A Magnetic Moment of 2.3 BM That Does Not Correspond to Any Whole Number of Electrons
Two iron(III) samples sit on the bench: deep red K3[Fe(CN)6] and very pale violet Fe(H2O)63. Both contain octahedral Fe(III). A student measures the magnetic moment of the red solid on a Gouy balance at room temperature and obtains μeff = 2.3 BM. She concludes that the cyanide complex must contain three unpaired electrons, because a single unpaired electron would give 1.73 BM and 2.3 is much closer to that region than to 5.9. A second student computes the spin-only values for 1 and 5 unpaired electrons and points out that no integral number of unpaired electrons reproduces 2.3 BM. The aqua complex behaves quite differently: its solutions are distinctly acidic (pH ≈ 2 at 0.1 mol L^-1), it is only faintly coloured, and its μeff is close to 5.9 BM. The discussion then turns to two cobalt(III) compounds, [Co(NH3)6]Cl3, which is diamagnetic, and K3[CoF6], for which μeff ≈ 4.9 BM.
1. Assign the d-electron configuration and number of unpaired electrons for each iron(III) complex, and explain why 2.3 BM does not imply three unpaired electrons. Is the first student's inference defensible?
Read the worked answer
Spin states. Fe(III) is d^5. In [Fe(CN)6]^3-, CN^- is a strong-field ligand: Δo exceeds the pairing energy, so the configuration is t2g^5 eg^0, low spin, one unpaired electron, spin-only μ = √(1 × 3) = 1.73 BM. In [Fe(H2O)6]^3+, water is a weak-field ligand, so the configuration is t2g^3 eg^2, high spin, five unpaired electrons, spin-only μ = √(5 × 7) = 5.92 BM - which is what is observed for the aqua complex.
Why 2.3 BM. The spin-only formula √(n(n+2)) assumes the orbital contribution is fully quenched, which requires a non-degenerate ground term. A low-spin d^5 ion has a single unpaired electron in the triply degenerate t2g set, so orbital angular momentum is not fully quenched; spin–orbit coupling mixes excited states into the ground state and the measured moment at 300 K can be anything from about 1.7 to 2.5 BM. In real salts of [Fe(CN)6]^3- values near 2.0–2.4 BM are routine. The value 2.3 BM is therefore fully consistent with one unpaired electron.
Verdict. The student's inference is not defensible. 2.3 BM lies between the spin-only values for 1 and 3 unpaired electrons, and there is no integral n that yields it exactly; treating √(n(n+2)) as an invertible formula is the error. The correct procedure is to use the spin-only value as a guide for high-spin/high-symmetry cases and to invoke orbital contribution when the number falls between plateaus. Independent evidence - the deep colour and the well-known position of CN^- in the spectrochemical series - supports low spin.
2. For [Co(NH3)6]^3+ and [CoF6]^3-, give the d configuration, the spin state, the predicted spin-only moment and the qualitative ligand-field energy-level diagram. Do the observed moments match?
Read the worked answer
Both are d^6 cobalt(III), octahedral. The only difference is Δo.
[Co(NH3)6]^3+. NH3 is a mid-to-strong-field ligand, and Co(III) has a high charge (Δo scales roughly with the metal's charge), so Δo > P. Configuration t2g^6 eg^0, all electrons paired, S = 0, μ(spin-only) = 0. Observed: diamagnetic, as expected.
[CoF6]^3-. F^- is at the weak-field end of the spectrochemical series, so Δo < P and the high-spin configuration t2g^4 eg^2 results, with four unpaired electrons. Spin-only μ = √(4 × 6) = √24 = 4.90 BM, matching the observed 4.9 BM closely (slight orbital contribution is usually present, so values of 4.9–5.2 BM are typical).
Energy-level picture. Draw the d-orbital splitting with t2g below and eg above, separated by Δo. For the ammine complex, Δo is large and all six electrons occupy t2g, so eg is empty. For the fluoro complex, Δo is small and the fourth electron pairs in t2g only at the cost of P; instead it enters eg, leaving two electrons in each eg orbital.
Exam-relevant caution. [CoF6]^3- is one of the very few genuine high-spin cobalt(III) complexes. Do not generalise from the familiar low-spin Co(III) ammine chemistry and assume all Co(III) is diamagnetic - the ligand decides.
3. A classmate writes: 'K3[CoF6] is less stable than [Co(NH3)6]^3+ purely because of the CFSE difference.' Evaluate this statement, using CFSE values and at least two factors the CFSE argument ignores.
Read the worked answer
The CFSE part is real but small in comparison. For an octahedral complex, CFSE = (−0.4 n_t2g + 0.6 n_eg)Δo, and where an electron is forced to pair against the field an extra +P appears.
- Low-spin d^6: CFSE = 6(−0.4Δo) + 2P = −2.4Δo + 2P.
- High-spin d^6: CFSE = 4(−0.4Δo) + 2(0.6Δo) = −0.4Δo.
The low-spin complex gains roughly 2.0Δo more than the high-spin one (offset by 2P), so CFSE does favour the ammine complex. But this is a difference of the order of a few hundred kJ mol^-1 at most, and Δo for NH3 versus F^- is the dominant variable inside that term.
What the CFSE argument ignores.
- Redox thermodynamics. Co^3+/Co^2+ has E° = +1.92 V in water; high-spin cobalt(III) is a powerful oxidant and will attack water or traces of reductant. Low-spin Co(III) is stabilised by the ligand field, but the underlying one-electron potential, not the CFSE number, determines whether the complex survives in solution.
- Kinetic inertness. Low-spin d^6 with a large ligand-field activation energy exchanges ligands extremely slowly, so [Co(NH3)6]^3+ can be isolated and handled; the fluoro complex is far more labile and reactive.
- Lattice and solvation energies. Both compounds are ionic solids; K3[CoF6] versus [Co(NH3)6]Cl3 differ in ion size, charge distribution and hydration enthalpy, none of which appears in a CFSE expression.
Verdict. The statement is half true and as written misleading. CFSE contributes to the extra stabilisation of the low-spin species, but calling it the sole cause ignores redox potential, kinetic lability, and lattice/solvation terms. In the aqua–iron case the same lesson appears from another angle: [Fe(H2O)6]^3+ is acidic (pKa roughly 2) because the high charge density of Fe(III) polarises the coordinated water, a Brønsted effect with no CFSE component.