NEET Chemistry Question (2026) — Solution

Step-by-step solution

The general electronic configuration of lanthanide ions Ln ^ 3+ is [ Xe ] 4f^ Z-57 , where Z is the atomic number. For Ho ^ 3+ (Z=67): Number of 4f electrons = 67 - 57 = 10 Configuration is [ Xe ] 4f^ 10 . Since the f-subshell has 7 orbitals, 10 electrons will fill as 3 pairs and 4 unpaired electrons. Number of unpaired electrons = 14 - 10 = 4. For Nd ^ 3+ (Z=60): Number of 4f electrons = 60 - 57 = 3 Configuration is [ Xe ] 4f^ 3 . Number of unpaired electrons = 3. For Ce ^ 3+ (Z=58): Number of 4f electrons = 58 - 57 = 1 Configuration is [ Xe ] 4f^ 1 . Number of unpaired electrons = 1. For Tb ^ 3+ (Z=65): Number of 4f electrons = 65 - 57 = 8 Configuration is [ Xe ] 4f^ 8 . Number of unpaired electrons = 14 - 8 = 6. Thus, Ho ^ 3+ has exactly four unpaired electrons. Answer: Ho ^ 3+