Which one of the following is a low spin complex?

The correct answer is option 1. \([Co(NH_3)_6]^{3+}\). 

To determine which of the given complexes is a low-spin complex, we need to evaluate the ligands and the metal ions involved. The spin state (low-spin or high-spin) is influenced by the ligand's position in the spectrochemical series and the oxidation state of the metal ion.

Low-spin complexes form when the ligands cause a large crystal field splitting energy \((\Delta )\), which forces electrons to pair up in the lower energy \(t_{2g}\) orbitals before occupying the higher energy \(e_g\) orbitals.

High-spin complexes form when the crystal field splitting energy is small, allowing electrons to occupy higher energy orbitals before pairing up.

Let us analyze each complex:

1. \([Co(NH_3)6]^{3+}\):

The central metal is \(Co^{3+}\) which has a \(3d^6\) configuration. Ammonia \((NH_3)\) is a strong field ligand, causing large crystal field splitting. With 6 electrons in the 3d orbitals and strong field ligands, it is likely to form a low-spin complex.

2. \([Fe(H_O)_6]^{2+}\):

The central metal is \(Fe^{2+}\) which has a \(3d^6\) configuration. Water \((H_2O)\) is a weak field ligand, causing small crystal field splitting. With a weak field ligand, this complex is expected to be high-spin.

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

The central metal is \(Co^{3+}\) which has a \(3d^6\) configuration. Fluoride \((F^-)\) is a weak field ligand, causing small crystal field splitting. With a weak field ligand, this complex is expected to be high-spin.

4. \([Mn(H_2O)_6]^{2+}\):

The central metal is \(Mn^{2+}\) which has a \(3d^5\) configuration. Water \((H_2O)\) is a weak field ligand. With a \(d^5\) configuration and weak field ligands, this is expected to be a high-spin complex.

Thus, the low-spin complex is \([Co(NH_3)_6]^{3+}\).