JEE Main 2021 · 31 August · morning · Single correct · +4 / −1
Which one of the following lanthanides exhibits +2 oxidation state with diamagnetic nature? (Given Z for Nd = 60, Yb = 70, La = 57, Ce = 58)
- (a) Nd
- (b) Yb
- (c) La
- (d) Ce
Answer and solution
Answer (b) Yb
Solution
Chemistry ma’am · 4 steps
Remove two electrons and look at the 4f shell: only a full 4f¹⁴ leaves no unpaired electron.
- Diamagnetic means. An ion is diamagnetic only when all its electrons are paired. So you need an M²⁺ ion with a completely empty or completely full subshell.
- Form the M2+ ion. Lanthanides lose the two 6s electrons first. Write each neutral atom's configuration, take away 6s², and count the unpaired electrons left.
Element Neutral atom M²⁺ ion Unpaired electrons Nd (60) [Xe]4f⁴ 6s² [Xe]4f⁴ 4, paramagnetic Yb (70) [Xe]4f¹⁴ 6s² [Xe]4f¹⁴, full 0, diamagnetic La (57) [Xe]5d¹ 6s² [Xe]5d¹ 1, paramagnetic Ce (58) [Xe]4f¹ 5d¹ 6s² [Xe]4f² 2, paramagnetic - Why Yb2+ is stable. A completely filled 4f¹⁴ shell is extra stable. That is why ytterbium readily shows the +2 state, just as europium does with its half-filled 4f⁷.
- So. Yb²⁺ (4f¹⁴) shows the +2 state and has no unpaired electrons, so the answer is (b) Yb.
The idea to keep. For lanthanide ions, strip the 6s electrons first, then check the 4f count. Only 4f⁰ or 4f¹⁴ gives a diamagnetic ion; anything in between is paramagnetic.
Answer from the NTA final answer key. Solution by PrepWiser.