Answer:
[tex]V(BrO_3)_3[/tex]
Explanation:
Hello,
In this case, vanadium is specified with +3 as its oxidation state, however it is stated that vanadium is combined with the bromate ion, which according to the IUPAC rules, accounts for bromine's last oxidation state, that is +5, in such a way, we build up the bromate ion forming the bromic acid shown below:
[tex]Br^{+5}+O^{-2}-->Br_2O_5\\Br_2O_5+H_2O-->HBrO_3[/tex]
Such bromate comes from the bromic acid's anion, this is [tex](BrO_3)^-[/tex], thus, the required formula of the vanadium (III) bromate turns out into:
[tex]V^{+3}(BrO_3)^-_3\\V(BrO_3)_3[/tex]
Best regards.
