Give the formula of a negative ion that would have the same number of electrons as each of the following positive ions. a. \(\mathrm{Na}^{+}\) c. \(\mathrm{Al}^{3+}\) b. \(\mathrm{Ca}^{2+}\) d. \(\mathrm{Rb}^{+}\)

First, we need to recall that the atomic number gives the number of protons in the nucleus. In a neutral atom, the number of protons equals the number of electrons. For a positive ion, the number of electrons is reduced by the number of positive charges. So, we will first find the atomic numbers of the given elements and then calculate the number of electrons in their ions. a. \(\mathrm{Na}^{+}\): Sodium (Na) has an atomic number of 11. In the neutral atom, there are 11 electrons. In the \(\mathrm{Na}^{+}\) ion, there are 11 - 1 = 10 electrons. b. \(\mathrm{Ca}^{2+}\): Calcium (Ca) has an atomic number of 20. In the neutral atom, there are 20 electrons. In the \(\mathrm{Ca}^{2+}\) ion, there are 20 - 2 = 18 electrons. c. \(\mathrm{Al}^{3+}\): Aluminum (Al) has an atomic number of 13. In the neutral atom, there are 13 electrons. In the \(\mathrm{Al}^{3+}\) ion, there are 13 - 3 = 10 electrons. d. \(\mathrm{Rb}^{+}\): Rubidium (Rb) has an atomic number of 37. In the neutral atom, there are 37 electrons. In the \(\mathrm{Rb}^{+}\) ion, there are 37 - 1 = 36 electrons.

Now, we will find a negative ion that has the same number of electrons as each given positive ion by adding the atomic number and the number of charges. a. For 10 electrons, we look for an element with atomic number 8 and a -2 charge: This is oxygen, and the ion we are looking for is \(\mathrm{O}^{2-}\). b. For 18 electrons, we look for an element with atomic number 16 and a -2 charge: This is sulfur, and the ion we are looking for is \(\mathrm{S}^{2-}\). c. For 10 electrons, we already found that the ion with the same number of electrons is \(\mathrm{O}^{2-}\). d. For 36 electrons, we look for an element with atomic number 34 and a -2 charge: This is selenium, and the ion we are looking for is \(\mathrm{Se}^{2-}\). So the respective negative ions with the same number of electrons as the given positive ions are: a. \(\mathrm{O}^{2-}\) b. \(\mathrm{S}^{2-}\) c. \(\mathrm{O}^{2-}\) d. \(\mathrm{Se}^{2-}\)