Problem 45 - Chemistry 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) \(\left[\mathrm{IrCl}_{6}\right]^{2-}\) (low spin), (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\), (f) \(\left[\mathrm{NiF}_{6}\right]^{4-}\)

Step-by-Step Solution

Answer

A short version of the answer is: (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\): Cr²⁺ ion with 3d⁴ configuration; high spin; energy-level diagram: ↑↓ ↑ ↑ ↑. (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\): Mn²⁺ ion with 3d⁵ configuration; high spin; energy-level diagram: ↑↑↑↑↑. (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{H}_{2} \mathrm{O}\right]^{2+}\): Ru²⁺ ion with 4d⁴ configuration; low spin; energy-level diagram: ↑↓ ↑↓ ↑. (d) \(\left[\mathrm{IrCl}_{6}\right]^{2-}\): Ir⁴⁺ ion with 5d⁵ configuration; low spin; energy-level diagram: ↑↓ ↑↓ ↑↓. (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\): Cr³⁺ ion with 3d³ configuration; low spin (assumed); energy-level diagram: ↑↓ ↑. (f) \(\left[\mathrm{NiF}_{6}\right]^{4-}\): Ni²⁺ ion with 3d⁸ configuration; high spin (assumed); energy-level diagram: ↑↓ ↑↓ ↑↑↑.

Step 1: Identify the oxidation state

Cr has an oxidation state of +2, as shown by the 2+ charge on the complex.

Step 2: Calculate the number of d-electrons

Cr has an atomic number of 24 (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹). In the +2 oxidation state, Cr loses 2 electrons, leaving it with 3d⁴ configuration.

Step 3: Draw the energy-level diagram and place electrons

Given there are four unpaired electrons, this complex is high spin. So, in an octahedral complex, the d-electrons fill as follows: ↑↓ ↑ ↑ ↑. The diagram will have three lower energy levels filled and one higher energy level filled, giving four unpaired electrons. (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (high spin)

Step 1: Identify the oxidation state

Mn has an oxidation state of +2, as indicated by the 2+ charge on the complex.

Step 2: Calculate the number of d-electrons

Mn has an atomic number of 25 (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²). In the +2 oxidation state, Mn loses 2 electrons from the 4s orbital, leaving it with 3d⁵ configuration.

Step 3: Draw the energy-level diagram and place electrons

Because this complex is high spin, we place the d-electrons in an octahedral complex as follows: ↑↑↑↑↑. The diagram will have three lower energy levels and two higher energy levels filled. All electrons are unpaired. (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{H}_{2} \mathrm{O}\right]^{2+}\) (low spin)

Step 1: Identify the oxidation state

Ru has an oxidation state of +2, as indicated by the 2+ charge on the complex.

Step 2: Calculate the number of d-electrons

Ru has an atomic number of 44 (1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d⁴). In the +2 oxidation state, Ru loses 2 electrons from the 5s orbital, leaving it with 4d⁴ configuration.

Step 3: Draw the energy-level diagram and place electrons

Since this complex is low spin, we place the d-electrons in an octahedral complex as follows: ↑↓ ↑↓ ↑. The diagram will have three lower energy levels completely filled and no electrons in the higher energy levels. (d) \(\left[\mathrm{IrCl}_{6}\right]^{2-}\) (low spin)

Step 1: Identify the oxidation state

Since each chloride contributes -1 and there are six of them, the oxidation state of Ir must be +4 to balance the 2- charge on the complex.

Step 2: Calculate the number of d-electrons

Ir has an atomic number of 77 (1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 5d⁷). In the +4 oxidation state, Ir loses 2 electrons from the 6s and 2 electrons from the 5d orbital, leaving it with 5d⁵ configuration.

Step 3: Draw the energy-level diagram and place electrons

As this complex is low spin, the d-electrons in the octahedral complex fill as follows: ↑↓ ↑↓ ↑↓. The diagram will have three lower energy levels completely filled and two electrons paired in one higher energy level. (e) \(\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}\)

Step 1: Identify the oxidation state

Cr has an oxidation state of +3, as shown by the 3+ charge on the complex.

