Lutetium(III) bromide is a crystalline compound made of one lutetium atom and three bromine atoms.[1] It takes the form of a white powder at room temperature. It is hygroscopic.[1] It is odorless.
Lutetium(III) bromide can be synthesized through the following reaction:[2]
2 Lu(s) + 3 Br2(g) → 2 LuBr3(s)
If burned, lutetium(III) bromide may produce hydrogen bromide and metal oxide fumes.[3]
Lutetium(III) bromide reacts to strong oxidizing agents.[3]
An experiment by T. Mioduski showed that the solubility of LuBr3 in tetrahydrofuran at 21-23 °C was 0.30 g per 100 ml of solution.[4]