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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

Amino acids can act as ligands toward transition metal ions. The simplest amino acid is glycine \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\right) .\) Draw a structure of the glycinate anion \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\right)\) acting as a bidentate ligand. Draw the structural isomers of the square planar complex \(\mathrm{Cu}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2}\right)_{2}\)

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Four different octahedral chromium coordination compounds exist that all have the same oxidation state for chromium and have \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{Cl}^{-}\) as the ligands and counterions. When 1 mole of each of the four compounds is dissolved in water, how many moles of silver chloride will precipitate upon addition of excess \(\mathrm{AgNO}_{3} ?\)
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