NEET 2026 (Phase 1) · ChemistryValence bondPrevious 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:} The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM. $\ce{Ti^2+}$ has the configuration $3d^2$, giving $n = 2$ unpaired electrons. $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.84$ BM. Hence the answer is option C.
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