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.

Let's find the oxidation state of iron (Fe) in each complex. In the complex \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\): Each CN ligand has a charge of -1, so the overall charge due to the 6 CN ligands is -6. Since the net charge of the complex is -4, the charge on iron must be: \[+2\ (\text{for balancing the overall charge})\] In the complex \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): Each H2O ligand has a charge of 0. Since the net charge of the complex is +2, the charge on iron must be: \[+2\ (\text{as given by the complex})\]

Now, let's analyze the electron configurations of iron. In both complexes, the oxidation state of iron is +2, which means it has lost 2 electrons from its neutral state of Fe (\(\text{atomic number}=26\)). The electron configuration of neutral iron is: \[[\text{Ar}] 3d^6 4s^2\] After losing 2 electrons, the electron configuration of Fe(II) will be: \[[\text{Ar}] 3d^6\]

Now we need to determine the number of unpaired electrons in the d-orbitals of both complexes. For \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\): The Fe(II) ion in this complex undergoes strong-field ligand (CN-) bonding, which results in the pairing of electrons in its d-orbitals. Its electron configuration becomes: \[[\text{Ar}] 3d^5 4s^1\] (Note that there is only one unpaired electron in the 4s orbital) For \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): The Fe(II) ion in this complex undergoes weak-field ligand (H2O) bonding, which does not result in the pairing of electrons in its d-orbitals. Its electron configuration remains: \[[\text{Ar}] 3d^6\] (Note that there are four unpaired electrons in the 3d orbitals)

We have determined the number of unpaired electrons in both complexes: - In \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\), there is only one unpaired electron in the 4s orbital. - In \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\), there are four unpaired electrons in the 3d orbitals. Since the \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) complex has more unpaired electrons, it is more likely to be paramagnetic.