Answer
$Ba^{2+}$, $Br^-$
$BaBr_2$
Work Step by Step
Barium (Ba) is from group 2A, so it should lose two electrons: $Ba^{2+}$
Bromine (Br) is from group 7A, so it should lose one electron: $Br^-$
The ionic compound formed by the two would be $BaBr_2$