Problem 46 - Chemistry: The Central Science

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 \(),(\mathrm{e})\left[\mathrm{PtBr}_{6}\right]^{2-},(\mathrm{f})\left[\mathrm{Ti}(\mathrm{en})_{3}\right]^{2+}\).

Step-by-Step Solution

Answer

The crystal-field energy level diagrams and electrons placement for the given complexes are the following: a. $[\text{VCl}_6]^{3-}$: High-spin octahedral; electron configuration $d_{xy}: 1 \, e^-$, $d_{xz}: 1 \, e^-$, $d_{yz}: 0 \, e^-$. b. $[\text{FeF}_6]^{3-}$: High-spin octahedral; electron configuration $d_{xy}: 1\, e^-$, $d_{xz}: 1 \, e^-$, $d_{yz}: 3\, e^-$. c. $[\text{Ru(bipy)}_3]^{3+}$: Low-spin octahedral; electron configuration $d_{xy}: 5\, e^-$, $d_{xz}: 0 \, e^-$, $d_{yz}: 0 \, e^-$. d. $[\text{NiCl}_4]^{2-}$: High-spin tetrahedral; electron configuration $d_{x^2-y^2}: 8\, e^-$. e. $[\text{PtBr}_6]^{2-}$: High-spin octahedral; electron configuration $d_{xy}: 2\, e^-$, $d_{xz}: 4\, e^-$, $d_{yz}: 0\, e^-$. f. $[\text{Ti(en)}_3]^{2+}$: Low-spin octahedral; electron configuration $d_{xy}: 2\, e^-$, $d_{xz}: 0\, e^-$, $d_{yz}: 0\, e^-$.

Step 1: Determine the ligand type and coordination geometry

: For each complex, we'll identify the ligand type (strong or weak) and coordination geometry (tetrahedral or octahedral). This can be done by observing the ligands and the overall structure of each complex: a. \([\text{VCl}_6]^{3-}\): Chloride, weak ligand; octahedral geometry. b. \([\text{FeF}_6]^{3-}\): Fluoride, weak ligand; octahedral geometry. c. \([\text{Ru}(\text{bipy})_3]^{3+}\): Bipyridine, strong ligand; octahedral geometry. d. \([\text{NiCl}_4]^{2-}\): Chloride, weak ligand; tetrahedral geometry. e. \([\text{PtBr}_6]^{2-}\): Bromide, weak ligand; octahedral geometry. f. \([\text{Ti}(\text{en})_3]^{2+}\): Ethylenediamine, strong ligand; octahedral geometry

Step 2: Determine the metal ion and its oxidation state

: For each complex, identify the metal ion and determine its oxidation state: a. \([\text{VCl}_6]^{3-}\): vanadium (V) ion; oxidation state: +3. b. \([\text{FeF}_6]^{3-}\): iron (Fe) ion; oxidation state: +3. c. \([\text{Ru}(\text{bipy})_3]^{3+}\): ruthenium (Ru) ion; oxidation state: +3. d. \([\text{NiCl}_4]^{2-}\): nickel (Ni) ion; oxidation state: +2. e. \([\text{PtBr}_6]^{2-}\): platinum (Pt) ion; oxidation state: +4. f. \([\text{Ti}(\text{en})_3]^{2+}\): titanium (Ti) ion; oxidation state: +2.

