Which of the following set represents only coloured ions?(1) Fe^3+ (2…

The correct answer is option 4. 1, 3 and 4.

Let us go through each ion in detail to understand why they are colored or colorless:

1. \( \text{Fe}^{3+} \) (Iron(III)):

The electronic configuration of \( \text{Fe}^{3+} \) is \( [\text{Ar}] 3d^5 \). In aqueous solution or in complexes, the d-orbitals of \( \text{Fe}^{3+} \) are split into different energy levels by the surrounding ligands. Electronic transitions (d-d transitions) between these split d-orbitals absorb specific wavelengths of light, causing the ion to appear colored. For \( \text{Fe}^{3+} \), the color can range from yellow to brown.

2. \( \text{Sc}^{3+} \) (Scandium(III)):

The electronic configuration of \( \text{Sc}^{3+} \) is \( [\text{Ar}] \). This configuration means there are no electrons in the d-orbitals. Without electrons in the d-orbitals, there are no d-d transitions possible, making \( \text{Sc}^{3+} \) colorless.

3. \( \text{Cu}^{2+} \) (Copper(II)):

The electronic configuration of \( \text{Cu}^{2+} \) is \( [\text{Ar}] 3d^9 \). In aqueous solution, the d-orbitals are split by the ligand field of the surrounding water molecules. The d-d transitions absorb light in the visible region, causing \( \text{Cu}^{2+} \) solutions to appear blue or blue-green.

4. \( \text{Cr}^{3+} \) (Chromium(III)):

The electronic configuration of \( \text{Cr}^{3+} \) is \( [\text{Ar}] 3d^3 \). The d-orbitals are split in the ligand field, allowing for d-d transitions. These transitions result in the absorption of light in the visible region, causing \( \text{Cr}^{3+} \) solutions to appear green or violet.

Based on this analysis:

\( \text{Fe}^{3+} \), \( \text{Cu}^{2+} \), and \( \text{Cr}^{3+} \) all have electronic configurations that allow for d-d transitions, resulting in colored ions. \( \text{Sc}^{3+} \) does not have d-electrons available for these transitions, making it colorless.

Therefore, the set that represents only colored ions is 1, 3 and 4.