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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 \(\mathrm{AgCl}\) will precipitate per mole of \(\mathrm{CrCl}_{3} * 6 \mathrm{H}_{2} \mathrm{O}\) dissolved in solution? (c) Crystals of anhydreus 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 \(\mathrm{Cl}\) is not 1:6?

Short Answer

Expert verified
The complex ion in hydrated chromium(III) chloride (\(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\)) is \(\mathrm{[Cr(H}_{2} \mathrm{O})_{4} \mathrm{Cl}_{2}]^{+}\). When treated with excess \(\mathrm{AgNO}_{3}(a q)\), 2 moles of \(\mathrm{AgCl}\) will precipitate per mole of \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) dissolved. The octahedral coordination geometry of chromium in anhydrous \(\mathrm{CrCl}_{3}\) can be maintained with a 1:3 Cr:Cl ratio by each \(\mathrm{Cr}^{3+}\) ion being coordinated to three \(\mathrm{Cl}^{-}\) ions and sharing three additional \(\mathrm{Cl}^{-}\) ions with neighboring \(\mathrm{Cr}^{3+}\) ions in the crystal lattice.

Step by step solution

01

(a) Write the complex ion

From the empirical formula, the hydrated chromium(III) chloride compound contains one \(\mathrm{Cr}^{3+}\) ion and three \(\mathrm{Cl}^{-}\) ions. In addition, there are six \(\mathrm{H}_{2} \mathrm{O}\) molecules as water of hydration. Since \(\mathrm{Cr}^{3+}\) typically forms octahedral complexes, it will bind to six ligands. In this case, four will be water molecules, and the other two will be \(\mathrm{Cl}^{-}\) ions. Thus, the complex ion is given by: \[\mathrm{[Cr(H}_{2} \mathrm{O})_{4} \mathrm{Cl}_{2}]^{+}\]
02

(b) Moles of AgCl precipitated

When the complex is treated with excess \(\mathrm{AgNO}_{3}(a q)\), the \(\mathrm{Ag}^{+}\) ions will react with the \(\mathrm{Cl}^{-}\) ions, forming \(\mathrm{AgCl}\) precipitate. As each mole of the complex ion contains two \(\mathrm{Cl}^{-}\) ions, two moles of \(\mathrm{AgCl}\) will be formed per mole of the complex. Therefore, the number of moles of \(\mathrm{AgCl}\) precipitated per mole of \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) dissolved in solution is 2.
03

(c) Coordination geometry explanation

In anhydrous \(\mathrm{CrCl}_{3}\), the \(\mathrm{Cr}^{3+}\) ions have an overall coordination number of 6, forming an octahedral complex. Although the empirical formula \(\mathrm{CrCl}_{3}\) suggests a 1:3 ratio of \(\mathrm{Cr}\) to \(\mathrm{Cl}\), we can still have an octahedral coordination geometry for the \(\mathrm{Cr}^{3+}\) ions. This is because each \(\mathrm{Cr}^{3+}\) ion is coordinated to three \(\mathrm{Cl}^{-}\) ions and shares three additional \(\mathrm{Cl}^{-}\) ions with neighboring \(\mathrm{Cr}^{3+}\) ions in the crystal lattice. In this arrangement, each \(\mathrm{Cr}^{3+}\) ion is octahedrally surrounded by six \(\mathrm{Cl}^{-}\) ions, thus preserving the octahedral coordination geometry.

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Key Concepts

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

Complex Ions
In coordination chemistry, complex ions arise when a central metal ion, like chromium, binds to several molecules or ions called ligands. In our exercise, the central metal ion is \(\mathrm{Cr}^{3+}\). This chromium ion forms a complex with ligands to achieve a more stable electron configuration.

For hydrated chromium(III) chloride, the formula is given as \(\mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\). This means that each chromium ion coordinates with six ligands. These are four water molecules and two chloride ions. The coordination happens because the water and chloride ions use their lone electron pairs to form bonds with the chromium ion.

