Questions: Consider the electron configuration of the ion to determine which ion shown below has an incorrect ionic charge? Ba^2+ Rb^+ Al^3- Se^2-

Solution Steps

To determine if the ionic charge is correct, we need to compare the electron configuration of each ion to its neutral atom.

1. Barium (Ba): Neutral barium has an atomic number of 56, so its electron configuration is: \[ \mathrm{Ba}: [Xe] 6s^2 \] For \(\mathrm{Ba}^{2+}\), it loses 2 electrons: \[ \mathrm{Ba}^{2+}: [Xe] \]

2. Rubidium (Rb): Neutral rubidium has an atomic number of 37, so its electron configuration is: \[ \mathrm{Rb}: [Kr] 5s^1 \] For \(\mathrm{Rb}^{+}\), it loses 1 electron: \[ \mathrm{Rb}^{+}: [Kr] \]

3. Aluminum (Al): Neutral aluminum has an atomic number of 13, so its electron configuration is: \[ \mathrm{Al}: [Ne] 3s^2 3p^1 \] For \(\mathrm{Al}^{3-}\), it gains 3 electrons: \[ \mathrm{Al}^{3-}: [Ne] 3s^2 3p^4 \]

4. Selenium (Se): Neutral selenium has an atomic number of 34, so its electron configuration is: \[ \mathrm{Se}: [Ar] 4s^2 3d^{10} 4p^4 \] For \(\mathrm{Se}^{2-}\), it gains 2 electrons: \[ \mathrm{Se}^{2-}: [Ar] 4s^2 3d^{10} 4p^6 \]

1. \(\mathrm{Ba}^{2+}\): The electron configuration \([Xe]\) is correct for \(\mathrm{Ba}^{2+}\), as it loses 2 electrons.
2. \(\mathrm{Rb}^{+}\): The electron configuration \([Kr]\) is correct for \(\mathrm{Rb}^{+}\), as it loses 1 electron.
3. \(\mathrm{Al}^{3-}\): The electron configuration \([Ne] 3s^2 3p^4\) is incorrect for \(\mathrm{Al}^{3-}\). Aluminum typically forms a \(3+\) ion, not a \(3-\) ion.
4. \(\mathrm{Se}^{2-}\): The electron configuration \([Ar] 4s^2 3d^{10} 4p^6\) is correct for \(\mathrm{Se}^{2-}\), as it gains 2 electrons.

Final Answer