Question

Which one of the following ions is diamagnetic in nature? 
 

• A. \({Ce}^{2+}\)
• B. \({Yb}^{2+}\)
• C. \({Eu}^{2+}\)
• D. \({Lu}^{2+}\)

Hint:

To determine whether an ion is diamagnetic, check if all its electrons are paired in its ground-state electron configuration. Elements with completely filled orbitals (like 4f\(^{14}\)) are generally diamagnetic.

Answer 1

The Correct Option is B

Diamagnetic substances have all their electrons paired, meaning they do not have unpaired electrons in their electronic configuration. Let's analyze each given ion:
- Ce\(^2+\) (Cerium): Cerium has an atomic number of 58, and in the Ce\(^2+\) state, it has unpaired electrons in the 4f orbitals, making it paramagnetic.
- Yb\(^2+\) (Ytterbium): Ytterbium has an atomic number of 70. In the Yb\(^2+\) state, its electron configuration is [Xe] 4f\(^{14}\). Since all 14 electrons in the f-orbital are paired, Yb\(^2+\) is diamagnetic.
- Eu\(^2+\) (Europium): Europium has an atomic number of 63. In Eu\(^2+\), its electron configuration is [Xe] 4f\(^{7}\), which contains unpaired electrons, making it paramagnetic.
- Lu\(^2+\) (Lutetium): Lutetium has an atomic number of 71. In Lu\(^2+\), its electron configuration is [Xe] 4f\(^{14}\) 5d\(^1\), which still contains an unpaired electron, making it paramagnetic.
Thus, the only diamagnetic ion among the given options is Yb\(^2+\).