The crystal structure of juabite, CaCu 10 (Te 4+ O 3 ) 4 (AsO 4 ) 4 (OH) 2 (H 2 O) 4 , triclinic, a 8.9903(7), b 10.1197(8), c 8.9959(7) A, α 102.654(1), β 92.432(1), γ 70.432(1)°, V 752.0(2) A 3 , space group P 1, Z = 1, has been solved by direct methods and refined by least-squares techniques on the basis of F 2 for 3358 unique reflections collected for a microcrystal using graphite-monochromated Mo K α X-radiation and a CCD area detector. The agreement index ( R 1) was 7.3%, calculated for 2288 unique observed [ F ≥ 4σ F ] reflections, and the goodness-of-fit ( S ) was 1.03. The structure determination has shown that the formula originally proposed for juabite is incorrect with regards to the oxidation state of Te, as well as the absence of Ca. The structure contains five unique Cu 2+ positions that are each in square pyramidal coordination. The two symmetrically distinct Te 4+ cations are in the usual one-sided coordination owing to the presence of a lone pair of electrons on the cation; there are three short Te 4+ –O bonds with lengths ~1.9 A in each polyhedron, as well as two or three longer bonds. The structure contains two unique As 5+ cations that are tetrahedrally coordinated by O 2− anions, and one Ca position that is octahedrally coordinated by O 2− anions. Juabite possesses a layered heteropolyhedral framework structure; layers parallel to (010) are weakly bonded in the [010] direction, resulting in a perfect {010} cleavage. Each layer contains all of the cation polyhedra of the structure, and involves two symmetrically identical sheets that contain four of the Cuϕ 5 square pyramids [ϕ: O 2− , (OH) − , H 2 O], both Te 4+ ϕ n polyhedra, and both AsO 4 tetrahedra. The sheets are parallel to (010), and chains of edge-sharing CaO 6 octahedra and Cuϕ 5 square pyramids extending parallel to [001] are sandwiched between adjacent symmetrically identical sheets, such that all anions contained within the chains are also linked to the sheets on either side, resulting in the heteropolyhedral layers.