The correct answer is option 2. \([CoF_6]^{3-}\).
A spin-free complex (high-spin complex) is one where the electrons are arranged to maximize the total spin, typically occurring with weak-field ligands.
- $[\text{Co}(\text{NH}_3)_6]^{3+}$: Ammonia ($\text{NH}_3$) is a strong-field ligand, creating a low-spin (spin-paired) complex.
- $\text{CoF}_6^{3-}$: Fluoride ($\text{F}^-$) is a weak-field ligand. In an octahedral field, the $\text{Co}^{3+}$ ion ($d^6$) arranges its electrons as $t_{2g}^4 e_g^2$, resulting in four unpaired electrons. This is a high-spin, spin-free complex.
- $[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$: Oxalate ($\text{C}_2\text{O}_4^{2-}$) acts as a relatively strong-field ligand, typically forming a low-spin complex.
- $[\text{Mn}(\text{CN})_6]^{3-}$: Cyanide ($\text{CN}^-$) is a very strong-field ligand, creating a low-spin complex.
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