NEET 2026 (Phase 1) · ChemistryPrevious Year Question
The calculated 'spin-only' magnetic moment of \ce{Ti^2+} ($3d^{2}$) is:
- A.5.92 BM
- B.3.87 BM
- C.2.84 BM✓
- D.4.90 BM
Correct Answer
(C) 2.84 BM
Solution & Explanation
\textbf{Answer:} (C) 2.84 BM \textbf{Solution:} Spin-only magnetic moment $\mu = \sqrt{n(n+2)}$ BM, where $n$ = number of unpaired electrons. \ce{Ti^2+} $= [\ce{Ar}]3d^{2}$ has 2 unpaired electrons, so $\mu = \sqrt{2\times4} = \sqrt{8} = 2.84$ BM.
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