Calculate the magnetic moment (spin-only) of a trivalent ion in aqueous solution if its atomic number is 22.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.73 BM
The spin-only magnetic moment depends only on the number of unpaired electrons, n:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)
The element with Z = 22 is titanium, \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \). The trivalent ion \( \ce{Ti^3+} \) is formed by losing the two 4s electrons and one 3d electron, which leaves \( \mathrm{3d^{1}} \): one unpaired electron.
\( \displaystyle \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\ \mathrm{BM} \)
So the magnetic moment of \( \ce{Ti^3+} \) in aqueous solution is 1.73 BM.
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 – …
