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Chapter 23: Problem 60

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
In conclusion, 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 ligand interactions are stronger in octahedral geometry due to the ligand-metal ion bond angles. The relationship between the crystal-field splitting energies in octahedral (Δ_oct) and tetrahedral (Δ_tet) geometries can be defined as: Δ_tet = \( \frac{4}{9} \) * Δ_oct The factor \( \frac{4}{9} \) indicates that the tetrahedral geometry has a smaller crystal-field splitting energy compared to the octahedral geometry.

Step by step solution

01

Understanding Tetrahedral and Octahedral Geometries

Tetrahedral geometry consists of a metal ion surrounded by four ligands, while octahedral geometry consists of a metal ion surrounded by six ligands. In both cases, the ligands interact differently with the metal ion's d-orbitals, leading to different crystal-field splitting energies.
02

Crystal-Field Splitting in Octahedral Geometry

In octahedral geometry, the d-orbitals of the metal ion are split into two groups: the t₂g orbitals (d(xy), d(xz), and d(yz)) and the e_g orbitals (d(x²-y²) and d(z²)). The t₂g orbitals are lower in energy, while the e_g orbitals are higher in energy. The crystal-field splitting energy in octahedral geometry (Δ_oct) is the energy difference between the t₂g and e_g orbitals.
03

Crystal-Field Splitting in Tetrahedral Geometry

In tetrahedral geometry, the d-orbitals are also split into two groups, but this time the e (d(xy), d(xz), and d(yz)) orbitals are lower in energy, and the t₂ orbitals (d(x²-y²) and d(z²)) are higher in energy. The crystal-field splitting energy in tetrahedral geometry (Δ_tet) is the energy difference between the e and t₂ orbitals.
04

Comparing Crystal-Field Splitting Energies

In general, the crystal-field splitting energy in octahedral geometry is larger than that in tetrahedral geometry. This is because the ligand interactions are stronger in octahedral geometry due to the ligand-metal ion bond angles (90° for octahedral vs 109.5° for tetrahedral). To make a comparison, we can define a relationship between Δ_oct and Δ_tet: Δ_tet = 4/9 * Δ_oct The factor 4/9 indicates that the crystal-field splitting energy is smaller for tetrahedral geometry than for octahedral geometry.
05

Conclusion

In conclusion, 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.

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Most popular questions from this chapter

Solutions of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}(\) both octahedral \()\) and \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) (tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less that that in octahedral ones, assign a color to each complex.

For each of the following molecules or polyatomic ions, draw the Lewis structure and indicate if it can act as a monodentate ligand, a bidentate ligand, or is unlikely to act as a ligand at all: (a) ethylamine, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) , (b) trimethylphosphine, \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3},\) (c) carbonate, \(\mathrm{CO}_{3}^{2-},\) \((\mathbf{d})\) ethane \(, \mathrm{C}_{2} \mathrm{H}_{6}.\)

Indicate the coordination number and the oxidation number of the metal for each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CdBr}_{4}\right]\) (c) \(\left[\mathrm{Pt}(\mathrm{en})_{3}\right]\left(\mathrm{ClO}_{4}\right)_{4}\) (d) \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right]^{+}\) (e) \(\mathrm{NH}_{4}\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{NCS})_{4}\right]\) (f) \(\left[\mathrm{Cu}(\mathrm{bipy})_{2} \mathrm{I}\right] \mathrm{I}\)

Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin Fe(Il) complex; deoxyhemoglobin, without the O \(_{2}\) molecule, is a high-spin complex. (a) Assuming that the coordination environment about the metal is octahedral, how many unpaired electrons are centered on the metal ion in each case? (b) What ligand is coordinated to the iron in place of \(\mathrm{O}_{2}\) in deoxyhemoglobin? (c) Explain in a general way why the two forms of hemoglobin have different colors (hemoglobin is red, whereas deoxyhemoglobin has a bluish cast. (d) \(\mathrm{A} 15\) -minute exposure to air containing 400 \(\mathrm{ppm}\) of CO causes about 10\(\%\) of the hemoglobin in the blood to be converted into the carbon monoxide complex, called carboxyhemoglobin. What does this suggest about the relative equilibrium constants for binding of carbon monoxide and \(\mathrm{O}_{2}\) to hemoglobin? (e) \(\mathrm{CO}\) is a strong-field ligand. What color might you expect carboxyhemoglobin to be?

Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right (\mathbf{b})\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]\) (d) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{ClO}_{4},(\mathbf{e})[\operatorname{Sr}(\mathrm{EDTA})]^{2-}\)
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