GATE 2026 CY – Question 54
The pair(s) of lanthanide ions whose ground state term symbols have same spin multiplicity (2S+1) and orbital angular momentum (L), but different spin-orbit coupling (J) is(are) (Given: Atomic number: Nd = 60; Pm = 61; Ho = 67; Er = 68)
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: (A) Pm3+ and Ho3+; (B) Nd3+ and Er3+
Explanation
The ground-state term of a $4f^n$ ion follows Hund's rules. Ions with $n$ and $14-n$ electrons (electron-hole pairs) have the same $S$ and $L$ but different $J$: for less than half-filled shells $J=|L-S|$, and for more than half-filled shells $J=L+S$.
| Ion | Configuration | $S$ | $L$ | Term | $J$ |
|---|---|---|---|---|---|
| Nd³⁺ | $f^3$ | 3/2 | 6 | $^4I$ | $9/2$ |
| Pm³⁺ | $f^4$ | 2 | 6 | $^5I$ | $4$ |
| Ho³⁺ | $f^{10}$ | 2 | 6 | $^5I$ | $8$ |
| Er³⁺ | $f^{11}$ | 3/2 | 6 | $^4I$ | $15/2$ |
- **Pm³⁺ and Ho³⁺:** same multiplicity (5) and $L$ (I), different $J$ (4 and 8). ✓
- **Nd³⁺ and Er³⁺:** same multiplicity (4) and $L$ (I), different $J$ (9/2 and 15/2). ✓
- **Pm³⁺ and Nd³⁺, Ho³⁺ and Er³⁺:** different $S$. ✗
Answer **A and B**.