Abstract
Phase relation could be elucidated between the two phases previously found out in BaPbO3 -BaCuO2 system by obtaining the single crystal product. Both phases have the molar ratio of Ba:Pb:Cu=1:0.725:0.10 and are basically isostructural to Ba4Pb3O10 which is a member of Ruddlesden-Popper series Ban+1PbnO3n+1(n=3). They can be described as Ba3.64Pb2.64Cu0.36O10_δ. The phase 2 was an oxidized form of the phase 1 with an intercalation of 0.6H2O. The phase 1 crystallized in orthorhombic Immm with a=0.430(4), b=0.429(8), c=3.026(7)nm and was a semiconductor. The phase 2 crystallized in tetragonal P4/mmm with a=0.430(5), c=1.715(2)nm and was a semimetallic conductor.
