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

To determine the number of electrons in each positive ion, we will subtract the charge of each ion from the atomic number of the element. a. \(\mathrm{Na}^{+}\) - Sodium (Na) has an atomic number of 11. Therefore, the number of electrons in \(\mathrm{Na}^+\) is 11 - 1 = 10 electrons. b. \(\mathrm{Ca}^{2+}\) - Calcium (Ca) has an atomic number of 20. Therefore, the number of electrons in \(\mathrm{Ca}^{2+}\) is 20 - 2 = 18 electrons. c. \(\mathrm{Al}^{3+}\) - Aluminum (Al) has an atomic number of 13. Therefore, the number of electrons in \(\mathrm{Al}^{3+}\) is 13 - 3 = 10 electrons. d. \(\mathrm{Rb}^{+}\) - Rubidium (Rb) has an atomic number of 37. Therefore, the number of electrons in \(\mathrm{Rb}^+\) is 37 - 1 = 36 electrons.

In this step, we will find elements that match the number of protons as the positive ions, but have a negative charge. a. Fluorine (F) has an atomic number of 9. To have 10 electrons, the charge on the fluorine ion will be -1. Hence, the negative ion is \(\mathrm{F}^-\). b. Sulfur (S) has an atomic number of 16. To have 18 electrons, the charge on the sulfur ion will be -2. Hence, the negative ion is \(\mathrm{S}^{2-}\). c. Neon (Ne) has an atomic number of 10. To have 10 electrons, the charge on the neon ion will be 0. Since neon is a noble gas, it does not form ions. Therefore, there is no negative ion that matches this condition. d. Bromine (Br) has an atomic number of 35. To have 36 electrons, the charge on the bromine ion will be -1. Hence, the negative ion is \(\mathrm{Br}^-\).

Finally, we will write the ion formulas for each of the elements we found in Step 2. a. \(\mathrm{F}^-\) b. \(\mathrm{S}^{2-}\) c. No suitable ion found d. \(\mathrm{Br}^-\)