Question

Consider that a \(\mathrm{d^6}\) metal ion (\(\mathrm{M^{2+}}\)) forms a complex with aqua ligands, and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilization energy of the complex is

• A. tetrahedral and \(-1.6\,\Delta_t + 1P\)
• B. octahedral and \(-2.4\,\Delta_0 + 2P\)
• C. tetrahedral and \(-0.6\,\Delta_t\)
• D. octahedral and \(-1.6\,\Delta_0\)

Hint:

4.90 BM means 4 unpaired electrons; match this with the CFSE expression of each option for a d6 ion.

Answer 1

The Correct Option is C

\(\mu = \sqrt{n(n+2)} = 4.90\) gives \(n(n+2) = 24\), so \(n = 4\) unpaired electrons for the \(\mathrm{d^6}\) ion.
Aqua is a weak field ligand, so the complex is high spin.
Tetrahedral \(\mathrm{d^6}\): \(e^3t_2^3\) has 4 unpaired electrons; CFSE \(= 3(-0.6) + 3(+0.4) = -0.6\,\Delta_t\).
Check the others: tetrahedral \(-1.6\Delta_t + 1P\) is \(e^4t_2^2\) (2 unpaired, 2.83 BM); octahedral \(-2.4\Delta_0 + 2P\) is \(t_{2g}^6\) (diamagnetic); octahedral \(-1.6\Delta_0\) corresponds to \(t_{2g}^4\), a d4 low-spin pattern, not d6.
Among the options, only tetrahedral with \(-0.6\,\Delta_t\) corresponds to 4 unpaired electrons.
Answer: (3) tetrahedral and \(-0.6\,\Delta_t\)