Step 1: Understanding the Concept:
The color of Lanthanoid ions is typically due to \(f-f\) transitions. For an ion to be colorless, it either has no electrons in \(f\)-orbitals (\(4f^{0}\)), a completely filled \(f\)-subshell (\(4f^{14}\)), or its electronic transitions do not fall in the visible range of the spectrum.
Step 2: Key Formula or Approach:
Examine the electronic configurations:
- \(La^{3+} (Z=57)\): \([Xe] 4f^{0}\) (Colorless due to absence of \(f\) electrons)
- \(Lu^{3+} (Z=71)\): \([Xe] 4f^{14}\) (Colorless due to filled \(f\) subshell)
- \(Gd^{3+} (Z=64)\): \([Xe] 4f^{7}\)
: Detailed Explanation:
1. \(Gd^{3+}\) has 7 unpaired electrons in its \(4f\) subshell (half-filled stability).
2. Although it has unpaired electrons, the energy required for \(f-f\) transitions in \(Gd^{3+}\) is very high because the transitions must occur from a stable half-filled configuration to a higher energy level.
3. These transitions occur in the ultraviolet (UV) region of the electromagnetic spectrum, not the visible region.
4. Therefore, to the human eye, the ion appears colorless.
Step 3: Final Answer:
The ion is \(Gd^{3+}\).