Step 2: Calculate the number of d-electrons

Cr has an atomic number of 24 (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹). In the +3 oxidation state, Cr loses 3 electrons, leaving it with 3d³ configuration.

Step 3: Draw the energy-level diagram and place electrons

As no specific spin configuration is given, assume low spin considering the en ligand is strong field. In an octahedral complex, the d-electrons fill as follows: ↑↓ ↑. The diagram will have two lower energy levels filled and no electrons in the higher energy levels. (f) \(\left[\mathrm{NiF}_{6}\right]^{4-}\)

Step 1: Identify the oxidation state

Since each fluoride contributes -1 and there are six of them, the oxidation state of Ni must be +2 to balance the 4- charge on the complex.

Step 2: Calculate the number of d-electrons

Ni has an atomic number of 28 (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²). In the +2 oxidation state, Ni loses 2 electrons from the 4s orbital, leaving it with 3d⁸ configuration.

Step 3: Draw the energy-level diagram and place electrons

As no specific spin configuration is given, assume high spin considering the F ligand is weak field. In an octahedral complex, the d-electrons fill as follows: ↑↓ ↑↓ ↑↑↑. The diagram will have three lower energy levels and two higher energy levels filled, giving two unpaired electrons.

Key Concepts

d-Electron Configuration

Understanding the d-electron configuration is crucial when exploring transition metal complexes. The d-orbitals can hold up to ten electrons, and their arrangement plays a pivotal role in the properties and color of the complexes.

When transition metals form complexes, they do so in specific oxidation states, impacting the remaining number of electrons in their d-orbitals. For example, a chromium(II) ion, noted as Cr²⁺, would have a d⁴ electron configuration since chromium normally has six d-electrons (3d⁶) in its neutral state, and two are removed due to the +2 charge.

Ligands, the molecules or ions attached to the metal in a complex, influence electron distribution among d-orbitals. Strong field ligands, such as water or ammonia, can cause electrons to pair up in lower energy d-orbitals, while weak field ligands, like halides, typically result in high spin configurations with more unpaired electrons.

Octahedral Complexes

In octahedral complexes, where a central metal ion is surrounded by six ligands positioned at the corners of an octahedron, the d-orbitals split into two sets of differing energy levels due to the electric field of the surrounding ligands.

The crystal-field splitting in an octahedral complex divides the five degenerate d-orbitals into two groups—three lower-energy t_2g and two higher-energy e_g orbitals. The pattern in which the d-electrons fill these orbitals gives insight into the magnetic and spectroscopic properties of the complex. A key to mastering complex formation is visualizing the energy-level diagram and accurately distributing the d-electrons according to the Hund's rule and the Pauli exclusion principle.

For instance, a high spin [Mn(H₂O)₆]²⁺ complex features five unpaired electrons arranged as ↑↑↑↑↑, occupying both the t_2g and e_g orbitals with the maximum number of unpaired electrons.

High Spin and Low Spin Complexes

High spin and low spin complexes refer to the two possible arrangements of d-electrons based on the strength of the ligand field due to the surrounding ligands. High spin complexes occur with weak field ligands and maximize the number of unpaired electrons, leading to magnetic complexes.

For example, [NiF₆]^4-, a high spin complex, has electrons in both t_2g and e_g orbitals since fluorides are considered to produce a weaker crystal field, resulting in a configuration of ↑↓ ↑↓ ↑↑↑, with two unpaired electrons.

Conversely, low spin complexes form with strong field ligands, which have a larger splitting energy that favors pairing of electrons in the lower t_2g orbitals, thus resulting in complexes with fewer unpaired electrons. Such low spin configurations are often non-magnetic. [IrCl₆]^2- illustrates this with its 5d⁵ electron configuration, creating a low spin arrangement with all d-electrons paired: ↑↓ ↑↓ ↑↓.