Cobalt Ammine Complexes: Charge, Magnetism and Oxidation State
A student prepares two cobalt ammine complexes by the air-oxidation route from CoCl2·6H2O, NH4Cl, concentrated NH3, and H2O2. Complex A is orange-yellow. A 0.0100 M solution of A has a molar conductance close to that of a 1:3 electrolyte. With excess AgNO3, 1.00 mol of A gives 3.00 mol AgCl. Complex B is purple. Its 0.0100 M solution has a molar conductance close to that of a 1:2 electrolyte, and 1.00 mol of B gives 2.00 mol AgCl. Both A and B are diamagnetic (μeff = 0 BM). The student expected paramagnetic Co(II) complexes because CoCl2·6H2O was the starting material. A colleague suggests that H2O2 oxidized the metal and that a paramagnetic complex could be isolated if the H2O2 were omitted. The student must assign formulas, explain the magnetic data, and decide whether omitting H2O2 alone is enough to isolate a paramagnetic Co(II) ammine complex.
1. Assign formulas to A and B from the conductance and AgCl data. Write their dissociation equations and explain why one chloride is not precipitated by AgNO3 in B.
Read the worked answer
A behaves as a 1:3 electrolyte and gives 3 mol AgCl per mol complex. Therefore all three chlorides are ionic and the cation carries a 3+ charge: A is [Co(NH3)6]Cl3.
Dissociation: [Co(NH3)6]Cl3 → [Co(NH3)6]3+ + 3Cl−.
B behaves as a 1:2 electrolyte and gives 2 mol AgCl per mol complex. Two chlorides are ionic; one chloride remains bound to cobalt and is not precipitated by AgNO3 at room temperature: B is [Co(NH3)5Cl]Cl2.
Dissociation: [Co(NH3)5Cl]Cl2 → [Co(NH3)5Cl]2+ + 2Cl−.
The coordinated Cl− is covalently bound in the inner coordination sphere. AgNO3 precipitates only free Cl− in solution; it does not readily remove a ligand from the inert Co(III) centre under these conditions. The number of ions in solution also explains the conductance order: four ions for A versus three ions for B, so A conducts more strongly at the same concentration.
2. Explain why both complexes are diamagnetic even though CoCl2·6H2O was the starting material. Calculate spin-only magnetic moments for Co(III) d6 and for Co(II) d7 to support the argument.
Read the worked answer
CoCl2·6H2O contains Co(II), d7, which is not diamagnetic. The H2O2 in the synthesis oxidizes Co(II) to Co(III). In an octahedral NH3 field, Co(III) d6 is strongly stabilized as low-spin t2g^6 eg^0. All six d electrons are paired, so S = 0 and the spin-only moment is: μ = √[n(n+2)] BM = √[0(0+2)] = 0 BM. That matches the observed diamagnetism of both A and B.
For comparison, Co(II) d7 can be high-spin t2g^5 eg^2 with three unpaired electrons: μ = √[3(5)] = √15 = 3.87 BM. A low-spin Co(II) d7 case would have t2g^6 eg^1, one unpaired electron, μ = √3 = 1.73 BM. Either Co(II) case would be paramagnetic, so the observed 0 BM rules out a simple Co(II) ammine complex and supports Co(III).
The oxidation state assignment is also consistent with charge: [Co(NH3)6]3+ requires Co(III), and [Co(NH3)5Cl]2+ requires Co(III) because NH3 is neutral and Cl− is −1.
3. Decide whether omitting H2O2 alone would allow isolation of a paramagnetic Co(II) ammine complex. Recommend practical changes and a check that the product is really Co(II).
Read the worked answer
Omitting H2O2 is not enough by itself. Co(II) ammine complexes are readily oxidized by dissolved oxygen in ammoniacal solution, especially when the solution is stirred or warmed. The brown/orange colour of Co(III) ammines can develop even without added peroxide. A paramagnetic Co(II) product may form initially but will not survive normal bench exposure to air.
Practical changes: use degassed or oxygen-free water and ammonia, carry out the preparation and filtration under an inert atmosphere (N2 or Ar), exclude air during drying, and keep the system cool. A reducing agent may be needed to suppress oxidation, but it must not itself coordinate strongly or precipitate cobalt. Excess ammonia should be controlled because Co(II) ammines are labile and can disproportionate or oxidize on work-up. Characterise the product quickly.
The key check is magnetic susceptibility: Co(II) high-spin should give about 3.9 BM, low-spin about 1.7 BM, while Co(III) low-spin gives 0 BM. Also check the charge by conductance and the number of ionic chlorides by AgNO3 precipitation. A product with μeff near 0 BM and a 1:3 electrolyte pattern is Co(III), not the desired Co(II) complex; a paramagnetic product with a different conductance pattern supports Co(II).