Question

Which of the following lanthanoid ion is cloured?

• A. \(\text{Lu}^{3+}\)
• B. \(\text{Sm}^{2+}\)
• C. \(\text{Yb}^{2+}\)
• D. \(\text{La}^{3+}\)

Hint:

To quickly find colorless lanthanoid ions, look for the extremes: \(\text{La}^{3+}\) (\(f^0\)), \(\text{Ce}^{4+}\) (\(f^0\)), \(\text{Lu}^{3+}\) (\(f^{14}\)), and \(\text{Yb}^{2+}\) (\(f^{14}\)).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The color of lanthanoid ions, whether in solid state or in aqueous solution, is primarily attributed to \(f-f\) transitions. These transitions absorb light in the visible region. For an \(f-f\) transition to occur, the ion must have partially filled \(f\)-orbitals (i.e., an \(f^1\) to \(f^{13}\) configuration).
Step 2: Key Formula or Approach:
The approach requires writing the ground-state electronic configuration of each lanthanoid ion and identifying if it has unpaired electrons in its \(4f\) subshell. Ions with \(f^0\) or \(f^{14}\) configurations will be colorless.
Step 3: Detailed Explanation:
Let's determine the electronic configuration of each given ion:
- A) \(\text{Lu}^{3+}\): Lutetium (\(Z = 71\)). Configuration of Lu is \([\text{Xe}] 4f^{14} 5d^1 6s^2\). For \(\text{Lu}^{3+}\), remove 3 electrons (two from \(6s\), one from \(5d\)): \([\text{Xe}] 4f^{14}\). The \(f\)-subshell is completely filled, so no \(f-f\) transitions can occur. It is colorless.
- B) \(\text{Sm}^{2+}\): Samarium (\(Z = 62\)). Configuration of Sm is \([\text{Xe}] 4f^6 6s^2\). For \(\text{Sm}^{2+}\), remove 2 electrons from \(6s\): \([\text{Xe}] 4f^6\). It has 6 unpaired electrons in the partially filled \(f\)-subshell, allowing for \(f-f\) transitions. It will be colored (typically blood-red).
- C) \(\text{Yb}^{2+}\): Ytterbium (\(Z = 70\)). Configuration of Yb is \([\text{Xe}] 4f^{14} 6s^2\). For \(\text{Yb}^{2+}\), remove 2 electrons from \(6s\): \([\text{Xe}] 4f^{14}\). The \(f\)-subshell is completely filled, so it is colorless.
- D) \(\text{La}^{3+}\): Lanthanum (\(Z = 57\)). Configuration of La is \([\text{Xe}] 5d^1 6s^2\). For \(\text{La}^{3+}\), remove 3 electrons: \([\text{Xe}] 4f^0\). The \(f\)-subshell is completely empty, so it is colorless.
Step 4: Final Answer:
\(\text{Sm}^{2+}\) has a partially filled f-orbital (\(4f^6\)) and is therefore the colored ion.