JEE Main 2021 · 31 August · evening · Single correct · +4 / −1
The Eu²⁺ ion is a strong reducing agent in spite of its ground state electronic configuration (outermost) : [Atomic number of Eu = 63]
- (a) 4f⁷6s²
- (b) 4f⁶
- (c) 4f⁷
- (d) 4f⁶6s²
Answer and solution
Answer (c) 4f⁷
Solution
Chemistry ma’am · 4 steps
Take the two 6s electrons off first: Eu²⁺ is left with a half-filled 4f⁷ shell.
- Neutral atom. Europium has Z = 63, so its configuration is [Xe] 4f⁷ 6s². That is 54 + 7 + 2 = 63 electrons.
- Remove two electrons. When a lanthanide forms a cation, it loses its outermost 6s electrons first. So Eu → Eu²⁺ removes both 6s electrons, which leaves [Xe] 4f⁷.
- Why in spite of. 4f⁷ is a half-filled shell, which is extra stable. Even so, Eu²⁺ readily gives up one more electron to become Eu³⁺ (4f⁶), because +3 is the common and most stable oxidation state for lanthanides. That is why Eu²⁺ acts as a strong reducing agent.
- So. The outermost configuration of Eu²⁺ is 4f⁷, option (c).
The idea to keep. For lanthanide ions, strip the 6s electrons before touching 4f. Then check whether the ion is f⁰, f⁷ or f¹⁴, and ask how far it is from the usual +3 state.
Answer from the NTA final answer key. Solution by PrepWiser.