NEET 2026 (Phase 1) · ChemistryMagnetismPrevious Year Question
The calculated 'spin-only' magnetic moment of \ce{Ti^2+} (\ce{3d^2}) is:
- A.(a) $5.92$ BM
- B.(b) $3.87$ BM
- C.(c) $2.84$ BM✓
- D.(d) $4.90$ BM
Correct Answer
(C) (c) $2.84$ BM
Solution & Explanation
\textbf{Answer:} (c) $2.84$ BM \textbf{Solution:} \ce{Ti^2+} has the configuration \ce{3d^2}, so it has $n=2$ unpaired electrons. Using the spin-only formula $\mu=\sqrt{n(n+2)}$: $\mu=\sqrt{2(2+2)}=\sqrt{8}=2.84$ BM. Hence the spin-only magnetic moment of \ce{Ti^2+} is $2.84$ BM.
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