 Solution

The N-bound isomer $C$ is the thermodynamically favored form; heating accelerates access to it. Formation of $B$ retains the original metal–oxygen attachment and is a plausible kinetic control outcome. Rearranging that attachment requires overcoming an activation free energy, especially at a substitutionally inert low-spin cobalt(III) center. Raising temperature speeds the linkage isomerization, allowing the more stable N-bound nitro ligand to accumulate. Its stronger ligand field and different covalent bonding help rationalize the stability, but a larger field splitting alone is not a complete calculation of the isomerization Gibbs free energy. The classical crystallographic comparison establishes the different attachments and thermodynamic direction.^[ref]