Question

Draw the crystal-field energy-level diagrams and show the placement of electrons for the following complexes: (a) \(\left[\mathrm{VCl}_{6}\right]^{3-}\), (b) \(\left[\mathrm{FeF}_{6}\right]^{3-}\) (a high-spin complex), (c) \(\left[\mathrm{Ru}(\mathrm{bipy})_{3}\right]^{3+}\) (a low-spin complex), (d) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) (tetrahedral), (e) \(\left[\mathrm{PtBr}_{6}\right]^{2-}\), (f) [Ti(en) \(\left._{3}\right]^{2+}\).

Short answer

The crystal-field energy-level diagrams and electron placements for the given complexes are as follows: (a) \([\text{VCl}_6]^{3-}\) (Octahedral): Electron configuration: \(t_{2g}^{2}e_{g}^{0}\) (b) \([\text{FeF}_6]^{3-}\) (Octahedral, high-spin): Electron configuration: \(t_{2g}^{3}e_{g}^{0}\) (c) \([\text{Ru}(\text{bipy})_3]^{3+}\) (Octahedral, low-spin): Electron configuration: \(t_{2g}^{4}e_{g}^{0}\) (d) \([\text{NiCl}_4]^{2-}\) (Tetrahedral): Electron configuration: \(e^{2}t_{2}^{4}\) (e) \([\text{PtBr}_6]^{2-}\) (Octahedral): Electron configuration: \(t_{2g}^{6}e_{g}^{0}\) (f) [\text{Ti}(\text{en})\(_3\)]^{2+} (Octahedral): Electron configuration: \(t_{2g}^{0}e_{g}^{0\)

Step-by-step solution

Identify oxidation state and d electron count

For each of the six complexes, identify the central metal ion, the oxidation state of that ion, and the number of d electrons it will have. We will use the given formulas to calculate this information. (a) \(\left[\mathrm{VCl}_{6}\right]^{3-}\) Metal ion: V (Vanadium) Oxidation state: +3 d electron count: \(5 - 3 = 2\) (b) \(\left[\mathrm{FeF}_{6}\right]^{3-}\) Metal ion: Fe (Iron) Oxidation state: +3 d electron count: \(6 - 3 = 3\) (c) \(\left[\mathrm{Ru}(\mathrm{bipy})_{3}\right]^{3+}\) Metal ion: Ru (Ruthenium) Oxidation state: +3 d electron count: \(7 - 3 = 4\) (d) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) Metal ion: Ni (Nickel) Oxidation state: +2 d electron count: \(8 - 2 = 6\) (e) \(\left[\mathrm{PtBr}_{6}\right]^{2-}\) Metal ion: Pt (Platinum) Oxidation state: +4 d electron count: \(10 - 4 = 6\) (f) [Ti(en) \(\left._{3}\right]^{2+}\) Metal ion: Ti (Titanium) Oxidation state: +2 d electron count: \(2 - 2 = 0\)

Determine the geometry of each complex

Using the information about the ligands provided in the exercise, determine the geometry of each complex. This will be either octahedral or tetrahedral. (a) \(\left[\mathrm{VCl}_{6}\right]^{3-}\): Octahedral (b) \(\left[\mathrm{FeF}_{6}\right]^{3-}\): Octahedral (c) \(\left[\mathrm{Ru}(\mathrm{bipy})_{3}\right]^{3+}\): Octahedral (d) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\): Tetrahedral (e) \(\left[\mathrm{PtBr}_{6}\right]^{2-}\): Octahedral (f) [Ti(en) \(\left._{3}\right]^{2+}\): Octahedral

