JEE Main 2020 · 6 September · morning · Single correct · +4 / −1
The lanthanoid that does NOT show +4 oxidation state is :
- (a) Tb
- (b) Dy
- (c) Ce
- (d) Eu
Answer and solution
Answer (d) Eu
Solution
Chemistry ma’am · 3 steps
Eu reaches the stable 4f⁷ shell at +2, so taking out more electrons would only break that shell.
- Usual rule. Every lanthanoid shows +3. Some also show +2 or +4, and this usually happens when that state gives an empty (4f⁰), half-filled (4f⁷) or full (4f¹⁴) f subshell.
Element Atom Ion with special f shell Shows +4? Ce [Xe]4f¹ 5d¹ 6s² Ce⁴⁺: 4f⁰ Yes Tb [Xe]4f⁹ 6s² Tb⁴⁺: 4f⁷ Yes Dy [Xe]4f¹⁰ 6s² Dy⁴⁺: 4f⁸, close to 4f⁷ Yes, in fluorides such as Cs₃DyF₇ Eu [Xe]4f⁷ 6s² Eu²⁺: 4f⁷ No, only +2 and +3 - Why not Eu. Europium already has a half-filled 4f⁷ shell once it loses its two 6s electrons. Reaching +4 would mean removing two electrons from that stable 4f⁷ shell, leaving 4f⁵. So Eu stops at +2 or +3.
- So. The lanthanoid that does not show +4 is Eu, option (d).
The idea to keep. For odd oxidation states of lanthanoids, write the f count of the ion. States giving 4f⁰, 4f⁷ or 4f¹⁴ occur. Eu and Yb favour +2, while Ce and Tb favour +4.
Answer from the NTA final answer key. Solution by PrepWiser.