoF}_6]^{3-} > [\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$$.