For each of the following ions, indicate the number of protons and electrons the ion contains. a. \(\mathrm{Ba}^{2+}\) b. \(\mathrm{Zn}^{2+}\) c. \(\mathrm{N}^{3-}\) d. \(\mathrm{Rb}^{+}\) e. \(\mathrm{Co}^{3+}\) f. \(\mathrm{Te}^{2-}\) g. \(\mathrm{Br}^{-}\)

Use the periodic table to find the atomic number of each element, which represents the number of protons. These are the atomic numbers: a. Ba (Barium) has an atomic number (Z) of 56. b. Zn (Zinc) has an atomic number (Z) of 30. c. N (Nitrogen) has an atomic number (Z) of 7. d. Rb (Rubidium) has an atomic number (Z) of 37. e. Co (Cobalt) has an atomic number (Z) of 27. f. Te (Tellurium) has an atomic number (Z) of 52. g. Br (Bromine) has an atomic number (Z) of 35.

Using the atomic number and the charge of the ion, calculate the number of electrons in the ion. a. Ba^(2+): 56 (number of protons) - 2 (charge) = 54 electrons b. Zn^(2+): 30 (number of protons) - 2 (charge) = 28 electrons c. N^(3-): 7 (number of protons) + 3 (since it's negative charge, we add) = 10 electrons d. Rb^(+): 37 (number of protons) - 1 (charge) = 36 electrons e. Co^(3+): 27 (number of protons) - 3 (charge) = 24 electrons f. Te^(2-): 52 (number of protons) + 2 (since it's negative charge, we add) = 54 electrons g. Br^(-): 35 (number of protons) + 1 (since it's negative charge, we add) = 36 electrons So, the number of protons and electrons for each ion are: a. \(\mathrm{Ba}^{2+}\): 56 protons, 54 electrons b. \(\mathrm{Zn}^{2+}\): 30 protons, 28 electrons c. \(\mathrm{N}^{3-}\): 7 protons, 10 electrons d. \(\mathrm{Rb}^{+}\): 37 protons, 36 electrons e. \(\mathrm{Co}^{3+}\): 27 protons, 24 electrons f. \(\mathrm{Te}^{2-}\): 52 protons, 54 electrons g. \(\mathrm{Br}^{-}\): 35 protons, 36 electrons