Question 7.PS.9: Atoms and Their Ions Complete this table....

Strategy and Explanation A neutral Se atom has the electron configuration [Ar] 3d^{10}4s^24p^4; Se is in Group 6A, so it gains two electrons in the 4p sublevel to form Se^{2-} and achieve the noble gas configuration of krypton, Kr (36 electrons). Barium is a Group 2A element and loses the two 6s electrons to acquire the electron configuration of xenon (54 electrons), the preceding noble gas. Therefore, Ba is [Xe] 6s^2 and Ba^{2+} is [Xe]. Bromine is in Group 7A and will gain one electron to form Br^-, which has the electron configuration of krypton, the next noble gas. An electron configuration of [Kr] 5s^1 indicates 37 electrons in the neutral atom, which is a rubidium atom, Rb. A neutral Rb atom loses the 5s^1 electron to form an Rb^+ ion, which has a [Kr] configuration. The [Ne] 3s^23p^3 configuration is for an element with 15 electrons, which is a phosphorus atom, P. By gaining three electrons, a neutral phosphorus atom becomes a P^{3-} ion with the [Ar] configuration.