Past exam of the chemistry B course of the University of Cambridge / 2025 / Paper 2 / B6 / b / i / Solution

$A=[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{OH}){]}^{2+}$, $B=[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{ONO}){]}^{2+}$, $C=[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{N}{\mathrm{O}}_2){]}^{2+}$, and the by-product is $\mathrm{HN}{\mathrm{O}}_2$. The first transformation is deprotonation of coordinated water, not loss of an ammonia ligand. The hydroxo ligand in $A$ is then nitrosated at oxygen, retaining its original Co–O attachment in the initially formed O-bound nitrito ligand. Atom and electric charge balance giveThe relative molecular mass of nitrous acid is approximately $1+14+2(16)=47$. Heating converts the O-bound nitrito ligand to the N-bound nitro ligand; these are linkage isomers, with Co remaining in oxidation state $+3$ throughout.