Question

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.

Short answer

\(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is more likely to be paramagnetic due to its 4 unpaired electrons, as opposed to \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\), which is diamagnetic with 0 unpaired electrons. This difference arises from the strong field ligand (CN) in \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) causing low-spin complex and the weak field ligand (H2O) in \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) leading to a high-spin complex.

Step-by-step solution

Determine the oxidation state of Fe in each complex ion

To find the unpaired electrons, we first need to find the oxidation state of Fe in both complex ions. For \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\): The overall charge of the complex ion is -4, and CN is a monodentate ligand with a charge of -1. Since there are 6 CN molecules, the total charge contributed by the ligands is -6. Thus, the oxidation state of Fe in this complex ion is +2. For \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): The overall charge of the complex ion is +2, and water (H2O) is a neutral ligand, not contributing any charge. Therefore, the oxidation state of Fe in this complex ion is also +2.

Use the electron configuration of Fe to determine the number of unpaired electrons

Fe has an electron configuration of [Ar] 3d6 4s2, which means that in its ground state, it has 6 electrons in the 3d orbital and 2 electrons in the 4s orbital. When Fe has an oxidation state of +2, it loses 2 electrons and its electron configuration becomes [Ar] 3d6 for both complex ions.

Determine unpaired electrons in the complexes

Next, we need to determine the number of unpaired electrons for each complex ion. For \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\): The electron configuration of Fe in this complex ion is [Ar] 3d6. CN is a strong field ligand, causing the energy difference between the d-orbitals to be large (greater crystal field splitting). This results in a low-spin complex where all six electrons of the Fe(II) ion will occupy the three lower-energy 3d orbitals, following the Aufbau principle, and pairing up with each other. Therefore, this complex will have 0 unpaired electrons and be diamagnetic. For \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): The electron configuration of Fe in this complex ion is also [Ar] 3d6. However, H2O is a weak field ligand, resulting in smaller energy differences between the d-orbitals (lower crystal field splitting). Consequently, the Fe(II) electrons will fill the five 3d orbitals before pairing up, following Hund's rule. This complex will have 4 unpaired electrons and be paramagnetic.

Conclusion

Given our findings, it's clear that \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is more likely to be paramagnetic, as it has 4 unpaired electrons while \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) will be diamagnetic with 0 unpaired electrons.

Key concepts

Oxidation State

The oxidation state of an element in a compound indicates the degree of oxidation and can help predict the element's behavior in reactions. Knowing this is essential for understanding the oxidation process. To determine the oxidation state of iron (Fe) in complex ions like \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) and \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\), you focus on each ligand's charge.

• In \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\), each cyanide (CN) ion carries a -1 charge. With six cyanides, the total ligand charge is -6, giving the Fe a +2 oxidation state when combined with the overall -4 charge of the complex.
• In contrast, \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) involves neutral water ligands. Therefore, the Fe oxidation state is simply +2, matching the complex's +2 charge.

Understanding oxidation states is crucial to assess how the electron configuration in each ion affects magnetic properties.

Electron Configuration

Electron configuration reflects the arrangement of electrons around an atom's nucleus, which plays a critical role in chemical bonding and properties. For iron (Fe), the electron configuration in its elemental form is [Ar] 3d\(^6\) 4s\(^2\).
When Fe is in a +2 oxidation state, it loses two electrons: typically those from the 4s orbital.

• Thus, the electron configuration becomes [Ar] 3d\(^6\) for both \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) and \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\).
• This configuration is essential to predicting the number of unpaired electrons, which determines if the substance will be paramagnetic or diamagnetic.

The loss and arrangement of these electrons will be influenced by factors such as the crystal field theory and the strength of the ligands involved.

Crystal Field Theory

Crystal Field Theory (CFT) explains how metal ion's d-orbitals are affected by surrounding ligands. It is foundational in predicting and understanding properties like magnetism in complex ions.

• In complexes like \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\), CN is considered a strong field ligand. It causes significant splitting of the d-orbitals, favoring electron pairing in lower-energy orbitals, leading to a low-spin state with fewer unpaired electrons.
• Conversely, water (H\(_2\)O) in \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) acts as a weak field ligand. This results in less splitting, encouraging a high-spin state wherein electrons occupy all five d-orbitals before any pairing occurs.

Understanding how ligands interact and split the d-orbitals underlies why some complexes exhibit paramagnetic properties, like \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\), while others do not.

Unpaired Electrons

Unpaired electrons in an atom or ion's d-orbitals are the driving force behind paramagnetism, a property where substances are attracted to magnetic fields.

• The presence of unpaired electrons results in unbalanced magnetic moments that align with external magnetic fields. This is the case in high-spin complexes like \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\), where weak field ligands lead to four unpaired electrons.
• In low-spin complexes like \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\), all d-orbital electrons pair due to the strong field ligand, resulting in a diamagnetic compound lacking unpaired electrons and magnetic properties.

Thus, to determine if a complex is paramagnetic, count the unpaired electrons—more typically equals paramagnetic; none suggests diamagnetic behavior.