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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 (螖_oct) and tetrahedral (螖_tet) geometries can be defined as: 螖_tet = \( \frac{4}{9} \) * 螖_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 t鈧俫 orbitals (d(xy), d(xz), and d(yz)) and the e_g orbitals (d(x虏-y虏) and d(z虏)). The t鈧俫 orbitals are lower in energy, while the e_g orbitals are higher in energy. The crystal-field splitting energy in octahedral geometry (螖_oct) is the energy difference between the t鈧俫 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 t鈧� orbitals (d(x虏-y虏) and d(z虏)) are higher in energy. The crystal-field splitting energy in tetrahedral geometry (螖_tet) is the energy difference between the e and t鈧� 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掳 for octahedral vs 109.5掳 for tetrahedral). To make a comparison, we can define a relationship between 螖_oct and 螖_tet: 螖_tet = 4/9 * 螖_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

Consider the following three complexes: (Complex 1) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SCN}\right]^{2+}\) (Complex 2) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]^{2+}\) (Complex 3) \(\mathrm{CoClBr} \cdot 5 \mathrm{NH}_{3}\) Which of the three complexes can have (a) geometric isomers, (b) linkage isomers, (c) optical isomers, (d) coordination-sphere isomers?

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

Write the names of the following compounds, using the standard nomenclature rules for coordination complexes: (a) \(\left[\mathrm{Rh}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (b) \(\mathrm{K}_{2}\left[\mathrm{TiCl}_{6}\right]\) (c) \(\mathrm{MoOCI}_{4}\) (d) \(\left[\operatorname{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\right] \mathrm{Br}_{2}\)

Four-coordinate metals can have either a tetrahedral or a square-planar geometry; both possibilities are shown here for \(\left[\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right] .\) (a) \(\mathrm{What}\) is the name of this molecule? (b) Would the tetrahedral molecule have a geometric isomer? (c) Would the tetrahedral molecule be diamagnetic or paramagnetic? (d) Would the square-planar molecule have a geometric isomer? (e) Would the square-planar molecule be diamagnetic or paramagnetic? (f) Would determining the number of geometric isomers help you distinguish between the tetrahedral and square-planar geometries? (g) Would measuring the molecule's response to a magnetic field help you distinguish between the two geometries? [Sections 23.4-23.6 ]

Which periodic trend is partially responsible for the observation that the maximum oxidation state of the transition-metal elements peaks near groups 7 \(\mathrm{B}\) and 8 \(\mathrm{B} ?\) (a) The number of valence electrons reaches a maximum at group 8 \(\mathrm{B} .\) (b) The effective nuclear charge increases on moving left across each period. (c) The radii of the transition-metal elements reach a minimum for group \(8 \mathrm{B},\) and as the size of the atoms decreases it becomes easier to remove electrons.
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