Question

Give the electron distribution in low-spin and/or high-spin complexes of (a) \(\mathrm{Ru}^{4+}\) (b) \(\mathrm{Pt}^{2+}\)

Short answer

In summary, the electron distribution for low-spin and high-spin complexes for the given ions is as follows: (a) Ru^4+ (same for both low-spin and high-spin complexes): \((t2g)^2 (e_g)^0\) (b) Pt^2+ (low-spin complex): \((t2g)^6 (e_g)^1\) Pt^2+ is difficult to have high-spin complexes because it forms square-planar complexes, where it is typically low-spin.

Step-by-step solution

Determine the electron configuration of the neutral elements

First, we need to determine the electron configuration of the neutral atoms of Ru (Ruthenium) and Pt (Platinum). For Ruthenium (Ru) with an atomic number of 44, the electron configuration is: \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^6\) For Platinum (Pt) with an atomic number of 78, the electron configuration is: \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \, 6s^2 \, 4f^{14} \, 5d^9\)

Determine the electron configuration of the given ions

The given ions are Ru^4+ and Pt^2+. We will remove the required number of electrons from the neutral atoms to get the electron configuration of these ions. For Ru^4+, remove 4 electrons from the neutral atom: \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^2\) For Pt^2+, remove 2 electrons from the neutral atom: \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \, 6s^2 \, 4f^{14} \, 5d^7\)

Determine electron distribution in low-spin and high-spin complexes

Using the crystal field theory, we will determine the electron distribution for low-spin and high-spin complexes. (a) \(\boldsymbol{Ru^{4+}}\) Electron configuration: \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^2\) For low-spin complexes, the electrons preferentially fill in the lower energy states (t2g orbitals), staying unpaired: Low-spin: \((t2g)^2 (e_g)^0\) For high-spin complexes, electrons preferentially remain unpaired, even in some higher-energy states (e_g orbitals): High-spin: \((t2g)^2 (e_g)^0\) As a strong field ligand, Ru^4+ does not form high-spin complexes, so the electron distribution remains the same for both low-spin and high-spin complexes. (b) \(\boldsymbol{Pt^{2+}}\) Electron configuration: \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \, 6s^2 \, 4f^{14} \, 5d^7\) For low-spin complexes, the electrons preferentially fill in the lower energy states (t2g orbitals), staying unpaired: Low-spin: \((t2g)^6 (e_g)^1\) For high-spin complexes, electrons preferentially remain unpaired, even in some higher-energy states (e_g orbitals): High-spin: \((t2g)^5 (e_g)^2\) For Pt^2+, it is difficult to have high-spin complexes because it forms square-planar complexes, where it is typically low-spin.

Key concepts

Electron Configuration

Understanding electron configuration is fundamental when studying coordination chemistry and crystal field theory. An atom's electron configuration lays the groundwork for how it will interact with other atoms when forming complexes. In complexes, metal ions are surrounded by ligands, which can alter the energy of d-orbitals, thus affecting the electron distribution.

For example, when iron (Fe) loses two electrons to form an Fe^{2+} ion, its electron configuration adjusts accordingly. This subtle shift can dramatically change the way this ion interacts with ligands in a complex. Instructors often use the analogy of a hotel with rooms (orbitals) being filled by guests (electrons) to simplify this concept for students. Just as there are rules for which rooms guests fill first in a hotel, there are rules for how electrons fill orbitals.

Low-Spin and High-Spin Complexes

Distinguishing between low-spin and high-spin complexes is essential in crystal field theory. These terms describe the distribution of electrons in a metal ion's d-orbitals in the presence of a crystal field—created by the ligands surrounding the metal ion. Whether a complex is low-spin or high-spin depends on the nature of the ligands; those causing a large splitting of d-orbital energies lead to low-spin complexes, while ligands causing a smaller splitting result in high-spin complexes.

This is relevant in magnetic properties and color. For instance, a low-spin complex might have all its electrons paired in the lower energy orbitals, minimizing repulsions and magnetic interactions. On the other hand, high-spin complexes will have more unpaired electrons, leading to stronger magnetic properties.

To help students visualize this, one might compare it to a party where some guests (electrons) prefer to sit closer together (low-spin) while others prefer to maintain distance and sit further apart (high-spin).

Ligand Field Stabilization

Ligand field stabilization is a key aspect of crystal field theory, directly impacting the stability and formation of complexes. Ligands create an electric field around the metal ion that splits the d-orbitals into different energy levels. In some complexes, this leads to lower energy arrangements of electrons, resulting in a more stabilized and preferred structure.

For instance, in an octahedral complex, ligands will generate a crystal field that splits the five d-orbitals into two sets with different energies—t2g of lower energy and eg of higher energy. This creates a stabilization energy for the complex, known as the ligand field stabilization energy (LFSE). Strongly electronegative ligands can increase LFSE, leading to more stable, low-spin configurations.

To illustrate LFSE in the classroom, it can be likened to a discount system; stronger field ligands offer a 'greater discount' on energy, making the 'purchase' of a low-energy configuration more 'cost-effective' for the metal ion.