## Q. 7.PS.9

Atoms and Their Ions

Complete this table.

 Neutral Atom Neutral Atom Electron Configuration Ion Ion Electron Configuration $Se$ __________ __________ $[Kr]$ $Ba$ __________ $Ba^{2+}$ __________ $Br$ __________ $Br^-$ __________ __________ $[Kr] 5s^1$ $Rb^+$ __________ __________ $[Ne] 3s^23p^3$ __________ $[Ar]$

## Verified Solution

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.

 Neutral Atom Neutral Atom Electron Configuration Ion Ion Electron Configuration $Se$ $[Ar] 3d^{10}4s^24p^4$ $Se^{2-}$ $[Kr]$ $Ba$ $[Xe] 6s^2$ $Ba^{2+}$ $[Xe]$ $Br$ $[Ar] 3d^{10}4s^24p^5$ $Br^-$ $[Kr]$ $Rb$ $[Kr] 5s^1$ $Rb^+$ $[Kr]$ $P$ $[Ne] 3s^23p^3$ $P^{3-}$ $[Ar]$