Reaction of the binucleating bridging ligands 3,3′,4,4′-tetrahydroxybiphenyl (H₄L¹) or 3,3″,4,4″-tetrahydroxy-p-terphenyl (H₄L²) with 2 equivalents of [Ru(bipy)₂Cl₂]·2H₂O (bipy = 2,2′-bipyridine) results in the formation of binuclear complexes [{Ru(bipy)₂}(µ-L){Ru(bipy)₂}]²⁺1(L = L¹) and 2(L = L²) in which the ligand L^4– has been oxidized to the semiquinone (sq) form L^2–. In each complex the co-ordinated catecholate fragment (cat) may be oxidised reversibly to the semiquinone and quinone (q) redox states, giving the five-membered redox series cat–cat, cat–sq, sq–sq, sq–q and q–q for the bridging ligands. In the sq–sq state the bridging ligands are necessarily planar due to the presence of double bonds between the aromatic rings; in the cat–cat and q–q states there is formally no double-bond character between the aromatic rings and they are free to adopt a twisted conformation. Spectroelectrochemical measurements confirm that in the cat–cat and q–q states, 1 and 2 behave approximately like their mononuclear counterparts [Ru(bipy)₂(cat)] and [Ru(bipy)₂(bq)]²⁺[H₂cat = catechol (benzene-1,2-diol); bq =o-benzoquinone]; for 1, the results also show that in the mixed-valence sq–q and sq–cat oxidation states the two halves of the ligand are electronically equivalent (i.e. valence delocalised), which is best explained by the bridging ligand retaining the planar conformation of the sq–sq state. The change from planar to twisted therefore occurs at the extremes of the redox series, on formation of the cat–cat and q–q oxidation states. This control of bridging ligand conformation with oxidation state may form the basis for a molecular switch.