Observation of a 1/3 magnetization plateau in Pb2Cu10O4(SeO3)4Cl7 arising from (Cu2+)7 clusters of corner-sharing (Cu2+)4 tetrahedra

A mixed-valence compound Pb2Cu10O4(SeO3)4Cl<sub<7 has a complex structure consisting of one nonmagnetic Cu+ (S = 0) ion and four nonequivalent magnetic Cu2+ (S = 1/2) ions. It exhibits antiferromagnetic ordering at TN = 10.2 K. At a temperature below TN, a sequence of spin-flop transition at Bspin-flop = 1.3 T and 1/3 plateau formation at Bspin-flip = 4.4 K is observed in the magnetization curve M(B). The 1/3 magnetization plateau persists at least up to 53.5 T. The spin exchanges of Pb2Cu10O4(SeO3)4Cl7 evaluated by performing energy-mapping analysis based on DFT+U calculations show that the magnetic properties of Pb2Cu10O4(SeO3)4Cl7 are described by the (Cu2+)7 cluster of corner-sharing (Cu2+)4 tetrahedra, and that each (Cu2+)7 cluster has a S = 3/2 spin arrangement in the ground state. The 1/3 magnetization plateau observed for Pb2Cu10O4(SeO3)4Cl7 is explained by the field-induced flip of every second (Cu2+)7 cluster within a unit cell.