Question

Which of the following is colorless?

• A. Eu³⁺
• B. Lu³⁺
• C. Nd³⁺
• D. Sm³⁺

Answer 1

The Correct Option is B

To determine which of the given ions is colorless, we need to understand the electronic configurations of these ions and their positions in the periodic table, particularly focusing on the presence of unpaired electrons. Color in transition and inner transition metal ions generally arises due to electronic transitions among d-orbitals or f-orbitals. Let's analyze each option:

1. Eu³⁺: Europium (Eu) has an atomic number of 63. Its electronic configuration is [Xe] 4f⁷ 6s². When it loses three electrons to form Eu³⁺, the electronic configuration becomes [Xe] 4f⁶. Since it still has unpaired electrons, Eu³⁺ will exhibit color.
2. Lu³⁺: Lutetium (Lu) has an atomic number of 71. Its electronic configuration is [Xe] 4f¹⁴ 5d¹ 6s². In the Lu³⁺ ion, the electronic configuration is [Xe] 4f¹⁴ because it loses the 5d and 6s electrons. The 4f subshell is completely filled, resulting in no unpaired electrons. Hence, Lu³⁺ is colorless due to the lack of f-f electronic transitions.
3. Nd³⁺: Neodymium (Nd) has an atomic number of 60. Its electronic configuration is [Xe] 4f⁴ 6s². In the Nd³⁺ ion, the configuration becomes [Xe] 4f³, containing unpaired electrons which result in color.
4. Sm³⁺: Samarium (Sm) has an atomic number of 62. Its electronic configuration is [Xe] 4f⁶ 6s². When Sm forms Sm³⁺, the configuration becomes [Xe] 4f⁵. The presence of unpaired electrons will also cause color to be exhibited.

Conclusion: Among the given options, Lu³⁺ is colorless because it has a completely filled 4f subshell with no unpaired electrons, preventing any electronic transitions that would result in color.