Lanthanoid ions with 4 f⁷ configuration are :
(A) Eu²⁺
(B) Gd³⁺
(C) Eu³⁺
(D) Tb³⁺
(E) Sm²⁺
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Count f electrons from atomic number: Ln³⁺ has Z - 57, and the 6s²-only ions lose just two.

1. Quick counting rule. La³⁺ (Z = 57) is 4f⁰, and each element after it adds one 4f electron. So a Ln³⁺ ion has Z - 57 electrons in 4f. A Ln²⁺ ion keeps one more, so it has Z - 56.

   Ion	Neutral atom	Electrons removed	Ion configuration
   (A) Eu²⁺	Z = 63: [Xe] 4f⁷ 6s²	both 6s	4f⁷
   (B) Gd³⁺	Z = 64: [Xe] 4f⁷ 5d¹ 6s²	both 6s and the 5d	4f⁷
   (C) Eu³⁺	Z = 63: [Xe] 4f⁷ 6s²	both 6s and one 4f	4f⁶
   (D) Tb³⁺	Z = 65: [Xe] 4f⁹ 6s²	both 6s and one 4f	4f⁸
   (E) Sm²⁺	Z = 62: [Xe] 4f⁶ 6s²	both 6s	4f⁶

2. Why these two. Eu²⁺ and Gd³⁺ are isoelectronic. Both reach the half-filled 4f⁷ shell, which is extra stable, and that is why Eu²⁺ is a common oxidation state for europium.
3. So. Only (A) Eu²⁺ and (B) Gd³⁺ are 4f⁷, which is option (d).

The idea to keep. For any lanthanoid ion, start from La³⁺ = 4f⁰ and count along: Ln³⁺ has Z - 57 f electrons. Half-filled 4f⁷ is why Eu²⁺, Gd³⁺ and Tb⁴⁺ exist.

Answer from the NTA final answer key. Solution by PrepWiser.