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For a given metal ion and set of ligands, is the crystal-field splitting energy larger for a tetrahedral or an octahedral geometry?

Short Answer

Expert verified
For a given metal ion and set of ligands, the crystal-field splitting energy is larger for an octahedral geometry than for a tetrahedral geometry. This is because the ligands in an octahedral geometry are closer to the metal ion and interact more strongly with the d-orbitals, resulting in a larger energy difference between the lower and higher energy d-orbitals. The magnitude of the crystal-field splitting energy for a tetrahedral geometry is typically about 4/9 times that of an octahedral geometry with the same metal ion and ligands.

Step by step solution

01

Understand crystal-field theory

Crystal field theory is a model used to explain the electronic structure and properties of transition metal complexes. In this theory, the interaction between a metal ion and the ligands is considered to be purely electrostatic. When ligands approach a metal ion, they interact with the metal's d-orbitals and create an uneven pattern of repulsion, leading to a splitting of the d-orbitals into higher and lower energy levels.
02

Compare the geometries

Tetrahedral geometry has four ligands surrounding the metal ion in a tetrahedral arrangement. In this geometry, the d-orbitals split into two sets: the lower energy d_xy, d_yz, and d_xz, and the higher energy d_x^2-y^2 and d_z^2 orbitals. Octahedral geometry has six ligands surrounding the metal ion in an octahedral arrangement. In this geometry, the d-orbitals also split into two sets: the lower energy d_x^2-y^2 and d_z^2 orbitals, and the higher energy d_xy, d_yz, and d_xz orbitals.
03

Compare crystal-field splitting energies

The crystal-field splitting energy is the energy difference between the lower and higher energy d-orbitals. The splitting energy for tetrahedral and octahedral geometries depends on the position and electrostatic effect of the ligands surrounding the metal ion. For tetrahedral geometry, the splitting is less intense because the ligands are farther from the metal ion and interact less with the d-orbitals compared to an octahedral geometry. The magnitude of the crystal-field splitting energy for a tetrahedral geometry is typically about 4/9 times that of an octahedral geometry for the same set of ligands and metal ion. Therefore, the crystal-field splitting energy is larger for an octahedral geometry than for a tetrahedral geometry for a given metal ion and set of ligands.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tetrahedral Geometry
Tetrahedral geometry is a common arrangement in coordination complexes where a central metal ion is surrounded by four ligands positioned at the corners of a tetrahedron. This particular geometry causes the ligands to distribute evenly around the metal ion, resulting in a specific kind of interaction with the metal's d-orbitals.

In a tetrahedral geometry, the splitting of the d-orbitals is unique. The orbitals split into two different sets:
  • The lower energy set: These are the orbitals labeled as \(d_{xy}, d_{yz}, ext{ and } d_{xz}\). Here, the electrons experience less repulsion from the ligand electrons.
  • The higher energy set: These orbitals include \(d_{x^2-y^2} ext{ and } d_{z^2}\). These orbitals sit more directly along the axes, where the ligands are aligned hence experiencing greater repulsion.
The geometry leads to less crystal-field splitting energy compared to other geometries, such as octahedral, because the spatial arrangement means that the ligands interact farther from the d-orbitals.
Octahedral Geometry
Octahedral geometry occurs when a metal ion is surrounded by six ligands, forming a symmetrical octahedron. This is one of the most prevalent and stable geometrical configurations in coordination chemistry.

Here’s how the d-orbitals split in an octahedral geometry:
  • The lower energy set comprises of the \(d_{x^2-y^2} ext{ and } d_{z^2}\) orbitals. These orbitals experience less direct interaction from the ligands because they lie between the axes, which aligns respectively.
  • The higher energy set: Involves the orbitals \(d_{xy}, d_{yz}, ext{ and } d_{xz}\). These orbitals are positioned such that they face the ligands directly along the planes, leading to a stronger repulsion.
Due to the manner of orbital splitting, the crystal-field splitting energy in octahedral complexes is heightened compared to that in tetrahedral ones. In simple terms, octahedral geometry induces greater energy differences between the d-orbitals due to more substantial ligand interaction.
Crystal-Field Splitting Energy
Crystal-field splitting energy, often denoted by \(\Delta\), is a pivotal concept in crystal field theory explaining how the presence of ligands leads to an energy differentiation in metal d-orbitals.

When comparing various geometries:
  • Tetrahedral geometry: Here, the ligands are positioned further from the metal ion's surface. As a result, the splitting of the d-orbitals is relatively minor, making the crystal-field splitting energy smaller. Typically, it is about 4/9 the size of the splitting energy in the octahedral arrangement for equivalent sets of metal ions and ligands.
  • Octahedral geometry: The ligands are positioned closer, interacting significantly with the d-orbitals, which leads to a larger splitting of these orbitals. This close proximity increases the crystal-field splitting energy.
In essence, the configuration of the metal-ligand complex and the spatial relationship of ligands significantly impacts the value of the crystal-field splitting energy, which ultimately determines the electronic properties and chemical reactivity of the complex.

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Most popular questions from this chapter

Which type of substance is attracted by a magnetic field, a diamagnetic substance or a paramagnetic substance?

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

Determine if each of the following metal complexes is chiral and therefore has an optical isomer: (a) square planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right],(\mathbf{b})\) octahedral \(\left[\mathrm{Ni (\mathrm{en})\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+},\) (c) octahedral cis-[V(en) \(\left._{2} \mathrm{ClBr}\right]\).

Indicate the coordination number and the oxidation number of the metal for each of the following complexes: (a) \(\mathrm{Na}_{2}[\mathrm{Co}(\mathrm{EDTA})]\) (b) \(\mathrm{KMnO}_{4}\) (c) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}\) (d) \(\mathrm{K}_{3} \mathrm{Fe}(\mathrm{CN})_{6}\) (e) \(\mathrm{Rh}\left(\mathrm{PPh}_{3}\right)_{3} \mathrm{Cl}\) (f) \(\mathrm{Zn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{NH}_{3}\right)_{2}\)

Write the names of the following compounds, using the standard nomenclature rules for coordination complexes: (a) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{NO}_{3}\) (b) \(\mathrm{Hg}\left[\mathrm{Co}(\mathrm{SCN})_{4}\right]\) (c) \(\left[\mathrm{Ru}\left(\mathrm{PPh}_{3}\right)_{3} \mathrm{Cl}_{3}\right]\) (d) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{CO}_{3}\right]_{2} \mathrm{SO}_{4}\)

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