Q.18 The magnetic moment of an octahedral Co(II) complex is approximately 4.0 μB
(atomic number of Co is 27). The CFSE for this complex, in Δo units, is ________.
CFSE of Octahedral Co(II) Complex with 4.0 μB Magnetic Moment: Detailed Calculation Guide
Co(II) in an octahedral complex shows a magnetic moment of 4.0 μB, indicating high-spin d⁷ configuration with three unpaired electrons. This guides the crystal field stabilization energy (CFSE) calculation in Δo units using standard crystal field theory principles.
Electron Configuration Analysis
Cobalt (atomic number 27) as Co²⁺ loses two 4s electrons, yielding d⁷ configuration. In octahedral fields, d orbitals split into t₂g (lower, -0.4 Δo each) and e_g (higher, +0.6 Δo each). A 4.0 μB moment matches spin-only formula μ = √[n(n+2)] ≈ 3.87 μB for n=3 unpaired electrons, confirming high-spin due to weak field ligands.
- High-spin d⁷ fills as t₂g⁵ e_g²: five t₂g electrons (three unpaired, two paired) and two unpaired e_g electrons, totaling three unpaired.
- Low-spin (t₂g⁶ e_g¹, one unpaired, μ≈2.8 μB) is ruled out by the observed moment.
CFSE Calculation
CFSE sums electron energies relative to barycenter: CFSE = [-0.4 × (t₂g electrons) + 0.6 × (e_g electrons)] Δo. For high-spin t₂g⁵ e_g²:
Final Answer
CFSE = [-0.4 × 5 + 0.6 × 2] Δo = [-2.0 + 1.2] Δo = -0.8 Δo
Pairing energy is irrelevant for CFSE, which excludes it. The value -0.8 reflects stabilization from splitting.
Common Options Explained
No explicit options given, but typical MCQ choices for such problems include:
| Option | Configuration | Status |
|---|---|---|
| -0.8 Δo | High-spin d⁷ (t₂g⁵ e_g²) | CORRECT |
| -1.2 Δo | Low-spin d⁷ or d⁶ high-spin | INCORRECT |
| -0.4 Δo | d³ or d⁸ high-spin | INCORRECT |
Exam Tips
- For Co(II) octahedral problems, always check μ first: 4.0 μB signals high-spin.
- Practice d⁷ diagrams: weak ligands prioritize unpaired electrons over pairing.
- This yields consistent -0.8 Δo across similar queries.
