The \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}\) ion is diamagnetic, but \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is paramagnetic. Explain.

First, write down the electronic configurations of the central metal ions, Cobalt (Co) and Iron (Fe), considering their respective oxidation states. For Co in \(\mathrm{Co}(\mathrm{NH}_3)_6^{3+}\), the oxidation state of Co is +3. For Fe in \(\mathrm{Fe}(H_2O)_6^{2+}\), the oxidation state of Fe is +2. The electronic configuration of Co and Fe in their neutral states are: Co: [Ar] 4s² 3d⁷ Fe: [Ar] 4s² 3d⁶ Now let's remove the appropriate number of electrons to obtain the electronic configurations of Co³⁺ and Fe²⁺: Co³⁺: [Ar] 3d⁶ Fe²⁺: [Ar] 3d⁴

Now we need to understand the ligand field splitting in the octahedral complexes. The ligands surrounding the metal ions in both complexes cause the d-orbitals to split into two sets of different energies: three orbitals of lower energy (t₂g) and two orbitals of higher energy (e_g). The strong field ligand, NH₃, in Co complex results in a larger splitting, represented by Δ₀ (crystal field splitting energy), causing electrons to first fill the lower energy t₂g orbitals. On the other hand, H₂O is a weaker field ligand in Fe complex leading to smaller splitting, represented by Δ₀, such that electrons might not necessarily fill the lower t₂g orbitals first.

Now, fill up the d-electrons with respect to the splitting of the ligand field and Hund's rule (keep electrons unpaired in degenerate orbitals until all orbitals are half-filled): Co³⁺: In the octahedral complex, the t₂g orbitals are completely filled and the e_g orbitals are empty: t₂g↑↓ t₂g↑↓ t₂g↑↓ e_g↑-- e_g↑-- Fe²⁺: In the octahedral complex, the t₂g orbitals are half-filled and one of the e_g orbitals has an unpaired electron: t₂g↑↓ t₂g↑-- t₂g↑-- e_g↑-- e_g↑--

Now, we can determine the magnetic properties of both complexes: In \(\mathrm{Co}(\mathrm{NH}_3)_6^{3+}\) complex, all the electrons are paired up with no unpaired electrons in its d-orbitals: t₂g↑↓ t₂g↑↓ t₂g↑↓ e_g↑-- e_g↑--. This complex is diamagnetic, having no unpaired electrons. In \(\mathrm{Fe}(H_2O)_6^{2+}\) complex, there are four unpaired electrons in the d-orbitals: t₂g↑↓ t₂g↑-- t₂g↑-- e_g↑-- e_g↑--. This complex is paramagnetic due to the presence of unpaired electrons. In summary, the \(\mathrm{Co}(\mathrm{NH}_{3})_{6}^{3+}\) ion is diamagnetic because there are no unpaired electrons in its d-orbitals, while \(\mathrm{Fe}(H_{2}O)_{6}^{2+}\) is paramagnetic due to the presence of four unpaired electrons in its d-orbitals.