Write electron configurations for the most stable ion formed by each of the elements Rb, Ba, Se, and I (when in stable ionic compounds).

For each element, we need to find the charge for the most stable ion formed when they are in stable ionic compounds. For Rb (Rubidium) and Ba (Barium): They belong to group 1 and group 2 respectively, so they are both alkali metals. Rb will lose its single valence electron, and Ba will lose its two valence electrons. This gives Rb a +1 charge and Ba a +2 charge. For Se (Selenium) and I (Iodine): They belong to group 16 and group 17 respectively, so they are both nonmetals. Se will gain two electrons, while I will gain one electron. This gives Se a -2 charge and I a -1 charge.

Rb: Its atomic number is 37, so the neutral atom has 37 electrons. When it loses one electron (to form Rb+), it will have 36 electrons. So the electron configuration of Rb+ is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1\] Ba: Its atomic number is 56, so the neutral atom has 56 electrons. When it loses two electrons (to form Ba2+), it will have 54 electrons. So the electron configuration of Ba2+ is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\] Se: Its atomic number is 34, so the neutral atom has 34 electrons. When it gains two electrons (to form Se2-), it will have 36 electrons. So the electron configuration of Se2- is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\] I: Its atomic number is 53, so the neutral atom has 53 electrons. When it gains one electron (to form I-), it will have 54 electrons. So the electron configuration of I- is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\] To summarize, the electron configurations of the most stable ions formed by the elements Rb, Ba, Se, and I are: Rb+: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1\) Ba2+: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) Se2-: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) I-: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\)