Question

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

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

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+: \([Ar]3 d^9\)

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.

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.

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.

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.