Give the electron configurations for the following ions: \(\mathrm{P}^{5+}, \mathrm{P}^{3-}, \mathrm{Sn}^{4+}, \mathrm{Se}^{2-}, \mathrm{I}^{-},\) and \(\mathrm{Ni}^{2+}\)

First, we need to use the periodic table to find the atomic number (number of protons) for each element. We can find the number of electrons by adding the charge to the atomic number for ions. - \(\mathrm{P}\) has an atomic number of 15. - \(\mathrm{P}^{5+}\) has 10 electrons (15 - 5). - \(\mathrm{P}^{3-}\) has 18 electrons (15 + 3). - \(\mathrm{Sn}\) has an atomic number of 50. - \(\mathrm{Sn}^{4+}\) has 46 electrons (50 - 4). - \(\mathrm{Se}\) has an atomic number of 34. - \(\mathrm{Se}^{2-}\) has 36 electrons (34 + 2). - \(\mathrm{I}\) has an atomic number of 53. - \(\mathrm{I}^{-}\) has 54 electrons (53 + 1). - \(\mathrm{Ni}\) has an atomic number of 28. - \(\mathrm{Ni}^{2+}\) has 26 electrons (28 - 2).

Now that we have the number of electrons, we can write the electron configurations for each ion using the Aufbau Principle, which means filling the lower-energy orbitals first. - \(\mathrm{P}^{5+}\) with 10 electrons: 1s\(^{2}\) 2s\(^{2}\) 2p\(^{6}\). - \(\mathrm{P}^{3-}\) with 18 electrons: 1s\(^{2}\) 2s\(^{2}\) 2p\(^{6}\) 3s\(^{2}\) 3p\(^{6}\). - \(\mathrm{Sn}^{4+}\) with 46 electrons: 1s\(^{2}\) 2s\(^{2}\) 2p\(^{6}\) 3s\(^{2}\) 3p\(^{6}\) 4s\(^{2}\) 3d\(^{10}\) 4p\(^{6}\) 5s\(^{2}\) 4d\(^{10}\). - \(\mathrm{Se}^{2-}\) with 36 electrons: 1s\(^{2}\) 2s\(^{2}\) 2p\(^{6}\) 3s\(^{2}\) 3p\(^{6}\) 4s\(^{2}\) 3d\(^{10}\) 4p\(^{6}\). - \(\mathrm{I}^{-}\) with 54 electrons: 1s\(^{2}\) 2s\(^{2}\) 2p\(^{6}\) 3s\(^{2}\) 3p\(^{6}\) 4s\(^{2}\) 3d\(^{10}\) 4p\(^{6}\) 5s\(^{2}\) 4d\(^{10}\) 5p\(^{6}\). - \(\mathrm{Ni}^{2+}\) with 26 electrons: 1s\(^{2}\) 2s\(^{2}\) 2p\(^{6}\) 3s\(^{2}\) 3p\(^{6}\) 4s\(^{2}\) 3d\(^{8}\). These are the electron configurations for the given ions.