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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 Cd²⁺ 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 Cu²+: \([Ar]3 d^9\)
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 (Cd²⁺) with an atomic number of 48. The electronic configuration of cadmium is: \[Cd^{2+}: [Kr] 4d^{10}\] The compound has ammonium (NH₃) as a ligand and two chloride (Cl⁻) 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 Cd²⁺ 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

Ethylenediaminetetraacetate (EDTA \(^{4-} )\) is used as a complexing agent in chemical analysis with the structure shown in Fig. \(21.7 .\) 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 virtually prevents the heavy metal ions from reacting with biochemical systems. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-(a q)} \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mol of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) 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? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Pb}(\mathrm{OH})_{2}=1.2 \times 10^{-15}\right]$

Will 0.10 mol of AgBr completely dissolve in 1.0 \(\mathrm{L}\) of 3.0 \(\mathrm{M} \mathrm{NH}_{3} ?\) The \(K_{\mathrm{sp}}\) value for \(\mathrm{AgBr}(s)\) is \(5.0 \times 10^{-13},\) and the overall formation constant for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7}\) , that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$

Write electron configurations for the following transition metal ions. a. \(\mathrm{Sc}^{3+}\) b. \(\mathrm{Ru}^{2+}, \mathrm{Ru}^{3+}\) c. \(\operatorname{Ir}^{+}, \operatorname{Ir}^{3+}\) d. \(M n^{2+}\)

Until the discoveries of Alfred Werner, it was thought that carbon had to be present in a compound for it to be optically active. Werner prepared the following compound containing \(\mathrm{OH}^{-}\) ions as bridging groups and separated the optical isomers. a. Draw structures of the two optically active isomers of this compound. b. What are the oxidation states of the cobalt ions? c. How many unpaired electrons are present if the complex is the low-spin case?

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction $$\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$$ is \(1.0 \times 10^{-5}\) a. Calculate the pH of a 0.10\(M\) solution of $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$ b. Will a 1.0\(M\) solution of cobalt(Il) nitrate have a higher or lower pH than a 1.0\(M\) solution of cobalt (III) nitrate? Explain. c. \(\mathrm{Co}^{3+}\) complex ions are generally low-spin cases, whereas \(\mathrm{Co}^{2+}\) complex ions are generally high-spin cases. Explain. If this is the situation, how many unpaired electrons are present in \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) and \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+} ?\)
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