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Chapter 21: Problem 14

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(II) complexes have a partially filled d-shell, allowing for d-d electronic transitions and causing them to be colored. However, copper(I) complexes have a full d-shell, preventing d-d transitions and making them colorless. The given complex, \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\), has a central cadmium ion with a completely filled d-shell, meaning it will not undergo d-d transitions, and is expected to be colorless.

Step by step solution

01

Electronic Configuration of Copper(I) and Copper(II) Complexes

Copper has an atomic number of 29, so its electron configuration is [Ar] 3d10 4s1 for the neutral Cu atom. In copper(I) compounds, copper loses one electron, forming a Cu(I) ion with the electron configuration [Ar] 3d10, leaving a full d-shell. In copper(II) compounds, copper loses two electrons, and the electron configuration for Cu(II) becomes [Ar] 3d9 with one electron removed from the d-shell.
02

d-d Electronic Transitions and Color in Complexes

In transition metal complexes, colors are observed due to the absorption of light that corresponds to electronic transitions between d-orbitals. These electronic transitions are typically called d-d transitions. When a complex absorbs light and undergoes a d-d transition, the absorbed light's complementary color is observed by the human eye. Complexes that do not absorb light in the visible spectrum appear colorless. For a d-d transition to occur, the metal ion must have a partially filled d-shell (i.e., at least one empty or partially filled d orbital). In the case of copper(I) complexes, the d-shell is full (3d10), so no d-d transitions are possible. This explains why copper(I) complexes are typically colorless. On the other hand, copper(II) complexes have the electron configuration [Ar] 3d9, with an empty or partially filled d orbital, allowing for d-d transitions to occur. As these transitions occur in the visible light region of the electromagnetic spectrum, they result in the color we observe in copper(II) complexes.
03

Predicting the Color of \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) Complex

Now, we need to determine if the given \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) complex will be colored. First, we need to examine the central metal ion, which is cadmium (Cd). Cadmium is a group 12 element with an electron configuration [Kr] 4d10 5s2 for the neutral atom. Upon forming the \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) complex, cadmium loses two electrons to form a Cd(II) ion. The electron configuration becomes [Kr] 4d10, with a completely filled d-shell. As we discussed earlier, a complex with a fully filled d-shell will not undergo d-d transitions, meaning that the \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) complex is expected to be colorless.

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Most popular questions from this chapter

Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis with the structure shown in Figure 21.7. Solutions of EDTA \(^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The complex ion essentially eliminates the heavy metal ions from reacting with biochemical systems. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is \(\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}\) Consider a solution with \(0.010 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to \(1.0 \mathrm{~L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing \(0.050\) \(M \mathrm{Na}_{4} \mathrm{EDTA} .\) Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution? \(\left(K_{\text {?? }}\right.\) for \(\left.\mathrm{Pb}(\mathrm{OH})_{2}=1.2 \times 10^{-15} .\right)\)

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Consider the following data: $$\begin{aligned} \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+} & & \mathscr{E}^{\circ}=1.82 \mathrm{~V} \\ \mathrm{Co}^{2+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+} & K &=1.5 \times 10^{12} \\ \mathrm{Co}^{3+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en}){ }^{3+} & K &=2.0 \times 10^{47} \end{aligned}$$ where en \(=\) ethylenediamine. a. Calculate \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{Co}(\mathrm{en})_{3}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+}$$ b. Based on your answer to part a, which is the stronger oxidizing agent, \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}(\mathrm{en})_{3}{ }^{3+}\) ? c. Use the crystal field model to rationalize the result in part b.

A metal ion in a high-spin octahedral complex has two more unpaired electrons than the same ion does in a low-spin octahedral complex. Name some possible metal ions for which this would be true.

The ferrate ion, \(\mathrm{FeO}_{4}{ }^{2-}\), is such a powerful oxidizing agent that in acidic solution, aqueous ammonia is reduced to elemental nitrogen along with the formation of the iron(III) ion. a. What is the oxidation state of iron in \(\mathrm{FeO}_{4}{ }^{2-}\), and what is the electron configuration of iron in this polyatomic ion? b. If \(25.0 \mathrm{~mL}\) of a \(0.243 \mathrm{M} \mathrm{FeO}_{4}^{2-}\) solution is allowed to react with \(55.0 \mathrm{~mL}\) of \(1.45 M\) aqueous ammonia, what volume of nitrogen gas can form at \(25^{\circ} \mathrm{C}\) and \(1.50 \mathrm{~atm}\) ?
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