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(II) compounds have an unpaired electron in their 3d9 configuration, which can get excited, causing the absorption of light and coloration. In contrast, copper(I) compounds have a completely filled 3d10 configuration, leaving no room for electron excitation and appearing colorless. The compound \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) also has a completely filled d orbital (4d10), similar to copper(I) compounds, with no unpaired electrons for excitation. Therefore, we can expect \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) to be colorless.

Step-by-step solution

Understanding copper(II) and copper(I) compounds

In order to explain the difference in coloration between copper(II) and copper(I) compounds, we need to understand their electronic configurations and how it affects their interactions with light. Both copper(II) and copper(I) have d orbitals with unpaired electrons. In copper(II) compounds, Cu has an electronic configuration of 3d9, while in copper(I) compounds it has an electronic configuration of 3d10.

Absorption of Light

When a compound absorbs light of a certain wavelength, it reflects wavelengths that were not absorbed. Our eyes perceive these reflected wavelengths as the color of the compound. Compounds with unpaired electrons can absorb light when electrons get excited from the ground state to an excited state by promoting to a higher energy level. In the case of copper(II) compounds, there is an unpaired electron in the 3d9 configuration that can get excited, which results in the absorption of light and the subsequent coloration of these compounds. Meanwhile, in the case of copper(I) compounds, the 3d10 configuration is completely filled, which means there is no room for an electron to get excited and transition to a higher energy level. Consequently, copper(I) compounds do not absorb light and appear colorless.

Examining the given compound: \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\)

Now, let's analyze the given compound, \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\), and determine if we can expect it to be colored or not. This compound contains cadmium, Cd, instead of copper. Cadmium has an electronic configuration of 4d10 5s2.

Predicting color of \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) with an explanation

Since cadmium in \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) has a completely filled d orbital, similar to copper(I) compounds, it does not have any unpaired electrons available for excitation. This indicates that the given compound will not absorb light and therefore, we can expect \(\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\mathrm{Cl}_{2}\) to be colorless.