Why is \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) diamagnetic while \(\left[\mathrm{CoF}_{6}\right]^{3-}\) is paramagnetic?

In order to determine the oxidation state of the cobalt (Co) ion, we need to analyze both complexes individually. In \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\), let the oxidation state of Co be x, and since ammonia (NH3) is a neutral ligand, we have: x + 0 = +3. Therefore, the oxidation state of Co in this complex is +3. In \(\left[\mathrm{CoF}_{6}\right]^{3-}\), let the oxidation state of Co be y. We know that the oxidation state of each fluoride (F) ion is -1. So we have: y - 6(-1) = -3. Therefore, the oxidation state of Co in this complex is also +3.

Since both complexes have Co in the +3 oxidation state, we need to determine the number of d-electrons in each complex. Cobalt (Co) has an atomic number of 27, which means its ground-state electronic configuration is \([Ar]4s^2\ 3d^7\). In a +3 oxidation state, Co loses 3 electrons, resulting in 6 d-electrons left in d-orbitals. Now we know that both complexes have a Co(III) ion with 6 d-electrons.

To analyze the energy level splitting in both complexes, we need to determine the coordination geometry around the central Co(III) ion. The ligands in both complexes are octahedral (6 ligands surrounding the metal ion), which results in the d-orbitals splitting into two sets: three lower-energy orbitals (\(t_{2g}\)) and two higher-energy orbitals (\(e_g\)).

Ammonia (NH3) is a strong field ligand, which results in a large crystal field splitting where the energy difference between the \(t_{2g}\) and \(e_g\) levels is significant. This means that the 6 d-electrons in \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) will fill the lower energy orbitals before occupying any of the higher energy orbitals, which results in the \(t_{2g}^6 e_g^0\) electronic configuration. Fluoride (F-) is a weak field ligand, which results in a small crystal field splitting where the energy difference between the \(t_{2g}\) and \(e_g\) levels is not significant enough to prevent electrons from occupying the higher energy orbitals. As a result, the 6 d-electrons in \(\left[\mathrm{CoF}_{6}\right]^{3-}\) will fill the orbitals according to Hund's rule, resulting in the \(t_{2g}^4 e_g^2\) electronic configuration.

Based on the electronic configurations obtained in Step 4, \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) has the configuration \(t_{2g}^6 e_g^0\), which means that all of its electrons are paired, and thus it is diamagnetic. On the other hand, \(\left[\mathrm{CoF}_{6}\right]^{3-}\) has the configuration \(t_{2g}^4 e_g^2\), which means that there are 2 unpaired electrons. Therefore, this complex is paramagnetic. In conclusion, \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is diamagnetic because all of its d-electrons are paired, while \(\left[\mathrm{CoF}_{6}\right]^{3-}\) is paramagnetic due to the presence of 2 unpaired d-electrons.