## Chemistry and Chemical Reactivity (9th Edition)

$Ba^{2+}$, $Br^-$ $BaBr_2$
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 $BaBr_2$