Diamagnetic Lanthanoid ions are :

Solution

Chemistry ma’am · 3 steps

Write the 4f count for each ion: only 4f⁰ or 4f¹⁴ leaves no unpaired electron.

1. What diamagnetic needs. A diamagnetic ion has no unpaired electrons. For lanthanoid ions the electrons that matter are the 4f electrons, so the 4f shell must be either empty (4f⁰) or completely full (4f¹⁴).
2. Find the 4f count. Start from the neutral atom's configuration and remove electrons in the order 6s, then 5d, then 4f. The 4f count left behind tells you the number of unpaired electrons, using Hund's rule over the 7 f orbitals.

   Ion	Atom	Ion configuration	Unpaired electrons
   La³⁺	[Xe] 5d¹ 6s²	[Xe] 4f⁰	0, diamagnetic
   Ce⁴⁺	[Xe] 4f¹ 5d¹ 6s²	[Xe] 4f⁰	0, diamagnetic
   Lu³⁺	[Xe] 4f¹⁴ 5d¹ 6s²	[Xe] 4f¹⁴	0, diamagnetic
   Nd³⁺	[Xe] 4f⁴ 6s²	[Xe] 4f³	3, paramagnetic
   Eu³⁺	[Xe] 4f⁷ 6s²	[Xe] 4f⁶	6, paramagnetic

3. So. Only option (a) has two ions with no unpaired electrons: La³⁺ and Ce⁴⁺ are both 4f⁰.

The idea to keep. For lanthanoid ions, check only the 4f count. 4f⁰ (La³⁺, Ce⁴⁺) and 4f¹⁴ (Yb²⁺, Lu³⁺) are diamagnetic; every other 4f count is paramagnetic.

Answer from the NTA final answer key. Solution by PrepWiser.