Spin free complex among the following is:

\([CoF_6]^{3-}\)

The correct answer is option 2. \([CoF_6]^{3-}\).

To determine which complex is "spin-free," we need to analyze the electronic configuration of the metal center in each complex and assess whether it has unpaired electrons. A spin-free (low-spin) complex is one where all electrons are paired, leading to no unpaired electrons.

Analysis of Each Complex

1. \([Co(NH_3)_6]^{3+}\)

Oxidation State: Co is in the \(+3\) oxidation state.

Electronic Configuration: Co in the \(+3\) oxidation state has an electronic configuration of \(d^6\) (since the neutral Co atom is \(3d^7 4s^2\)).

Geometry: This is an octahedral complex (since NH\(_3\) is a neutral ligand and there are six ligands).

Field Strength: NH\(_3\) is a weak field ligand, so it does not cause a strong splitting of the d-orbitals.

Spin State: In a weak field ligand environment, \(d^6\) in an octahedral field usually results in a high-spin configuration with 4 unpaired electrons.

2. \([CoF_6]^{3-}\)

Oxidation State: Co is in the \(+3\) oxidation state.

Electronic Configuration: Co in the \(+3\) oxidation state has \(d^6\) configuration.

Geometry: This is an octahedral complex (since there are six fluorine ligands).

Field Strength: Fluoride is a very weak field ligand, leading to a high-spin configuration.

Spin State: In a high-spin environment with weak field ligands, \(d^6\) in an octahedral field will have unpaired electrons.

3. \([Co(C_2O_4)_3]^{3-}\)

Oxidation State: Co is in the \(+3\) oxidation state.

Electronic Configuration: Co in the \(+3\) oxidation state has \(d^6\) configuration.

Geometry: This complex is octahedral (oxalate is a bidentate ligand, and there are three of them).

Field Strength: Oxalate is a moderate field ligand, but the complex may still be high-spin.

Spin State: With moderate field ligands, this complex may or may not be low-spin. However, it is more likely to be high-spin given the field strength.

4. \([Mn(CN)_6]^{3-}\)

Oxidation State: Mn is in the \(+3\) oxidation state.

Electronic Configuration: Mn in the \(+3\) oxidation state has \(d^4\) configuration.

Geometry: This is an octahedral complex.

Field Strength: Cyanide is a strong field ligand.

Spin State: Strong field ligands like cyanide cause a large splitting of the d-orbitals, leading to a low-spin configuration. For \(d^4\) in a strong field environment, this complex would typically have all electrons paired (low-spin).

Conclusion

Among the given complexes, the one that is "spin-free" (i.e., has all electrons paired and is in a low-spin state) is: 4. \([Mn(CN)_6]^{3-}\).

This complex, due to the strong field nature of cyanide ligands, results in a low-spin configuration where all the d-electrons are paired, making it spin-free.