Questions: Magnetism of Coordination Compounds

Question 1

A complex has a measured magnetic moment of 4.9 BM. How many unpaired electrons does it have, according to the spin-only formula?

• A. 3 unpaired electrons (μ = 3.87 BM)
• B. 4 unpaired electrons (μ = 4.90 BM)
• C. 5 unpaired electrons (μ = 5.92 BM)
• D. 2 unpaired electrons (μ = 2.83 BM)

Answer: (B)

The spin-only formula μ = √(n(n+2)) gives: n=1 → 1.73, n=2 → 2.83, n=3 → 3.87, n=4 → 4.90, n=5 → 5.92 BM. A measured value of 4.9 BM matches n=4 almost exactly. This measurement would be consistent with, for example, a high-spin d⁶ octahedral complex (t₂g⁴ eg², four unpaired) or a d⁴ high-spin complex (t₂g³ eg¹, four unpaired). Distinguishing between these requires additional information like the metal identity and oxidation state.

Question 2

[Fe(CN)₆]⁴⁻ is diamagnetic while [Fe(H₂O)₆]²⁺ is paramagnetic with four unpaired electrons, even though both contain Fe²⁺ (d⁶). This difference arises because CN⁻ is a strong-field ligand that forces all electrons into the t₂g set.

• T. True
• F. False

Answer: True

Fe²⁺ has six d-electrons. With CN⁻ (strong field, large Δ_oct), the splitting energy exceeds the pairing energy: all six electrons pair in the t₂g orbitals (t₂g⁶ eg⁰), giving zero unpaired electrons and diamagnetic behavior. With H₂O (weak field, small Δ_oct), the pairing energy exceeds Δ: electrons spread across both t₂g and eg sets following Hund's rule (t₂g⁴ eg², four unpaired electrons), giving paramagnetic behavior. This is the same d⁶ ion with the same total electron count — the ligand environment alone determines the magnetism.

Question 3

The spin-only formula always accurately predicts the magnetic moment of any coordination compound.

• T. True
• F. False

Answer: False

The spin-only formula ignores the orbital angular momentum contribution to the magnetic moment. For first-row transition metals, orbital contributions are largely quenched by the crystal field, so the spin-only formula works reasonably well. But for second- and third-row metals (where spin-orbit coupling is stronger), and for first-row ions where the ground state has an orbital degeneracy (like T terms in octahedral fields), the orbital contribution becomes significant and measured moments deviate from spin-only values. Lanthanide complexes require a completely different approach (the full J-based formula) because spin-orbit coupling dominates.

Question 4

Explain the phenomenon of spin crossover and describe the conditions under which a coordination compound might switch between high-spin and low-spin states.

Model answer: Spin crossover occurs when the crystal field splitting energy Δ is close to the electron pairing energy P, so that the high-spin and low-spin states are nearly degenerate. In this regime, external perturbations — changes in temperature, pressure, or light irradiation — can shift the equilibrium between the two states. Raising temperature favors the high-spin state (higher entropy due to more unpaired electrons and longer metal-ligand bonds). Increasing pressure favors the low-spin state (shorter bonds, smaller volume). The phenomenon is most common for d⁴ through d⁷ octahedral complexes with intermediate-field ligands, particularly Fe²⁺ (d⁶) and Fe³⁺ (d⁵) complexes with nitrogen-donor ligands. Spin-crossover compounds are of interest for molecular switches, sensors, and displays because the spin state change produces measurable changes in color, magnetic moment, and crystal volume.

The spin-crossover transition can be gradual (smooth thermal equilibrium) or abrupt (cooperative, with hysteresis), depending on crystal packing and intermolecular interactions. Abrupt transitions with hysteresis are the most technologically interesting because they provide bistability — the compound has memory of its thermal history.