Calculate the magnetic moment of the Fe2+ ion using the spin-only formula.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4.90 BM
The magnetic moment of a transition metal ion comes mainly from its unpaired electrons. The spin-only formula is:
\( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM}\quad(n=\text{number of unpaired electrons}) \)
To find n, write the configuration of the ion. The 4s electrons are removed before the 3d electrons:
\( \displaystyle \ce{Fe}\ (Z=26):\ [\ce{Ar}]\,3d^{6}\,4s^{2} \)
\( \displaystyle \ce{Fe^2+}:\ [\ce{Ar}]\,3d^{6} \)
The six d electrons first occupy the five 3d orbitals singly (Hund’s rule), and the sixth electron pairs up. So \( \ce{Fe^{2+}} \) has 4 unpaired electrons:
\( \displaystyle \mu=\sqrt{4(4+2)} \)
\( \displaystyle \mu=\sqrt{24} \)
\( \displaystyle \mu=4.90\ \mathrm{BM} \)
So the spin-only magnetic moment of \( \ce{Fe^{2+}} \) is 4.90 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 – …
