Question

Predict: (a) whether the square-planar complex ion \(\left[\mathrm{Cu}(\mathrm{py})_{4}\right]^{2+}\) is diamagnetic or paramagnetic (b) whether octahedral \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) or tetrahedral \(\left[\mathrm{FeCl}_{4}\right]^{-}\) has the greater number of unpaired electrons.

Short answer

(a) The square planar complex ion \(\left[\mathrm{Cu}(\mathrm{py})_{4}\right]^{2+}\) is diamagnetic. (b) The octahedral \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) complex has more unpaired electrons than the tetrahedral \(\left[\mathrm{FeCl}_{4}\right]^{-}\) complex.

Step-by-step solution

Determine electron configuration of the central metal ion for Cu complex ion

First, we need to identify the charge and electron configuration of the central copper ion, Cu. As a metal in the transition series, Cu typically has an electron configuration of [Ar]4s2 3d9. However, in this complex ion, Cu is in its Cu2+ ion state, where it loses 2 electrons to become [Ar]3d9-2 = [Ar]3d7. However, copper is an exception and its stable electron configuration is actually [Ar]3d10 4s1. After losing two electrons, the copper ion will have an electron configuration of [Ar]3d9.

Predict magnetic properties of Cu complex ion

The complex ion is square planar, meaning the d-orbitals will be split. In a square planar complex, the electrons will fill the lower energy d-orbitals completely before occupying the higher energy d-orbitals. As a result, since all the d-orbitals are filled there are no unpaired electrons, meaning the complex will be diamagnetic.

Determine electron configuration of the central metal ion for Mn and Fe complex ions

The Mn ion in this complex is in its Mn3+ state and the Fe ion is in its Fe- state. The electron configurations are [Ar]4s2 3d5 for Mn and [Ar]4s2 3d6 for Fe. The Mn3+ ion will have lost 3 electrons from the 4s and 3d orbitals, leading to a configuration of [Ar]3d4, while the Fe- ion will have gained an electron, leading to a configuration of [Ar]3d7.

Predict number of unpaired electrons in Mn and Fe complex ions

For the octahedral Mn complex, the d-orbitals will be split into two groups, with the 3 lower energy orbitals filled first. This results in 4 unpaired electrons. For the tetrahedral Fe complex, the d-orbitals will be split with the higher energy orbitals filled first. This results in 3 unpaired electrons.

Key concepts

Electron Configuration

In transition metal complexes, understanding the electron configuration of the central metal ion is crucial. This configuration determines many of the chemical and physical properties of the metal complex. Let's take a closer look at the copper ion in the square-planar complex \( \left[\mathrm{Cu}(\mathrm{py})_{4}\right]^{2+} \). Copper in its neutral state has the electron configuration \([\text{Ar}] \ 3d^{10} 4s^1\). However, when it forms a \(\text{Cu}^{2+}\) ion, it loses two electrons, resulting in \([\text{Ar}] \ 3d^9\). It’s important to note that copper and similar transition metals often have unconventional electron configurations due to stability preferences.

• Neutral copper: \([\text{Ar}] \ 3d^{10} 4s^1\)
• Copper in \(\text{Cu}^{2+}\): \([\text{Ar}] \ 3d^9\)

Similarly, when predicting electron configurations for other metals like Mn and Fe in complexes, consider their oxidation states. Mn in \( \left[\mathrm{Mn} (\mathrm{CN})_{6}\right]^{3-}\) exists as \(\text{Mn}^{3+}\), with a configuration of \([\text{Ar}] \ 3d^4\). Fe in \( \left[\mathrm{FeCl}_{4}\right]^-\) exists as \(\text{Fe}^{-}\), gaining an electron to become \([\text{Ar}] \ 3d^7\). Understanding these configurations lays the groundwork for predicting other properties like magnetism.

Magnetic Properties

Magnetic properties of a complex are deeply linked to its electron configuration and arrangement. A key aspect to consider is the effect of the ligand field, especially in geometries such as square planar or tetrahedral. In a square planar complex like \( \left[\mathrm{Cu}(\mathrm{py})_{4}\right]^{2+} \), the d-orbitals split into different energy levels.

• The lower energy orbitals fill completely before the higher ones.
• With all these filled, no unpaired electrons remain, rendering the complex diamagnetic.

Octahedral and tetrahedral arrangements differ in splitting patterns:

• In an octahedral field, the d-orbitals are split into t₂g and eₙg levels. Lower levels fill first, often resulting in paired and unpaired configurations.
• Tetrahedral fields split with reversed priority, often leading to fewer unpaired electrons due to quicker filling of lower energy orbitals.

In summary, the geometry and configuration interplay of electrons within the metal complex influence its magnetic property, determining if it is paramagnetic, with unpaired electrons, or diamagnetic, with paired electrons.

Unpaired Electrons

The presence of unpaired electrons in transition metal complexes often dictates their magnetic behavior. When examining how many unpaired electrons exist, it's tied directly to how the d-orbitals are occupied by electrons.

• In the octahedral \( \left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\), the Mn \(\text{Mn}^{3+}\) ion has an electron configuration of \([\text{Ar}] \ 3d^4\).The splitting of the d-orbitals fills lower energy levels first, typically resulting in a higher number of unpaired electrons—in this case, four.
• Conversely, for the tetrahedral \( \left[\mathrm{FeCl}_{4}\right]^{-}\), Fe \(\text{Fe}^{-}\) also has unpaired electrons but fewer, with a configuration of \([\text{Ar}] \ 3d^7\).Here, three unpaired electrons are present due to how the d-orbitals split and fill at lower field strengths.

Thus, the number of unpaired electrons arises from both the specific electron configuration and the geometrical shape of the complex, influencing magnetic properties and behavior.