Magnetic moment of a divalent ion in aqueous solution of an element with atomic number 25 is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5.92 BM
The spin only magnetic moment depends on the number of unpaired electrons, n:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)
The element with atomic number 25 is manganese. Beyond the argon core it has 25 − 18 = 7 electrons: Mn is [Ar]3d54s2. The divalent ion is formed by losing the two 4s electrons, so \( \ce{Mn^{2+}} \) is [Ar]3d5. By Hund’s rule the five electrons occupy the five 3d orbitals singly:
3d: [↑ ] [↑ ] [↑ ] [↑ ] [↑ ]
So n = 5 unpaired electrons:
\( \displaystyle \mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\ \mathrm{BM} \)
Note 1: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Magnetic Properties – …
Note 2: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | 28-05 S2 | – | – | 20-08 S1 10-08 S1 |