The number of stereo isomers possible for $$[Co(ox)_2(Br)(NH_3)]^{2-}$$ is ______.
[ox = oxalate]

The complex $$[Co(ox)_2(Br)(NH_3)]^{2-}$$ has cobalt as the central metal ion with two bidentate oxalate (ox) ligands, one bromide ion, and one ammonia molecule. The coordination number is $$2 \times 2 + 1 + 1 = 6$$, so the geometry is octahedral.

For an octahedral complex of the type $$[M(AA)_2(B)(C)]$$ where AA is a bidentate ligand and B, C are monodentate ligands, we need to consider geometric and optical isomerism.

First, the two monodentate ligands (Br and $$NH_3$$) can be either cis or trans to each other. In the trans arrangement, the two oxalate ligands occupy the remaining four positions in one plane. This gives one trans isomer. The trans isomer has a plane of symmetry and is optically inactive, so it counts as 1 isomer.

In the cis arrangement, the two monodentate ligands are adjacent. This cis isomer lacks a plane of symmetry and is therefore optically active, existing as a pair of non-superimposable mirror images (enantiomers). This gives 2 more isomers (d and l forms).

Therefore, the total number of stereoisomers is $$1 + 2 = 3$$.