Give three examples of ions that have an electron configuration of \(n d^{8}(n=3,4,5, \ldots).\)

We are looking for elements with \(nd^8\) electron configuration. As the given values of n are 3, 4, and 5, let's find elements having 3d^8, 4d^8, and 5d^8 electron configurations. 3d^8: This corresponds to an element with 3d subshell having 2 less electrons than the completely filled 3d subshell, which has 10 electrons. The 3d^10 electron configuration belongs to Zinc (Zn) with atomic number 30. So, the element with 3d^8 electron configuration would have 2 less electrons than Zinc, making it an element with atomic number 28, which is Nickel (Ni). 4d^8: This corresponds to an element with 4d subshell having 2 less electrons than the completely filled 4d subshell. The 4d^10 electron configuration belongs to Cadmium (Cd) with atomic number 48. So, the element with 4d^8 electron configuration would have 2 less electrons than Cadmium, making it an element with atomic number 46, which is Palladium (Pd). 5d^8: This corresponds to an element with 5d subshell having 2 less electrons than the completely filled 5d subshell. The 5d^10 electron configuration belongs to Hafnium (Hf) with atomic number 72. So, the element with 5d^8 electron configuration would have 2 less electrons than Hafnium, making it an element with atomic number 70, which is Ytterbium (Yb).

Now, we have identified the neutral elements with the desired electron configurations. Let's find three examples of ions from these elements or other elements by considering the charge of the ion and the difference in the number of electrons. 1. Nickel (Ni): Nickel, a transition metal, generally forms Ni^2+ ions by losing two 4s electrons. As the electron configuration of neutral Nickel is [Ar] 3d^8 4s^2, when it loses 2 electrons to form Ni^2+ ion, it will have an electron configuration of [Ar] 3d^8, which is one of the desired electron configurations. 2. Palladium (Pd): Palladium mostly forms Pd^4+ ions by losing four electrons from its 4d and 5s orbitals. As the electron configuration of neutral Palladium is [Kr] 4d^8 5s^2, when it loses 4 electrons to form Pd^4+ ion, it will have an electron configuration of [Kr] 4d^8, which fulfills our requirement. 3. Platinum (Pt): Platinum has an electron configuration of [Xe] 4f^14 5d^9 6s^1. When it forms Pt^2+ ion by losing both the 6s electron and one 5d electron, the electron configuration becomes [Xe] 4f^14 5d^8, which matches our required electron configuration.

The three examples of ions with the given electron configurations are: 1. Ni^2+ ion with a 3d^8 electron configuration 2. Pd^4+ ion with a 4d^8 electron configuration 3. Pt^2+ ion with a 5d^8 electron configuration