Write the electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Fe}^{2+}\), (b) \(\mathrm{V}^{3+}\), (c) \(\mathrm{Ni}^{2+}\), (d) \(\mathrm{Pt}^{2+}\), (e) \(\mathrm{Ge}^{2-}\), (f) \(\mathrm{Ba}^{2+}\).

We use periodic table to find the atomic number(Z) of each given element and then write their corresponding electron configurations.

We remove/add the required number of electrons as per their given charges (loss of electrons for positive ions and gain of electrons for negative ions).

If the electron configuration of an ion matches with the electron configuration of noble gas element, then it has a noble gas configuration. Now let's use these steps for the given ions: (a) \(\mathrm{Fe}^{2+}\): Step 1: Electron configuration of neutral Fe (Z=26) is \([Ar] 4s^2 3d^6\). Step 2: Removing 2 electrons from Fe atom to form Fe²⁺ ion. Its configuration is: \([Ar] 3d^6\). Step 3: The electron configuration does not match any noble gas, so Fe²⁺ does not have a noble-gas configuration. (b) \(\mathrm{V}^{3+}\): Step 1: Electron configuration of neutral V (Z=23) is \([Ar] 4s^2 3d^3\). Step 2: Removing 3 electrons from V atom to form V³⁺ ion. Its configuration is: \([Ar] 3d^0\). Step 3: The electron configuration matches Argon, so V³⁺ has a noble-gas configuration. (c) \(\mathrm{Ni}^{2+}\): Step 1: Electron configuration of neutral Ni (Z=28) is \([Ar] 4s^2 3d^8\). Step 2: Removing 2 electrons from Ni atom to form Ni²⁺ ion. Its configuration is: \([Ar] 3d^8\). Step 3: The electron configuration does not match any noble gas, so Ni²⁺ does not have a noble-gas configuration. (d) \(\mathrm{Pt}^{2+}\): Step 1: Electron configuration of neutral Pt (Z=78) is \([Xe] 6s^2 4f^{14} 5d^9\). Step 2: Removing 2 electrons from Pt atom to form Pt²⁺ ion. Its configuration is: \([Xe] 4f^{14} 5d^7\). Step 3: The electron configuration does not match any noble gas, so Pt²⁺ does not have a noble-gas configuration. (e) \(\mathrm{Ge}^{2-}\): Step 1: Electron configuration of neutral Ge (Z=32) is \([Ar] 4s^2 3d^{10} 4p^2\). Step 2: Adding 2 electrons to Ge atom to form Ge²⁻ ion. Its configuration is: \([Ar] 4s^2 3d^{10} 4p^4\). Step 3: The electron configuration does not match any noble gas, so Ge²⁻ does not have a noble-gas configuration. (f) \(\mathrm{Ba}^{2+}\): Step 1: Electron configuration of neutral Ba (Z=56) is \([Xe] 6s^2\). Step 2: Removing 2 electrons from Ba atom to form Ba²⁺ ion. Its configuration is: \([Xe]\). Step 3: The electron configuration matches Xe, so Ba²⁺ has a noble-gas configuration. So, the ions with noble-gas configurations are V³⁺ and Ba²⁺.