Question

The lanthanide ion that would show colour is-

• A. ${Sm^{3+} }$
• B. ${La^{3+} }$
• C. ${Lu^{3+} }$
• D. ${Gd^{3+} }$

Answer 1

The Correct Option is A

To determine which lanthanide ion would show color, we must consider the electronic configuration of each option and the concept of f-f electron transitions which cause coloration in lanthanide ions.

1. {La^{3+}}: Lanthanum ion loses 3 electrons and has the electronic configuration of {[Xe]4f^0}. With no electrons in the 4f orbital, there are no f-f transitions possible, hence it is colorless.
2. {Lu^{3+}}: Lutetium ion also loses 3 electrons, resulting in {[Xe]4f^{14}}. With a completely filled 4f orbital, no f-f transitions are possible, resulting in a colorless ion.
3. {Gd^{3+}}: Gadolinium ion has the configuration {[Xe]4f^7}. Even though it has unpaired electrons which may suggest possible transitions, the half-filled nature provides greater symmetry and less energy difference between levels, leading to weak or no visible coloration.
4. {Sm^{3+}}: Samarium ion has the configuration {[Xe]4f^5}. This ion has unpaired electrons in the 4f orbital, which permits f-f electronic transitions resulting in the manifestation of color.

Therefore, among the given ions, {Sm^{3+}} has an electronic configuration that allows f-f transitions, making it the ion that exhibits color.