Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cd}^{2+}\) (b) \(\mathrm{P}^{3-}\) (c) \(Z r^{4+}\) (d) \(\mathrm{Ru}^{3+}\), (e) \(\mathrm{As}^{3-},(\mathrm{f}) \mathrm{Ag}^{+}\)

: 1. For Neutral \(\mathrm{Cd}\): \(\mathrm{Cd}\) (Cadmium) has an atomic number of 48. Therefore, an uncharged \(\mathrm{Cd}\) atom has 48 protons and 48 electrons. The electron configuration of \(\mathrm{Cd}\) is [Kr]\(\,5s^2\,4d^1$$\,0\). 2. For \(\mathrm{Cd}^{2+}\) ion: \(\mathrm{Cd}^{2+}\) has lost 2 electrons. Therefore, the electron configuration becomes [Kr]\(\,4d^1$$\,0\).

: 1. For Neutral \(\mathrm{P}\): \(\mathrm{P}\) (Phosphorus) has an atomic number of 15. Therefore, an uncharged \(\mathrm{P}\) atom has 15 protons and 15 electrons. The electron configuration of \(\mathrm{P}\) is [Ne]\(\,3s^2\,3p^3\). 2. For \(\mathrm{P}^{3-}\) ion: \(\mathrm{P}^{3-}\) has gained 3 electrons. Therefore, the electron configuration becomes [Ne]\(\,3s^2\,3p^6\).

: 1. For Neutral \(Z r\): \(Z r\) (Zirconium) has an atomic number of 40. Therefore, an uncharged \(Z r\) atom has 40 protons and 40 electrons. The electron configuration of \(Z r\) is [Kr]\(\,5s^2\,4d^2\). 2. For \(Z r^{4+}\) ion: \(Z r^{4+}\) has lost 4 electrons. Therefore, the electron configuration becomes [Kr].

: 1. For Neutral \(\mathrm{Ru}\): \(\mathrm{Ru}\) (Ruthenium) has an atomic number of 44. Therefore, an uncharged \(\mathrm{Ru}\) atom has 44 protons and 44 electrons. The electron configuration of \(\mathrm{Ru}\) is [Kr]\(\,5s^2\,4d^6\). 2. For \(\mathrm{Ru}^{3+}\) ion: \(\mathrm{Ru}^{3+}\) has lost 3 electrons. Therefore, the electron configuration becomes [Kr]\(\,4d^3\).

: 1. For Neutral \(\mathrm{As}\): \(\mathrm{As}\) (Arsenic) has an atomic number of 33. Therefore, an uncharged \(\mathrm{As}\) atom has 33 protons and 33 electrons. The electron configuration of \(\mathrm{As}\) is [Ar]\(\,4s^2\,3d^1$$\,0\,4p^3\). 2. For \(\mathrm{As}^{3-}\) ion: \(\mathrm{As}^{3-}\) has gained 3 electrons. Therefore, the electron configuration becomes [Ar]\(\,4s^2\,3d^1$$\,0\,4p^6\).

: 1. For Neutral \(\mathrm{Ag}\): \(\mathrm{Ag}\) (Silver) has an atomic number of 47. Therefore, an uncharged \(\mathrm{Ag}\) atom has 47 protons and 47 electrons. The electron configuration of \(\mathrm{Ag}\) is [Kr]\(\,5s^2\,4d^9\). 2. For \(\mathrm{Ag}^{+}\) ion: \(\mathrm{Ag}^{+}\) has lost 1 electron. Therefore, the electron configuration becomes [Kr]\(\,4d^1$$\,0\). Now, let's identify which ions have noble-gas configurations: - \(\mathrm{Cd}^{2+}\): [Kr]\(\,4d^1$$\,0\) (No) - \(\mathrm{P}^{3-}\): [Ne]\(\,3s^2\,3p^6\) (Yes, matches \(\mathrm{Ar}\)) - \(Z r^{4+}\): [Kr] (Yes) - \(\mathrm{Ru}^{3+}\): [Kr]\(\,4d^3\) (No) - \(\mathrm{As}^{3-}\): [Ar]\(\,4s^2\,3d^1$$\,0\,4p^6\) (Yes, matches \(\mathrm{Kr}\)) - \(\mathrm{Ag}^{+}\): [Kr]\(\,4d^1$$\,0\) (No) So, the ions with noble-gas configurations are \(\mathrm{P}^{3-}\), \(Z r^{4+}\), and \(\mathrm{As}^{3-}\).