The coordination complex \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) forms colorless, diamagnetic crystals that melt at \(90^{\circ} \mathrm{C}\) . (a) What is the oxidation number of chromium in this compound? (b) Given that \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) is diamagnetic, what is the electron configuration of chromium in this compound? (c) Given that \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) is colorless, would you expect CO to be a weak-field or strong-field ligand? (d) Write the name for \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) using the nomenclature rules for coordination compounds.

Short Answer

Expert verified
In summary, the oxidation number of chromium in \(\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\) is 0. The electron configuration of chromium in the complex is [Ar] 4s0 3d6, with all electrons paired, making it diamagnetic. CO is a strong-field ligand, as the complex is colorless with no unpaired electrons. The coordination compound is named hexacarbonylchromium.

Step by step solution

01

Find the Oxidation Number of Chromium

To find the oxidation number of chromium, we first need to figure out the charge of the ligand, CO. Since CO is a neutral ligand, it has a charge of 0. As the complex has no overall charge, the oxidation number of chromium is also 0. So, the oxidation number of chromium in this compound is 0.
02

Determine the Electron Configuration of Chromium

Originally, the electron configuration of chromium is [Ar] 4s1 3d5. Since the oxidation state of chromium in the complex is 0, there is no loss or gain of electrons. Hence, the electron configuration of chromium in this complex is the same as its original configuration, which is [Ar] 4s1 3d5. Considering that the complex is diamagnetic (no unpaired electrons), and each CO ligand donates a pair of electrons to the chromium atom, the 6 CO ligands will cause the Cr atom to promote 1 electron from the 4s orbital to the 3d orbital. The new electron configuration will be [Ar] 4s0 3d6 (paired electrons).
03

Determine if CO is Weak-Field or Strong-Field Ligand

As the given coordination complex is colorless, it means that there is no unpaired electron left to absorb visible light and create a color. This indicates that CO is a strong-field ligand, as it leads to pairing of electrons in the chromium's orbitals, leaving no unpaired electrons.
04

Name the Coordination Compound

Using the nomenclature rules for coordination compounds, we name the complex as follows: 1. Name the ligand(s) first, using alphabetical order: hexacarbonyl 2. Name the central metal atom: chromium Since the oxidation number of chromium is 0, we do not use roman numerals to denote its oxidation state. The name of the complex becomes: hexacarbonylchromium

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Key Concepts

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

Oxidation Number
Understanding the oxidation number of elements in compounds like coordination complexes is fundamental for grasping their chemistry. The oxidation number represents the charge that an atom would have if the compound was composed of ions. It is pivotal for predicting reactivity, understanding electron transfer, and writing chemical formulas correctly.

In the coordination compound \[\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\], the oxidation number of chromium is determined by considering the charge of the ligands bonded to it. Since carbonyl (CO) is a neutral ligand, it has an oxidation number of 0. The complex itself doesn't carry an overall charge; thus, chromium's oxidation number is 0. This means there are no net electron transfers from the chromium to the ligands, keeping its electron configuration unchanged.
Electron Configuration
The electron configuration tells us how electrons are distributed within the orbitals of an atom. For transition metals, electron configurations can offer insight into the atom's chemical behaviors, including bonding and magnetism.

The electron configuration of an isolated chromium atom is \[\text{[Ar]} 4s^1 3d^5\]. However, in the complex \[\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\], chromium maintains its electrons because its oxidation number is 0. The complex is noted to be diamagnetic, which means all electrons are paired. This implies that one electron from the 4s orbital is promoted to the 3d orbital, leading to a new configuration of \[\text{[Ar]} 4s^0 3d^6\] with all paired electrons. This configuration explains the observed diamagnetic property of the compound.
Ligand Field Theory
Ligand field theory aids in understanding the color and magnetic properties of coordination compounds. It is an adaptation of molecular orbital theory that looks at the effect ligands have on the d-orbitals of a transition metal ion.

In the context of \[\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\], we observe that the compound is colorless and all its electrons are paired, which is indicative of a strong-field ligand. Carbon monoxide is, indeed, a strong-field ligand and causes a larger splitting in the chromium's d-orbitals. This prevents electrons from being unpaired and promotes a low-spin configuration, where electrons pair up in the lower energy orbitals, hence no electron transition is possible to absorb visible light, resulting in the compound being colorless.
Chemical Nomenclature
The rules for naming chemical compounds, including coordination complexes, allows for the clear communication of a compound's composition. The nomenclature involves systematic naming of the ligands followed by the central metal with its oxidation state when it is not zero.

For \[\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]\], the ligands are named first followed by the central metal. Since the oxidation number of chromium is 0, it is not indicated in the compound's name. Thus, the correct nomenclature for the coordination complex is hexacarbonylchromium, reflecting the six CO ligands (hexacarbonyl) and the chromium center without the need for numeric oxidation state indicators.

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Most popular questions from this chapter

(a) In early studies it was observed that when the complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br}\) was placed in water, the electrical conductivity of a 0.05\(M\) solution changed from an initial value of 191 \(\mathrm{ohm}^{-1}\) to a final value of 374 \(\mathrm{ohm}^{-1}\) over a period of an hour or so. Suggest an explanation for the observed results.(See Exercise 23.69 for relevant comparison data.) (b) Write a balanced chemical equation to describe the reaction. (c) \(A 500\)-mL solution is made up by dissolving 3.87g of the complex. As soon as the solution is formed, and before any change in conductivity has occurred, a 25.00-mL portion of the solution is titrated with 0.0100 \(\mathrm{M} \mathrm{AgNO}_{3}\) solution. What volume of AgNO \(_{3}\) solution do you expect to be required to precipitate the free \(\operatorname{Br}^{-}(a q) ?(\mathbf{d})\) Based on the response you gave to part (b), what volume of \(\mathrm{AgNO}_{3}\) solution would be required to titrate a fresh 25.00 -mL sample of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br}\) after all conductivity changes have occurred?

Determine if each of the following complexes exhibits geometric isomerism. If geometric isomers exist, determine how many there are. (a) [ Rh(bipy) \((o-\) phen \()_{2} ]^{3+},\) \((\mathbf{b})\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}(\mathrm{bipy}) \mathrm{Br}\right]^{2+},(\mathbf{c})\) square-planar \(\left[\mathrm{Pd}(\mathrm{en})(\mathrm{CN})_{2}\right].\)

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) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{ClO}_{4},(\mathbf{e})[\operatorname{Sr}(\mathrm{EDTA})]^{2-}\)

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 ]

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 \(, \mathrm{C}_{2} \mathrm{H}_{6}.\)

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