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The complexes \(\left[\mathrm{CrBr}_{6}\right]^{3-}\) and \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) are both known. (a) Draw the \(d\) -orbital energy-level diagram for octahedral \(\mathrm{Cr}(\mathrm{III})\) complexes. (b) What gives rise to the colors of these complexes? (c) Which of the two complexes would you expect to absorb light of higher energy?

Short Answer

Expert verified
The d-orbital energy-level diagram for octahedral Cr(III) complexes shows three spin-up electrons in the lower-energy t₂g orbitals and empty e_g orbitals. The colors of these complexes arise from electronic transitions between the d-orbitals when light is absorbed. Comparing the ligand-field splitting energies for both complexes, we find that [Cr(NH₃)₆]³⁺ absorbs light of higher energy due to a larger energy gap between the t₂g and e_g orbitals, as NH₃ causes a larger splitting than Br⁻ according to the spectrochemical series.

Step by step solution

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(a) d-orbital energy-level diagram for octahedral Cr(III) complexes

To draw the d-orbital energy-level diagram for octahedral Cr(III) complexes, we must first know the electron configuration of Cr(III). Chromium has an atomic number of 24, and its ground state electron configuration is [Ar] 3d^5 4s^1. In the Cr(III) state, it loses 3 electrons, resulting in the following electron configuration: [Ar] 3d^3. In an octahedral complex, the d-orbitals split into two sets due to the crystal field: the t₂g set (dxy, dyz, dxz) and the e_g set (dz^2, dx^2-y^2). The t₂g orbitals are lower in energy than the e_g orbitals. The Cr(III) ion has 3 electrons to distribute among the d-orbitals. These electrons will fill the t₂g orbitals first, following Hund's rule. The resulting d-orbital energy-level diagram is as follows: (1) t₂g: ↑↑↑ (three spin-up electrons) (2) e_g : (empty)
02

(b) Cause of the colors of these complexes

The colors of these complexes arise from the electronic transitions between the d-orbitals. When light is absorbed, an electron from a lower energy d-orbital (t₂g) is promoted to a higher energy d-orbital (e_g). The energy difference between these orbitals (∆E) corresponds to the energy of the absorbed photon and depends on the ligands surrounding the metal ion. Since visible light covers a range of energies (and hence colors), the absorbed light's energy (color) will determine the color of the complex as perceived by the human eye. The complementary color of the absorbed light will be observed as the color of the complex.
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(c) Comparing the energy of absorbed light for both complexes

To determine which of the two complexes would absorb light of higher energy, we must compare the ligand-field splitting energies in both cases. Different ligands cause different degrees of splitting, which is summarized in the spectrochemical series, a list of ligands ordered by their ability to split d-orbitals: I^- < Br^- < Cl^- < F^- < OH^- < H_2O < NH_3 < en < NO_2^- < CN^- In the given complexes, the Cr ion is coordinated to Br^- in the first complex and NH_3 in the second complex. According to the spectrochemical series, NH_3 causes a larger splitting than Br^-. A larger splitting means that the energy gap between the t₂g and e_g orbitals will be greater in the [Cr(NH₃)₆]³⁺ complex compared to the [CrBr₆]³⁻ complex. Therefore, the [Cr(NH₃)₆]³⁺ complex would absorb light of higher energy, as the electronic transitions in this complex require a higher energy to promote an electron from the t₂g orbitals to the e_g orbitals.

