Question

The correct set of ions (aqueous solution) with the same colour from the following is:

• A. \( V^{2+} \), \( Cr^{3+} \), \( Mn^{3+} \)
• B. \( Zn^{2+} \), \( V^{3+} \), \( Fe^{3+} \)
• C. \( Ti^{4+} \), \( V^{4+} \), \( Mn^{2+} \)
• D. \( Sc^{3+} \), \( Ti^{3+} \), \( Cr^{2+} \)

Hint:

The colour of transition metal ions in aqueous solutions is determined by the d-electron configuration and the ligand field around the ion. Ions that are part of the same group or period often exhibit similar colours due to similar electronic transitions.

Answer 1

The Correct Option is A

To determine the correct set of ions in an aqueous solution that have the same color, we need to consider the typical colors of transition metal ions. Transition metals often exhibit characteristic colors due to d-d transitions of electrons. Let's analyze each ion in the provided options:

1. \(V^{2+}\), \(Cr^{3+}\), \(Mn^{3+}\):
   • \(V^{2+}\): Pale violet or lavender color 
   • \(Cr^{3+}\): Typically green in aqueous solutions, but can appear violet as well depending on the ligands present.
   • \(Mn^{3+}\): Often appears pale purple.
2. \(Zn^{2+}\), \(V^{3+}\), \(Fe^{3+}\):
   • \(Zn^{2+}\): Colorless because it's not a transition metal.
   • \(V^{3+}\): Typically green.
   • \(Fe^{3+}\): Usually yellow or brown in water.
3. \(Ti^{4+}\), \(V^{4+}\), \(Mn^{2+}\):
   • \(Ti^{4+}\): Colorless (not a transition metal).
   • \(V^{4+}\): Typically blue.
   • \(Mn^{2+}\): Pale pink.
4. \(Sc^{3+}\), \(Ti^{3+}\), \(Cr^{2+}\):
   • \(Sc^{3+}\): Colorless.
   • \(Ti^{3+}\): Purple.
   • \(Cr^{2+}\): Blue.

Based on the above reasoning, the correct answer is the first option, where \(V^{2+}\), \(Cr^{3+}\), and \(Mn^{3+}\) can appear in similar shades (pale violet, lavender, and purple). Therefore, the correct set is:

\(V^{2+}\), \(Cr^{3+}\), \(Mn^{3+}\)

Answer 2

Transition metals and their ions often exhibit characteristic colors due to d-d transitions where electrons in the d-orbitals absorb specific wavelengths of light and move to higher energy levels. The color observed is complementary to the color of light absorbed. Here is the breakdown of each ion:

Set 1: \( V^{2+} \), \( Cr^{3+} \), \( Mn^{3+} \)

These ions are known to produce similar colors in solution. Specifically, they appear in shades of violet or blue-green. The colors arise from specific d-d transitions unique to each ion, but similar enough to result in perceived color similarity:

• \( V^{2+} \): Blue or violet
• \( Cr^{3+} \): Often green
• \( Mn^{3+} \): Typically red-violet

Set 2: \( Zn^{2+} \), \( V^{3+} \), \( Fe^{3+} \)

\( Zn^{2+} \) has a completely filled d-orbital, resulting in a colorless solution, thus not matching the others which may appear yellow to brown.

Set 3: \( Ti^{4+} \), \( V^{4+} \), \( Mn^{2+} \)

\( Ti^{4+} \) usually produces a colorless solution due to its electronic configuration, whereas the other ions have visible colors.

Set 4: \( Sc^{3+} \), \( Ti^{3+} \), \( Cr^{2+} \)

\( Sc^{3+} \) is colorless in solution due to its lack of d-electrons, and thus, this set does not have similar colors either.

Thus, the correct set of ions that produce similarly colored aqueous solutions is \( V^{2+} \), \( Cr^{3+} \), \( Mn^{3+} \), due to their ability to exhibit complementary colored transitions in solutions.