Which of the following will form coloured solution in aqueous medium?

\(Cu^{2+}\)

The correct answer is option 2. \(Cu^{2+}\).

To understand why \(Cu^{2+}\) forms a coloured solution while the other ions do not, we need to look at their electronic configurations and the phenomenon of d-d transitions in transition metal ions. This involves understanding the structure of the d-orbitals and how they interact with light in an aqueous medium.

The color of a solution formed by a transition metal ion is largely due to the d-d transitions of electrons between energy levels in the d-orbitals. When light hits the solution, electrons in lower-energy d-orbitals get excited to higher-energy d-orbitals by absorbing specific wavelengths of light. The remaining wavelengths are reflected or transmitted, which results in the colour we observe.

In transition metal ions:

If there are unpaired electrons in the d-orbitals, d-d transitions can occur, leading to the absorption of light and the formation of a colored solution.

If the d-orbitals are either completely filled or completely empty, no d-d transitions can occur, and the solution will be colourless.

Analysis of Each Ion:

1. \(Sc^{3+}\):

Electronic Configuration: \([Ar] 3d^0\).

In the case of \(Sc^{3+}\), the 3d orbital is completely empty because all d-electrons are lost when scandium forms the +3 oxidation state. Since there are no d-electrons, there are no possible d-d transitions. As a result, \(Sc^{3+}\) does not form a colored solution. The solution is colourless.

2. \(Cu^{2+}\):

Electronic Configuration: \([Ar] 3d^9\).

\(Cu^{2+}\) has one unpaired electron in its 3d orbital. When \(Cu^{2+}\) is in an aqueous solution, it forms a coordination complex with water molecules, and this complex creates a splitting of the d-orbital energy levels (due to crystal field theory). The unpaired electron can undergo a d-d transition, moving from a lower-energy d-orbital to a higher-energy one when visible light is absorbed. This absorption of light in the visible spectrum gives rise to a blue or blue-green colour in the solution. Hence,\(Cu^{2+}\) forms a coloured solution.

3. \(Cd^{2+}\):

Electronic Configuration: \([Kr] 4d^{10}\).

\(Cd^{2+}\) has a completely filled d-orbital \((4d^{10})\). Since all of the d-orbitals are filled, there are no available lower-energy d-orbitals for electrons to move to. This means that no d-d transitionn can occur, and as a result, \(Cd^{2+}\) forms a colourless solution in water.

4. \(Zn^{2+}\):

Electronic Configuration: \([Ar] 3d^{10}\).

Like \(Cd^{2+}\), \(Zn^{2+}\) also has a completely filled 3d orbital. All 10 electrons occupy the 3d orbitals, so there are no empty orbitals for d-d transitions to occur. Since there are no d-d transitions, \(Zn^{2+}\) forms a colourless solution.

Thus, \(Cu^{2+}\) is the only ion that will form a colored solution due to its partially filled d-orbitals and the presence of d-d transitions.