Give three examples of +2 ions that have an electron configuration of $n d^{10}(n=3,4,5 \ldots)$

An ion with an electron configuration of \(n d^{10}\) has 10 electrons in its d subshell. Given that the value of n ranges from 3 to 5, we are looking at elements whose d subshell can be filled with 10 electrons after losing two electrons to form a +2 ion.

For this step, we need to look at the periodic table and examine elements within the d-block, also known as the transition metals. Transition metals usually form positive ions by losing their valence electrons from the s-subshell before losing the d electrons. Here, we are looking for the elements that will achieve a fully filled d-subshell after losing two s electrons. Candidate 1: Zinc (Zn, atomic number 30) Zinc's electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2\). Upon losing 2 electrons (from the 4s subshell), Zinc's configuration becomes: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\), which is the \(n d^{10}\) configuration with n = 3. Candidate 2: Cadmium (Cd, atomic number 48) Cadmium's electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2\). Upon losing 2 electrons (from the 5s subshell), Cadmium's configuration becomes: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10}\), which is the \(n d^{10}\) configuration with n = 4. Candidate 3: Mercury (Hg, atomic number 80) Mercury's electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^{14} 5d^{10} 6s^2\). Upon losing 2 electrons (from the 6s subshell), Mercury's configuration becomes: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 4f^{14} 5d^{10}\), which is the \(n d^{10}\) configuration with n = 5.

We found three examples of elements that, when forming a +2 ion, reach the \(n d^{10}\) electron configuration: 1. Zinc ion (Zn+2) with electron configuration: \(3d^{10}\) 2. Cadmium ion (Cd+2) with electron configuration: \(4d^{10}\) 3. Mercury ion (Hg+2) with electron configuration: \(5d^{10}\)