Question:

The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion (\(\mathrm{M^{z+}}\)) are \(-0.8\,\Delta_0\) and 3.87 BM, respectively. Identify (\(\mathrm{M^{z+}}\)) :

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\(\mu = 3.87\) BM means 3 unpaired electrons. Aqua is a weak-field ligand (high spin); check which \(d^n\) gives \(-0.8\Delta_0\).
Updated On: Nov 4, 2026
  • \(\mathrm{V^{3+}}\)
  • \(\mathrm{Cr^{3+}}\)
  • \(\mathrm{Mn^{4+}}\)
  • \(\mathrm{Co^{2+}}\)
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The Correct Option is D

Solution and Explanation

\(\mu = \sqrt{n(n+2)} = 3.87\) gives \(n = 3\) unpaired electrons.
\(\mathrm{V^{3+}}\) (\(d^2\)): 2 unpaired, \(\mu = 2.83\) BM, CFSE \(= -0.8\Delta_0\); rejected on \(\mu\).
\(\mathrm{Cr^{3+}}\) and \(\mathrm{Mn^{4+}}\) (\(d^3\)): \(t_{2g}^3\), 3 unpaired, CFSE \(= 3(-0.4) = -1.2\Delta_0\); rejected on CFSE.
\(\mathrm{Co^{2+}}\) (\(d^7\)), weak field \(\mathrm{H_2O}\): \(t_{2g}^5e_g^2\), 3 unpaired; CFSE \(= 5(-0.4)+2(+0.6) = -0.8\Delta_0\). Both conditions satisfied.
Answer: (4) \(\mathrm{Co^{2+}}\)
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