The lanthanide ion having four unpaired electrons is: (Atomic numbers: Ce =58, Nd =60, Tb =65, Ho =67)
- A.Nd³⁺
- B.Ce³⁺
- C.Tb³⁺
- D.Ho³⁺✓
Correct Answer
(D) Ho³⁺
Solution & Explanation
For a lanthanide ion Ln³⁺ the electronic configuration is [Xe]4f^n; find n, then count unpaired electrons in the 7 f-orbitals by Hund's rule (fill singly first up to 7, then pair). Ce³⁺ (Z=58): 4f¹ ⇒ 1 unpaired. Nd³⁺ (Z=60): 4f³ ⇒ 3 unpaired. Tb³⁺ (Z=65): 4f⁸ ⇒ 7 singly-filled then 1 paired ⇒ 6 unpaired. Ho³⁺ (Z=67): 4f¹⁰ ⇒ 7 orbitals each get one electron, then 3 pair up, leaving 7 − 3 = 4 unpaired. The ion with exactly four unpaired electrons is Ho³⁺. Answer: option D.
✅ We had this exact concept in our bank before the exam — Strong match, 88% match.
That's the Predicted Batch — it trains you on the exact high-probability concepts NEET repeats every year. 98% of the 2026 paper (176/180) was in our bank before the exam.
Master these concepts + 50,000+ NEET PYQs in the app — free, built for NEET 2027.
Start the Predicted Batch — Free →