Werner Complexes and Nickel(II) Geometry
A student prepares cobalt ammine complexes from CoCl2·6H2O and ammonia in air, and two nickel complexes. Sample A analyses as CoCl3·5NH3 (M = 250.4 g mol⁻¹); 0.150 g of A treated with excess AgNO3 gives 0.172 g AgCl (M = 143.3 g mol⁻¹), and A has a magnetic moment of 0.0 BM. Sample B analyses as CoCl3·4NH3 (M = 233.4 g mol⁻¹); 0.150 g of B gives 0.0921 g AgCl, and B is also diamagnetic. Different crystallisations of B give a violet form and a green form, both with the same analysis. Sample C is (Et4N)2[NiCl4] and sample D is K2[Ni(CN)4]; their magnetic moments are 2.83 BM and 0.0 BM respectively. The student must assign structures without relying on conductivity alone, because both A and B are electrolytes.
1. Use the AgCl data and magnetic moment to deduce the structure of A. Include the oxidation state, hybridization and expected geometry.
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Moles of A: (0.150/250.4 = 5.99\times10^{-4}) mol. Moles of AgCl: (0.172/143.3 = 1.20\times10^{-3}) mol. The ratio is 2.00, so two chloride ions are ionic and are precipitated by AgNO3; the third chloride is covalently bound in the coordination sphere. The formula is therefore [Co(NH3)5Cl]Cl2.
Charge balance on the complex cation: two free Cl⁻ give +2 overall; five NH3 are neutral and the inner Cl⁻ is −1, so Co must be +3. Co(III) is d⁶. With NH3 and one Cl⁻, the complex is low-spin t2g⁶, giving zero unpaired electrons and μ = 0.0 BM, as observed. The hybridization is d²sp³ (inner-orbital), and the geometry is octahedral with five NH3 ligands and one Cl ligand. The two outside chlorides are the ones responsible for the AgCl precipitate; this is why conductivity alone would only show that the complex is a 1:2 electrolyte, not which chloride is bound.
2. Sample B gives one mole of AgCl per mole and is diamagnetic, but it appears in violet and green forms. Deduce its structure and explain how the two forms can be distinguished, including which colour corresponds to which isomer.
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Moles of B: (0.150/233.4 = 6.43\times10^{-4}) mol. Moles of AgCl: (0.0921/143.3 = 6.43\times10^{-4}) mol. The ratio is 1.00, so one chloride is ionic and two are bound. The formula is [Co(NH3)4Cl2]Cl. Co is again +3 (one free Cl⁻ gives +1; two inner Cl⁻ and four neutral NH3 require Co = +3), so the d⁶ ion is low-spin and diamagnetic. Both the AgCl test and the magnetic moment are therefore consistent with two geometrical isomers: cis, with the two Cl ligands adjacent, and trans, with them opposite.
The two forms cannot be distinguished by AgCl count or magnetism, because both give one free Cl⁻ and both are diamagnetic. They can be distinguished by dipole moment: the cis isomer is polar, while the centrosymmetric trans isomer has zero dipole moment. They also differ in colour: the cis isomer is violet and the trans isomer is green in this classic cobalt(III) series. Spectroscopically, the cis isomer shows two Co–Cl stretching bands and the trans isomer one; X-ray diffraction would be definitive.
3. Explain the magnetic moments of C and D and deduce the geometry and hybridization of each.
Read the worked answer
In C, [NiCl4]²⁻, nickel is +2, i.e. d⁸. Chloride is a weak-field ligand, so the tetrahedral splitting Δt is small and the d⁸ configuration remains high-spin with two unpaired electrons. The spin-only moment is
[\mu = \sqrt{n(n+2)} = \sqrt{2\times4} = 2.83\ \text{BM},]
which matches the observation. The geometry is tetrahedral and the hybridization is sp³.
In D, [Ni(CN)4]²⁻, nickel is also +2 (d⁸), but CN⁻ is a strong-field ligand. The large splitting pairs all eight d electrons, giving μ = 0.0 BM. The complex is square planar with dsp² hybridization. The key judgement is that the magnetic moment alone does not prove geometry; it becomes decisive when combined with ligand-field strength. A tetrahedral Ni(II) complex is paramagnetic, whereas a square-planar d⁸ Ni(II) complex is diamagnetic, so C and D cannot be swapped.