Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Al}\) and \(\mathrm{Cl} \quad\) c. \(\mathrm{Sr}\) and \(\mathrm{F}\) b. \(\mathrm{Na}\) and \(\mathrm{O} \quad\) d. \(\mathrm{Ca}\) and \(\mathrm{Se}\)

Since all the elements provided are metals and non-metals, we can know their usual ionic charges by observing their groups in the periodic table. Metals tend to lose electrons and form positive ions (cations), while the non-metals tend to gain electrons and form negative ions (anions). a. Al: Aluminium is in group 13 and forms +3 cation, Al⁺³. Cl: Chlorine is in group 17 and forms -1 anion, Cl⁻¹. b. Na: Sodium is in group 1 and forms +1 cation, Na⁺¹. O: Oxygen is in group 16 and forms -2 anion, O⁻². c. Sr: Strontium is in group 2 and forms +2 cation, Sr⁺². F: Fluorine is in group 17 and forms -1 anion, F⁻¹. d. Ca: Calcium is in group 2 and forms +2 cation, Ca⁺². Se: Selenium is in group 16 and forms -2 anion, Se⁻².

We need to balance the positive and negative charges in each compound by finding the least common multiple (LCM) of the charges, and then adjusting the number of ions so that their charges are equal. a. Al⁺³ and Cl⁻¹: The LCM of +3 and -1 is 3, so 1 Al⁺³ ion should combine with 3 Cl⁻¹ ions to form AlCl₃. b. Na⁺¹ and O⁻²: The LCM of +1 and -2 is 2, so 2 Na⁺¹ ions should combine with 1 O⁻² ion to form Na₂O. c. Sr⁺² and F⁻¹: The LCM of +2 and -1 is 2, so 1 Sr⁺² ion should combine with 2 F⁻¹ ions to form SrF₂. d. Ca⁺² and Se⁻²: The charges of Ca⁺² and Se⁻² are equal, so 1 Ca⁺² ion should combine with 1 Se⁻² ion to form CaSe.

We can name the compounds using the names of the positively charged ion followed by the negatively charged ion with an -ide suffix. a. AlCl₃: Aluminium Chloride b. Na₂O: Sodium Oxide c. SrF₂: Strontium Fluoride d. CaSe: Calcium Selenide