Step 3: Identify the appropriate energy-level diagram and fill electron

: a. \([\text{VCl}_6]^{3-}\): Vanadium(+3) has electronic configuration as \([Ar] 3d^2\). As the geometry is octahedral and ligands are weak, it is a high-spin complex. Filling the electron in energy level diagram: d(xy) __ d(xz) __ d(yz) __ d(x2-y2) __ d(z2) | | | ↓ ↓ ↓ 1 e^⁻- 1 e^⁻- 0 e^⁻- b. \([\text{FeF}_6]^{3-}\): Iron(+3) has electronic configuration as \([Ar] 3d^5\). Given, FeF6^{3-} is a high-spin complex so the energy level diagram and filling the electron: d(xy) __ d(xz) __ d(yz) __ d(x2-y2) __ d(z2) | | | ↓ ↓ ↓ 1 e^⁻- 1 e^⁻- 3 e^⁻- c. \([\text{Ru(bipy)_3}]^{3+}\): Ruthenium(+3) has electronic configuration as \([Kr] 4d^5\]. Given, Ru(bipy)_3^{3+} is a low-spin complex, because bipyridine is a strong ligand, and the energy level diagram and filling the electron: d(xy) __ d(xz) __ d(yz) __ d(x2-y2) __ d(z2) | | | ↓ ↓ ↓ 5 e^⁻- 0 e^⁻- 0 e^⁻- d. \([\text{NiCl}_4]^{2-}\): Nickel(+2) has electronic configuration as [Ar] 3d^8. The geometry is tetrahedral and ligands are weak. Filling the electrons in the energy-level diagram: d(x2-y2) __ d(z2) __d(xy) __ d(xz) __ d(yz) | ↓ 8 e^⁻- e. \([\text{PtBr}_6]^{2-}\): Platinum(+4) has electronic configuration as \([Xe] 4f^{14}5d^6\). Given, PtBr6^{2-} is a high-spin complex and geometry is octahedral. Filling the electron in energy-level diagram: d(xy) __ d(xz) __ d(yz) __ d(x2-y2) __ d(z2) | | | ↓ ↓ ↓ 2 e^⁻- 4 e^⁻- 0 e^⁻- f. \([\text{Ti(en)}_3]^{2+}\): Titanium(+2) has electronic configuration as \([Ar] 3d^2\). Given, Ti(en)_3^{2+} is a low-spin complex due to strong ligand, geometry is octahedral. Filling the electron in energy-level diagram: d(xy) __ d(xz) __ d(yz) __ d(x2-y2) __ d(z2) | ↓ 2 e^⁻-

Key Concepts

Coordination Complexes

Coordination complexes are fascinating structures formed by a central metal atom or ion bonded to surrounding molecules or ions, known as ligands. These ligands donate electron pairs to the metal center, creating a coordinate covalent bond. The nature and number of these ligands determine the geometry and properties of the complex. Common geometries include octahedral, tetrahedral, and square planar. In an octahedral complex, the metal is at the center of an octahedron, coordinated by six ligands. For example, \[ \text{[FeF}_6\text{]}^{3-} \] is an octahedral complex. In tetrahedral coordination, four ligands surround the metal, as seen in \[ \text{[NiCl}_4\text{]}^{2-} \]. Coordination complexes play a vital role in fields such as catalysis, biochemistry, and industrial chemistry, illustrating the versatility and importance of these compounds.

Electron Configuration

Electron configuration is crucial for understanding the behavior of coordination complexes. It describes the distribution of electrons in the atomic or molecular orbitals of the central metal ion. When the oxidation state and ligand arrangement of a metal are known, we can determine its electron configuration. For vanadium in \[ \text{[VCl}_6\text{]}^{3-} \], the configuration is \[ \text{[Ar] 3d}^{2} \], as it has lost three electrons. Similarly, iron in \[ \text{[FeF}_6\text{]}^{3-} \] has a configuration of \[ \text{[Ar] 3d}^{5} \], reflecting its higher oxidation state. This information helps predict the magnetic properties and reactivity of the complexes. Electron configurations also influence the color of these complexes due to the d-d transitions and ligand-to-metal charge transfer.

Ligand Field

The ligand field theory helps explain how different ligands affect the energy levels of the d-orbitals in a coordination complex. When ligands approach a metal ion, they interact with its d-orbitals, splitting their energy levels. The extent of this splitting depends on the strength of the ligand field. Strong field ligands, like bipyridine in \[ \text{Ru(bipy)}_3^{3+} \], create a large energy gap between the split d-orbitals, often leading to "low-spin" complexes where electrons pair up in lower energy orbitals. Weak field ligands, such as fluoride and chloride in \[ \text{[FeF}_6\text{]}^{3-} \] and \[ \text{[VCl}_6\text{]}^{3-} \], result in a smaller gap, leading to "high-spin" complexes where electrons remain unpaired to minimize electron repulsion.

Oxidation States

Understanding oxidation states is crucial for knowing the electron count and overall charge balance in coordination complexes. The oxidation state of a metal ion is the hypothetical charge it would hold if all its ligands were removed. Calculating this requires considering the charges of the ligands and the overall charge of the complex. In \[ \text{[VCl}_6\text{]}^{3-} \], each chloride ion has a charge of \(-1\), with the metal vanadium in a +3 oxidation state. For \[ \text{[PtBr}_6\text{]}^{2-} \], platinum is in a +4 state, with bromide ions each carrying a \(-1\) charge. Identifying the oxidation state helps predict reactivity and provides insight into the type of bonding and electronic structure within the complex.