The resulting complex is denoted as \([\mathrm{Cr(H}_{2}\mathrm{O})_{4}\mathrm{Cl}_{2}]^{+}\). This notation indicates not only the number and type of ligands involved but also the charge on the entire complex. Understanding complex ions is crucial as they play pivotal roles in many chemical processes, including catalysis and color formation in compounds.
Octahedral Geometry
Octahedral geometry is a common geometry for complex ions, especially involving transition metals like chromium. In this form, the central metal ion is surrounded by six ligands at the corners of an octahedron, which is a geometrical shape with eight faces.

In the context of our chromium compound, the \(\mathrm{Cr}^{3+}\) ion in its hydrated form connects with six ligands. Specifically, these are four water molecules and two chloride ions, forming a stable octahedral complex. Each ligand occupies a vertex around the central metal, maintaining equal spacing to minimize repulsion between electron pairs.

This octahedral arrangement results in specific chemical properties. For example, the stability of the complex, its solubility in different solvents, and its color depend significantly on the spatial arrangement of the ligands. This geometry is favored because it allows a uniform distribution of the donor atoms around the central ion, minimizing the energy of the system.
Chromium Compounds
Chromium, particularly in its +3 oxidation state, forms interesting and varied compounds. The hydrated form of chromium(III) chloride is green, while the anhydrous form is violet. These color variations are due to differences in the coordination of chromium ions and the ligands in their environment.

In hydrated chromium(III) chloride, the presence of water as ligands leads to a greener hue. Conversely, the violet color in anhydrous chromium(III) chloride arises because the crystal lattice arrangement changes. Chromium in its anhydrous form still maintains an octahedral geometry by coordinating directly with chloride ions. However, it involves shared chloride ions between adjacent chromium centers, contributing to a different electronic environment and, thus, a color change.

Chromium compounds demonstrate the intriguing nature of transition metals, where small changes in ligand identity or connectivity can lead to significant changes in chemical behavior. This property is fundamental in understanding the complex behavior of transition metal chemistry.

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Most popular questions from this chapter

Carbon monoxide, \(\mathrm{CO}\), is an important ligand in coordination chemistry. When \(\mathrm{CO}\) is reacted with nickel metal the product is \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\), which is a toxic, pale yellow liquid. (a) What is the oxidation number for nickel in this compound? (b) Given that \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) is diamagnetic molecule with a tetrahedral geometry, what is the electron configuration of nickel in this compound? (c) Write the name for \(\left[\mathrm{Nu}(\mathrm{CO})_{4}\right]\) using the nomenclature rules for coordination compounds.

Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin \(\mathrm{Fe}\) (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 \(400 \mathrm{ppm}\) 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?

Draw the structure for \(\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_{2}\) and use it to answer the following questions: (a) What is the coordination number for platinum in this complex? (b) What is the coordination geometry? (c) What is the oxidation state of the platinum? (d) How many unpaired electrons are there? [Sections \(23.2\) and 23.6]

The lanthanide contraction explains which of the following periodic trends? (a) The atomic radii of the transition metals first decrease and then increase when moving horizontally across each period. (b) When forming ions the transition metals lose their valence s orbitals before their valence \(d\) orbitals. (c) The radii of the period 5 transition metals (Y-Cd) are very similar to the radii of the period 6 transition metals (Lu-Hg).

Draw the crystal-field energy-level diagrams and show the placement of \(d\) electrons for each of the following: (a) \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) (four unpaired electrons), (b) \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (high spin), (c) \(\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+}\) (low spin), (d) \(\left[\mathrm{IrCl}_{6}\right]^{2-}\) (low spin), (c) \(\left[\mathrm{Cr}(\mathrm{cn})_{3}\right]^{1+}\), (f) \(\left[\mathrm{NiF}_{6}\right]^{4-}\).

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