Draw energy-level diagrams and place electrons

Now that we have determined the geometry and d electron count for each complex, we can draw the energy-level diagrams and place the electrons accordingly. (a) \(\left[\mathrm{VCl}_{6}\right]^{3-}\) (Octahedral, 2 d electrons): Place 2 electrons in the \(t_{2g}\) level. [Electron configuration: \(t_{2g}^{2}e_{g}^{0}\)] (b) \(\left[\mathrm{FeF}_{6}\right]^{3-}\) (Octahedral, 3 d electrons, high spin): Place 3 electrons in the \(t_{2g}\) level. [Electron configuration: \(t_{2g}^{3}e_{g}^{0}\)] (c) \(\left[\mathrm{Ru}(\mathrm{bipy})_{3}\right]^{3+}\) (Octahedral, 4 d electrons, low spin): Place 4 electrons in the \(t_{2g}\) level. [Electron configuration: \(t_{2g}^{4}e_{g}^{0}\)] (d) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) (Tetrahedral, 6 d electrons): Place 2 electrons in the \(e\) level and 4 electrons in the \(t_{2}\) level. [Electron configuration: \(e^{2}t_{2}^{4}\)] (e) \(\left[\mathrm{PtBr}_{6}\right]^{2-}\) (Octahedral, 6 d electrons): Place 6 electrons in the \(t_{2g}\) level. [Electron configuration: \(t_{2g}^{6}e_{g}^{0}\)] (f) [Ti(en) \(\left._{3}\right]^{2+}\) (Octahedral, 0 d electrons): No d electrons to place. [Electron configuration: \(t_{2g}^{0}e_{g}^{0}\)]

Key concepts

Crystal-Field Energy-Level Diagrams

Understanding crystal-field energy-level diagrams is crucial for chemistry students, especially when analyzing transition metal complexes. These diagrams display the energy separation between d orbitals when a transition metal ion is surrounded by ligands, forming a coordination complex.

In an octahedral field, the five d orbitals split into two groups: the lower-energy 't_2g' orbitals (d_xy, d_xz, d_yz) and the higher-energy 'e_g' orbitals (d_z², d_x²-y²). On the other hand, in a tetrahedral field, the e orbitals are lower in energy compared to the higher-energy 't₂' orbitals. These differences in energy levels stem from how the electron cloud of the metal ion interacts with the electric fields produced by the surrounding ligands.

For step-by-step solution improvement, let's take complex (a) \(\left[\mathrm{VCl}_{6}\right]^{3-}\) from our exercise, which is an octahedral complex with a vanadium ion having 2 d electrons. The crystal-field energy-level diagram for this complex shows the e_g orbitals at a higher level than the t_2g orbitals, and the 2 d electrons would occupy the t_2g orbitals according to Hund's rule, which mandates that each electron occupies a separate orbital before any pairing occurs.

Electron Configuration in Complexes

The pattern of electron configuration within a complex is fundamental to understanding its chemical behaviour. Electron configuration refers to how d electrons of the central metal ion are distributed among the available orbitals in the presence of a crystal field.

In high-spin and low-spin configurations, the number of unpaired electrons can vary even if the complexes have identical metal ions and charges. This is due to the strength of the ligands present, characterized as either strong-field (resulting in low-spin complexes) or weak-field (resulting in high-spin complexes).

To improve clarity in the step-by-step solution, let's consider complex (c) \(\left[\mathrm{Ru}(\mathrm{bipy})_{3}\right]^{3+}\), a low-spin octahedral complex. The strong-field ligand 'bipyridine' causes higher pairing energy, so the 4 d electrons completely fill the t_2g orbitals before any occupy the e_g orbitals (electron configuration: t_2g⁴e_g⁰). This contrasts with a high-spin complex (b) \(\left[\mathrm{FeF}_{6}\right]^{3-}\), where the weak-field fluoride ligands result in a setup where electrons fill higher energy e_g orbitals to maintain more unpaired spins (electron configuration: t_2g³e_g⁰).

Ligand Field Stabilization Energy

Ligand Field Stabilization Energy (LFSE) is the energy by which a complex becomes stabilized due to the splitting of d orbitals in a crystal field. The LFSE is a crucial concept for predicting the stability, color, and magnetic properties of coordination complexes.

LFSE can be calculated by considering the number of electrons in t_2g and e_g orbitals, the extent of the crystal-field splitting (\(\Delta\)), and whether the complex is high-spin or low-spin. It's important to note that LFSE is typically negative, signifying stabilization.

For an optimized solution, take the octahedral complex (d) \(\left[\mathrm{NiCl}_{4}\right]^{2-}\), which is a tetrahedral complex and thus has an inverted crystal field with respect to an octahedral complex. Its LFSE will be calculated using the t₂ and e orbitals, considering four electrons are present in the t₂ set, which is the higher energy set in this geometry, resulting in a less stabilized configuration compared to if it were octahedral. Reflecting on this energy consideration enables students to better appreciate why certain geometries and configurations result in more stable complexes.