Question:

Consider the hydrated ions of \(\mathrm{Ti^{2+}}\), \(\mathrm{V^{2+}}\), \(\mathrm{Ti^{3+}}\), and \(\mathrm{Sc^{3+}}\). The correct order of their spin-only magnetic moments is :

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Find the number of unpaired electrons n for each ion and use \(\mu=\sqrt{n(n+2)}\) BM.
Updated On: Nov 5, 2026
  • \(\mathrm{Sc^{3+}} \lt \mathrm{Ti^{3+}} \lt \mathrm{V^{2+}} \lt \mathrm{Ti^{2+}}\)
  • \(\mathrm{Sc^{3+}} \lt \mathrm{Ti^{3+}} \lt \mathrm{Ti^{2+}} \lt \mathrm{V^{2+}}\)
  • \(\mathrm{Ti^{3+}} \lt \mathrm{Ti^{2+}} \lt \mathrm{Sc^{3+}} \lt \mathrm{V^{2+}}\)
  • \(\mathrm{V^{2+}} \lt \mathrm{Ti^{2+}} \lt \mathrm{Ti^{3+}} \lt \mathrm{Sc^{3+}}\)
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The Correct Option is B

Solution and Explanation

Electronic configurations: \(\mathrm{Sc^{3+}}\): \(3d^0\), n = 0, \(\mu=0\). \(\mathrm{Ti^{3+}}\): \(3d^1\), n = 1, \(\mu=\sqrt3=1.73\) BM. \(\mathrm{Ti^{2+}}\): \(3d^2\), n = 2, \(\mu=\sqrt8=2.83\) BM. \(\mathrm{V^{2+}}\): \(3d^3\), n = 3, \(\mu=\sqrt{15}=3.87\) BM.
Order: \(\mathrm{Sc^{3+}} \lt \mathrm{Ti^{3+}} \lt \mathrm{Ti^{2+}} \lt \mathrm{V^{2+}}\).
Answer: (2) Sc3+ < Ti3+ < Ti2+ < V2+
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