Which of the following ions will exhibit colour in aqueous solution?

Ti³⁺ (Z=22)

The correct answer is option 2. Ti³⁺ (Z=22).

The color in transition metal ions and some f-block elements in aqueous solutions is due to the presence of unpaired electrons in their d or f orbitals. These unpaired electrons can absorb certain wavelengths of visible light and get excited to higher energy levels. The specific wavelengths absorbed depend on the electron configuration and the crystal field environment created by the surrounding ligands (water molecules in this case).

Analysis of Each Ion

1. \(La^{3+}\) (Lanthanum, Z=57):

The electronic Configuration of Lanthanum is \([Xe] 5d^1 6s^2\)

\(La^{3+}\) = \([Xe]\)

In the \(La^{3+}\) ion, the configuration is simply the xenon core \(([Xe])\) with no electrons in the \(d\) or \(f\) orbitals. Since there are no unpaired electrons, there are no \(d-d\) transitions or \(f-f\) transitions possible. Thus, \(La^{3+}\) is colorless in an aqueous solution.

2. \(Ti^{3+}\) (Titanium, Z=22)

The electronic Configuration of Titanium is \([Ar] 3d^2 4s^2\)

\(Ti^{3+}\)= \([Ar] 3d^1\)

\(Ti^{3+}\) has one electron in the \(3d\) orbital.

The presence of this single unpaired electron in the \(d\) orbital allows for \(d-d\) transitions when light is absorbed. This leads to the ion being colored in an aqueous solution.

3. \(Lu^{3+}\) (Lutetium, Z=71)

The electronic Configuration of Lutetium is \([Xe] 4f^{14} 5d^1 6s^2\)

\(Lu^{3+}\) = \([Xe] 4f^{14}\)

In the \(Lu^{3+}\) ion, the 4f subshell is completely filled with 14 electrons, and the d and s orbitals are empty. Since the \(4f\) subshell is completely filled and there are no unpaired electrons, \(Lu^{3+}\) cannot exhibit any \(f-f\) transitions. Therefore, \(Lu^{3+}\) is colorless in an aqueous solution.

4. \(Sc^{3+}\) (Scandium, Z=21)

The electronic Configuration of Scandium is \([Ar] 3d^1 4s^2\)

\(Sc^{3+}\) = \([Ar]\)

\(Sc^{3+}\) has no electrons in the d or f orbitals.

Since there are no unpaired electrons, \(Sc^{3+}\) cannot exhibit any \(d-d\) transitions. Thus, \(Sc^{3+}\) is colorless in an aqueous solution.

Summary

\(La^{3+}\): No unpaired electrons, colorless.

\(Ti^{3+}\): One unpaired electron in the 3d orbital, colored.

\(Lu^{3+}\): No unpaired electrons, colorless.

\(Sc^{3+}\): No unpaired electrons, colorless.

Therefore, the ion that will exhibit color in an aqueous solution is: \(Ti^{3+}\) (Z=22)

This is due to the presence of one unpaired electron in the \(3d\) orbital, which allows for \(d-d\) transitions and thus results in the ion being colored.