Match List I with List II
| List I | List II |
| A. \(Fe^{2+}\) | I. 2.84 BM |
| B. \(Ni^{2+}\) | II. 0.00 BM |
| C. \(Mn^{2+}\) | III. 4.90 BM |
| D. \(Zn^{2+}\) | IV. 5.92 BM |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-I, C-IV, D-II
Spin-only magnetic moment: μ = √[n(n + 2)] BM, where n = number of unpaired electrons.
A. \( \ce{Fe^2+} \) — This matches III. 4.90 BM. Fe (Z = 26) [Ar] 3d6 4s2; \( \ce{Fe^2+} \) 3d6, n = 4: μ = √24 = 4.90 BM.
B. \( \ce{Ni^2+} \) — This matches I. 2.84 BM. Ni (Z = 28) [Ar] 3d8 4s2; \( \ce{Ni^2+} \) 3d8, n = 2: μ = √8 = 2.84 BM.
C. \( \ce{Mn^2+} \) — This matches IV. 5.92 BM. Mn (Z = 25) [Ar] 3d5 4s2; \( \ce{Mn^2+} \) 3d5, n = 5: μ = √35 = 5.92 BM.
D. \( \ce{Zn^2+} \) — This matches II. 0.00 BM. Zn (Z = 30) [Ar] 3d10 4s2; \( \ce{Zn^2+} \) 3d10, n = 0.
NCERT (2026-27 edition):
Magnetic Properties – …
