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

To begin, we need to find the atomic numbers of each given element, which indicates the number of protons and electrons in their neutral state. The respective atomic numbers are: P: 15 Sn: 50 Se: 34 I: 53 Ni: 28

Next, we need to write the electron configurations for the neutral atoms. The electron configuration shows the distribution of electrons in the various orbitals/sub-shells around the nucleus: P: \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^3\) Sn: \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^2\) Se: \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^4\) I: \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^5\) Ni: \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^8\)

Now, we'll adjust the electron configurations based on the charges of the ions. A positive charge means that the ion has lost electrons, while a negative charge means that the ion has gained electrons. With this in mind, we'll adjust the electron configurations as follows: P\(^{5+}\): remove 5 electrons from the neutral P configuration => \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^3 \rightarrow 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^{\cancelto{0}{3}}\) P\(^{3-}\): add 3 electrons to the neutral P configuration => \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^3 \rightarrow 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^{6}\) Sn\(^{4+}\): remove 4 electrons from the neutral Sn configuration => \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^2 \rightarrow 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^{\cancelto{0}{2}}\) Se\(^{2-}\): add 2 electrons to the neutral Se configuration => \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^4 \rightarrow 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\) I\(^{-}\): add 1 electron to the neutral I configuration => \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^5 \rightarrow 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^6\) Ni\(^{2+}\): remove 2 electrons from the neutral Ni configuration => \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^8 \rightarrow 1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^{\cancelto{0}{2}}\ 3d^8\)

The electron configurations for the given ions are as follows: P\(^{5+}\): \(1s^2\ 2s^2\ 2p^6\ 3s^2\) P\(^{3-}\): \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\) Sn\(^{4+}\): \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\) Se\(^{2-}\): \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\) I\(^{-}\): \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^{10}\ 4p^6\ 5s^2\ 4d^{10}\ 5p^6\) Ni\(^{2+}\): \(1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^8\)