Question

Compounds of copper(II) are generally colored, but compounds of copper(I) are not. Explain. Would you expect $\mathbf{Cd}{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}$ to be colored? Explain.

Short answer

Copper (II) compounds are coloured but copper (I) not coloured.

$\mathrm{Cd}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2$compound is not coloured.

Step-by-step solution

Step:1

In copper (II) compounds copper has 3d⁹electronic configuration so it has one unpaired electron, so it isblue coloured. But in copper (I)copper has 3d¹⁰ electronic configuration so it has no unpaired electrons. All electrons are paired in copper (I) so it iscolourless.

Step:2

In$\mathrm{Cd}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2$compound cadminum is in +2 oxidation state. Cd⁺² has 4d¹⁰electronic configuration so it has no unpaired electrons. So it is colourless.

The colour of the transition metal ions arises from the excitation of electrons from the d orbitals of lower energy to the d orbitals of higher energy (d-d transitions). Light energy for such a transition is available in visible region.