Question

Which of the following lanthanoid ions is diamagnetic ?
(At. nos. Ce = 58, Sm = 62, Eu = 63, Yb = 70)

• A. Ce²⁺
• B. Sm²⁺
• C. Eu²⁺
• D. Yb²⁺

Answer 1

The Correct Option is D

To determine which of the given lanthanoid ions is diamagnetic, we need to consider their electronic configurations. Diamagnetism occurs when all electrons are paired in an atom or ion. Let us evaluate each option step-by-step:

1. Cerium (Ce²⁺):
   • The atomic number of Ce is 58. The neutral atom electronic configuration is [Xe] 4f¹ 5d¹ 6s².
   • For Ce²⁺, two electrons are removed. The resulting configuration is [Xe] 4f².
   • The 4f shell is not full, hence electrons are unpaired, making Ce²⁺ paramagnetic.
2. Samarium (Sm²⁺):
   • The atomic number of Sm is 62. The neutral atom electronic configuration is [Xe] 4f⁶ 6s².
   • For Sm²⁺, two electrons are removed, resulting in the configuration [Xe] 4f⁶.
   • The 4f⁶ configuration has unpaired electrons, thus Sm²⁺ is paramagnetic.
3. Europium (Eu²⁺):
   • The atomic number of Eu is 63. The neutral atom electronic configuration is [Xe] 4f⁷ 6s².
   • For Eu²⁺, two electrons are removed, leading to the configuration [Xe] 4f⁷.
   • The 4f⁷ configuration is half-filled, which means unpaired electrons are present. Hence, Eu²⁺ is paramagnetic.
4. Ytterbium (Yb²⁺):
   • The atomic number of Yb is 70. The neutral atom electronic configuration is [Xe] 4f¹⁴ 6s².
   • For Yb²⁺, two electrons are removed, giving the configuration [Xe] 4f¹⁴.
   • The 4f¹⁴ configuration is fully filled, meaning all electrons are paired.
   • Thus, Yb²⁺ is diamagnetic.

Conclusion: Among the given options, Yb²⁺ is the only ion that is diamagnetic, as it has all paired electrons with a completely filled 4f subshell. Therefore, the correct answer is Yb²⁺.