JEE Main 2022 · 27 June · evening · Single correct · +4 / −1

The 'f' orbitals are half and completely filled, respectively in lanthanide ions :
[Given : Atomic no. Eu, 63; Sm, 62; Tm, 69; Tb, 65; Yb, 70; Dy, 66]

  1. (a) Eu²⁺ and Tm²⁺
  2. (b) Sm²⁺ and Tm³⁺
  3. (c) Tb⁴⁺ and Yb²⁺
  4. (d) Dy³⁺ and Yb³⁺
Answer and solution

Answer (c) Tb⁴⁺ and Yb²⁺

Solution

Chemistry ma’am · 3 steps

Strip the 6s electrons first, then the 4f, and look for exactly 7 and exactly 14 f electrons.

  1. Neutral atom. Most lanthanide atoms are [Xe] 4fⁿ 6s². Xenon has 54 electrons, so 4f plus 6s together hold Z - 54 electrons.
  2. Forming the ion. The two 6s electrons leave first. Any further electrons come out of 4f. A lanthanide ion keeps no 5d or 6s electrons, so the rule below works for every ion.

    number of 4f electrons in Lnⁿ⁺ = Z - 54 - n

    OptionFirst ionSecond ionHalf (7), then full (14)?
    (a)Eu²⁺: 63 - 54 - 2 = 7Tm²⁺: 69 - 54 - 2 = 13no, the second ion is one short of 14
    (b)Sm²⁺: 62 - 54 - 2 = 6Tm³⁺: 69 - 54 - 3 = 12no
    (c)Tb⁴⁺: 65 - 54 - 4 = 7Yb²⁺: 70 - 54 - 2 = 14yes
    (d)Dy³⁺: 66 - 54 - 3 = 9Yb³⁺: 70 - 54 - 3 = 13no
  3. So. Tb⁴⁺ is 4f⁷ (half filled) and Yb²⁺ is 4f¹⁴ (completely filled), in that order. The answer is (c).

The idea to keep. For any lanthanide ion, the 4f count is Z - 54 - charge. A count of 7 or 14 explains the unusual stable states: Eu²⁺, Gd³⁺ and Tb⁴⁺ have 7, and Yb²⁺ has 14.

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

Study d & f Block Elements

Asked about 170 times in JEE Main since 2020. This question is from 2022.