Problem 61 - Chemistry The Central Science

Question

Draw the crystal-field energy-level diagrams and show the placement of \(d\) electrons for each of the following: (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (four unpaired electrons), (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (high spin), (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{H}_{2} \mathrm{O}\right]^{2+}\) (low spin), (d) \(\left[\operatorname{Ir} \mathrm{Cl}_{6}\right]^{2-}\) (low spin), (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+},(\mathrm{f})\left[\mathrm{NiF}_{6}\right]^{4-}\)

Step-by-Step Solution

Answer

In summary, the crystal-field energy-level diagrams for each complex are as follows: a) [Cr(H2O)6]2+: 3d electrons placed as ↑, ↑, ↑, 0, 0. b) [Mn(H2O)6]2+: 3d electrons placed as ↑, ↑, ↑, ↑, ↑. c) [Ru(NH3)5H2O]2+: 4d electrons placed as ↑↓, ↑↓, ↑↓, ↑, 0. d) [IrCl6]2-: 5d electrons placed as ↑↓, ↑↓, ↑↓, ↑, 0. e) [Cr(en)3]3+: 3d electrons placed as ↑↓, ↑, 0, 0, 0. f) [NiF6]4-: 3d electrons placed as ↑↓, ↑↓, ↑↓, ↑↓, ↑↑.

Step 1: a) [Cr(H2O)6]2+

: 1. Determine the oxidation state of the Cr atom: Since water has a net charge of 0, the charge on the complex is solely due to the Cr atom. Therefore, the oxidation state of Cr is +2. 2. Determine the number of \(d\) electrons: In the ground state, Cr has an electron configuration of [Ar] \(3d^5 4s^1\). Since the oxidation state of Cr is +2, we must remove two electrons, resulting in a \(3d^3\) configuration. 3. Crystal-field energy-level diagram: As water is a weak field ligand, the system will be high-spin. Following Hund's rule, we will place the three \(d\) electrons as follows:↑ , ↑ , ↑ , 0 , 0.

Step 2: b) [Mn(H2O)6]2+

: 1. Determine the oxidation state of the Mn atom: The oxidation state of Mn is +2. 2. Determine the number of \(d\) electrons: In the ground state, Mn has an electron configuration of [Ar] \(3d^5 4s^2\). After removing two electrons, we are left with a \(3d^5\) configuration. 3. Crystal-field energy-level diagram: As the system is high-spin and water is a weak-field ligand, we will distribute the \(d\) electrons as follows:↑ , ↑ , ↑ , ↑ , ↑.

Step 3: c) [Ru(NH3)5H2O]2+

: 1. Determine the oxidation state of the Ru atom: The oxidation state of Ru is +2. 2. Determine the number of \(d\) electrons: In the ground state, Ru has an electron configuration of [Kr] \(4d^7 5s^1\). After removing two electrons, we are left with a \(4d^6\) configuration. 3. Crystal-field energy-level diagram: As the system is low-spin and NH3 is a strong-field ligand, we will distribute the \(d\) electrons as follows:↑↓ , ↑↓ , ↑↓ , ↑ , 0.

Step 4: d) [IrCl6]2-

: 1. Determine the oxidation state of the Ir atom: The oxidation state of Ir is +3. 2. Determine the number of \(d\) electrons: In the ground state, Ir has an electron configuration of [Xe] \(4f^{14} 5d^7 6s^2\). After removing three electrons, we are left with a \(5d^6\) configuration. 3. Crystal-field energy-level diagram: As the system is low-spin and Cl- is a strong-field ligand, we will distribute the \(d\) electrons as follows:↑↓ , ↑↓ , ↑↓ , ↑ , 0.

Step 5: e) [Cr(en)3]3+

: 1. Determine the oxidation state of the Cr atom: The oxidation state of Cr is +3. 2. Determine the number of \(d\) electrons: In this case, we are left with a \(3d^2\) configuration. 3. Crystal-field energy-level diagram: As the system has only two \(d\) electrons and en is a strong-field ligand, we will distribute the \(d\) electrons as follows:↑↓ , ↑ , 0 , 0 , 0.

Step 6: f) [NiF6]4-

: 1. Determine the oxidation state of the Ni atom: The oxidation state of Ni is +2. 2. Determine the number of \(d\) electrons: In the ground state, Ni has an electron configuration of [Ar] \(3d^8 4s^2\). After removing two electrons, we are left with a \(3d^8\) configuration. 3. Crystal-field energy-level diagram: As the system has eight \(d\) electrons and F- is a strong-field ligand, we will distribute the \(d\) electrons as follows:↑↓ , ↑↓ , ↑↓ , ↑↓ , ↑↑.

Key Concepts

Ligand Field Theory

Ligand Field Theory (LFT) is an extension of Crystal Field Theory (CFT) and provides a more comprehensive understanding of how ligands interact with transition metal complexes. While CFT primarily considers the electrostatic interaction between ligands and metal ions, LFT includes covalent bonding aspects as well which makes it applicable to a wider range of complexes.
When ligands approach a transition metal ion, they affect the energy of the metal's d-orbitals. This is because different ligands create varying degrees of splitting in these d-orbitals, depending on their strength as a field (referred to as either strong-field or weak-field ligands).
**In simple terms:**

• Weak-field ligands lead to smaller splitting of d-orbitals, frequently resulting in high-spin configurations where electrons remain unpaired.
• Strong-field ligands cause larger splitting, often resulting in low-spin states where electrons pair up in the lower energy orbitals.

By understanding Ligand Field Theory, you can determine if a complex has high-spin or low-spin electron configurations, which directly affects its magnetic and spectroscopic properties.

Electron Configuration

Electron Configuration is a way to describe the distribution of electrons in an atom or molecule's orbitals. For transition metals, knowing the electron configuration is essential as it helps to understand their chemical behaviors.
In the context of ligand field theory and transition metal complexes, electron configuration determines how d-electrons are distributed among the energy levels that get split upon ligand interaction.
**To determine the electron configuration:**

• First, find the electron configuration of the neutral metal atom.
• Determine the oxidation state in the complex which tells us how many electrons have been removed.
• Adjust the electron count based on the metal’s oxidation state in the given complex.
• Use ligand field theory to determine the arrangement of these electrons in split d-orbitals.

Importantly, this concept helps predict the magnetic properties and color of coordination compounds. High-spin complexes, with unpaired electrons, show paramagnetism, while low-spin ones, with paired electrons, tend to be diamagnetic.

Transition Metal Complexes

Transition metal complexes are formed by the interaction of transition metal ions with surrounding molecules or ions, known as ligands. These complexes play crucial roles in various chemical processes and have diverse applications.
One of the unique features of these complexes is the variability in oxidation states and the ability to form colorful compounds. The nature of the ligands and the type of complexation affect the properties such as magnetism and color.
**Understanding Transition Metal Complexes involves:**

• Identifying the central metal ion and its oxidation state.
• Understanding the geometry of the complex which could be octahedral, tetrahedral, or square planar.
• Determining the type and strength of the ligands which affects the electron distribution according to ligand field theory.

By examining these factors, the properties and reactivity of transition metal complexes can be better understood, allowing us to predict their behavior in different chemical environments. This understanding is foundational for various fields such as bioinorganic chemistry, catalysis, and materials science.