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Chapter 21: Problem 16

Compounds of copper(II) are generally colored, but compounds of copper(I) are not. Explain. Would you expect \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) to be colored? Explain.

Short Answer

Expert verified
Copper(I) compounds are generally colorless because their d-orbital is completely filled, which means there are no d-d transitions possible. Copper(II) compounds, with a partially filled d-orbital, exhibit color due to d-d transitions when visible light is absorbed. In \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\), the central Cd虏鈦� ion has a filled d-orbital, and no d-d transitions are possible, making the compound colorless.

Step by step solution

01

Electronic Configuration of Copper Ions

First, let's examine the electronic configurations of copper(II) and copper(I) ions considering copper's atomic number is 29: Copper atom: \([Ar]3 d^{10} 4 s^1\) Copper(I) ion, Cu(I) or Cu+: \([Ar]3 d^{10}\) Copper(II) ion, Cu(II) or Cu虏+: \([Ar]3 d^9\)
02

Understand Color in Compounds

The color in compounds is generally caused by the absorption of visible light as an electron is excited from a lower-energy orbital to a higher-energy orbital. In transition metal complexes, this usually involves a ligand-to-metal charge transfer (LMCT) or a d-d transition within the metal's d-orbitals.
03

Analyze Copper Ions and Their Color

In copper(I) ion complexes, the d-orbital is completely filled, with no possibility of d-d transitions. As d-d transitions are responsible for the color in transition metal compounds, no d-d transitions mean no color for the copper(I) compounds. On the other hand, copper(II) ions have one less electron in the d-orbital, leaving it partially filled. When the visible light is absorbed, a d-d transition can occur as electrons can be excited from a lower-energy d-orbital to a higher-energy d-orbital. These d-d transitions produce color in copper(II) compounds.
04

Analyze the Given Compound \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\)

Now, let's examine the compound \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\). Here, the central metal ion is cadmium (Cd虏鈦�) with an atomic number of 48. The electronic configuration of cadmium is: \[Cd^{2+}: [Kr] 4d^{10}\] The compound has ammonium (NH鈧�) as a ligand and two chloride (Cl鈦�) ions as counterions. The ammonia acts as a ligand with electron-pair donation to the central metal ion. However, the chloride ions do not participate in ligand-metal interactions.
05

Determine Color of the Given Compound

As we analyze the electronic configuration of the Cd虏鈦� ion, it has a filled d-orbital, meaning that there are no d-d transitions possible for this ion. Thus, there is no visible light absorption, and the compound \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) is expected to be colorless.

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Most popular questions from this chapter

The compound \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6} \mathrm{Cl}_{2}\) is green, whereas \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) is violet. Predict the predominant color of light absorbed by each compound. Which compound absorbs light with the shorter wavelength? Predict in which compound \(\Delta\) is greater and whether \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{NH}_{3}\) is the stronger field ligand. Do your conclusions agree with the spectrochemical series?

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in water, the colors of the solutions are red, yellow, green, and blue. Further experiments reveal that two of the complex ions are paramagnetic with four unpaired electrons and the other two are diamagnetic. What can be deduced from this information about the four coordination compounds?

\(\mathrm{CoCl}_{4}^{2-}\) forms a tetrahedral complex ion and \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) forms an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the \(d\) orbital splitting diagrams? a. \(\mathrm{CoCl}_{4}^{2-}\) is an example of a strong-field case having two unpaired electrons. b. Because \(\mathrm{CN}^{-}\) is a weak-field ligand, \(\mathrm{Co}(\mathrm{CN})_{6}^{3-}\) will be a low-spin case having four unpaired electrons.

a. In the absorption spectrum of the complex ion \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) there is a band corresponding to the absorption of a photon of light with an energy of \(1.75 \times 10^{4} \mathrm{cm}^{-1} .\) Given \(1 \mathrm{cm}^{-1}=1.986 \times 10^{-23} \mathrm{J},\) what is the wavelength of this photon? b. The \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle in \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) is predicted to be \(180^{\circ} .\) What is the hybridization of the N atom in the \(\mathrm{NCS}^{-}\) ligand when a Lewis acid-base reaction occurs between \(\mathrm{Cr}^{3+}\) and \(\mathrm{NCS}^{-}\) that would give a \(180^{\circ} \mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle? \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) undergoes substitution by ethylenediamine (en) according to the equation $$\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}+2 \mathrm{en} \longrightarrow \mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}+4 \mathrm{NCS}^{-}$$ Does \(\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit geometric isomerism? Does \(\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit optical isomerism?

Why are \(\mathrm{CN}^{-}\) and \(\mathrm{CO}\) toxic to humans?
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