JEE Main 2021 · 25 July · evening · Single correct · +4 / −1

The spin only magnetic moments (in BM) for free Ti³⁺, V²⁺ and Sc³⁺ ions respectively are :
(At.No. Sc : 21, Ti : 22, V : 23)

  1. (a) 3.87, 1.73, 0
  2. (b) 1.73, 3.87, 0
  3. (c) 1.73, 0, 3.87
  4. (d) 0, 3.87, 1.73
Answer and solution

Answer (b) 1.73, 3.87, 0

Solution

Chemistry ma’am · 2 steps

Write each ion's 3d configuration, count the unpaired electrons, then put n into the spin-only formula.

  1. Remove 4s first. A transition metal loses its 4s electrons before its 3d electrons. So subtract the charge from the atomic number and fill only 3d with what is left after the argon core.

    μ_(spin only) = √n(n+2) BM

    IonElectronsConfigurationUnpaired n√n(n+2)
    Ti³⁺22 - 3 = 19[Ar] 3d¹1√3 = 1.73
    V²⁺23 - 2 = 21[Ar] 3d³3√15 = 3.87
    Sc³⁺21 - 3 = 18[Ar] 3d⁰00
  2. So. In the order asked, the moments are 1.73, 3.87 and 0 BM, which is option (b).

The idea to keep. For a 3d ion, take the 4s electrons off first and count unpaired 3d electrons. Remember n = 1, 2, 3, 4, 5 give 1.73, 2.83, 3.87, 4.90, 5.92 BM.

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 2021.