1. Find oxidation states and electronic configurations

• In $\mathrm{[FeF_6]^{3-}}$:

  Let oxidation state of Fe be $x$. $x+6(-1)=-3 \Rightarrow x=+3$ So, $\mathrm{Fe^{3+}}$ has configuration: $\mathrm{Fe}: [Ar]3d^64s^2 \Rightarrow Fe^{3+}: [Ar]3d^5$

• In $\mathrm{[CoF_6]^{3-}}$:

  Let oxidation state of Co be $x$. $x+6(-1)=-3 \Rightarrow x=+3$ So, $\mathrm{Co^{3+}}$ has configuration: $\mathrm{Co}: [Ar]3d^74s^2 \Rightarrow Co^{3+}: [Ar]3d^6$

• In $\mathrm{[Co(C_2O_4)_3]^{3-}}$:

  Oxalate $\mathrm{(C_2O_4)^{2-}}$ is a bidentate ligand with charge $-2$. $x+3(-2)=-3 \Rightarrow x=+3$ Again metal is $\mathrm{Co^{3+}} = 3d^6$$

2. Determine high spin / low spin nature

• $\mathrm{F^-}$ is a weak field ligand, so:

  • $\mathrm{[FeF_6]^{3-}}$ is high spin
  • $\mathrm{[CoF_6]^{3-}}$ is high spin

• Oxalate $\mathrm{(C_2O_4)^{2-}}$ is stronger than $\mathrm{F^-}$, and with $\mathrm{Co^{3+}}$ (which has large $\Delta_o$), the complex $\mathrm{[Co(C_2O_4)_3]^{3-}}$ is low spin.

3. Count unpaired electrons

(i) $\mathrm{[FeF_6]^{3-}}$

• Metal ion: $\mathrm{Fe^{3+}} = d^5$
• High spin octahedral: $t_{2g}^3 e_g^2$
• Number of unpaired electrons, $n=5$

(ii) $\mathrm{[CoF_6]^{3-}}$

• Metal ion: $\mathrm{Co^{3+}} = d^6$
• High spin octahedral: $t_{2g}^4 e_g^2$
• Number of unpaired electrons, $n=4$

(iii) $\mathrm{[Co(C_2O_4)_3]^{3-}}$

• Metal ion: $\mathrm{Co^{3+}} = d^6$
• Low spin octahedral: $t_{2g}^6 e_g^0$
• Number of unpaired electrons, $n=0$

4. Calculate spin-only magnetic moment order

Spin-only magnetic moment is: $\mu = \sqrt{n(n+2)}\ \text{BM}$

So,

• For $n=5$: $\mu = \sqrt{5(7)}=\sqrt{35}$
• For $n=4$: $\mu = \sqrt{4(6)}=\sqrt{24}$
• For $n=0$: $\mu = 0$

Hence, $\mathrm{[FeF_6]^{3-} > [CoF_6]^{3-} > [Co(C_2O_4)_3]^{3-}}$

5. Match with options

This corresponds to: $\boxed{\text{C}}$

6. Comparison with stored correct answer

Stored correct answer: $\mathrm{C}$

Derived answer: $\mathrm{C}$

They agree.