Identify \(a+2\) cation that has the following ground state electron configurations: (a) \([\mathrm{Ne}]\) (b) \([\mathrm{Ar}] 3 d^{9}\) (c) \([\mathrm{Xe}] 4 f^{14} 5 d^{10} 6 s^{2}\)

Refer to the periodic table and find the element with the electron configuration \([\mathrm{Ne}]\). In this case, Neon's electron configuration is the same as the ground state electron configuration in question. The electron configuration of neutral Neon (Ne) is \(1s^2 2s^2 2p^6\), which is also represented as \([\mathrm{Ne}]\). For an \(a+2\) cation state, we need to consider removing two electrons from the outermost shell. However, considering the full electron configuration of neon, there are no outer electrons for the \(a+2\) cation. Thus, there is no valid \(a+2\) cation for case (a).

First, let's find the element with the given ground state electron configuration. The electron configuration of neutral Argon (Ar) is \(1s^2 2s^2 2p^6 3s^2 3p^6\), which can be represented as \([\mathrm{Ar}]\). Now, consider adding the \(3d^9\) configuration to argon, making the electron configuration \([\mathrm{Ar}] 3d^9\). This results in the element, Copper (Cu) with a neutral electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\). For an \(a+2\) cation, we need to remove two electrons from the outermost shell. In this case, we will remove the 4s electrons and then remove 1 of the 3d electrons. Doing this, the electron configuration of the \(a+2\) cation for copper will be \([\mathrm{Ar}] 3d^{9}\). The \(a+2\) cation for case (b) is \(Cu^{2+}\).

First, let's find the element with the given ground state electron configuration. The electron configuration of neutral Xenon (Xe) is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6\), which can be represented as \([\mathrm{Xe}]\). Now, consider adding the \(4f^{14}\), \(5d^{10}\), and \(6s^2\) configurations to xenon, making the electron configuration \([\mathrm{Xe}] 4f^{14} 5d^{10} 6s^2\). This results in the element, Mercury (Hg) with a neutral electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^{14} 5d^{10} 6s^2\). For an \(a+2\) cation, we need to remove two electrons from the outermost shell. In this case, we will remove the 6s electrons. Doing this, the electron configuration of the \(a+2\) cation for mercury will be \([\mathrm{Xe}] 4f^{14} 5d^{10}\). The \(a+2\) cation for case (c) is \(Hg^{2+}\).