Calculated value of magnetic moment (BM) of an aqueous solution of \( Mn^{2+} \) is .........
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5.92
The magnetic moment of a transition metal ion comes mainly from the spin of its unpaired electrons. The spin-only formula gives it in Bohr magnetons (BM) from the number of unpaired electrons, n:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)
Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Mn^2+} \): \( \mathrm{3d^{5}} \). In aqueous solution (with water as a weak field ligand) the five 3d electrons occupy the five d orbitals singly, so n = 5:
\( \displaystyle \mu = \sqrt{5(5+2)} = \sqrt{35} \)
\( \displaystyle \mu \approx 5.92\ \mathrm{BM} \)
Note: 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 – …
