NEETChemistryd and f Block Elements
Match List-I with List-II. List-I (Element) List-II (Anomalous state and configuration) (A) Cerium (Ce) (I) +4 ( 4f⁷ ) (B) Europium (Eu) (II) +2 ( 4f¹⁴ ) (C) Terbium (Tb) (III) +4 ( 4f⁰ ) (D) Ytterbium (Yb) (IV) +2 ( 4f⁷ ) Choose the correct answer from the options given below:
Options
- A(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
- B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- C(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
- D(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
Correct answer
B. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Step-by-step solution
The electronic configurations of the given lanthanoids and their stable anomalous oxidation states are as follows: Cerium (Ce, Z=58 ): [Xe] 4f¹ 5d¹ 6s² . It loses four electrons to form Ce⁴⁺ with a stable noble gas core configuration [Xe] 4f⁰ . Europium (Eu, Z=63 ): [Xe] 4f⁷ 6s² . It loses two electrons to form Eu²⁺ with a stable half-filled configuration [Xe] 4f⁷ . Terbium (Tb, Z=65 ): [Xe] 4f⁹ 6s² . It loses four electrons to form Tb⁴⁺ with a stable half-filled configuration [Xe] 4f⁷ . Ytterbium (Yb, Z=70 ): [Xe]