metal-organic compounds
NbBr5 reacts with three equivalents of Me3SiNPh2 in ether to give the title compound, [NbBr2(C12H10N)3]. C4H8O. Nb(NPh2)3Br2 has trigonal bipyramidal geometry with axial bromides ligands.
Supporting information
Crystallographic Information File (CIF) |
CCDC reference: 126993