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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) Contains datablocks global, TOZ |
CCDC reference: 126993
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