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.

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.

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.

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.