How about the various electron pairs provided by adding O atoms to the complex? I'd assume that this would provide a sort of "cushion" whereby a negative charge would be stabilised. That'd make me say that 1 is incorrect.
I'd also assume that extra Br would have the same effect, so for this reason I would say 3 is correct.
Another thing to consider is that Sb really is much more stable as a free atom than P would be, and so would form less stable anion than P, Using this reasoning i'd have to say that 2 is incorrect.
Please correct me if I am wrong...