JEE Main 2021 · 26 February · morning · Single correct · +4 / −1
Which one of the following lanthanoids does not form MO₂? [M is lanthanoid metal]
- (a) Nd
- (b) Dy
- (c) Yb
- (d) Pr
Answer and solution
Answer (c) Yb
Solution
Chemistry ma’am · 4 steps
MO₂ means the metal is in the +4 state, so ask which of these four can lose a fourth electron.
- Read the formula. In MO₂, two O²⁻ ions carry a total charge of -4. So the metal must be M⁴⁺. The question is really asking which lanthanoid does not show the +4 oxidation state.
- Known +4 lanthanoids. +3 is the common state for all lanthanoids. Only a few also show +4: Ce, Pr, Nd, Tb and Dy. The oxides PrO₂, NdO₂ and DyO₂ are known in the solid state.
Element Configuration Extra state besides +3 Forms MO₂? Nd [Xe]4f⁴ 6s² +4 (and +2) yes Dy [Xe]4f¹⁰ 6s² +4 yes Yb [Xe]4f¹⁴ 6s² +2 only (4f¹⁴, full) no Pr [Xe]4f³ 6s² +4 yes - Why not Yb. Yb can lose just its two 6s electrons to give Yb²⁺ with a completely filled 4f¹⁴ shell, and this stability is why +2 appears. Going the other way to +4 would mean pulling two electrons out of that almost full 4f shell, which needs too much energy.
- So. Ytterbium shows +2 and +3 but never +4, so it does not form MO₂. The answer is (c).
The idea to keep. Match the formula to an oxidation state first. For lanthanoids, remember the lists: +4 for Ce, Pr, Nd, Tb, Dy; +2 for Eu and Yb. Yb sits near 4f¹⁴, so it goes down to +2, never up to +4.
Answer from the NTA final answer key. Solution by PrepWiser.