JEE MainChemistryd and f Block Elements
Match List I with List II: List I (Ion) List II (Property and Configuration) (A) Ce ⁴⁺ (I) Reductant, f¹⁴ configuration (B) Eu ²⁺ (II) Oxidant, f⁰ configuration (C) Tb ⁴⁺ (III) Reductant, f⁷ configuration (D) Yb ²⁺ (IV) Oxidant, f⁷ configuration Choose the correct answer from the options given below:
Options
- A(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
- B(A)-(II), (B)-(IV), (C)-(III), (D)-(I)
- C(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- D(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Correct answer
A. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Step-by-step solution
The most stable oxidation state for all lanthanoids is +3 . Ions in the +4 state will tend to gain an electron to reach +3 , thus acting as oxidants. Ions in the +2 state will tend to lose an electron to reach +3 , thus acting as reductants. (A) Cerium (Ce, Z=58 ) is [ Xe ] 4f¹ 5d¹ 6s² . Ce ⁴⁺ has an f⁰ configuration. It acts as an oxidant to become Ce ³⁺ . Matches (II). (B) Europium (Eu, Z=63 ) is [ Xe ] 4f⁷ 6s² . Eu ²⁺ has an f⁷ configuration. It acts as a reductant to become Eu ³⁺ . Matches (III). (C) Terbium (T