Which is more likely to be paramagnetic, \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{4-}\) or \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+} ?\) Explain.

For \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{4-}\) complex: Each cyanide ligand (CN) carries a charge of -1. Since there are six of them, the total charge on the ligands is -6. The overall charge of the complex is -4, so the charge on the iron atom must be +2 to balance the charge. Thus, the oxidation state of iron in this complex is +2. For \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) complex: Each water ligand (H2O) is neutral, so the total charge on the ligands is 0. The overall charge of the complex is +2, which is solely due to the iron atom. Thus, the oxidation state of iron in this complex is also +2.

In both complexes, iron has an oxidation state of +2. The electron configuration of the neutral iron atom (Fe) is [Ar] 3d6 4s2. When it loses 2 electrons to form Fe2+ ion, its electron configuration becomes [Ar] 3d6.

For \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{4-}\) complex: Cyanide (CN) is a strong-field ligand, meaning it can cause the splitting of the d-orbitals in such a way that the energy difference between the lower and higher energy orbitals is large. Therefore, in this complex, the electrons will preferentially occupy the lower energy orbitals resulting in a low-spin electron configuration. Low-spin electron configuration for d6: \(t_{2g}^{6} e_{g}^{0}\) For \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) complex: Water (H2O) is a weak-field ligand, meaning it does not cause a large splitting of the d-orbitals. In this scenario, electrons occupy the d orbitals according to Hund's rule, resulting in a high-spin electron configuration. High-spin electron configuration for d6: \(t_{2g}^{4} e_{g}^{2}\)

Paramagnetic compounds have unpaired electrons in their electron configuration. In the case of the \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{4-}\) complex, the low-spin electron configuration has all of its electrons paired (no unpaired electrons): \(t_{2g}^{6} e_{g}^{0}\). In the case of the \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) complex, the high-spin electron configuration has two unpaired electrons: \(t_{2g}^{4} e_{g}^{2}\). Since the \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) complex has unpaired electrons, it is more likely to be paramagnetic compared to the \(\mathrm{Fe}(\mathrm{CN})_{6}{ }^{4-}\) complex, which does not have any unpaired electrons.