The magnetic moment of a trivalent ion (Z=27) in its aqueous solution is calculated as
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4.90 BM
The spin-only magnetic moment of an ion depends on its number of unpaired electrons, n:
\( \displaystyle \mu=\sqrt{n(n+2)}\ \text{BM} \)
The element with Z = 27 is cobalt. When it forms an ion, the 4s electrons are removed first:
\( \displaystyle \ce{Co}:\ [\text{Ar}]\,3d^{7}\,4s^{2} \)
\( \displaystyle \ce{Co^3+}:\ [\text{Ar}]\,3d^{6} \)
In 3d6, five electrons occupy the five d orbitals singly and the sixth pairs up, leaving 4 unpaired electrons.
\( \displaystyle \mu=\sqrt{4(4+2)}=\sqrt{24}=4.90\ \text{BM} \)
Shortcut: The spin-only values for n = 1 to 5 are 1.73, 2.84, 3.87, 4.90 and 5.92 BM, roughly n + 1 in each case.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Magnetic Properties – …
