Nitro versus Nitrito Linkage Isomerism in a Cobalt(III) Complex
A student prepares [Co(NH3)5(NO2)]Cl2 by two routes. Product A is obtained by heating [Co(NH3)5Cl]Cl2 with NaNO2 at 60 °C. Product B is obtained by reacting [Co(NH3)5(OH2)]3+ with NaNO2 at 0 °C. Both analyze correctly for the formula and behave as 1:2 electrolytes in water, with molar conductance near 250 S cm^2 mol^-1. Their UV-vis spectra are nearly identical. IR spectra differ: A has strong bands at 1430, 1320, and 825 cm^-1; B has bands at 1460, 1060, and 840 cm^-1. A is thermodynamically more stable and does not isomerize on heating, whereas B converts to A in the solid state over time. The N-bound isomer is known to be more stable. The student must assign the structures and explain the observations. The student has no X-ray facility.
1. Assign A and B as the nitro (N-bound) or nitrito (O-bound) isomers. Use the IR data to justify your assignment, identifying the key vibrational modes.
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A is the nitro (N-bound) isomer; B is the nitrito (O-bound) isomer. For N-bound nitrite, the local symmetry is C2v and the two N-O bonds are equivalent by resonance; IR shows ν_asym(NO2) near 1430 cm^-1, ν_sym(NO2) near 1320 cm^-1, and δ(ONO) near 825 cm^-1. For O-bound nitrite, the N-O bonds are inequivalent: a terminal N=O stretch appears near 1460 cm^-1, a Co-O-N stretch near 1060 cm^-1, and a bending mode near 840 cm^-1. The 1060 cm^-1 band is diagnostic of the O-bound nitrito group; it is absent in A. The thermodynamic stability of A also supports the nitro assignment, because the Co-N bond is stronger than Co-O and the N-bound form is the more stable linkage isomer.
2. Explain why product B forms at 0 °C while A forms at 60 °C. Is this thermodynamic or kinetic control? Relate to the reaction mechanism.
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The reactions are substitution reactions at Co(III), which is kinetically inert and typically proceeds by dissociative or interchange mechanisms. At 0 °C, the reaction is under kinetic control: nitrite attacks through its more nucleophilic oxygen atom, giving the O-bound nitrito complex B with a lower activation barrier. At 60 °C, the system has enough energy to reach the thermodynamic product A, where nitrite binds through nitrogen, forming the stronger Co-N bond. The solid-state conversion of B to A on standing is also consistent with B being the kinetic product and A the thermodynamic product. Thus B is kinetically favored, A thermodynamically favored; heating does not simply accelerate the same path but allows isomerization to the more stable linkage.
3. The molar conductance of both complexes is similar. Does conductivity distinguish the linkage isomers? What additional experiment would you propose to confirm the assignment?
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No. Both complexes are 1:2 electrolytes because the chloride ions are outside the coordination sphere; linkage isomerism changes only the inner-sphere donor atom (N vs O), not the number of free ions. Conductivity therefore cannot distinguish nitro from nitrito. IR is already diagnostic, but a confirmatory experiment could be far-IR spectroscopy: the Co-N stretch of the nitro complex appears at a different frequency from the Co-O stretch of the nitrito complex. Alternatively, 15N NMR or X-ray crystallography would definitively locate the donor atom. A simple chemical test is also possible: the nitrito complex may react with acid or undergo photochemical isomerization differently, but IR remains the most accessible evidence.