Problem 57 - General Chemistry

Question

Using crystal field theory, sketch the energy-level diagram for the \(d\) orbitals in an octahedral field; then fill in the electrons for the metal ion in each of the following complexes. How many unpaired electrons are there in each case? a. \(\left[\mathrm{V}(\mathrm{CN})_{6}\right]^{3-}\) b. \(\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{4-}\) (high-spin) C. \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}\) (low-spin)

Step-by-Step Solution

Answer

a: 0 unpaired electrons; b: 3 unpaired electrons; c: 0 unpaired electrons.

Step 1: Understanding Crystal Field Splitting

In an octahedral field, the metal ion's five degenerate \(d\) orbitals split into two groups due to the electrostatic interactions with the six ligands arranged symmetrically around the metal ion. The higher energy group contains the \(e_g\) orbitals (\(d_{x^2-y^2}\), \(d_{z^2}\)), and the lower energy group contains the \(t_{2g}\) orbitals (\(d_{xy}\), \(d_{yz}\), \(d_{xz}\)). This energy split is known as the crystal field splitting energy, denoted by \(\Delta_o\).

Step 2: Filling Electrons - Compound a

For \([\text{V(CN)}_6]^{3-}\), the metal ion is \(\text{V}^{3+}\) which has an electron configuration of \([\text{Ar}]3d^2\). CN- is a strong field ligand, resulting in a low-spin configuration. Electrons fill the lower energy \(t_{2g}\) orbitals first:- \(t_{2g}\): 2 electrons, both paired - \(e_g\): 0 electronsThus, there are no unpaired electrons.

Step 3: Filling Electrons - Compound b

For \([\text{Co}({\text{C}_2\text{O}_4})_3]^{4-}\), the metal ion is \(\text{Co}^{2+}\) which has an electron configuration of \([\text{Ar}]3d^7\). \({\text{C}_2\text{O}_4}^{2-}\) is a moderate field ligand, and the problem indicates a high-spin configuration. Electrons fill as follows:- \(t_{2g}\): 3 electrons, all unpaired - \(e_g\): 4 electrons, having 2 unpairedTotal unpaired electrons: 3.

Step 4: Filling Electrons - Compound c

For \([\text{Mn(CN)}_6]^{3-}\), the metal ion is \(\text{Mn}^{3+}\) which has an electron configuration of \([\text{Ar}]3d^4\). Since CN- is a strong field ligand, it results in a low-spin configuration. Electrons fill the lower energy \(t_{2g}\) orbitals first:- \(t_{2g}\): 4 electrons, all paired- \(e_g\): 0 electronsThus, there are no unpaired electrons.

Key Concepts

Octahedral Field

In crystal field theory, an octahedral field is one where six ligands symmetrically surround a metal ion, typically forming an octahedron. This symmetrical arrangement impacts the energy levels of the metal ion's five degenerate (i.e., equal energy) d orbitals. These interactions cause the orbitals to split in energy, creating two distinct groups. Understanding this octahedral field is crucial for predicting the behavior of metal ions in various complexes.

• Six ligands create a strong electrostatic field around the metal ion.
• Splitting results in two different energy levels.
• This splitting varies with the strength of the ligands.

The precise energy difference between these groups in the octahedral field is vital for understanding electronic structure and chemical properties. This difference is known as the crystal field splitting energy, \( \,\Delta_o \)).

Energy-Level Diagram

In an octahedral field, the energy-level diagram reveals how d orbitals split into two groups: \(t_{2g}\) and \(e_g\). This energy separation is due to the symmetrical arrangement of ligands that create differences in electrostatic interactions with the orbitals. It is essential to understand these differences when predicting how electrons occupy these orbitals.

• \(t_{2g}\) orbitals (\(d_{xy}, d_{yz}, d_{xz}\)) are lower in energy.
• \(e_g\) orbitals (\(d_{x^2-y^2}, d_{z^2}\)) are higher in energy.

The difference in energy, \( \,\Delta_o \), is crucial in determining whether a complex is high-spin or low-spin, affecting electron arrangements and chemical properties.

d Orbitals

The d orbitals are the primary focus when examining how a metal ion interacts with surrounding ligands in a complex. In an octahedral field, these orbitals split into two sets, affecting how the electrons are distributed. Each d orbital has a distinctive shape impacting how it interacts with the ligands around it.Key points about d orbitals in octahedral fields:

• Five d orbitals, each with unique lobes and directional properties.
• Splitting into \(t_{2g}\) and \(e_g\) sets is due to spatial arrangement.
• Crucial for determining electron configuration of metal ion.

This differentiation impacts magnetic properties and ultimately influences how a complex behaves chemically. Understanding these properties aids in predicting how a metal ion will interact in various chemical contexts.

Unpaired Electrons

Unpaired electrons in a metal complex can greatly influence its magnetic properties. In the context of an octahedral field, the spin state — high-spin or low-spin — determines the number of unpaired electrons present. High-spin complexes tend to have more unpaired electrons, whereas low-spin complexes have fewer. Considerations for unpaired electrons:

• High-spin: ligands with weaker field strength allow for more unpaired electrons.
• Low-spin: strong field ligands like CN- pair electrons in lower energy orbitals, reducing unpaired counts.
• Directly affects magnetic properties like paramagnetism or diamagnetism.

This is pivotal for understanding the reactivity and color characteristics of different metal complexes, as these properties are deeply linked to the electron arrangement within the d orbitals.