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Key Concepts

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

d-Orbital Energy-Level Diagram
One of the essential aspects of understanding transition metal complexes is the splitting of the d-orbitals in different geometric arrangements. In the case of octahedral complexes, the d-orbitals experience an interaction with ligands, leading to energy level splitting. This phenomenon is deeply connected to the electron configuration of the metal center.
For chromium(III), which has the electron configuration \ ext{[Ar] 3d}^3\, we observe that in an octahedral field, the five degenerate d-orbitals split into two groups: the lower-energy \(t_{2g}\) orbitals \((d_{xy}, d_{yz}, d_{xz})\) and the higher-energy \(e_g\) orbitals \((d_{z^2}, d_{x^2-y^2})\).
In general:
  • The \(t_{2g}\) set is lower in energy because these orbitals point between the axes where ligands are located.
  • The \(e_g\) set is higher in energy as these orbitals point directly at the ligands along the octahedral axes.
In an octahedral Cr(III) complex, you will place the three d-electrons in the \(t_{2g}\) orbitals, according to the Aufbau principle and Hund's rule, which minimizes electron repulsion.
Octahedral Coordination
When it comes to coordination chemistry, octahedral complexes are among the most studied due to their prevalence in transition metal chemistry. In an octahedral complex, six ligands symmetrically surround a central metal ion. This arrangement influences both the geometry and the electronic properties of the metal.
The nature of the ligands and the geometry contribute significantly to the properties of the complex:
  • The six ligands coordinate in a way that they maximize their distance from each other, forming an octahedral shape.
  • This arrangement creates two different environments for the d-orbitals, leading to the crystal field splitting observed.
In terms of symmetry, octahedral complexes have a high degree of symmetry, which affects vibrational and optical properties. These complexes provide insights into the electronic transitions that depend on ligand interactions and geometry.
Ligand-Field Splitting
Ligand-field splitting is a core concept in transition metal chemistry, explaining the behavior of electrons in split energy levels within a complex. This splitting dictates many properties including color and reactivity. The size of \(\Delta\) (the energy gap between \(t_{2g}\) and \(e_g\) orbitals) is influenced by the ligands surrounding the metal – described by the spectrochemical series.
The spectrochemical series is a ranking that shows a ligand's ability to cause d-orbital splitting:
  • Weak field ligands like \(I^-\) and \(Br^-\) cause small splitting \(\Delta\).
  • Strong field ligands such as \(NH_3\) and \(CN^-\) lead to significant splitting.
In the exercise, we compared the complexes \([\text{CrBr}_6]^{3-}\) and \([\text{Cr(NH}_3\text{)}_6]^{3+}\). \(NH_3\), being a stronger field ligand than \(Br^-\), results in a larger \(\Delta\), leading to the absorption of higher energy light with \([\text{Cr(NH}_3\text{)}_6]^{3+}\). This principle is an essential explanation for the difference in coloration and absorption properties of complexes with different ligands.

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Most popular questions from this chapter

The \(E^{\circ}\) values for two low-spin iron complexes in acidic solution are as follows: $$ \begin{aligned} \left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}(a q)+\mathrm{e}^{-} \rightleftharpoons \\ \left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{2+}(a q) & E^{\circ}=1.12 \mathrm{~V} \end{aligned} $$ $$ \begin{aligned} \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}(a q)+\mathrm{e}^{-} \rightleftharpoons & \\ &\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}(a q) \quad E^{\circ}=0.36 \mathrm{~V} \end{aligned} $$ (a) Is it thermodynamically favorable to reduce both Fe(III) complexes to their Fe(II) analogs? Explain. (b) Which complex, \(\left[\mathrm{Fe}(o \text { -phen })_{3}\right]^{3+}\) or \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-},\) is more difficult to reduce? (c) Suggest an explanation for your answer to (b).

Crystals of hydrated chromium(III) chloride are green, have an empirical formula of \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O},\) and are highly soluble, (a) Write the complex ion that exists in this compound. (b) If the complex is treated with excess \(\mathrm{AgNO}_{3}(a q)\), how many moles of AgCl will precipitate per mole of \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) dissolved in solution? (c) Crystals of anhydrous chromium(III) chloride are violet and insoluble in aqueous solution. The coordination geometry of chromium in these crystals is octahedral, as is almost always the case for \(\mathrm{Cr}^{3+}\). How can this be the case if the ratio of \(\mathrm{Cr}\) to Cl is not \(1: 6 ?\)

Explain why the transition metals in periods 5 and 6 have nearly identical radii in each group.

Which periodic trend is partially responsible for the observation that the maximum oxidation state of the transition-metal elements peaks near groups 7 and \(8 ?(\mathbf{a})\) The number of valence electrons reaches a maximum at group 8. (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 \(8,\) and as the size of the atoms decreases it becomes easier to remove electrons.

Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin Fe(II) complex; deoxyhemoglobin, without the \(\mathrm{O}_{2}\) 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) A 15-minute exposure to air containing 400 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) CO is a strong-field ligand. What color might you expect carboxyhemoglobin to be?

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