How many unpaired electrons are in the following ions? (a) \(\mathrm{Al}^{3+}\) (b) \(\mathrm{Cl}^{-}\) (c) \(\mathrm{Sr}^{2+}\) (d) \(\mathrm{Ag}^{+}\)

For each ion, we will remove or add electrons to obtain the correct electron configuration. (a) \(\mathrm{Al}^{3+}\): Remove 3 electrons from \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^1}\). The resulting configuration is \(\mathrm{1s^2 2s^2 2p^6}\). (b) \(\mathrm{Cl}^{-}\): Add 1 electron to \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^5}\). The resulting configuration is \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6}\). (c) \(\mathrm{Sr}^{2+}\): Remove 2 electrons from \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2}\). The resulting configuration is \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6}\). (d) \(\mathrm{Ag}^{+}\): Remove 1 electron from \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1 4d^{10}}\). The resulting configuration is \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^{10}}\).

For each ion, we will count the number of unpaired electrons in their orbitals. (a) \(\mathrm{Al}^{3+}\): In the configuration \(\mathrm{1s^2 2s^2 2p^6}\), all electrons are paired. Therefore, there are 0 unpaired electrons. (b) \(\mathrm{Cl}^{-}\): In the configuration \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6}\), all electrons are paired. Therefore, there are 0 unpaired electrons. (c) \(\mathrm{Sr}^{2+}\): In the configuration \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6}\), all electrons are paired. Therefore, there are 0 unpaired electrons. (d) \(\mathrm{Ag}^{+}\): In the configuration \(\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^{10}}\), all electrons are paired. Therefore, there are 0 unpaired electrons. So, in all given ions, there are 0 unpaired electrons.